An Introduction to Chemical Science
by R.P. Williams, A.M.,
PREFACE, BY R.P. WILLIAMS TABLE OF CONTENTS AN INTRODUCTION TO CHEMICAL SCIENCE APPENDIX TEXTBOOK ADVERTISEMENTS THAT APPEARED IN THE ORIGINAL EDITION INFO ABOUT THIS E-TEXT EDITION
PREFACE, BY R.P. WILLIAMS
The object held constantly in view in writing this book has been to prepare a suitable text-book in Chemistry for the average High School,—one that shall be simple, practical, experimental, and inductive, rather than a cyclopaedia of chemical information.
For the accomplishment of this purpose the author has endeavored to omit superfluous matter, and give only the most useful and interesting experiments, facts and theories.
In calling attention, by questions, and otherwise, to the more important phenomena to be observed and facts to be learned, the best features of the inductive system have been utilized. Especially is the writing of equations, which constitute the multum in parvo of chemical knowledge, insisted upon. As soon as the pupil has become imbued with the spirit and meaning of chemical equations, he need have little fear of failing to understand the rest. To this end Chapters IX., XI., and XVI. should be studied with great care.
In the early stages of the work the equations may with advantage be memorized, but this can soon be discontinued. Whenever symbols are employed, pupils should be required to give the corresponding chemical names, or, better, both names and symbols.
The classification of chemical substances into acids, bases and salts, and the distinctions and analogies between each of these classes, have been brought into especial prominence. The general relationship between the three classes, and the general principles prevailing in the preparation of each, must be fully understood before aught but the merest smattering of chemical science can be known.
Chapters XV.-XXI. should be mastered as a key to the subsequent parts of the book.
The mathematical and theoretical parts of Chemistry it has been thought best to intersperse throughout the book, placing each where it seemed to be especially needed; in this way, it is hoped that the tedium which pupils find in studying consecutively many chapters of theories will be avoided, and that the arrangement will give an occasional change from the discussion of facts and experiments to that of principles. In these chapters additional questions should be given, and the pupil should be particularly encouraged to make new problems of his own, and to solve theta.
It is needless to say that this treatise is primarily designed to be used in connection with a laboratory. Like all other text- books on the subject, it can be studied without such an accessory; but the author attaches very little value to the study of Chemistry without experimental work. The required apparatus and chemicals involve but little expense, and the directions for experimentation are the result of several years' experience with classes as large as are to be found in the laboratory of any school or college in the country.
During the present year the author personally supervises the work of more than 180 different pupils in chemistry. This enables him not only to assure himself that the experiments of the book are practical, but that the directions for performing them are ample. It is found advisable to perform most of the experiments, with full explanation, in presence of the class, before requiring the pupils either to do the work or to recite the lesson. In the laboratory each pupil has a locker under his table, furnished with apparatus, as specified in the Appendix. Each has also the author's "Laboratory Manual," which contains on every left-hand page full directions for an experiment, with observations to be made, etc. The right-hand page is blank, and on that the pupil makes a record of his work. These notes are examined at the time, or subsequently, by the teacher, and the pupil is not allowed to take the book from the laboratory; nor can he use any other book on Chemistry while experimenting. By this means he learns to make his own observations and inferences.
For the benefit of the science and the added interest in the study, it is earnestly recommended that teachers encourage pupils to fit up laboratories of their own at home. This need not at first entail a large outlay. A small attic room with running water, a very few chemicals, and a little apparatus, are enough to begin with; these can be added to from time to time, as new material is wanted. In this way the student will find his love for science growing apace.
While endeavoring, by securing an able corps of critics, and in all other ways possible, to reduce errors to a minimum, the author disclaims any pretensions to a work entirely free from mistakes, holding himself alone responsible for any shortcomings, and trusting to the leniency of teachers and critics.
The manuscript has been read by Prof. Henry Carmichael, Ph.D., of Boston, and to his broad and accurate scholarship, as well as to his deep personal interest in the work, the author is indebted for much valuable and original matter. The following persons have generously read the proof, as a whole or in part, and made suggestions regarding it, and to them the author would return his thanks, as well as acknowledge his obligation: Prof. E. J. Bartlett, Dartmouth College, N.H.; Prof. F. C. Robinson, Bowdoin College, Me.; Prof. H. S. Carhart, Michigan University; Prof. B. D. Halsted, Iowa Agricultural College; Prof. W. T. Sedgwick, Institute of Technology, Boston; Pres. M. E. Wadsworth, Michigan Mining School; Prof. George Huntington, Carleton College, Minn.; Prof. Joseph Torrey, Iowa College; Mr. C. J. Lincoln, East Boston High.School; Mr. W. H. Sylvester, English High School, Boston; Mr. F. W. Gilley, Chelsea, Mass., High School; the late D. S. Lewis, Chemist of the Boston Gas Works, and others.
R. P. W.
BOSTON, January 3, 1888.
TABLE OF CONTENTS
THE METRIC SYSTEM.
DIVISIBILITY OF MATTER.
MOLECULES AND ATOMS.
ELEMENTS AND BINARIES.
Symbols.—Names.—Coefficients.—Exponents.—Table of elements
To prepare and cut glass, etc.
Preparation.—Properties.—Combustion of carbon; sulphur; phosphorus; iron.
UNION BY WEIGHT
Meaning of equations—Problems
Preparation—Allotropic forms: diamond, graphite, amorphous carbon, coke, mineral coal.—Carbon a reducing agent, a decolorizer, disinfectant, absorber of gases
Poles of attraction—Radicals
ELECTRO-CHEMICAL RELATION OF ELEMENTS
Deposition of silver; copper; lead—Table of metals and non- metals, and discussion of their differences
Decomposition of water and of salts—Conclusions CHAPTER XIV.
UNION BY VOLUME.
Avogadro's law and its applications.
ACIDS AND BASES.
Characteristics of acids and bases.—Anhydrides.—Naming of acids.—Alkalies
Preparation from acids and bases.—Naming of salts.—Occurrence
Preparation and tests.—Bromhydric, iodhiydric, and fluorhydric acids.—Etching glass
Preparation, properties, tests, and uses.—Aqua regia: preparation and action
Preparation, tests, manufacture, and importance.-Fuming sulphuric acid
Preparation of bases.—Formation, preparation, tests, and uses of ammonia.
Preparation and properties.—Potassium hydrate and calcium hydrate
OXIDES OF NITROGEN.
Nitrogen monoxide, dioxide, trioxide, tetroaide, pentoxide.
LAWS OF DEFINITE AND OF MULTIPLE PROPORTION, and their application
CARBON PROTOXIDE and water gas.
Preparation and tests.—Oxidation in the human system.—Oxidation in water.—Deoxidation in plants
Description, preparation, and test
CHEMISTRY OF THE ATMOSPHERE.
Constituents of the air.—Air a mixture.—Water, carbon dioxide, and other ingredients of the atmosphere
THE CHEMISTRY OF WATER.
Distillation of water.—Three states.—Pure water, sea-water, river-water, spring-water CHAPTER XXIX.
THE CHEMISTRY OF FLAME.
Candle flame.—Bunsen flame.—Light and heat.—Temperature of combustion.—Oxidizing and reducing flames.—Combustible and supporter.—Explosive mixture of gases.—Generalizations
Preparation.—Chlorine water.—Bleaching properties.— Disinfecting power.—A supporter of combustion.—Sources and uses
Preparation.—Tests.—Iodo-starch paper.—Occurrence.—Uses.— Fluorine
Comparison.—Acids, oxides, and salts
VAPOR DENSITY AND MOLECULAR WEIGHT.
Gaseous weights and volumes.—Vapor density defined.—Vapor density of oxygen
Definition.—Atomic weight of oxygen.—Molecular symbols.— Molecular and atomic volumes CHAPTER XXXVI.
DIFFUSION AND CONDENSATION OF GASES.
Diffusion of gases.—Law of diffusion.—Cause.—Liquefaction and solidification of gases
Separation.—Crystals from fusion.—Allotropy.—Solution.— Theory of Allotropy.—Occurrence and purification.—Uses.—- Sulphur dioxide
Preparation.—Tests.—Combustion.—Uses.—An analyzer of metals.- -Occurrence and properties
Solution and combustion.—Combustion under water.—Occurrence.— Sources.—Preparation of phosphates and phosphorus.—- Properties.—Uses.—Matches.—Red phosphorus.—-Phosphene
Separation.—Tests.—Expert analysis.—Properties and occurrence.— Atomic volume.—Uses of arsenic trioxide
SILICON, SILICA, AND SILICATES.
Comparison of silicon and carbon.—Silica.—Silicates.—Formation of silica.
GLASS AND POTTERY.
Glass an artificial silicate.—Manufacture.—Importance.— Porcelain and pottery.
METALS AND THEIR ALLOYS.
Comparison of metals and non-metals.—Alloys.—Low fusibility. — Amalgams
SODIUM AND ITS COMPOUNDS.
Order of derivation.—Occurrence and preparation of sodium chloride; uses.—Sodium sulphate: manufacture and uses. —Sodium carbonate: occurrence, manufacture, and uses.— Sodium: preparation and uses.—Sodium hydrate: preparation and use.— Hydrogen sodium carbonate.—Sodium nitrate
POTASSIUM AND AMMONIUM.
Occurrence and preparation of potassium.—Potassium chlorate and cyanide.—Gunpowder.—Ammonium compounds
Calcium carbonate.—Lime and its uses.—Hard water.—Formation of caves.—Calcium sulphate
MAGNESIUM, ALUMINIUM, AND ZINC.
Occurrence and preparation of magnesium.—Compounds of aluminium: reduction; properties, and uses.—Compounds, uses, and reduction of zinc CHAPTER XLVIII.
IRON AND ITS COMPOUNDS.
Ores of iron.—Pig-iron.—Steel.—Wrought-iron.—Properties. — Salts of iron.—Change of valence and of color
LEAD AND TIN.
Distribution of lead.—Poisonous properties.—Some lead compounds.— Tin
COPPER, MERCURY, AND SILVER.
Occurrence and uses of copper.—Compounds and uses of mercury.— Occurrence, reduction, and salts of silver
PLATINUM AND GOLD.
Methods of obtaining, and uses
CHEMISTRY OF ROCKS.
Classification.—Composition.—Importance of siliceous rocks.— Soils.—Minerals.—The earth's interior.—Percentage of elements
Comparison of organic and inorganic compounds.—Molecular differences.—Synthesis of organic compounds.—Marsh-gas. series.—-Alcohols.—Ethers.—Other substitution products. — Olefines and other series.
Source, preparation, purification, and composition.—Natural gas
Fermented and distilled liquors.—Effect on the system.—Affinity for water.—Purity
OILS, FATS, AND SOAPS.
Sources and kinds of oils and fats.—Saponification.—Manufacture and action of soap.—Glycerin, nitro-glycerin, and dynamite. — Butter and oleomargarine.
Sugars.—Glucose.—Starch.—Cellulose.—Gun-cotton.—Dextrin. — Zylonite
CHEMISTRY OF FERMENTATION.
Ferments.—Alcoholic, acetic, and lactic fermentation.— Putrefaction.—Infectious diseases
CHEMISTRY OF LIFE.
Growth of minerals and of organic life.—Food of plants and of man.—Conservation of energy and of matter
The La Place theory—Theory of evolution—New theory of chemistry
GAS VOLUMES AND WEIGHTS.
Quantitative experiments with oxygen and hydrogen—Problems
AN INTRODUCTION TO CHEMICAL SCIENCE
THE METRIC SYSTEM.
1. The Metric System is the one here employed. A sufficient knowledge of it for use in the study of this book may be gained by means of the following experiments, which should be performed at the outset by each pupil.
Experiment 1.—Note the length of 10 cm. (centimeters) on a metric ruler, as shown in Figure 1. Estimate by the eye alone this distance on the cover of a book, and then verify the result. Do the same on a t.t. (test-tube). Try this several times on different objects till you can carry in mind a tolerably accurate idea of 10 cm. About how many inches is it?
In the same way estimate the length of 1 cm, verifying each result. How does this compare with the distance between two blue lines of foolscap? Measure the diameter of the old nickel five- cent piece.
Next, try in the same way 5 cm. Carry each result in mind, taking such notes as may be necessary.
Experiment 2.—Into a graduate, shown in Figure 2, holding 25 or 50 cc. (cubic centimeters) put 10 cc. of water; then pour this into a t.t. Note, without marking, what proportion of the latter is filled; pour out the water, and again put into the t.t. the same quantity as nearly as can be estimated by the eye. Verify the result by pouring the water back into the graduate. Repeat several times until your estimate is quite accurate with a t.t. of given size. If you wish, try it with other sizes. Now estimate 1 cc. of a liquid in a similar way. Do the same with 5 cc.
A cubic basin 10 cm on a side holds a liter. A liter contains 1,000 cc. If filled with water, it weighs, under standard conditions, 1,000 grams. Verify by measurement.
Experiment 3.—Put a small piece of paper on each pan of a pair of scales. On one place a 10 g. (gram) weight. Balance this by placing fine salt on the other pan. Note the quantity as nearly as possible with the eye, then remove. Now put on the paper what you think is 10 g. of salt. Verify by weighing. Repeat, as before, several times. Weigh 1 g., and estimate as before. Can 1 g. of salt be piled on a one-cent coin? Experiment with 5 g.
5. Resume—Lengths are measured in centimeters, liquids in cubic centimeters, solids in grams. In cases where it is not convenient to measure a liquid or weigh a solid, the estimates above will be near enough for most experiments herein given. Different solids of the same bulk of course differ in weight, but for one gram what can be piled on a one-cent piece may be called a sufficiently close estimate. The distance between two lines of foolscap is very nearly a centimeter. A cubic centimeter is seen in Figure 1. Temperatures are recorded in the centigrade scale.
WHAT CHEMISTRY IS.
6. Divisibility of Matter.
Experiment 4.—Examine a few crystals of sugar, and crush them with the fingers. Grind them as fine as convenient, and examine with a lens. They are still capable of division. Put 3 g. of sugar into a t.t., pour over it 5 cc. of water, shake well, boil for a minute, holding the t.t. obliquely in the flame, using for the purpose a pair of wooden nippers (Fig. 3). If the sugar does not disappear, add more water. When cool, touch a drop of the liquid to the tongue. Evidently the sugar remains, though in a state too finely divided to be seen. This is called a solution, the sugar is said to be soluble in water, and water to be a solvent of sugar.
Now fold a filter paper, as in Figure 4, arrange it in a funnel (Fig. 5), and pour the solution upon it, catching what passes through, which is called the filtrate, in another t.t. that rests in a receiver (Fig. 5). After filtering, notice whether any residue is left on the filter paper. Taste a drop of the filtrate. Has sugar gone through the filter? If so, what do you infer of substances in solution passing through a filter? Save half the filtrate for Experiment 5, and dilute the other half with two or three times its own volume of water. Shake well, and taste.
(Fig 5.) We might have diluted the sugar solution many times more, and still the sweet taste would have remained. Thus the small quantity of sugar would be distributed through the whole mass, and be very finely divided.
By other experiments a much finer subdivision can be made. A solution of.00000002 g. of the red coloring matter, fuchsine, in 1 cc. of alcohol gives a distinct color.
Such experiments would seem to indicate that there is no limit to the divisibility of matter. But considerations which we cannot discuss here lead to the belief that such a limit does exist; that there are particles of sugar, and of all substances, which are incapable of further division without entirely changing the nature of the substance. To these smallest particles the name molecules is given.
A mass is any portion of a substance larger than a molecule; it is an aggregation of molecules.
A molecule is the smallest particle of a substance that can exist alone.
A substance in solution may be in a more finely divided state than otherwise, but it is not necessarily in its ultimate state of division.
7. A Chemical Change.—Cannot this smallest particle of sugar, the molecule, be separated into still smaller particles of something else? May it not be a compound body, and will not some force separate it into two or more substances? The next experiment will answer the question.
Experiment 5.—Take the sugar solution saved from Experiment 4, and add slowly 4 cc.of strong sulphuric acid. Note any change of color, also the heat of the t.t. Add more acid if needed.
A substance entirely different in color and properties has been formed. Now either the sugar, the acid, or the water has undergone a chemical change. It is, in fact, the sugar. But the molecule is the smallest particle of sugar possible. The acid must have either added something to the sugar molecules, or subtracted something from them. It was the latter. Here, then, is a force entirely different from the one which tends to reduce masses to molecules. The molecule has the same properties as the mass. Only a physical force was used in dissolving the sugar, and no heat was liberated. The acid has changed the sugar into a black mass, in fact into charcoal or carbon, and water; and heat has been produced. A chemical change has been brought about.
From this we see that molecules are not the ultimate divisions of matter. The smallest sugar particles are made up of still smaller particles of other things which do not resemble sugar, as a word is composed of letters which alone do not resemble the word. But can the charcoal itself be resolved into other substances, and these into still others, and so on? Carbon is one of the substances from which nothing else has been obtained. There are about seventy others which have not been resolved. These are called elements; and out of them are built all the compounds— mineral, vegetable, and animal—which we know.
8. An element is a chemically indivisible substance, or one from which nothing else can be extracted.
A compound is a substance which is made up of elements united in exact proportions by a force called chemism, or chemical affinity.
A mixture is composed of two or more elements or compounds blended together, but not held by any chemical attraction.
To which of these three classes does sugar belong? Carbon? The solution of sugar in water?
Carbon is an element; we call its smallest particle an atom.
An atom is the smallest particle of an element that can enter into combination. Atoms are indivisible and usually do not exist alone. Both elements and compounds have molecules.
The molecule of an element usually contains two atoms; that of a compound may have two, or it may have hundreds. For a given compound the number is always definite.
Chemism is the force that binds atoms together to form molecules. The sugar molecule contains atoms, forty-five in all, of three different elements: carbon, hydrogen, and oxygen. That of salt has two atoms: one of sodium, one of chlorine. Should we say "an atom of sugar"? Why? Of what is a mass of sugar made up? A molecule? A mass of carbon? A molecule? Did the chemical affinity of the acid break up masses or molecules? In this respect it is a type of all chemical action. The distinction between physics and chemistry is here well shown. The molecule is the unit of the physicist, the atom that of the chemist. However large the masses changed by chemical action, that action is always on the individual molecule, the atoms of which are separated. If the molecule were an indivisible particle, no science of chemistry would be possible. The physicist finds the properties of masses of matter and resolves them into molecules, the chemist breaks up the molecule and from its atoms builds up other compounds.
Analysis is the separation of compounds into their elements.
Synthesis is the building up of compounds from their elements.
Of which is the sugar experiment an example? Metathesis is an exchange of atoms in two different compounds; it gives rise to still other compounds.
A chemical change may add something to a substance, or subtract something from it, or it may both subtract and add, making a new substance with entirely different properties. Sulphur and carbon are two stable solids. The chemical union of the two forms a volatile liquid. A substance may be at one time a solid, at another a liquid, at another a gas, and yet not undergo any chemical change, because in each case the chemical composition is identical.
State which of these are chemical changes: rusting of iron, falling of rain, radiation of heat, souring of milk, evaporation of water, decay of vegetation, burning of wood, breaking of iron, bleaching of cloth. Give any other illustrations that occur to you.
Chemistry treats of matter in its simplest forms, and of the various combinations of those simplest forms.
MOLECULES AND ATOMS.
9. Molecules are Extremely Small.—It has been estimated that a liter of any gas at 0 degrees and 760 mm. pressure contains 10^24 molecules, i.e. one with twenty-four ciphers.
Thomson estimates that if a drop of water were magnified to the size of the earth, and its molecules increased in the same proportion, they would be larger than fine shot, but not so large as cricket balls.
A German has recently obtained a deposit of silver two-millionths of a millimeter thick, and visible to the naked eye. The computed diameter of the molecule is only one and a half millionths of a millimeter.
By a law of chemistry there is the same number of molecules in a given volume of every gas, if the temperature and pressure are the same. Hence, all gaseous molecules are of the same size, including, of course, the surrounding space. They are in rapid motion, and the lighter the gas the more rapid the motion. This gives rise to diffusion. See page 114.
10. We Know Nothing Definite of the Form of Molecules.—In this book they will always be represented as of the same size, that of two squares. A molecule is itself composed of atoms,—from two to several hundred. The size of the atom of most elements we represent by one square.11. Atoms.—If the gaseous molecules be of the same size, it is clear that either the atoms themselves must be condensed, or the spaces between them must be smaller than before. We suppose the latter to be the case, and that they do not touch one another, the same thing being true of molecules. Atoms composing sugar must be crowded nearer together than those of salt. These atoms are probably in constant motion in the molecule, as the latter is in the mass. If we regard this square as a mass of matter, the dots may represent molecules; if we call it a molecule, the dots may be called atoms, though many molecules have no more than two or three atoms.
The following experiments illustrate the union of atoms to form molecules, and of elements to form compounds.
12. Union of Atoms.
Experiment 6.—Mix, on a paper, 5 g. of iron turnings, and the same bulk of powdered sulphur, and transfer them to an ignition tube, a tube of hard glass for withstanding high temperatures. Hold the tube in the flame of a burner till the contents have become red-hot. After a minute break it by holding it under a jet of water. Put the contents into an evaporating-dish, and look for any uncombined iron or sulphur. Both iron and sulphur are elements. Is this an example of synthesis or of analysis? Why? Is the chemical union between masses of iron and sulphur, or between molecules, or between atoms? Is the product a compound, an element, or a mixture?
Experiment 7.—Try the same experiment, using copper instead of iron. The full explanation of these experiments is given on page 13.
ELEMENTS AND BINARIES.
13. About Seventy Different Elements are now recognized, half of which have been discovered within little more than a century. These differ from one another in (1) atomic weight, (2) physical and chemical properties, (3) mode of occurrence, etc. Page 12 contains the most important elements.
The symbol of an element is usually the initial letter or letters of its Latin name, and stands for one atom of the element. C is the symbol for carbon, and represents one atom of it. O means one atom of oxygen.[The symbols of elements will also be used in this book to stand for an indefinite quantity of them; e.g. O will be used for oxygen in general as well as for one atom. The text will readily decide when symbols have a definite meaning, and when they are used in place of words.] Write, explain, and memorize the symbols of the elements in heavy type.
14. The Atomic Weight of an element is the weight of its atom compared with that of hydrogen. H is taken as the standard because it has the least atomic weight. The atomic weight of O is 16, which means that its atom weighs 16 times as much as the H atom. Every symbol, then, stands for a definite weight of the element, i.e. its atomic weight, as well as for its atom.
How much bromine by weight does Br stand for? What do these symbols mean—As, Na, N, P? If O represents one atom, how much does O2 or 2 O stand for? How much by weight? Most elements have two atoms in the molecule. How many molecules in 6 H? 10 N? S8? I20?
The symbol of a compound is formed by writing in succession the symbols of the elements of which it is composed. How many atoms in the following molecules, and how many of each element: C2H60? HNO3? PbSO4? MgCl2? (Hg2(NO3)2?)
15. The Simplest Compounds are Binaries.—A binary is a substance composed of two elements; e.g. common salt, which is a compound of sodium and chlorine. Its symbol is NaCl, its chemical name sodium chloride. The ending ide is applied to the last name of binaries. How many parts by weight of Na and of Cl in NaCl? What is the molecular weight, i.e. the weight of its molecule? Name KCl. How many atoms in its molecule? Parts by weight of each element? Molecular weight? Does the symbol stand for more than one molecule? How many molecules in 4 NaCl? How many atoms of Na and of Cl? Name these: HCl, NaBr, NaI, KBr, AgCl, AgI, HBr, HI, HF, HgO, ZnO, ZnS, MgO, CaO. Compute the proportion by weight of each element in the last three.
A coefficient before the symbol of a compound includes all the elements of the symbol, and shows the number of molecules. How many in these: 6 KBr? 3 Sn0? 12 NaCl? How many atoms of each element in the above?
An exponent, always written below, applies only to the element after which it is written, and shows the number of atoms. Explain these: AuCl3, ZnCl2, Hg2Cl2.
Write symbols for four molecules of sodium bromide, one of silver iodide (always omit coefficient one), eight of potassium bromide, ten of hydrogen chloride; also for one molecule of each of these: hydrogen fluoride, potassium iodide, silver chloride.
In all the above cases the elements have united atom for atom. Some elements will not so unite. In CaCl2 how many atoms of each element? Parts by weight of each? Give molecular weight. Is the size of the molecule thereby changed? Name these, give the number of atoms of each element in the molecule, and the proportion by weight, also their molecular weights: AuCl3, ZnCl2, MnCl2, Na2O, K2S, H3P, H4C.
Principal Elements. Name. Sym. At. Wt. Valence. Vap.D. At.Vol. Mol.Vol. State. Aluminium Al 27. II, IV ... ... ... Solid Antimony Sb 120. III, V. ... ... ... " Arsenic As 75. III, V 150. " Barium Ba 137. II ... ... ... " Bismuth Bi 210. III, V ... ... ... " Boron B 11. III ... ... ... " Bromine Br 80. I, (V) 80. Liquid Cadmium Cd 112. II 56. Solid Calcium Ca 40. II ... ... ... " Carbon C 12. (II), IV ... ... ... " Chlorine Cl 35.5 I, (V) 35.5 Gas Chromium Cr 52. (II),IV,VI ... ... ... Solid Cobalt Co 59. II, IV ... ... ... Gas Copper Cu 63. I, II ... ... ... " Fluorine F 19. I, (V) ... ... ... Gas Gold Au 196. (I), III ... ... ... Solid Hydrogen H 1. I 1. Gas Iodine I 127. I, (V) 127. ... ... Solid Iron Fe 56. II,IV,(VI) ... ... ... " Lead Pb 206. II, IV ... ... ... " Lithium Li 7. I ... ... ... " Magnesium Mg 24. II ... ... ... " Manganese Mn 55. II, IV, VI ... ... ... " Mercury Hg 200. I, II 100. Liquid Nickel Ni 59. II, IV ... ... ... Solid Nitrogen N 14. (I),III,V 14. Gas Oxygen O 16. II 16. " Phosphorus P 31. (I),III, V 62. Solid Platinum Pt 197. (II), IV ... ... ... " Potassium K 39. I ... ... ... " Silicon Si 28. IV ... ... ... " Silver Ag 108. I ... ... ... " Sodium Na 23. I ... ... ... " Strontium Sr 87. II ... ... ... " Sulphur S 32. II,IV,(VI) 32(96) " Tin Sn 118. II, IV ... ... ... " Zinc Zn 65. II 32.5 "
If more than one atom of an element enters into the composition of a binary, a prefix is often used to denote the number. SO2 is called sulphur dioxide, to distinguish it from SO3, sulphur trioxide. Name these: CO2, SiO2, MnO2. The prefixes are: mono or proto, one; di or bi, two; tri or ter, three; tetra, four; pente, five; hex, six; etc. Diarsenic pentoxide is written, As2O5. Symbolize these: carbon protoxide, diphosphorus pentoxide, diphosphorus trioxide, iron disulphide, iron protosulphide. Often only the prefix of the last name is used.
16. An Oxide is a Compound of Oxygen and Some Other Element, as HgO. What is a chloride? Define sulphide, phosphide, arsenide, carbide, bromide, iodide, fluoride.
In Experiment 6, where S and Fe united, the symbol of the product was FeS. Name it. How many parts by weight of each element? What is its molecular weight? To produce FeS a chemical union took place between each atom of the Fe and of the S. We may express this reaction, i.e. chemical action, by an equation:—
Iron + Sulphur = Iron Sulphide Or, using symbols Fe + S = FeS Using atomic weights, 56 32 = 88.
These equations are explained by saying that 56 parts by weight of iron unite chemically with 32 parts by weight of sulphur to produce 88 parts by weight of iron sulphide. This, then, indicates the proportion of each element which combines, and which should be taken for the experiment. If 56 g. of Fe be used, 32 g. of S should be taken. If we use more than 56 parts of Fe with 32 of S, will it all combine? If more than 32 of S with 56 of Fe? There is found to be a definite quantity of each element in every chemical compound. Symbols would have no meaning if this were not so.
Write and explain the equation for the experiment with copper and sulphur, using names, symbols, and weights, as above.
17. To Break Glass Tubing.
Experiment 8.—Lay the tubing on a flat surface, and draw a sharp three-cornered file two or three times at right angles across it where it is to be broken, till a scratch is made. Take the tube in the hands, having the two thumbs nearly opposite the scratch, and the fingers on the other side. Press outward quickly with the thumbs, and at the same time pull the hands strongly apart, and the tubing should break squarely at the scratch.
To break large tubing, or cut off bottles, lamp chimneys, etc., first make a scratch as before; then heat the handle of a file, or a blunt iron—in a blast-lamp flame by preference—till it is red-hot, and at once press it against the scratch till the glass begins to crack. The fracture can be led in any direction by keeping the iron just in front of it. Re-heat the iron as often as necessary.
18. To Make Ignition-Tubes.
Experiment 9.—Hold the glass tubing between the thumb and forefinger of each hand, resting it against the second finger. Heat it in the upper flame, slowly at first, then strongly, but heat only a very small portion in length, and keep it in constant rotation with the right hand. Hold it steadily, and avoid twisting it as the glass softens. The yielding is detected by the yellow flame above the glass and by an uneven pressure on the hands. Pull it a little as it yields, then heat a part just at one side of the most softened portion. Rotate constantly without twisting, and soon it can be separated into two closed tubes. No thread should be attached; but if there be one, it can be broken off and the end welded. The bottom can be made more symmetrical by heating it red-hot, then blowing, gradually, into the open end, this being inserted in the mouth. The parts should be annealed by holding above the flame for a short time, to cool slowly.
For hard glass—Bohemian—or large tubes, the blast-lamp or blowpipe is needed. In the blast-lamp air is forced out with illuminating gas. This gives a high degree of heat. Bulbs can be made in the same way as ignition-tubes, and thistle-tubes are made by blowing out the end of a heated bulb, and rounding it with charcoal.
19. To Bend Glass Tubing.
Experiment 10.—Hold the tube in the upper flame. Rotate it so as to heat all parts equally, and let the flame spread over 3 or 4 cm. in length. When the glass begins to yield, without removing from the flame slowly bend it as desired. Avoid twisting, and be sure to have all parts in the same plane; also avoid bending too quickly, if you would have a well-rounded joint. Anneal each bend as made. Heated glass of any kind should never be brought in contact with a cool body. For making O, H, etc., a glass tube — delivery-tube—50 cm. long should have three bends, as in Figure 6. The pupil should first experiment with short pieces of glass, 10 or 15 cm. long. An ordinary gas flame is the best for bending glass.
20. To Cut Glass.
Experiment 11.—Lay the glass plate on a flat surface, and draw a steel glass-cutter—revolving wheel—over it, holding this against a ruler for a guide, and pressing down hard enough to scratch the glass. Then break it by holding between the thumb and fingers, having the thumbs on the side opposite to the scratch, and pressing them outward while bending the ends of the glass inward. The break will follow the scratch.
Holes can be bored through glass and bottles with a broken end of a round file kept wet with a solution of camphor in oil of turpentine.
21. To Perforate Corks.
Experiment 12.—First make a small hole in the cork with the pointed handle of a round—rat-tail—file. Have the hole perpendicular to the surface of the cork. This can be done by holding the cork in the left hand and pressing against the larger surface, or upper part, of the cork, with the file in the right hand. Only a mere opening is made in this way, which must be enlarged by the other end of the file. A second or third file of larger size may be employed, according to the size of the hole to be made, which must be a little smaller than the tube it is to receive, and perfectly round.
22. To Obtain Oxygen.
Experiment 13.—Take 5 g. of crystals of potassium chlorate (KClO3) and, without pulverizing, mix with the same weight of pure powdered manganese dioxide (MnO2). Put the mixture into a t.t., and insert a d.t.—delivery-tube—having the cork fit tightly. Hang it on a r.s.—ring-stand,— as in Figure 7, having the other end of the d.t.
under the shelf, in a pneumatic trough, filled with water just above the shelf. Fill three or more receivers—wide-mouthed bottles—with water, cover the mouth of each with a glass plate, invert it with its mouth under water, and put it on the shelf of the trough, removing the plate. No air should be in the bottles. Have the end of the d.t. so that the gas will rise through the orifice. Hold a lighted lamp in the hand, and bring the flame against the mixture in the t.t. Keep
the lamp slightly in motion, with the hand, so as not to break the t.t. by over-heating in one place. Heat the mixture strongly, if necessary. The upper part of the t.t. is filled with air: allow this to escape for a few seconds; then move a receiver over the orifice, and fill it with gas. As soon as the lamp is taken away, remove the d.t. from the water. The gas contracts, on cooling, and if not removed, water will be drawn over, and the t.t. will be broken. Let the t.t. hang on the r.s. till cool.
With glass plates take out the receivers, leaving them covered, mouth upward (Fig. 8), with little or no water inside. When cool, the t.t. may be cleaned with water, by covering its mouth with the thumb or hand, and shaking it vigorously.
What elements, and how many, in KClO3? In Mn02? It is evident that each of these compounds contains O. Why, then, could we not have taken either separately, instead of mixing the two? This could have been done at a sufficiently high temperature. Mu02 requires a much higher temperature for dissociation, i.e. separation into its elements, than KClO3, while a mixture of the two causes O to come off from KClO3 at a lower temperature than if alone. It is not known that Mn02 suffers any change.
Each molecule of potassium chlorate undergoes the following change:—
Potassium Chlorate = Potassium Chloride + Oxygen KClO3 = KCl + 3 O.
Is this analysis or synthesis? Complete the equation, by using weights, and explain it. Notice whether the right- hand member of the equation has the same number of atoms as the left. Has anything been lost or gained? What element has heat separated? Does the experiment show whether O is very soluble in water? How many grams of O are obtainable from 122.58 g. KCIO3? PROPERTIES.
23. Combustion of Carbon.
OXYGEN Experiment 14.—Examine the gas in one of the receivers. Put a lighted splinter into the receiver, sliding along the glass cover. Remove it, blow it out, and put in again while glowing. Is it re-kindled? Repeat till it will no longer burn. Is the gas a supporter of combustion? How did the combustion compare with that in air? Is it probable that air is pure O? Why did the flame at last go out? Has the O been destroyed, or chemically united with something else?
Wood is in part C. CO2 is formed by the combustion; name it. The equation is C + 2O = CO2. Affix the names and weights. Is CO2 a supporter of combustion? Note that when C is burned with plenty of O, CO2 is always formed, and that no matter how great the conflagration, the union is atom by atom. Combustion, as here shown, is only a rapid union of O with some other substance, as C or H.
24. Combustion of Sulphur.
Experiment 15.—Hollow out one end of a piece of electric-light pencil, or of crayon, 3 cm. long, and attach it to a Cu wire (Fig. 9). Put into this a piece of S as large as a pea, ignite it by holding in the flame, and then hold it in a receiver of O. Note the color and brightness of the flame, and compare with the same in the air. Also note the color and odor of the product. The new gas is SO2. Name it, and write the equation for its production from S and O. How do you almost daily perform a similar experiment? Is the product a supporter of combustion?
25. Combustion of Phosphorus.
Experiment 16.—With forceps, which should always be used in handling this element, put a bit of P, half as large as the S above,into the crayon, called a deflagrating-spoon. Heat another wire, touch it to the P, and at once lower the latter into a receiver of O. Notice the combustion, the color of the flame and of the product. After removing, be sure to burn every bit of P by holding it in a flame, as it is liable to take fire if left. The product of the combustion is a union of what two elements? Is it an oxide? Its symbol is P2O5. Write the equation, using symbols, names, and weights. Towards the close of the experiment, when the O is nearly all combined, P2O3 is formed, as it is also when P oxidizes at a low temperature. Name it and write the equation.
26. Combustion of Iron.
Experiment 17.—Take in the forceps a piece of iron picture-cord wire 6 or 8cm long, hold one end in the flame for an instant, then dip it into some S. Enough S will adhere to be set on fire by holding it in the flame again. Then at once dip it into a receiver of O with a little water in the bottom. The iron will burn with scintillations. Is this analysis or synthesis? What elements combine? A watch-spring, heated to take out the temper, may be used, but picture-wire is better.
The product is Fe3O4. Write the equation. How much Fe by weight in the formula? How much O? What per cent by weight of Fe in the compound? Multiply the fractional part by 100. What per cent of 0? Whatper cent of C0 .is C? O2? Find the percentage composition of SO2. P2O5.
From the last five experiments what do you infer of the tendency of O to unite with other elements?
27. Oxygen is a Gas without Color, Odor, or Taste.
It is chemically a very active element; that is, it unites with almost everything. Fluorine is the only element with which it will not combine. When oxygen combines with a single element, what is the compound called? We have found that O makes up a certain portion of the air; later, we shall see how large the proportion is. Its tendency to combine with almost everything is a reason for the decay, rust, and oxidation of so many substances, and for conflagrations, great and small. New compounds are thusformed, of which O constitutes one factor. Water, H2O, is only a chemical union of O and H. Iron rust, Fe2O3 and H2O, is composed of O, Fe, and water. The burning of wood or of coal gives rise to carbon dioxide, CO2, and water. Decay of animal and vegetable matter is hastened by this all-pervading element. O forms a portion of all animal and vegetable matter, of almost all rocks and minerals, and of water. It is the most abundant of all elements, and makes up from one-half to two- thirds of the earth's surface. Compute the proportion of it, by weight, in water, H2O. It is the union of O in the air with C and H in our blood that keeps up the heat of the body and supports life. See page 81.
There are many ways of preparing this element besides the one given above. It may be obtained from water (Experiment 38) and from many other compounds, e.g. by heating mercury oxide, HgO.
Experiment 18.—Fasten a piece of electric-light pencil, or of crayon, to a wire, as in Experiment 15, and bend the wire so it will reach half-way to the bottom of a receiver. Using forceps, put into the crayon a small piece of phosphorus. Pass the wire up through the orifice in the shelf of a p.t. (pneumatic trough), having water at least l cm. above the shelf. Heat another wire, touch it to the P, and quickly invert an empty receiver over the P, having the mouth under water, so as to admit no air (Fig. 10). Let the P burn as long as it will, then remove the wire and the crayon, letting in no air. Note the color of the product, and leave till it is tolerably clear, then remove the receiver with a glass plate, leaving the water in the bottom.
Do the fumes resemble those of Experiment 16? Does it seem likely (Fig 10.) that part of the air is O? Why a part only? Find what proportion of the receiver is filled with water by measuring the water with a graduate; then fill it with water and measure that; compute the percentage which the former is of the latter. What proportion of the air, then, is O? What was the only means of escape for the P2O6, and P2O2 formed? These products are solids. Are they soluble in water? Compute the percentage composition, always by weight, of P2O2 and P2O5.
The gas left in the receiver is evidently not O. Experiment 19 will prove this conclusively, and show the properties of the new gas.
Experiment 19.—When the white cloud has disappeared, slide the plate along, and insert a burning stick; try one that still glows.
See whether the P and S on the end of a match will burn. Is the gas a supporter of combustion? Since it does not unite with C, S, or P, is it an active or a passive element? Compare it with O. Air is about 14 1/2 times as heavy as H. Which is heavier, air or N? See page 12. Air or O?
Write out the chief properties, physical and chemical, of N, as found in this experiment.
30. Inactivity of N.—N will scarcely unite chemically except on being set free from compounds. It has, however, an intense affinity for boron, and will even go through a carbon crucible to unite with it. It is not combined with O in the air; but the two form a mixture (page 86), of which N makes up four-fifths, its use being to dilute the O. What would be the effect, in case of a fire, if air were pure O? What effect on the human system?
Growing plants need a great deal of N, but they are incapable of making use of that in the air, on account of the chemical inactivity of the element. Their supply comes from compounds in earth, water, and air. By reason of its inertness N is very easily set free from its compounds. For this reason it is a constituent of most explosives, as gunpowder, nitro-glycerine, dynamite, etc. These solids, by heat or concussion, are suddenly changed to gases, which thereby occupy much more space, causing an explosion.
Nitrogen exists in many compounds, such as the nitrates; but the great source of it all is the atmosphere. See page 85.
Experiment 20.—Prepare apparatus as for making O. Be sure that the cork perfectly fits both d.t. and t.t., or the H will escape. Cover 5 g. granulated Zn, in the t.t., with 10 cc. H2O, and add 5 cc. chlorhydric acid, HCl. Adjust as for O (Fig. 7), except that no heat is to be applied. If the action is not brisk enough, add more HCl. Collect several receivers of the gas over water, adding small quantities of HCl when necessary. Observe the black floating residuum; it is carbon, lead, etc. With a glass plate remove the receivers, keeping them inverted (Fig. 11), or the H will escape.
32. The Chemical Change is as follows:—
Zinc + hydrogen chloride = zinc chloride + hydrogen.
Zn + 2 HCl = ZnCl2 + 2H.
Complete by adding the weights, and explain. Notice that the water does not take part in the change; it is added to dissolve the ZnCl2 formed, and thus keep it from coating the Zn and preventing further action of the acid. Note also that Zn has simply changed places with H, one atom of the former having driven off two atoms of the latter. The H, having nothing to unite with, is set free as a gas, and collected over water. Of course Zn must have a stronger chemical affinity for Cl than H has, or the change could not have taken place. Why one Zn atom replaces two H atoms will be explained later, asfar as an explanation is possible. This equation, should be studied carefully, as a type of all equations. The left-hand member shows what were taken, i.e. the factors; the right-hand shows what were obtained, i.e. the products. H2SO4 might have been used instead of HCl. In that case the reaction, or equation, would have been: —
Zinc + hydrogen sulphate = zinc sulphate + hydrogen.
Zn + H2SO4 = ZnSO4 + 2H.
Iron might have been used instead of zinc, in which case the reactions would have been:—
Iron + hydrogen chloride = iron chloride + hydrogen.
Fe + 2 HCl = FeCl2 + 2 H.
Iron + hydrogen sulphate = iron sulphate + hydrogen.
Fe + H2SO4 = FeSO4 + 2 H.
Write the weights and explain the equations. The latter should be memorized.
Experiment 21.—Lift with the left hand a receiver of H, still inverted, and insert a burning splinter with the right (Fig. 12). Does the splinter continue to burn? Does the gas burn? If so, where? Is the light brilliant? Note the color of the flame. Is there any explosion? Try this experiment with several receivers. Is the gas a supporter of combustion? i.e. will carbon burn in it? Is it combustible? i.e. does it burn? If so, it unites with some part of the air. With what part?34. Collecting H by Upward Displacement.
Experiment 22.—Pass a d.t. from a H generator to the top of a receiver or t.t. (Fig. 13). The escaping H being so much lighter than air will force the latter down. To obtain the gas unmixed with air, the d.t. should tightly fit a cardboard placed under the mouth of the receiver. When filled, the receiver can be removed, inverted as usual, and the gas tested. In this and other experiments for generating H, a thistle-tube, the end of which dips under the liquid, can be used for pouring in acid, as in Figure 13.
35. Philosopher's Lamp and Musical Flame.
Experiment 23.—Fit to a cork a piece of glass tubing 10 or 15 cm. long, having the outer end drawn out to a point with a small opening, and insert it in the H generator. Before igniting the gas at the end of the tube take the, precaution to collect a t.t. of it by upward displacement, and bring this in contact with a flame. If a sharp explosion ensues, air is not wholly expelled from the generator, and it would be dangerous to light the gas. When no sound, or very little, follows, light the escaping gas. The generation of H must not be too rapid, neither should the t.t. be held under the face, as the cork is liable to be forced out by the pressure of H. A safety-tube, similar to the thistle- tube above, will prevent this. This apparatus is called the "philosopher's lamp." Thrust the flame into a long glass tube 1- 1/2 to 3 cm. in diameter, as shown in Figure 14, and listen for a musical note.
36. Product of Burning H in Air.
Experiment 24.—Fill a tube 2 or 3 cm. in diameter with calcium chloride, CaCl2, and connect one end with a generator of H (Fig. 15). At the other end have a philosopher's lamp-tube.Observing the usual precautions, light the gas and hold over it a receiver, till quite a quantity of moisture collects. All water was taken from the gas by the dryer, CaCl2. What is, therefore, the product of burning H in air? Complete this equation and explain it: 2H + O = ? Figure 16 shows a drying apparatus arranged to hold CaCl2.
[Fig. 15][Fig. 16]
37. Explosiveness of H.
Experiment 25. — Fill a soda-water bottle of thick glass with water, invert it in a pneumatic trough, and collect not over 1/4 full of H. Now remove the bottle, still inverted, letting air in to fill the other 3/4. Mix the air and H by covering the mouth of the bottle with the hand, and shaking well; then hold the mouth of the bottle, slightly inclined, in a flame. Explain the explosion which follows. If 3/4 was air, what part was O? What use did the N serve? Note any danger in exploding H mixed with pure O. What proportions of O and H by volume would be most dangerously explosive? What proportion by weight?
By the rapid union of the two elements, the high temperature suddenly expanded the gaseous product, which immediately contracted; both expansion and contraction produced the noise of explosion.
38. Pure H Is a Gas without Color, Odor, or Taste.
—It is the lightest of the elements, 14 1/2 times as light asair. It occurs uncombined in coal-mines, and some other places, but the readiness with which it unites with other elements, particularly O, prevents its accumulation in large quantities. It constitutes two-thirds of the volume of the gases resulting from the decomposition of water, and one-ninth of the weight. Compute the latter from its symbol. It is a constituent of plants and animals, and some rocks. Considering the volume of the ocean, the total amount of H is large. It can be separated from H2O by electrolysis, or by C, as in the manufacture of water gas.
When burned with O it forms H2O. Pure O and H when burning give great heat, but little light. The oxy-hydrogen blow-pipe (Fig. 17) is a device for producing the highest temperatures of combustion. It has O in the inner tube and H in the outer. Why would it not be better the other way? These unite at the end, and are burned, giving great heat. A piece of lime put into the flame gives the brilliant Drummond or calcium light.
Chapter IX. UNION BY WEIGHT.
39. In the Equation —
Zn + 2 HCl = ZnCl2 + 2 H 65 + 73 = 136 + 2
65 parts by weight of Zn are required to liberate 2 parts by weight of H; or, by using 65 g Zn with 73 g HCl, we obtain 2 g H. If twice as much Zn (130 g) were used, 4 g H could be obtained, with, of course, twice as much HCl. With 260 g. Zn, how much H could be liberated? A proportion may be made as follows:—
Zn given : Zn required :: H given : H required. 65 : 260 :: 2 : x.
[footnote: Given, as here used, means the weight called for by the equation; required means that called for by the question.]
Solving, we have 8 g H.
How much H is obtainable by using 5 g Zn, as in the experiment?
To avoid error in solving similar problems, the best plan is as follows:—
Zn + 2HCl = ZnCl2 + 2 H 65:5::2:x 65 2 65 x = 10 5 x x = 10/65 = 2/13 Ans. 2/13 g.
The equation should first be written; next, the atomic or molecular weights which you wish to use, and only those, to avoid confusion; then, on the third line, the quantity of the substance to be used, with underneath the substance wanted. The example above will best how this. This plan will prevent the possibility of error. The proportion will then be:—
a given : a required :: b given : b required.
How much Zn is required to produce 30 g. H?
Zn + 2HCl = ZnCl2 + 2H 2:30::65:x 65 2 2x = 1950 x 30 x = 975 Ans. 975 g. Zn.
(1) How much Zn is necessary for 14 g. H?
(2) How many pounds of Zn are necessary for 3 pounds of H?
(3) How many grams of H from 17 g. of Zn?
(4) How many tons of H from 1/2 ton of Zn?
Suppose we wish to find how much chlorhydric acid—pure gas— will give 12 g. H. The question involves only HCl and H. Arrange as follows:—
Zn + 2HCl = ZnCl2 + 2 H H giv. : H req. :: HCl giv. : HCl req. 73 2 2 : 12 :: 73 x x 12 2x=876 x=438 Ans. 438 g. HCl.
(1) How much HCl is needed to produce 100 g. H?
(2) How much H in 10 g. HCl?
(3) How much ZnCl2 is formed by using 50 g. HCl? The question is now between HCl and ZnCl2.
Zn + 2HC1 = ZnCl2 + 2H 73 136 Arrange the proportion, and solve. 50 x
Suppose we have generated H by using H2S04: the equation is Zn + H2S04 = ZnSO4 + 2 H. There is the same relation as before between the quantities of Zn and of H, but the H2S04 and ZnS04 are different.
How much H2SO4 is needed to generate 12 g. H?
Zn + H2SO4 = ZnS04 + 2 H 98 2 Make the proportion, and solve x 12
(1) How much H in 200 g. H2S04?
(2) How much ZnS04 is produced from 200 g. H2S04? (3) How much H2S04 is needed for 7 1/2 g H? (4) How much Zn will 40 g. H2SO4 combine with? (5) How much Fe will 40 g. H2SO4 combine with? (6) How much H can be obtained by using 75 g Fe?
These principles apply to all reactions. Suppose, for example, we wish to get l0 g. of O: how much KClO3 will it be necessary to use? The reaction is:—
KClO3 = KCl + O3 48 : 10 :: 122.5 : x 122.5 48 x 10 Ans. 25.5+ g. KClO3.
The pupil should be required to make up problems of his own, using various reactions, and to solve them.
Examine graphite, anthracite coal, bituminous coal, cannel coal, wood, gas carbon, coke.
40. Preparation of C.
Experiment 26.—Hold a porcelain dish or a plate in the flame of a candle, or of a Bunsen burner with the openings at the bottom closed. After a minute examine the deposit. It is carbon, i.e. lamp- black or soot, which is a constituent of gas, or of the candle. Open the valve at the base of the Bunsen burner, and hold the deposit in the flame. Does the C gradually disappear? If so, it has been burned to CO2. C + 2 O = CO2. Is C a combustible element?
Experiment 27.—Ignite a splinter, and observe the combustion and the smoke, if any. Try to collect some C in the same way as before.
With plenty of O and high enough temperature, all the C is burned to CO2, whether in gas, candle, or wood. CO2 is an invisible gas. The porcelain, when held in the flame, cools the C below the point at which it burns, called the kindling-point, and hence it is deposited. The greater part of smoke is unburned carbon.
Experiment 28.—Hold an inverted dry t.t. or receiver over the flame of a burning candle, and look for any moisture (H2O). What two elements are shown by these experiments to exist in the candle? The same two are found in wood and in gas. Experiment 29.—Put into a small Hessian crucible (Fig. 18) some pieces of wood 2 or 3 cm long, cover with sand, and heat the crucible strongly. When smoking stops, cool the crucible, remove the contents, and examine the charcoal. The gases have been driven off from the wood, and the greater part of what is left is C.
Experiment 30.—Put 1 g. of sugar into a porcelain crucible, and heat till the sugar is black. C is left. See Experiment 5. Remove the C with a strong solution of sodium hydrate (page 208).
41. Allotropic Forms.—Carbon is peculiar in that it occurs in at least three allotropic, i.e. different, forms, all having different properties. These are diamond, graphite, and amorphous —not crystalline—carbon. The latter includes charcoal, lamp- black, bone-black, gas carbon, coke, and mineral coal. All these forms of C have one property in common; they burn in O at a high temperature, forming CO2. This proves that each is the element C, though it is often mixed with some impurities.
Allotropy, or allotropism, is the quality which an element often has of appearing under various forms, with different properties. The forms of C are a good illustration.
42. Diamond is the purest C; but even this in burning leaves a little ash, showing that it is not quite pure. It is a rare mineral, found in India, South Africa, and Brazil, and is the hardest and most highly refractive to light of all minerals. Boron is harder. [Footnote: B, not occurring free, is not a mineral.] When heated in the electric arc, at very high temperatures, diamond swells and turns black. 43. Graphite, or Plumbago, is One of the Softest Minerals.—It is black and infusible, and oxidizes only at very high temperatures, higher than the diamond. It contains from 95 to 98 per cent C. Graphite is found in the oldest rock formations, in the United States and Siberia. It is artificially formed in the iron furnace. Graphite is employed for crucibles where great heat is required, for a lubricant, for making metal castings, and, mixed with clay, for lead-pencils. It is often called black-lead.
44. Amorphous Carbon comprises the following varieties.
Charcoal is made by heating wood, for a long time, out of contact with the air. The volatile gases are thus driven off from the wood; what is left is C, and a small quantity of mineral matter which remains as ash when the coal is burned.
45. Lamp-black is prepared as in Experiment 26, or by igniting turpentine (C1OH16), naphtha, and various oils, and collecting the C of the smoke. It is used for making printers' ink, India ink, etc. A very pure variety is obtained from natural gas.
Bone-black, or animal charcoal, is obtained by distilling bones, i.e. by heating them in retorts into which no air is admitted. The C is the charred residue.
Gas Carbon is formed in the retorts of the gas-house. See page 182. It is used to some extent in electrical work.
46. Coke is the residue left after distilling soft coal. It is tolerably pure carbon, with some ash and a little volatile matter. It burns without flame. 47. Mineral Coal is fossilized wood or other vegetable matter. Millions of years ago trees and other vegetation covered the earth as they do to-day. In certain places they slowly sank, together with the land, into the interior of the earth, were covered with sand, rock, and water, and heated from the earth's interior. A slow distillation took place, which drove off some of the gases, and converted vegetable matter into coal. All the coal dug from the earth represents vegetable life of a former period. Millions of years were required for the transformation; but the same change is in progress now, where peat beds are forming from turf.
Coal is found in all countries, the largest beds being in the United States. From the nature of its formation, coal varies much in purity.
Anthracite, or hard coal, is purest in carbon, some varieties having from 90 to 95 per cent. This represents most complete distillation in the earth; i.e. the gases have mostly been driven off. It is much used in New England.
48. Bituminous, or soft coal, crocks the hands, and burns rapidly with much flame and smoke. The greater part of the coal in the earth is bituminous. It represents incomplete distillation. Hence, by artificially distilling it, illuminating gas is made. See page 180. It is far less pure C than anthracite.
49. Cannel Coal is a variety of bituminous coal which can be ignited like a candle. This is because so many of the gases are still left, and it shows cannel to be less pure C than bituminous coal.
50. Lignite, Peat, Turf, etc., are still less pure varieties of C. Construct a table of the naturally occurring forms of this element, in the order of their purity. Carbon forms the basis of all vegetable and animal life; it is found in many rocks, mineral oils, asphaltum, natural gas, and in the air as CO2.
51. C a Reducing Agent.
Experiment 31.—Put into a small ignition-tube a mixture of 4 or 5 g. of powdered copper oxide (CuO), with half its bulk of powdered charcoal. Heat strongly for ten or fifteen minutes. Examine the contents for metallic copper. With which element of CuO has C united? The reaction may be written: Cu0 + C = CO + Cu. Complete and explain.
A Reducing, or Deoxidizing, Agent is a substance which takes away oxygen from a compound. C is the most common and important reducing agent, being used for this purpose in smelting iron and other ores, making water-gas, etc.
An Oxidizing Agent is a substance that gives up its O to a reducing agent. What oxidizing agent in the above experiment?
52. C a Decolorizer.
Experiment 32.—Put 3 or 4 g. of bone-black into a receiver, and add 10 or 15 cc.of cochineal solution. Shake this thoroughly, covering the bottle with the hand. Then pour the whole on a filter paper, and examine the filtrate. If all the color is not removed, filter again. What property of C is shown by this experiment? Any other coloring solution may be tried.
The decolorizing power of charcoal is an important characteristic. Animal charcoal is used in large quantities for decolorizing sugar. The coloring matter is taken out mechanically by the C, there being no chemical action. 53. C a Disinfectant.
Experiment 33.—Repeat the previous experiment, adding a solution of H2S3 i.e. hydrogen sulphide, in water, instead of cochineal solution. See page 120. Note whether the bad odor is removed. If not, repeat.
Charcoal has the property of absorbing large quantities of many gases. Ill-smelling and noxious gases are condensed in the pores of the C; O is taken in at the same time from the air, and these gases are there oxidized and rendered odorless and harmless. For this reason charcoal is much used in hospitals and sick-rooms, as a disinfectant. This property of condensing O, as well as other gases, is shown in the experiment below.
54. C an Absorber of Gases and a Retainer of Heat.
Experiment 34.—Put a piece of phosphorus of the size of a pea, and well dried, on a thick paper. Cover it well with bone-black, and look for combustion after a while. O has been condensed from the air, absorbed by the C, and thus communicated to the P. Burn all the P at last.
55. The Symbols NaCl and MgCl2 differ in two ways.—What are they? Let us see why the atom of Mg unites with two Cl atoms, while that of Na takes but one. If the atoms of two elements attract each other, there must be either a general attraction all over their surfaces, or else some one or more points of attraction. Suppose the latter to be true, each atom must have one or more poles or bonds of attraction, like the poles of a magnet. Different elements differ in their number of bonds. Na has one, which may be written graphically Na-; Cl has one, -Cl. When Na unites with Cl, the bonds of each element balance, as follows: Na-Cl. The element Mg, however, has two such bonds, as Mg= or -Mg-. When Mg unites with Cl, in order to balance, or saturate, the bonds, it is evident that two atoms of Cl must be used, as Cl-Mg-Cl, or MgCl2.
A compound or an element, in order to exist, must have no free bonds. In organic chemistry the exceptions to this rule are very numerous, and, in fact, we do not know that atoms have bonds at all; but we can best explain the phenomena by supposing them, and for a general statement we may say that there must be no free bonds. In binaries the bonds of each element must balance.
56. The Valence, Quantivalence, of an Element is its Combining Power Measured by Bonds.—H, having the least number of bonds, one, is taken as the unit. Valence has always to be taken into account in writing the symbol of a compound. It is often written above and after the elements [i.e. written like an exponent], as K^I, Mg^II.
An element having a valence of one is a monad; of two, a dyad; three, a triad; four, tetrad; five, pentad; six, hexad, etc. It is also said to be monovalent, di- or bivalent, etc. This theory of bonds shows why an atom cannot exist alone. It would have free or unused bonds, and hence must combine with its fellow to form a molecule, in case of an element as well as in that of a compound. This is illustrated by these graphic symbols in which there are no free bonds: H-H, O=O, N[3-bond symbol]N, C[4-bond symbol]C. A graphic symbol shows apparent molecular structure.
After all, how do we know that there are twice as many Cl atoms in the chloride of magnesium as in that of sodium? The compounds have been analyzed over and over again, and have been found to correspond to the symbols MgCl2 and NaCl. This will be better understood after studying the chapter on atomic weights. In writing the symbol for the union of H with O, if we take an atom of each, the bonds do not balance, H-=O, the former having one; the latter, two. Evidently two atoms of H are needed, as H-O-H, or
H = O , or H2O. In the union of Zn and O, each has two bonds; H
hence they unite atom with atom, Zn = O, or ZnO.
Write the grapbic and the common symbols for the union of H^I and Cl^I; of K^I and Br^I; Ag^I and O^II; Na^I and S^II; H^I and P^III. Study valences. It will be seen that some elements have a variable quantivalence. Sn has either 2 or 4; P has 3 or 5. It usually varies by two for a given element, as though a pair of bonds sometimes saturated each other;. e.g. Sn, a quantivalence of 4, and Sn=, a quantivalence of 2. There are, therefore, two oxides of tin, SnO and SnO2, or Sn=O and OSnO. Write symbols for the two chlorides of tin; two oxides of P; two oxides of arsenic.
The chlorides of iron are FeCl2 and Fe2Cl6. In the latter, it might be supposed that the quantivalence of Fe is 3, but the graphic symbol shows it to be 4. It is called a pseudo-triad, or false triad. Cr and Al are also pseudo-triads.
Cl Cl Cl Fe Fe Cl Cl Cl
Write formulae for two oxides of iron; the oxide of Al.
57. A Radical is a Group of Elements which has no separate existence, but enters into combination like a single atom; e.g. (NO3) in the compounds HNO3 or KNO3; (SO4) in H2SO4. In HNO3 the radical has a valence of 1, to balance that of H, H-NO3). In H2SO4, what is the valence of (SO4)? Give it in each of these radicals, noting first that of the first element: K(NO3), Na2(SO4), Na2(CO3), K(ClO3), H3(PO4), Ca3(PO4)2, Na4(SiO4).
Suppose we wish to know the symbol for calcium phosphate. Ca and PO4 are the two parts. In H3(PO4) the radical is a triad, to balance H3. Ca is a dyad, Ca==(P04). The least common multiple of the bonds (2 and 3) is 6, which, divided by 2 (no. Ca bonds), gives 3 (no. Ca atoms to be taken). 6 / 3 (no. (PO4) bonds) gives 2 (no. PO4 radicals to be taken). Hence the symbol Ca3(P04)2. Verify this by writing graphically.
Write symbols for the union of Mg and (SO4), Na and (PO4), Zn and (NO3), K and (NO3), K and (SO4), Mg and (PO4), Fe and (SO4) (both valences of Fe), Fe and (NO3), taking the valences of the radicals from HNO3, H2SO4, H3PO4.
ELECTRO-CHEMICAL RELATION OF ELEMENTS.
58. Examine untarnished pieces of iron, silver, nickel, lead, etc.; also quartz, resin, silk, wood, paper. Notice that from the first four light is reflected in a different way from that of the others. This property of reflecting light is known as luster. Metals have a metallic luster which is peculiar to themselves; and this, for the present, may be regarded as their chief characteristic. Are they at the positive or negative end of the list? See page 43. How is it with the non-metals? This arrangement has a significance in chemistry which we must now examine. The three appended experiments show how one metal can be withdrawn from solution by a second, this second by a third, the third by a fourth, and so on. For expedition, three pupils can work together for the three following experiments, each doing one, and examining the results of the others.
59. Deposition of Silver.
Experiment 35.—Put a ten-cent Ag coin into an evaporating-dish, and pour over it a mixture of 5 cc. HNO3 and 10 cc. H2O. Warm till all, or nearly all, the Ag dissolves. Remove the lamp. 3 Ag + 4 HNO3 = 3 AgNO3 + 2 H2O + NO. Then add 10 cc. H2O, and at once put in a short piece of Cu wire, or a cent. Leave till quite a deposit appears, then pour off the liquid, wash the deposit thoroughly, and remove it from the coin. See whether the metal resembles Ag. 2 AgNO3 + Cu =?60. Deposition of Copper.
Experiment 36.—Dissolve a cent or some Cu turnings in dilute HNO3, as in Experiment 35, and dilute the solution. 3 Cu + 8 HN09 - 3 Cu (NOA+4 H2O+2 NO.)
Then put in a clean strip of Pb, and set aside as before, examining the deposit finally. Cu(NO3), + Pb - ?
61. Deposition of Lead.
Experiment 37.—Perform this experiment in the same manner as the two previous ones, dissolving a small piece of Pb, and using a strip of Zn to precipitate the Pb. 3 Pb + 8 HNO3 - 3 Pb (NO4)2 + 4 Ha0 + 2 NO. Pb (NO3) 2 + Zn = ? h.
62. Explanation. -These experiments show that Cu will replace Ag in a solution of AgNO3, that Pb will replace and deposit Cu from a similar compound, and that Zn will deposit Pb in the same way. They show that the affinity of Zn for (NO3) is stronger than either Ag, Cu, or Pb. We. express this affinity by saying that Zn is the most positive of the four metals, while Ag is the most nega- tive. Cu is positive to Ag, but negative to Pb and Zn. Which of the four elements are positive to Pb, and which negative? Mg would withdraw Zn from a similar solution, and be in its turn withdrawn by Na. The table on page 43 is founded on this relation. A given element is positive to every element above it in the list, and negative to all below it.
Metals are usually classed as positive, non-metals as negative. Each in union with O and 1=I gives rise to a very important class of compounds,=—the negative to acids, the positive to bases.
In the following, note whether the positive or the negative element is written first:—HCl, Na20,-As2S3, -MgBr2, Ag2S. Na2SO4 is made up of two parts, Na2 being positive, the radical SO4 negative. Like elements, radicals are either positive or negative. In the following, separate the positive element from the negative radical by a vertical line: Na2CO3, NaNO3, ZnSO4, KClO3.
The most common positive radical is NH4, ammonium, as in NH4Cl. It always deports itself as a metal. The commonest radical is the negative OH, called hydroxyl, from hydrogen- oxygen. Take away H from the symbol of water, H-O-H, and hydroxyl —(OH) with one free bond is left. If an element takes the place of H, i.e. unites with OH, the compound is called a hydrate. KOH is potassium hydrate. Name NaOH, Ca(OH)2, NH4OH, Zn(OH)2, Al2(OH)6. Is the first part of each symbol above positive or negative?
H has an intermediate place in the list. It is a constituent of both acids and bases, and of the neutral substance, water.
Negative or Non-Metallic Elements. Acid-forming with H(usually OH).
Oxygen Sulphur Nitrogen Fluorine Chlorine Bromine Iodine Phosphorus Arsenic Carbon Silicon Hydrogen
Positive or Metallic Elements. Base-forming with OH.
Gold Platinum Mercury Silver Copper Tin Lead Iron Zinc Aluminium Magnesium Calcium Sodium Potassium
The following experiment is to be performed only by the teacher, but pupils should make drawings and explain.
63. Decomposition of Water.
Experiment 38.—Arrange "in series" two or more cells of a Bunsen battery (Physics, page 164), [References are made in this book to Gage's Introduction to Physical Science.] and attach the terminal wires to an electrolytic apparatus (Fig. 19) filled with water made slightly acid with H2SO4. Construct a diagram of the apparatus, marking the Zn in the liquid +, since it is positive, and the C, or other element, -. Mark the electrode attached to the Zn -, and that attached to the C +; positive electricity at one end of a body commonly implies negative at the other. Opposites attract, while like electricities repel each other. These analogies will aid the memory. At the + electrode is the - element of H2O, and at the - electrode the + element. Note, page 43, whether H or O is positive with reference to the other, and write the symbol for each at the proper electrode. Compare the diagram with the apparatus, to verify your conclusion. Why does gas collect twice as fast at one electrode as at the other? What does this prove of the composition of water? When filled, test the gases in each tube, for O and H, with a burning stick. Electrical analysis is called electrolysis.
If a solution of NaCl be electrolyzed, which element will go to the + pole? Which, if the salt were K2SO4? Explain these reactions in the electrolysis of that salt. K2SO4 = K2 + S03 + O. SO4 is unstable, and breaks up into SO3 and O. Both K and SO3 have great affinity for water. K2 + 2 H2O = 2 KOH + H2. S03 + H2O = H2SO4.
The base KOH would be found at the - electrode, and the acid H2SO4 at the + electrode.
The positive portion, K, uniting with H2O forms a base; the negative part, S03, with H2O forms an acid. Of what does this show a salt to be composed?
64. Conclusions.—These experiments show (1) that at the + electrode there always appears the negative element, or radical, of the compound, and at the - electrode the positive element; (2) that these elements unite with those of water, to make, in the former case, acids, in the latter, bases; (3) that acids and bases differ as negative and positive elements differ, each being united with O and H, and yet producing compounds of a directly opposite character; (4) that salts are really compounded of acids and bases. This explains why salts are usually inactive and neutral in character, while acids and bases are active agents. Thus we see why the most positive or the most negative elements in general have the strongest affinities, while those intermediate in the list are inactive, and have weak affinities; why alloys of the metals are weak compounds; why a neutral substance, like water, has such a weak affinity for the salts which it holds in solution; and why an aqueous solution is regarded as a mechanical mixture rather than a chemical compound. In this view, the division line between chemistry and physics is not a distinct one. These will be better understood after studying the chapters on acids, bases and salts.
UNION BY VOLUME.
66. Avogadro's Law of Gases.—Equal volumes of all gases, the temperature and pressure being the same, have the same number of molecules. This law is the foundation of modern chemistry. A cubic centimeter of O has as many molecules as a cubic centimeter of H, a liter of N the same number as a liter of steam, under similar conditions. Compare the number of molecules in 5 l. of N2O with that in 10 l. Cl. 7 cc. vapor of I to 6 cc. vapor of S. The half-molecules of two gases have, of course, the same relation to each other, and in elements the half-molecule is usually the atom.
The molecular volumes—molecules and the surrounding space—of all gases must therefore be equal, as must the half-volumes. Notice that this law applies only to gases, not to liquids or solids. Let us apply it to the experiment for the electrolysis of water. In this we found twice as much H by volume as O. Evidently, then, steam has twice as many molecules of H as of O, and twice as many half-molecules, or atoms. If the molecule has one atom of O, it must have two of H, and the formula will be H2O.
Suppose we reverse the process and synthesize steam, which can be done by passing an electric spark through a mixture of H and O in a eudiometer over mercury; we should need to take twice as much H as O. Now when 2 cc. of H combine thus with 1 cc. of O, only 2 cc.of steam are produced. Three volumes are condensed into two volumes, and of course three molecular volumes into two, three atomic volumes into two. This may be written as follows:—
H + H + O = H2O.
This is a condensation of one-third.
If 2 l. of chlorhydric acid gas be analyzed, there will result 1 l. of H and 1 l. of Cl. The same relation exists between the molecules and the atoms, and the reaction is:—
HCl = H + Cl.
Reverse the process, and 1 l. of H unites with 1 l. of Cl to produce 2 l. of the acid gas; there is no condensation, and the symbol is HCl. In seven volumes HCl how many of each constituent?
The combination of two volumes of H with one volume of S is found to produce two volumes of hydrogen sulphide. Therefore two atoms of H combine with one of S to form a molecule whose symbol is H2S.
H + H + S = H2S.
What is the condensation in this case?
(1) How many liters of S will it take to unite with 4 l. of H? How much H2S will be formed?
(2) How many liters of H will it take to combine with 5 l. of S? How much H2S results?
(3) In 6 l. H2S how many liters H, and how much S? Prove.
(4) In four volumes H2S how many volumes of each constituent?
(5) If three volumes of H be mixed with two volumes of S, so as to make H2S, how much will be formed? How much of either element will be left? An analysis of 2 cc. of ammonia gives 1 cc. N and 3 cc. H. The symbol must then be NH3, the reaction,—
NH3 = N + H + H + H.
What condensation in the synthesis of NH3?
In 12 cc. NH3 how many cubic centimeters of each element? In 2 1/2 cc? How much H by volume is required to combine with nine volumes of N? How many volumes of NH3 are produced?
In elements that have not been weighed in the gaseous state, as C, the evidence of atomic volume is not direct, but we will assume it. Thus two volumes of marsh gas would separate into one of C and four of H. What is its symbol and supposed condensation? Two volumes of alcohol vapor resolve into two of C, six of H, and one of O. What is its symbol? its condensation?
The symbol itself of a compound will usually show what its condensation is; e.g. HCl, HBr, HF, etc., have two atoms; hence there will be no shrinkage. In H2O, SO2, CO2, the molecule has three atoms condensed into the space of two, or one-third shrinkage. In NH3 four volumes are crowded into the space of two, a condensation of one-half.
P, As, Hg, Zn, have exceptional atomic volumes.
ACIDS AND BASES.
66. What Acids Are.
Experiment 39.—Pour a few drops of chlorhydric acid, HCl, into a clean evaporating-dish. Add 5 cc. H2O, and stir. Touch a drop to the tongue, noting the taste. Dip into it the end of a piece of blue litmus paper, and record the result. Thoroughly wash the dish, then pour in a few drops of nitric acid, HNO3, and 5 cc. H2O, and stir. Taste, and test with blue litmus. Test in the same way sulphuric acid, H2SO4. Name two characteristics of an acid. In a vertical line write the formulae of the acids above. What element is common to them all? Is the rest of the formula positive or negative?