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A compound contains 6.0 $\mathrm{g}$ of carbon and 1.0 $\mathrm{g}$ of

hydrogen, and has a molar mass of 42.0 $\mathrm{g} / \mathrm{mol}$ . What are

the compound's percent composition, empirical formula, and molecular formula?

the molecular formula is C3H6

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Carleton College

Drexel University

University of Kentucky

Numerade Educator

to determine all of these. We first need to take the masses that are given to us 6 g of carbon and 1 g of hydrogen to calculate first. Then the percent carbon will take 6 g of carbon, divided by the total mass, which would be 7 g if we contain. If it contains 6 g of carbon and 1 g of hydrogen, and we get when we divide these two and multiplied by 100 we get 85.7% carbon. Because hydrogen is the only other element than 100 minus 85.7% carbon would give us percent hydrogen of 14.3. Then, to determine the empirical formula, we need to convert the grams of each element into moles. So we'll start with 6 g of carbon and convert 6 g of carbon by dividing by the molar mass of carbon in tow. Moles of carbon. It would get 60.4996 moles carbon. Then we'll take 1 g of hydrogen divided by the molar mass of hydrogen, 1.794 g, and we will get 0.99 to 1 moles of hydrogen, then What we'll do is we will divide the by the smallest when we divide by the smallest number that will then tell us the mole relationship one mole of carbon for every about two moles of hydrogen. So the empirical formula is going to be C H two. Now to determine the molecular formula, we need to determine the molar mass of the empirical formula. It's going to be one mole of carbon 12.1 plus two times the molar mass of hydrogen given us 14.3 g per mole. Now, to get the molecular formula from the empirical formula, we need to divide the empirical formula into the molar mass of the known compound. So we'll take the molar mass of the empirical formula divided into the given known molar mass of a compound that will then tell us what the sub script multiplier will be. Will then take that subscript multiplier and multiply it by every sub script in order to convert the empirical formula into the molecular formula. When we do that, we get a molecular formula of C three h six, so percent carbon is 85.7%. Hydrogen is 14.3. The empirical formula is C H. Two. On the molecular formula is C three, h six

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