Chapter 53

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?

PROBLEMS.

(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.