Law of Definite Proportions - Online Tutor, Practice Problems & Exam Prep

Law of Definite Proportions

Let's examine the law of definite proportions. In 1799, the French chemist Joseph L. Proust originated the law of definite proportions. Because of its immense contributions to it, it's sometimes referred to simply as Proust's law. But it also goes by the law of constant composition. The law itself uses what we call mass ratios. These mass ratios are fractions or proportions of elements by mass. The whole concept behind the law of definite proportions is that different samples of a pure chemical compound always contain the same proportions of elements by mass. So let's say, for example, that I take a sample from New York City and I suspect that it's CO₂, and I take a sample in another city, London, England, and I suspect that's CO₂. If we're following the law of definite proportions, both samples should have the same mass ratio. Now for the mass ratio, we're going to place the element with the larger mass on top.

Now, going back to our whole idea of CO₂ in two different cities. Remember, CO₂ itself is composed of 1 carbon and 2 oxygens. And we know that if we look at the periodic table, the atomic mass of carbon is 12.01 grams per mole, and the mass of oxygen is around 16 grams per mole. When you multiply the number of each element by its atomic mass, we'll see how much of it contributes to the overall mass of CO₂. So carbon itself contributes 12.01 grams total, and oxygen contributes 32 grams total. Now using the mass ratio, we place the larger mass on top, which is the 32 grams of oxygen, and the smaller mass on the bottom. When we divide those two numbers, it gives me approximately 2.66. What that 2.66 is telling me is that we have 2.66 oxygen for every one carbon. This would be our mass ratio. And if I were examining two samples of CO₂, one from New York City and one from London, if they both were indeed CO₂, they both should give me back the same exact mass ratio. That's what the law of definite proportion hinges on. If we know the mass ratio of a given sample, and we're examining it against an unknown sample, I use this law to determine if they're the same sample. Alright. Now we get the idea behind the law of definite proportions, so let's move on. Let's talk about calculations and different types of problems associated with the law of definite proportions.

Law of Definite Proportions Example 1

So in this example question, it says, 2 unknown compounds are examined. Compound A contains 2 grams of hydrogen and 32 grams of oxygen. Compound B contains 15 grams of hydrogen and 120 grams of oxygen. Do compounds A and B represent the same compound? Alright. So they're having us examine 2 different samples. Remember, if we're following the law of definite proportions, if they represent the same compound, then they should give us the same mass ratio. So we'd say here that for them to be the same, mass ratio A of compound A must equal the mass ratio of compound B. Alright. So remember, we place the larger mass on top, and here we have 32 grams of oxygen versus 2 grams of hydrogen in compound A. So that would be 32 grams O2 grams H. When we do that, that's going to give us 16 as our mass ratio. And then for compound B, we put the larger mass again on top, so 120 grams of oxygen divided by 15 grams of hydrogen. When we divide those two, that gives me a mass ratio of 8. Now again, these values are really saying that I have 16, grams of oxygen to 1 gram of hydrogen, and this is really saying that I have 8 grams of oxygen per 1 gram of hydrogen. Now we see that their mass ratios are definitely not the same, and since their mass ratios are not equal to one another, they cannot represent the same compound. So here we would say, no. Compounds A and B do not represent the same compound because their mass ratios are not equal to one another. So remember, this is how we utilize the law of different proportions to determine if different samples represent the same compound.

Law of Definite Proportions Example 2

Now remember that we can utilize mass ratios and proportions to determine if different samples represent the same compound. In addition to this, you can determine the unknown amount of an element if you know the mass ratio and the mass of the other element. Now again, we're utilizing proportions in order to do this. If we take a look at this example question, it says, a compound contains only calcium and fluorine. A sample of the compound is determined to contain 2 grams of calcium and 1.90 grams of fluorine. According to the law of definite proportions, how much calcium should another sample of this compound contain if it possesses 2.85 grams of fluorine? So remember, following the law of definite proportions, we know mass ratio A must equal mass ratio B. For mass ratio A, we know that we have 2 grams of calcium to 1.90 grams of fluorine. We place the larger mass on top and the smaller mass on the bottom. And we know that if they represent the same compound, then this ratio has to equal here we have 2.85 grams of fluorine, and we don't know the grams of calcium. That's what we're looking for. We've just established proportions and that's going to help us find out what x is. So remember, cross multiply 1.90 times x, and then we're going to multiply 2 times 2.85. So when we do that, we're going to get 1.90 x = 5.70 . Divide both sides now by 1.90 in order to isolate x, and when we do that we're gonna see that x = 3.00 grams, and that would be grams of calcium. So setting up these proportions helps us to realize the answer would have to be option So just remember, we can utilize our ratios in order to determine if our samples represent the same compound, or in this instance, to actually determine the mass of one of the element components. So just utilize these steps, and you'll be able to figure out anything dealing with the law of definite proportions.

A 7.74 g sample of HCN is found to contain 0.287 g of H and 4.01 g N. Find the mass of carbon in a sample of HCN with a mass of 3.43 g.

• A. 1.28 g C
• B. 1.53 g C
• C. 1.70 g C
• D. 2.01 g C