Law of Multiple Proportions

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The law of multiple proportions says: If two elements can form different chemical compounds , the masses of one element, which combine with a constant mass of the other element, are in the ratio of small whole numbers . John Dalton formulated the law in 1808, based on the law of constant proportions . It supported Dalton's atomic hypothesis, which was very controversial at the time .

Examples

Sulfur oxides

If, under certain conditions, sulfur is allowed to react with oxygen , 32 grams of sulfur (due to the molar mass of 32 g / mol corresponding to 1 mol) are also needed for 32 grams of oxygen (due to the molar mass of 16 g / mol corresponding to 2 mol), so both without a remaining excess of one of the reactants react completely with one another. To make another possible combination of sulfur and oxygen, you need 48 grams of oxygen for 32 grams of sulfur. The masses of oxygen, which in both cases react with 32 grams of sulfur each, behave as 32:48 or, in short, as 2: 3, while the stoichiometric ratio in the second reaction is 1: 3. The first compound is sulfur dioxide SO 2 , the second compound is sulfur trioxide SO 3 .

This observation could easily be explained with Dalton's atomic hypothesis, according to which chemical elements consist of identical atoms, while chemical compounds are the combination of the atoms of two or more elements in certain numerical proportions. Apparently the second compound mentioned contains one and a half times as many oxygen atoms as the first, so that the mathematically possible formulas for the resulting compounds can be derived from the observation described

  • S x O 2 and S x O 3 or
  • S x O 4 and S x O 6
  • and so on

narrow down.

Hydrogen oxides

To produce water in a chemical reaction , you need two grams of hydrogen for every 16 grams of oxygen . When producing another substance, you find that one gram of hydrogen is required for every 16 grams of oxygen. Assuming that the formula H 2 O for water is known, this information alone limits the possible formulas for the unknown substance to HO or H 2 O 2 . A closer examination shows that it is hydrogen peroxide with the formula H 2 O 2 .

Nitrogen oxides

In the five nitrogen oxides N 2 O, NO, N 2 O 3 , NO 2 and N 2 O 5, there are 16, 32, 48, 64 and 80 grams of oxygen for every 28 grams of nitrogen (N 2 ). These oxygen masses are in the ratio 16: 32: 48: 64: 80 or, in short, 1: 2: 3: 4: 5.

See also

Web links

literature

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Individual evidence

  1. a b L. Pauling: General Chemistry. Dover Publications, New York 1988, ISBN 0-486-65622-5 , p. 18.