The combustion of 4.16 grams of a compound which contains only...

Question

The combustion of 4.16 grams of a compound which contains only C,H,O and F yields 7.7 g CO₂ and 2.52 g H₂O. Another sample of the compound with a mass of 3.63 g is found to contain 0.58 g F. What is the empirical formula of the compound?

Accepted Answer

The combustion of 4.16 grams of a compound, which contains only carbon, hydrogen, oxygen, and fluorine, yields 7.7 grams of carbon dioxide and 2.52 grams of water. Another sample of the compound with a mass of 3.63 grams is found to contain 0.58 grams of fluorine. What is the empirical formula of the compound? Alright. So our compound itself contains carbon, hydrogen, oxygen, and fluorine. From the amount of carbon dioxide and of water, we know that we can determine the grams of hydrogen and the grams of carbon. So we're gonna have 7.7 grams of c02, and we have here 2.52 grams of water. We're going to change those into grams of c and grams of h respectively. So we've seen this before with other combustion analysis types of questions. We're gonna say 1 mole of c02 weighs 44.01 gram c02. So that's the weight of carbon and and the 2 oxygens together. Then we're gonna convert moles of c o two into just moles of c. And if we know the moles of c, then we can determine the grams of c. So one mole of c weighs 12.01 grams. So when we do that, we're gonna get 2.1012 grams of c. Now let's do it for water. One mole of water weighs 18.016 grams. When we take into account, it's made up of 2 hydrogens and 1 oxygen. One mole of h two o has within it 2 moles of h, because we can see that there's 2 hydrogens from the subscript here. Then we're going to say that 1 mole of h weighs 1.008 grams according to our periodic table. So we've determined the grams of water as 0.2820 grams of hydrogen. At this point, what we normally do is we say that the total mass of our sample, which in this case is made up of carbon, hydrogen, oxygen, and fluorine, we'd subtract out the amount of carbon and the amount of hydrogen, and that would help me determine the grams of oxygen remaining. In this case, we can't do it because we also have fluorine involved. So before we can determine the amount of oxygen remaining, we have to also determine the grams of fluorine present. So how do we do that? Well, in the second sentence, it tells us another sample of the same compound when it has a mass of 3.63 grams. It contains 0.58 grams of fluorine. What's gonna help us to get to the grams of fluorine is something called the law of definite proportions. Now the law of definite proportions basically says that if you have the same sample or compound, it doesn't matter if you have it in 1 sample or in multiple samples. If they're the same compound, they'll have the same mass ratio. So what I'm saying here is, we know that when we have 3.63 grams of our compound, there are 0.58 grams of fluorine involved. And if we're dealing with law of definite proportions, when we have this 4.16 gram sample of the same compound, we can determine the grams of fluorine by setting up a proportion and making them equal to one another. Okay? So the law of definite proportions is the key to helping us determine the amount of fluorine. Those two proportions have to equal each other, the 2 ratios, because they both represent the same compound. So the ratio of flooring to the compound's weight will remain constant. Alright. So we're going to cross multiply, so 3.63 times x and then 0.58 times 4.16. When we do that, we're gonna have 3.63x equals 0.58 times 4.16. Divide both sides here by 3.63, and when we do that we'll find x. X here will represent the mass of fluorine. So when we do that, we're gonna get 0.66 47 grams of fluorine. So with that number, we can add it here. So we can subtract out from our total mass of 4.16 the grams of carbon, hydrogen, and fluorine, and what's left would have to represent the grams of oxygen. So here, oxygen's mass will come out to 1.1121 grams oxygen. So at this point, we have the grams of each of the elements. So then it becomes a simple, few steps in order to find the empirical formula of the compound. We're gonna change each one of these grams into moles. So let's write them all down here and then we're gonna have oxygen, and I'm writing in the order that they're stated within this within the question. C h o f, so that's the order I'm writing them down here. So we're gonna convert each of these grams into moles and we do that by dividing them all by their atomic masses. So one mole of c here, 1 mole of h here, 1 mole of oxygen, and 1 mole of fluorine. Plug in their atomic masses from the periodic table and that'll give us the moles of each. Oxygen weighs 16 and then 4 in on the periodic table is 19 grams. So here, grams cancel out and I'll have moles of each one of these elements at the end. So I'll have here 0.1750 moles of c, I'll have here 0.2798 moles of h, I have here, let's see, 0.0695 moles of o, and then finally I'm gonna have 0.0350 moles of f. Remember the steps for determining empirical formula. Now that I have the moles of each one, I divide them all by the smallest mole answer, and that's gonna give me their ratios. So all of them get divided by point 0350. So when we have point 053 0.0350, that's gonna give me what? That's gonna give me 1 fluorine, 2 oxygens, it's gonna give me 8 hydrogens and 5 carbons. All of them are whole numbers, so using this information, I can find my empirical formula, which comes out as c5h802f. So this one was a little bit different. We had to employ the law of definite proportions to just help us find fluorine. But other than that, it's pretty similar to other types of combustion analysis questions. You find the amounts of the elements in grams, you deduct it from the sample, which usually helps you find a missing element. Once you have the grams of all of them, convert them into moles, divide all the moles by the smallest moles, and the whole number ratios that you get will be part of the empirical formula.

Struggling with General Chemistry?

The combustion of 4.16 grams of a compound which contains only C,H,O and F yields 7.7 g CO₂ and 2.52 g H₂O. Another sample of the compound with a mass of 3.63 g is found to contain 0.58 g F. What is the empirical formula of the compound?

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Struggling with General Chemistry?

The combustion of 4.16 grams of a compound which contains only C,H,O and F yields 7.7 g CO2 and 2.52 g H2O. Another sample of the compound with a mass of 3.63 g is found to contain 0.58 g F. What is the empirical formula of the compound?

Watch next

Combustion Analysis practice set

The combustion of 4.16 grams of a compound which contains only C,H,O and F yields 7.7 g CO₂ and 2.52 g H₂O. Another sample of the compound with a mass of 3.63 g is found to contain 0.58 g F. What is the empirical formula of the compound?