(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 H98 2 | Make the proportion, and solvex 12
Solve:—
(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 O=Sn=O. 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.
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).
OxygenSulphurNitrogenFluorineChlorineBromineIodinePhosphorusArsenicCarbonSiliconHydrogen
Positive or Metallic Elements.Base-forming with OH.
GoldPlatinumMercurySilverCopperTinLeadIronZincAluminiumMagnesiumCalciumSodiumPotassium
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 acidH2SO4 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.
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:—
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.
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,—
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.
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?
67. An Acid is a substance composed of H and a negative element or radical. It has usually a sour taste, and turns blue litmus red. Litmus is a vegetable extract obtained from lichens in Southern Europe. Acids have the same action on many other vegetable pigments. Are the following acid formulae, and why? H2SO3, HBr, HNO2, H3PO3, H4SiO4. Most acids have O as well as H. Complete the symbols for acids in the following list, and name them, from the type given:—
HCl, chlorhydric acid. HN03, nitric acid.?Br, ? ?Cl? ??I, ? ?Br? ??F, ? ?I? ?H3PO4, phosphoric acid. H3PO3, phosphorous acid.?As? ? ?As? ?
Complete these equations:—
H2SO3 - H2O = ? | 2 HN03 - H2O = ?H2SO4 - H2O = ? | 2 HNO2 - H2O = ?H2CO3 - H2O = ? | 2 H3AsO4 - 3 H2O = ?
Are the products in each case metallic or non-metallic oxides?They are called anhydrides. Notice that each is formed by thewithdrawal of water from an acid. Reverse the equations; as, SO3+ H2O = ?
68. An Anhydride is what remains after water has been removed from an acid; or, it is the oxide of a non- metallic element, which, united with water, forms an acid. SO2 is sulphurous anhydride, SO2 sulphuric anhydride, the ending ic meaning more O, or negative element, than ous. Name the others above.
Anhydrides were formerly called acids,—anhydrous acids, in distinction from hydrated ones, as CO2 even now is often called carbonic acid.
Experiment 40.—Hold a piece of wet blue litmus paper in the fumes of SO2, and note the acid test. Try the same with dry litmus paper.
Experiment 41.—Burn a little S in a receiver of air containing 10 cc. H2O, and loosely covered, as in the O experiment. Then shake to dissolve the SO2. H2O + SO2 = H2SO3. Apply test paper.
69. Naming Acids.—Compare formulae H2SO3 and H2SO4. Of two acids having the same elements, the name of the one with least O, or negative element, ends in ous, the other in ic. H2SO3 is sulphurous acid, H2SO4, sulphuric acid. Name H3PO4 and H3PO3; H3AsO3 and H3ASO4; HNO2 and HNO3.
If there are more than two acids in a series, the prefixes hypo, less, and per, more, are used. The following is such a series: HClO, HClO2, HClO3, HClO4.
HClO3 is chloric acid; HClO2, chlorous; HClO, hypochlorous; HClO4 perchloric. Hypo means less of the negative element than ous; per means more of the negative element than ic. Name: H3PO4 (ic), H3PO3, H3PO2. Also HBrO (HBrO2 does not exist), HBrO3 (ic), HBrO4.
What are the three most negative elements? Note their occurrence in the three strongest and most common acids. Hereafter note the names and symbols of all the acids you see.
70. What Bases Are.
Experiment 42.—Put a few drops of NH4OH into an evaporating- dish. Add 5 cc. H2O, and stir. Taste a drop. Dip into it a piece of red litmus paper, noting the effect. Cleanse the dish, and treat in the same way a few drops NaOH solution, recording the result. Do the same with KOH. Acid stains on the clothing, with the exception of those made by HNO3, maybe removed by NH4OH. H2SO4, however, rapidly destroys the fiber of the cloth.
Name two characteristics of a base. In the formulae of those bases, what two common elements? Name the radical. Compare those symbols with the symbol for water, HOH. Is (OH) positive or negative? Is the other part of each formula positive or negative? What are two constituents, then, of a base? Bases are called hydrates. Write in a vertical line five positive elements. Note the valence of each, and complete the formula for its base. Affix the names. Can you see any reason why the three bases above given are the strongest?
Taking the valences of Cr and Fe, write symbols for two sets of hydrates, and name them. Try to recognize and name every base hereafter met with.
A Base is a substance which is composed of a metal, or positive radical, and OH. It generally turns red litmus blue, and often has an acrid taste.
An Alkali is a base which is readily soluble in water. The three principal alkalies are NH4OH, KOH, and NaOH.
Alkali Metals are those which form alkalies. Name three.
An Alkaline Reaction is the turning of red litmus blue.
An Acid Reaction is the turning of blue litmus red.
Experiment 43.—Pour 5 cc. of a solution of litmus in water, into a clean t.t. or small beaker. Pour 2 or 3 cc. of HCl into an evaporating-dish, and the same quantity of NH4OH into another dish. Take a drop of the HCl on a stirring-rod and stir the litmus solution with it. Note the acid reaction. Clean the rod, and with it take a drop (or more if necessary) of NH4OH, and add this to the red litmus solution, noting the alkaline reaction. Experiment in the same way with the two other principal acids and the two other alkalies.
Litmus paper is commonly used to test these reactions, and hereafter whenever the term LITMUS is employed in that sense, the test-paper should be understood. This paper can be prepared by dipping unglazed paper into a strong aqueous solution of litmus.
71. Acids and Bases are usually Opposite in Character.—When two forces act in opposition they tend to neutralize each other. We may see an analogy to this in the union of the two opposite classes of compounds, acids and bases, to form salts.
72. Neutralization.
Experiment 44.—Put into an evaporating-dish 5 cc. of NaOH solution. Add HCl to this from a t.t., a few drops at a time, stirring the mixture with a glass rod (Fig. 20), and testing it with litmus paper, until the liquid is neutral, i.e. will not turn the test paper from blue to red, or red to blue. Test with both colors. If it turns blue to red, too much acid has been added; if red to blue, too much base. When it is very nearly neutral, add the reagent, HCl or NaOH, a drop at a time with the stirring-rod. It must be absolutely neutral to both colors. Evaporate the water by heating the dish over asbestus paper, wire gauze, or sand, in an iron plate (Fig. 21) till the residue becomes dry and white. Cool the residue, taste, and name it. The equation is: HCl + NaOH = NaCl + HOH or H2O. Note which elements, positive or negative, change places. Why was the liquid boiled? The residue is a type of a large class of compounds, called salts.
(Fig. 20) (Fig. 21)
Experiment 45. — Experiment in the same way with KOH solution and H2SO4, applying the same tests. H2SO4 + 2 KOH = K2SO4 + 2 HOH. What is the solid product?
Experiment 46.—Neutralize NH4OH with HNO3, evaporate, apply the tests, and write the equation. Write equations for the combination of NaOH and H2SO4; NaOH and HNO3; KOH and HCl; KOH and HNO3; NH4OH and HCl; NH4OH and H2SO4. Describe the experiment represented by each equation, and be sure you can perform it if asked to do so. What is the usual action of a salt on litmus? How is a salt made? What else is formed at the same time? Have all salts a saline taste? Does every salt contain a positive element or radical? A negative?
73. A Salt is the product of the union of a positive and a negative element or radical; it may be made by mixing a base and an acid.
The salt KI represents what acid? What base, or hydrate? Write the equation for making KI from its acid and base. Describe the experiment in full. Classify, as to acids, bases, or salts: KBr, Fe(OH)2, HI, NaBr, HNO2, Al2(OH)6, KClO3, HClO3, H2S, K2S, H2S03, K2SO4, Ca(OH)2, CaCO3, NaBr03, CaSO4, H2CO3, K2CO3, Cu(OH)2, Cu(NO3)2, PbSO4, H3P04, Na2P04. In the SALTS above, draw a light vertical line, separating the positive from the negative part of the symbol. Now state what acid each represents. What base. Write the reaction in the preparation of each salt above from its acid and base; then state the experiment for producing it.
74. Naming Salts.—(NO3) is the nitrate radical; KNO3 is potassium nitrate. From what acid? (NO2) is the nitrite radical; KN02 is potassium nitrite. From what acid? Note that the endings of the acids are OUS and IC; also that the names of their salts end in ITE and ATE. From which acid—IC or OUS—is the salt ending in ATE derived? That ending in ITE?
Name these salts, the acids from which they are derived, and the endings of both acids and salts: NaNO3, NaNO2, K2SO4, K2SO3, CaSO4, CaSO3, KClO3, KClO2, KClO, KClO4 (use prefixes HYPO and PER, as with acids), Ca3(PO4)2, Ca3(P03)2, CuSO4, CuSO3, AgNO3, Cu(NO3)2. FeS, FeS2, are respectively FERROUS SULPHIDE and FERRIC SULPHIDE. Name: HgCl, HgCl2, FeCl2, Fe2Cl6, FeSO4, Fe2(SO4)3.75. Acid Salts.—Write symbols for nitric, sulphuric, phosphoric acids. How many H atoms in each? Replace all the H in the symbol of each with Na, and name the products. Again, in sulphuric acid replace one atom of H with Na; then in phosphoric replace first one, then two, and finally three H atoms with Na. HNaSO4 is hydrogen sodium sulphate; HNa2P04 is hydrogen di-sodium phosphate. Name the other salts symbolized. Name HNaNH4P04. Though these products are all salts, some contain replaceable H, and are called acid salts. Those which have all the H replaced by a metal are normal salts. Name and classify, as to normal or acid salts: Na2CO3, HNaCO3, K2SO4, HKSO4, (NH4)2SO4, HNH4SO4, Na3P04, HNa2P04, H2NaP04.
The BASICITY of an acid is determined by the number of replaceable H atoms in its molecule. It is called MONOBASIC if it has one; DIBASIC if two; TRI- if three, etc. Note the basicity of each acid named above. How many possible salts of H2SO4 with Na? Of H3P04 with Na? Which are normal and which acid? What is the basicity of H4Si04?
Some normal, as well as acid, salts change litmus. Na2CO3, representing a strong base and a weak acid, turns it blue. There are other modes of obtaining salts, but this is the only one which we sball consider.
76. Salts Occur Abundantly in Nature, such as NaCl, MgSO4, CaCO3. Acids and bases are found in small quantities only. Why is this? Why are there not springs of H2SO4 and NH4OH? We have seen that acids and bases are extremely active, have opposite characters, and combine to form relatively inactive salts. If they existed in the free state, they would soon combine by reason of their strong affinities. This is what in all ages of the world has taken place, and this is why salts are common, acids and bases rare. Active agents rarely exist in the free state in large quantities. Oxygen seems to be an exception, but this is because there is a superabundance of it. While vast quantities are locked up in compounds in rocks, water, and salts of the earth, much remains with which there is nothing to combine.
77. We have seen that salts are made by the union of acids and bases. Can these last be obtained from salts?
78. Preparation of HCl.
Experiment 47.—Into a flask put 10 g. coarse NaCl, and add 20 cc. H2SO4. Connect with Woulff bottles [Woulff bottles may be made by fitting to wide-mouthed bottles corks with three holes, through which pass two delivery tubes, and a central safety tube dipping into the liquid, as in Figures 22 and 23.] partly filled with water, as in Figure 22. One bottle is enough to collect the HCl; but in that case it is less pure, since some H2SO4 and other impurities are carried over. Several may be connected, as in Figure 23. The water in the first bottle must be nearly saturated before much gas will pass into the second. Heat the mixture 15 or 20 minutes, not very strongly, to prevent too much foaming. Notice any current in the first bottle. NaCl + H2SO4 = HNaSO4 + HCl. Intense heat would have given: 2NaCl + H2SO4 = Na2SO4 + 2HCl. Compare these equations with those for HNO3. In which equation above is H2SO4 used most economically? Both reactions take place when HCl is made on the large scale.
(Fig. 22)
79. Tests. Experiment 48.—(1) Test with litmus the liquid in each Woulffbottle. (2) Put a piece of Zn into a t.t. and cover it with liquid from the first bottle. Write the reaction, and test the gas. (3) To 2 cc.solution AgNO3 in a t.t. add 2 cc.of the acid. Describe, and write the reaction. Is AgCl soluble in water? (4) Into a t.t. pour 5 cc.Pb(NO3)2 solution, and add the same amount of prepared acid. Give the description and the reaction. (5) In the same way test the acid with Hg2(NO3)2 solution, giving the reaction. (6) Drake a little HCl in a t.t., and bring the gas escaping from the d.t. in contact with a burning stick. Does it support the combustion of C? (7) Hold a piece of dry litmus paper against it. [figure 23] (8) Hold it over 2 cc.of NH4OH in an evaporating-dish. Describe, name the product, and write the reaction. (3), (4), (5), (8), are characteristic tests for this acid.
80. Chlorhydric, Hydrochloric or Muriatic, Acid is a Gas.—As used, it is dissolved, in water, for which it has great affinity. Water will hold, according to temperature, from 400 to 500 times its volume of HCl. Hundreds of thousands of tons of the acid are annually made, mostly in Europe, as a bye-product in Na2CO3 manufacture. The gas is passed into towers through which a spray of water falls; this absorbs it. The yellow color in most commercial HCl indicates impurities, some of which are Fe, S, As, and organic matter. As, S, etc., come from the pyrites used in making H2SO4. Chemically pure (C.P.) acid is freed from these, and is without color. The gas may be dried by passing it through a glass tube holding CaCl2 (Fig. 16) and collecting it over mercury.
The muriatic acid of commerce consists of about two- thirds water by weight. HCl can also be made by direct union of its constituents.81. Uses.—HCl is used to make Cl, and also bleaching- powder. Its use as a reagent in the laboratory is illustrated by the following experiment:— Experiment 49.—Put into a t.t. 2 cc. AgNO3 solution, add 5 cc. H2O, then add slowly HCl so long as a ppt. (precipitate) is formed. This ppt. is AgCl. Now in another t.t. put 2 cc. Cu(NO3)2, solution, add 5 cc. H2O, then a little HCl. No ppt. is formed. Now if a solution of AgNO3 and a solution of Cu(NO3)2 were mixed, and HCl added, it is evident that the silver would be precipitated as chloride of silver, while the copper would remain in solution. If now this be filtered, the silver will remain on the filter paper, while in the filtrate will be the copper. Thus we shall have performed an analysis, or separated one metal from another. Perform it. Note, however, that any soluble chloride, as NaCl, would produce the same result as HCl.
82. NaCl, being the most abundant compound of Cl, is the source of commercial HCl. KCl treated in the same way would give a like product. Theoretically HBr and HI might be made in the same way from NaBr and NaI, but the affinity of H for Br and I is weak, and the acids separate into their elements, when thus prepared.
83. To make HI.
Experiment 50.—Drop into a t.t. three or four crystals of I, and add 10 cc. H2O. Hold in the water the end of a d.t. from which H2S gas is escaping. Observe any deposit, and write the reaction.
84. Preparation and Action.
Experiment 51.—Put 3 or 4 g. powdered CaF2, i.e. fluor spar or fluorite, into a shallow lead tray, e.g. 4x5 cm, and pour over it 4 or 5 cc. H2SO4. A piece of glass large enough to cover this should previously be warmed and covered on one side with a very thin coat of beeswax. To distribute itevenly, warm the other side of the glass over a flame. When cool, scratch a design (Fig. 24) through the wax with a sharp metallic point. Lay the glass, film side down, over the lead tray. Warm this five minutes or more by placing it high over a small flame (Fig. 25) to avoid melting the wax. Do not inhale the fumes. Take away the lamp, and leave the tray and glass where it is not cold, for half an hour or more. Then remove the wax and clean the glass with naphtha or benzine. Look for the etching.
Two things should have occurred: (1) the generation of HF. Write the equation for it. (2) Its etching action on glass. In this last process HF acts on SiO2 of the glass, forming H2O and SiF4. Why cannot HF be kept in glass bottles?
A dilute solution of HF, which is a gas, may be kept in gutta percha bottles, the anhydrous acid in platinum only; but for the most part, it is used as soon as made, its chief use being to etch designs on glass-ware. Glass is also often etched by a blast of sand (SiO2).
Notice the absence of O in the acids HF, HCI, HBr, HI, and that each is a gas. HF is the only acid that will dissolve or act appreciably on glass.
85. Preparation. Experiment 52.—To 10 g. KNO3 or NaNO3, in a flask, add 15 cc. H2SO4. Securely fasten the cork of the d.t., as HNO3 is likely to loosen it, and pass the other end to the bottom of a t.t. held deep in a bottle of water (Fig. 26). Apply heat, and collect 4 or 5 cc.of the liquid. The usual reaction is: KNO3 + H2SO4 = HKSO4 + HNO3. With greater heat, 2 KNO3 + H2SO4 = K2SO4 + 2HNO3. Which is most economical of KNO3? Of H2SO4? Instead of a flask, a t.t. may be used if desired (Fig. 27).
86. Properties and Tests.
Experiment 53.—(1) Note the color of the prepared liquid. (2) Put a drop on the finger; then wash it off at once. (3) Dip a quill or piece of white silk into it; then wash off the acid. What color is imparted to animal substances? (4) Add a little to a few bits of Cu turnings, or to a Cu coin. Write the equation. (5) To 2 cc.indigo solution, add 2 cc. HNO3. State the leading properties of HNO3, from these tests.
87. Chemically Pure HNO3 is a Colorless Liquid.— The yellow color of that prepared in Experiment 52 is due to liquid NO2 dissolved in it. It is then called fuming HNO3, and is very strong. NO2 is formed at a high temperature.
Commercial or ordinary HNO3, is made from NaNO3, this being cheaper than KNO3; it is about half water.
88. Uses. HNO3 is the basis of many nitrates, as AgNO3, used for photography, Ba(NO3)2 and Sr(NO3)2 for fire-works, and others for dyeing and printing calico; it is employed in making aqua regia, sulphuric acid, nitro-glycerine, gun-cotton, aniline colors, zylonite, etc.
Enough experiments have been performed to answer the question whether some acids can be prepared from their salts. H2SO4 is not so made, because no acid is strong enough to act on its salts. In making HCl, HNO3, etc., sulphuric acid was used, being the strongest.
89. Preparation and Action. Experiment 54.—Into a t.t. put 2 cc. HNO3, and 14 qcm. of either Au leaf or Pt. Warm in a flame. If the metal is pure, no action takes place. Into another tube put 6 cc. HCl and add a similar leaf. Heat this also. There should be no action. Pour the contents of one t.t. into the other. Note the effect. Which is stronger, one of the acids, or the combination of the two? Note the odor. It is that of Cl. 3HCl + HNO3 = NOCl + 2H2O + Cl2. This reaction is approximate only. The strength is owing to nascent chlorine, which unites with Au. Au + 3Cl = AuCl3. If Pt be used, PtCl4 is produced. No other acid except nitro-hydrochloric will dissolve Au or Pt; hence the ancients called it aqua regia, or king of liquids. It must be made as wanted, since it cannot be kept and retain its strength.
90. Preparation.
Experiment 55.—Having fitted a cork with four or five perforations to a large t.t., pass a d.t. from three of these to three smaller t.t., leaving the others open to the air, as in Figure 28. Into one t.t. put 5 cc. H2O, into another 5 g. Cu turnings and 10 cc. H2SO4, into the third 5 g. Cu turnings and 10 cc. dilute HNO3, half water. Hang on a ring stand, and slowly heat the tubes containing H2O and H2SO4. Notice the fumes that pass into the large t.t.
Trace out and apply to Figure 28 these reactions:—
(1) Cu + 2 H2SO4 = CuSO4 + 2 H2O + SO2.
(2) 3 Cu + 8 HNO3 = 3 Cu(NO3)2+ 4 H2O + 2 NO.
(4) comes from combining the gaseous products in (1), (2), (3). In (3), NO takes an atom of O from the air, becoming NO2, and at once gives it up, to the H2SO3 (H2O + SO2), making H2SO4, and again goes through the same operation of taking up O and passing it along. NO is thus called a carrier of O. It is a reducing agent, while NO2 is an oxidizing agent. This is a continuous process, and very important, since it changes useless H2SO3 into valuable H2SO4. If exposed to the air, H2SO3 would very slowly take up O and become H2SO4.
Instead of the last experiment, this may be employed if preferred: Burn a little S in a receiver. Put into an evaporating-dish, 5 cc. HNO3, and dip a paper or piece of cloth into it. Hang the paper in the receiver of SO2, letting no HNO3 drop from it. Continue this operation till a small quantity of liquid is found in the bottle. The fumes show that HNO3 has lost O. 2 HNO3 + SO2 = H2SO4 + 2 NO2.
91. Tests for H2SO4.
Experiment 56.—(1) Test the liquid with litmus. (2) Transfer it to a t.t., and add an equal volume of BaCl2 solution. H2SO4 + BaCl2 = ? Is BaSO4 soluble? (3) Put one drop H2SO4 from the reagent bottle in 10 cc. H2O in a clean t.t., and add 1 cc. BaCl2 solution. Look for any cloudiness. This is the characteristic test for H2SO4 and soluble sulphates, and so delicate that one drop in a liter of H2O can be detected. (4) Instead of H2SO4, try a little Na2SO4 solution. (5) Put two or three drops of strong H2SO4 on writing-paper, and evaporate, high over a flame, so as not to burn the paper. Examine it when dry. (6) Put a stick into a t.t. containing 2 cc. H2SO4, and note the effect. (7) Review Experiment 5. (8) Into an e.d. pour 5 cc. H2O, and then 15 cc. H2SO4. Stir it meantime with a small t.t. containing 2 or 3 cc. NH4OH, and notice what takes place in the latter; also note the heat of the e.d.
The effects of (5), (6), (7), and (8) are due to the intense affinity which H2SO4 has for H2O. So thirsty is it that it even abstracts H and O from oxalic acid in the right proportion to form H2O, combines them, and then absorbs the water.
92. Affinity for Water.—This acid is a desiccator or dryer, and is used to take moisture from the air and prevent metallic substances from rusting. In this way it dilutes itself, and may increase its weight threefold. In diluting, the acid must always be poured into the water slowly and with stirring, not water into the acid, since, as H2O is lighter than H2SO4, heat enough may be set free at the surface of contact to cause an explosion. Contraction also takes place, as may be shown by accurately measuring each liquid in a graduate, before mixing, and again when cold. The mixture occupies less volume than the sum of the two volumes. For the best results the volume of the acid should be about three times that of the water.
93. Sulphuric Acid made on a Large Scale involves the same principles as shown in Experiment 55, excepting that S02 is obtained by burning S or roasting FeS2 (pyrite),
[Fig. 29.]
and HNO3 is made on the spot from NaNO3 and H2SO4. SO2 enters a large leaden chamber, often 100 to 300 feet long, and jets of steam and small portions of HNO3 are also forced in. The "chamber acid" thus formed is very dilute, and must be evaporated first in leaden pans, and finally in glass or platinum retorts, since strong H2SO4, especially if hot, dissolves lead. See Experiment 124. Study Figure 29, and write the reactions. 2 HNO3 breaks up into 2 NO2, H2O, and O. 94. Importance.—Sulphuric acid has been called, next to human food, the most indispensable article known. There is hardly a product of modern civilization in the manufacture of which it is not directly or indirectly used. Nearly a million tons are made yearly in Great Britain alone. It is the basis of all acids, as Na2CO3 is of alkalies. It is the life of chemical industry, and the quantity of it consumed is an index of a people's civilization. Only a few of its uses can be stated here. The two leading ones are the reduction of Ca3(PO4)2 for artificial manures and the sodium carbonate manufacture. Foods depend on the productiveness of soils and on fertilizers, and thus indirectly our daily bread is supplied by means of this acid; and from sodium carbonate glass, soap, saleratus, baking- powders, and most alkalies are made directly or indirectly. H2SO4 is employed in bleaching, dyeing, printing, telegraphy, electroplating, galvanizing iron and wire, cleaning metals, refining Au and Ag, making alum, blacking, vitriols, glucose, mineral waters, ether, indigo, madder, nitroglycerine, gun- cotton, parchment, celluloid, etc., etc.
95. Nordhausen or Fuming Sulphuric Acid, H2S207 used in dissolving indigo and preparing coal-tar pigments, is made by distilling FeSO4. 4FeSO4 + H2O = H2S207 + 2Fe203 + 2S02. This was the original sulphuric acid. It is also formed when S03 is dissolved in H2SO4. When exposed to the air, S03 escapes with fuming.
96. Preparation of Bases.—We have seen that many acids are made by acting on a salt of the acid required, with a stronger acid. This is the direct way. The following experiments will show that bases may be prepared in a similar way by acting on salts of the base required with other bases, which we may regard as stronger than the ones to be obtained.
97. Preparation of NH4OH and NH3.
Experiment 57.—Powder 10 g. ammonium chloride, NH4Cl, in a mortar and mix with 10 g. calcium hydrate, Ca(OH)2; recently slaked lime is the best. Cover with water in a flask, and connect with Woulff bottles, as for making HCl (Fig. 22); heat the flask for fifteen minutes or more. The experiment may be tried on a smaller scale with a t.t. if desired.
The reaction is: 2NH4Cl + Ca(OH)2 = CaCl2 + 2NH4OH. NH4OH is broken up into NH3, ammonia gas, and water. NH4OH = NH3 + H2O. These pass over into the first bottle, where the water takes up the NH3, for which it has great affinity. One volume of water at 0° will absorb more than 1000 volumes of NH3. Thus NH4OH may be called a solution of NH3, in H2O. Write the reaction.
Experiment 58.—Powder and mix 2 or 3 g. each of ammonium nitrate, NH4NO3, and Ca(OH)2; put them into a t.t., and heat slowly. Note the odor. 2NH4NO3 + Ca(OH)2 = ?
98. Tests.
Experiment 59.—(1) Generate a little of the gas in a t.t., and note the odor. (2) Test the gas with wet red litmus paper. (3) Put a little HCl into an e.d., and pass over it the fumes of NH3 from a d.t. Note the result, and write the equation. (4) Fill a small t.t. with the gas by upward displacement; then, while still inverted, put the mouth of the t.t. into water. Explain the rise of the water. (5) How might NH4Cl be obtained from the NH4OH in the Woulff bottles? (6) Test the liquid in each bottle with red litmus paper. (7) Add some from the first bottle to 5 or 10 cc. of a solution of FeSO4 or FeCl2, and look for a ppt. State the reaction.
99. Formation.—Ammonia, hartshorn, exists in animal and vegetable compounds, in salts, and, in small quantities, in the atmosphere. Rain washes it from the atmosphere into the soil; plants take it from the soil; animals extract it from plants. Coal, bones, horns, etc., are the chief sources of it, and from them it is obtained by distillation. It results also from decomposing animal matter. NH3 can be produced by the direct union of N and H, only by an electric discharge or by ozone. It may be collected over Hg like other gases that are very soluble in water.
100. Uses. —Ammonium hydrate, NH4OH, and ammonia, NH3, are used in chemical operations, in making artificial ice, and to some extent in medicine; from them also may be obtained ammonium salts. State what you would put with NH4OH to obtain (NH4)2SO4. To obtain NH4NO3. The use of NH4OH in the laboratory may be illustrated by the following experiment:—
Experiment 60.—Into a t.t. put 10 cc. of a solution of ferrous sulphate, FeSO4. Into another put 10 cc. of sodium sulphate solution, Na2SO4. Add a little NH4OH to each. Notice a ppt. in the one case but none in the other. If solutions of these two compounds were mixed, the metals Fe and Na could be separated by the addition of NH4OH, similar to the separation of Ag and Cu by HCl. Try the experiment.
101. Preparation.
Experiment 61.—Dissolve 3 g. sodium carbonate, Na2CO3, in 10 or 15 cc. H2O in an e.d., and bring it to the boiling-point. Then add to this a mixture of 1 or 2 g. calcium hydrate, Ca(OH)2, in 5 or 10cc. H2O. It will not dissolve. Boil the whole for five minutes. Then pour off the liquid which holds NaOH in solution. Evaporate if desired. This is the usual mode of preparing NaOH.
The reaction is Na2CO3 + Ca(OH)2 = 2NaOH + CaCO3. The residue is Ca(OH)2 and CaCO3; the solution contains NaOH, which can be solidified by evaporating the water. Sodium hydrate is an ingredient in the manufacture of hard soap, and for this use thousands of tons are made annually, mostly in Europe. It is an important laboratory reagent, its use being similar to that of ammonium hydrate. Exposed to the air, it takes up water and CO2, forming a mixture of NaOH and Na2CO3. It is one of the strongest alkalies, and corrodes the skin.
Experiment 62.—Put 20 cc. of H2O in a receiver. With the forceps take a piece of Na, not larger than half a pea, from the naphtha in which it is kept, drop it into the H2O, and at once cover the receiver loosely with paper or cardboard. Watch the action, as the Na decomposes H2O. HOH + Na = NaOH + H. If the water be hot the action is so rapid that enough heat is produced to set the H on fire. That the gas is H can be shown by putting the Na under the mouth of a small inverted t.t., filled with cold water, in a water-pan. Na rises to the top, and the t.t. fills with H, which can be tested. NaOH dissolves in the water.102. Properties.
Experiment 63.—(1) Test with red litmus paper the solutions obtained in the last two experiments. (2) To 5cc.of alum solution, K2A12(SO4)4, add 2cc.of the liquid, and notice the color and form of the ppt.
103. KOH is made in the Same Way as NaOH.
Describe the process in full (Experiment 61), and give the equation.
Experiment 64.—Drop a small piece of K into a receiver of H2O, as in Experiment 62. The K must be very small, and the experiment should not be watched at too close a range. The receiver should not be covered with glass, but with paper. The H burns, uniting with O of the air. The purple color is imparted by the burning, or oxidation of small particles of K. Write the equation for the combustion of each.
H2O might be considered the symbol of an acid, since it is the union of H and a negative element; or, if written HOH, it might be called a base, since it has a positive element and the (OH) radical. It is neutral to litmus, and on this account might be called a salt. It is better, however, to call it simply an oxide.
Potassium hydrate, caustic potash, is employed for the manufacture of soft soap. As a chemical reagent its action is almost precisely like that of caustic soda, though it is usually considered a stronger base, as K is a more electro-positive element than Na.
104. Calcium Hydrate, the Most Common of the Bases, is nearly as important to them as H2SO4 is to acids. Since it is used to make the other bases, it might be called the strongest base; as H2SO4 is often called the strongest acid. The strength of an acid or base, however,depends on the substance to which it is applied, as well as on itself, and for most purposes this one is classified as a weaker base than the three previously described.
Sulphuric acid, the most useful of the acids, is not made directly from its salts, but has to be synthesized. Calcium hydrate is also made by an indirect process, as follows:
CaCO3, i.e. limestone, marble, etc., is burnt in kilns with C, a process which separates the gas, CO2, according to the reaction: CaCO3 = CaO + CO2. CaO is unslaked lime, or quick-lime. On treating this with water, slaked lime, Ca(OH)2 is formed, with generation of great heat. CaO + H2O = Ca(OH)2. Its affinity for H2O is so great that it takes the latter from the air, if exposed.
Experiment 65.—Saturate some unslaked lime with water, in an e.d., and look for the results stated above, leaving it as long as may be necessary.
105. Resume.—From the experiments in the last few chapters on the three divisions of chemical compounds, acids, bases and salts, we have seen (1) that acids and bases are the chemical opposites of each other; (2) that salts are formed by the union of acids and bases; (3) that some acids can be obtained from their salts by the action of a stronger acid; (4) that some bases can be got from salts by the similar action of other bases; (5) that the strongest acids and bases, as well as others, may be obtained in an indirect way by synthesis.
106. There are five oxides of N, only two of which are important.
107. Preparation.
Experiment 66.—Put into a flask, holding 200cc, lOg of ammonium nitrate, NH4NO3; heat it over wire gauze or asbestus in an iron plate, having a d.t. connected with a large t.t., which is held in a receiver of water, and from this t.t., another d.t. passing into a pneumatic trough, so as to collect the gas over water (Fig. 30). Have all the bearings tight. The reaction is NH4NO3 = 2H2O + N2O. The t.t. is for collecting the H2O.
[Fig. 30.]
Note the color of the liquid in the t.t.; taste a drop, and test it with litmus. If the flask is heated too fast, some NO is formed, and this taking O from the air makes NO2, which liquefies and gives an acid reaction and a red color. Some NH4NO3 is also liable to be carried over.
108. Properties.
Experiment 67.—Test the gas in the receiver with a burning stick and a glowing one, and compare the combustion with that in O. N20may also be tested with S and P, if desired. N is set free in each case. Write the reactions.
Nitrogen monoxide or protoxide, the nitrous oxide of dentists, when inhaled, produces insensibility to pain,— anaesthesia,— and, if continued, death from suffocation. Birds die in half a minute from breathing it. Mixed with one-fourth O, and inhaled for a minute or two, it produces intoxication and laughter, and hence is called laughing gas. As made in Experiment 66, it contains Cl and NO, as impurities, and should not be breathed.
109. Preparation.
Experiment 68.—Into a t.t. or receiver put 5g Cu turnings, add 5 cc. H2O and 5 cc. HNO3. Collect the gas like H, over water. 3Cu + 8HNO3 = ? What two products will be left in the generator? Notice the color of the liquid. This color is characteristic of Cu salts. Notice also the red fumes of NO2.
110. Properties.
Experiment 69.—Test the gas with a burning stick, admitting as little air as possible. Test it with burning S. NO is not a supporter of C and S combustion. Put a small bit of P in a deflagrating-spoon, and when it is vigorously burning, lower it into the gas. It should continue to burn. State the reaction. What combustion will NO support? Note that NO is half N, while N2O is two-thirds N, and account for the difference in supporting combustion.
NITROGEN TETROXIDE (NO2 or N2O4).
111. Preparation.
Experiment 70.—Lift from the water-pan a receiver of NO, and note the colored fumes. They are NO2, or N2O4, nitrogen tetroxide. NO + O = NO2. Is NO combustible? What is the source of O in the experiment?OXIDES OF NITROGEN.
112. Preparation.
Experiment 71.—Put into a t.t. 1 g. of starch and 1 cc. of HNO3.Heat the mixture for a minute. The red fumes are N2O3 and NO2.
Nitrogen pentoxide, N2O5, is an unimportant solid. United with water it forms HNO3. N2O5 + H2O = 2HNO3.
113. Weight and Volume.—We have seen that water contains two parts of H by volume to one part of O; or, by weight, two parts of H to sixteen of O. These proportions are invariable, or no symbol for water would be possible. Every compound in the same way has an unvarying proportion of elements.
114. Law of Definite Proportion.—In a given compound the proportion of any element by weight, or, if a gas, by volume is always constant. Apply the law, by weight and by volume, to these: HCl, NH3, H2S, N2O.
There is another law of equal importance in chemistry, which the compounds of N and O well illustrate.
Weight. Volume.N. O. N. O.Nitrogen protoxide N2O 28 16 2 1Nitrogen dioxide N2O2 28 32 2 2Nitrogen trioxide. N2O3 28 48 2 3Nitrogen tetroxide N2O4 28 64 2 4Nitrogen pentoxide N2O5 28 80 2 5
Note that the proportion of O by weight is in each case a multiple of the first, 16. Also that the proportion by volume of O is a multiple of that in the first compound. In this example the N remains the same. If that had varied in the different compounds, it would also havevaried by a multiple of the smallest proportion. This is true in all compounds.
115. Law of Multiple Proportion.—Whenever one element combines with another in more than one proportion, it always combines in some multiple, one or more, of its least combining weight, or, if a gas, of its least combining volume.
The least combining weight of an element is its atomic weight; and it is this fact of a least combining weight that leads us to believe the atom to be indivisible.
Apply the law in the case of P2O, P2O3, P2O5; in HClO, HClO2, HClO3, HClO4, arranging the symbols, weights, and volumes in a table, as above.
The volumetric proportions of each element in the oxides of nitrogen are exhibited below.
++ _ = __ N + N + O = N2O
+++= __ N + N + O + O = N2O2
++++ _ = __ N + N + O + O + O = N2O3
+++++= __ N + N + O + O + O + O = N2O4
++++++ _ = __ N + N + O + O + O + O + O = N2O5
116. Preparation.
Experiment 72.—Put into a flask, of 200 cc., 5 g. of oxalic acid crystals, H2C2O4, and 25 cc. H2SO4. Have the d.t. pass into a solution of NaOH in a Woulff bottle (Fig. 31), and collect the gas over water. Heat the flask slowly, and avoid inhaling the gas.
117. Tests.
Experiment 73.—Remove a receiver of the gas, and try to light the latter with a splinter. Is it combustible, or a supporter of (C) combustion? What is the color of the flame? When the combustion ceases, shake up a little lime water with the gas left in the receiver. What gas has been formed by the combustion, as shown by the test? See page 80. Give the reaction for the combustion.