Law of Multiple Proportions Example Problem

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Helmenstine, Anne Marie, Ph.D. "Law of Multiple Proportions Example Problem." ThoughtCo, Mar. 24, 2017, thoughtco.com/law-of-multiple-proportions-problem-609564. Helmenstine, Anne Marie, Ph.D. (2017, March 24). Law of Multiple Proportions Example Problem. Retrieved from https://www.thoughtco.com/law-of-multiple-proportions-problem-609564 Helmenstine, Anne Marie, Ph.D. "Law of Multiple Proportions Example Problem." ThoughtCo. https://www.thoughtco.com/law-of-multiple-proportions-problem-609564 (accessed October 18, 2017).
The law of multiple proportions states the ratio of types of atoms in a compound is fixed and can be broken into whole numbers.
The law of multiple proportions states the ratio of types of atoms in a compound is fixed and can be broken into whole numbers. JGI/Tom Grill / Getty Images

This is a worked example chemistry problem using the Law of Multiple Proportions.

Example Law of Multiple Proportions Problem

Two different compounds are formed by the elements carbon and oxygen. The first compound contains 42.9% by mass carbon and 57.1% by mass oxygen. The second compound contains 27.3% by mass carbon and 72.7% by mass oxygen. Show that the data are consistent with the Law of Multiple Proportions.

Solution

The Law of Multiple Proportions is the third postulate of Dalton's atomic theory. It states that the masses of one element which combine with a fixed mass of the second element are in a ratio of whole numbers.

Therefore, the masses of oxygen in the two compounds that combine with a fixed mass of carbon should be in a whole-number ratio. In 100 g of the first compound (100 is chosen to make calculations easier) there are 57.1 g O and 42.9 g C. The mass of O per gram C is:

57.1 g O / 42.9 g C = 1.33 g O per g C

In the 100 g of the second compound, there are 72.7 g O and 27.3 g C. The mass of oxygen per gram of carbon is:

72.7 g O / 27.3 g C = 2.66 g O per g C

Dividing the mass O per g C of the second (larger value) compound:

2.66 / 1.33 = 2

Which mean that the masses of oxygen that combine with carbon are in a 2:1 ratio. The whole-number ratio is consistent with the Law of Multiple Proportions.

Tips for Solving Law of Multiple Proportions Problems

  • While the ratio in this example problem worked out to be exactly 2:1, it's more likely chemistry problems and real data will give you ratios that are close, but not whole numbers. If you ratio came out like 2.1:0.9, then you'd know to round to the nearest whole number and work from there. If you got a ratio more like 2.5:0.5, then you could be pretty certain you had the ratio wrong (or your experimental data was spectacularly bad, which happens too). While 2:1 or 3:2 ratios are most common, you could get 7:5, for example, or other unusual combinations.
  • The law works the same way when you work with compounds containing more than two elements. To make the calculation simple, choose a 100-gram sample (so you're dealing with percentages), and then divide the largest mass by the smallest mass. This isn't critically important -- you can work with any of the numbers -- but it helps to establish a pattern for solving this type of problem.
  • The ratio won't always be obvious! It takes practice to recognize ratios.
  • In the real world, the law of multiple proportions doesn't always hold. The bonds formed between atoms are more complex than what you learn about in a 101 chemistry class. Sometimes whole number ratios don't apply. In a classroom setting, you need to get whole numbers, but remember there may come a time when you'll get a pesky 0.5 in there (and it will be correct)!