Learn the toughest concepts covered in your GOB - General, Organic, and Biological Chemistry class with step-by-step video tutorials and practice problems.

Atoms and the Periodic Table

"Different samples of a pure chemical compound always contain the same proportions of elements by mass."

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Law of Definite Proportions

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Let's examine the law definite proportions. In 17 99 the French chemist Joseph L. Promised originated the law of definite proportions because of its immense contributions to it. It's sometimes referred to simply ask Proust law, but it also goes by the law of constant or definite 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 a lot of definite proportions is that different samples of a cure chemical compound always contained the same proportion development by mass. So let's say, for example, that I take ah sample from New York City, and I suspect that it's CO two and I take a sample of in another city, London, England. And I suspect that CO two, if we're following the law of different proportions, both samples should have the same mass ratio. Now, for the mass ratio, we're gonna place the element with the larger mass on top, not going back to our whole idea of co two in two different cities. Remember, Seal to itself is composed of one carbon and two oxygen, and we know that if we look at the periodic table. The atomic massive carbon is 12.1 g per mole in the massive, oxygen is around 16 g per mole. When you multiply the number of each element by atomic Mass, we'll see how much of it contributes to the old raw mass of CO. Two. So carbon itself contributes 12.1 g total, and oxygen contributes 32 g total. Not using the mass ratio, we place the larger mass on top, which is the 32 g of oxygen in the smaller mass on the bottom. When we divide those two numbers, that gives me approximately 2 66. What that 2.66 is telling me is that we have 2. oxygen for everyone. Carbon. This would be our mass ratio. And if our examining two samples of CO. 21 from New York City and one from London if they both were indeed co. Two, 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 the 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.

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Law of Definite Proportions Example 1

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So in this example, questions as to unknown compounds are examined. Compound A contains 2 g of hydrogen and 32 g of oxygen. Compound B contains 15 g hydrogen and 120 g off oxygen. New compounds A and B represent the same compound. All right, so they're having us examined two 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 off compound beat. All right, so remember we place the larger mass on top, and here we have 32 g of oxygen versus 2 g of hydrogen compound A. So that would be 32 g. Oh, divided by 2 g h. So when we do that, that's gonna give us 16 as our mass ratio. And then for Compound B, we put the larger mass again on top. So 120 g oxygen divided by g hydrogen. When we divide those two, that gives me a mass ratio of eight. Now, again these values were really saying that I have 16 g off oxygen toe, 1 g of hydrogen, and this is really saying that I have 8 g of oxygen per 1 g 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 different proportions to determine if different samples represent the same compound.

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Law of Definite Proportions Example 2

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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 Onley, calcium and Florrie Ah sample of the compound is determined to contain g of calcium at 1.90 g of flooring according to the law of definite proportions. How much calcium should another sample of this compound contain if it possesses 2.85 grands of foreign? So remember, following the law of definite proportions, we know mass ratio a must equal mass ratio be for mass ratio. A. We know that we have 2 g of calcium toe, 1.90 g of flooring. 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 g of flooring and we don't know the grams of calcium. That's what we're looking for. We just established proportions, and that's gonna help us find out what X is. So remember, Cross multiply 1.90 times X and then we're gonna multiply two times 2.85 So when we do that, we're going to get 1.90 X equals 5.70 Divide both sides now by 1.90 in orderto isolate X. And when we do that, we're going to see that X equals 3. g, and that would be grams of calcium. So setting up these proportions helps us to realize that the answer would have to be options. See? 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 law of definite proportions.

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Problem

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

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