A substance containing 2.04% H, 32.6% S, and 65.3% O was found to have a molecular weight of 98. In these 98 parts of the substance there are 98 × 2.04% = 2 parts of H, 98 × 32.6% = 32 parts of S, and 98 × 65.3% = 64 parts of O. If the molecule weighs 98, the hydrogen atoms present must together weigh 2, the sulphur atoms 32, and the oxygen atoms 64. Dividing these figures by the respective atomic weights of the three elements, we have, for H, 2 ÷ 1 = 2 atoms; for S, 32 ÷ 32 = 1 atom; for O, 64 ÷ 16 = 4 atoms. Hence the formula is H2SO4.
We have, then, this general procedure: Find the percentage composition of the substance and also its molecular weight. Multiply the molecular weight successively by the percentage of each element present, to find the amount of the element in the molecular weight of the compound. The figures so obtained will be the respective parts of the molecular weight due to the several atoms. Divide by the atomic weights of the respective elements, and the quotient will be the number of atoms present.
Avogadro's hypothesis and chemical calculations. This law simplifies many chemical calculations.
1. Application to volume relations in gaseous reactions. Since equal volumes of gases contain an equal number of molecules, it follows that when an equal number of gaseous molecules of two or more gases take part in a reaction, the reaction will involve equal volumes of the gases. In the equation
C2H2O4 = H2O + CO2 + CO,
since 1 molecule of each of the gases CO2 and CO is set free from each molecule of oxalic acid, the two substances must always be set free in equal volumes.
Acetylene burns in accordance with the equation
2C2H2 + 5O2 = 4CO2 + 2H2O.
Hence 2 volumes of acetylene will react with 5 volumes of oxygen to form 4 volumes of carbon dioxide and 2 volumes of steam. That the volume relations may be correct a gaseous element must be given its molecular formula. Thus oxygen must be written O2 and not 2O.
2. Application to weights of gases. It will be recalled that the molecular weight of a gas is determined by ascertaining the weight of 22.4 l. of the gas. This weight in grams is called the gram-molecular weight of a gas. If the molecular weight of any gas is known, the weight of a liter of the gas under standard conditions may be determined by dividing its gram-molecular weight by 22.4. Thus the gram-molecular weight of a hydrochloric acid gas is 36.458. A liter of the gas will therefore weigh 36.458 ÷ 22.4 = 1.627 g.