Ethylene glycol, commonly used as automobile antifreeze, contains only carbon, hydrogen, and oxygen. Combustion analysis of a 23.46 mg sample yields 20.42 mg of H2O and 33.27 mg of CO2. What is the empirical formula of ethylene glycol? What is its molecular formula if it has a molecular weight of 62.0?

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Hi everyone today, we have a question telling us that 0.487 g sample of quiNINE was combusted and produced 1.321 g of carbon dioxide 0.325g of water and 0. - one g of nitrogen. The molar mass of quiNINE is 324 g per mole. And we want to find its molecular formula if it only contains carbon, hydrogen, oxygen and nitrogen. So we first want to find our masses. So we'll start here with carbon dioxide, we want to find the mass of carbon. So we're going to start with 1.32, 1 g of carbon dioxide. We're going to multiply that by one mole of carbon dioxide Divided by .01g of carbon dioxide. That's carbon dioxides molar mass Times one mole of carbon Over one mole of carbon dioxide Times the molar mass of carbon, which is 12. And that is per one mold. So our grams of carbon docks that are gonna cancel out our moles of carbon dioxide. Going to cancel out our moles of carbon are going to cancel out leaving us with grams of carbon and that equals 0. g of carbon. Next we want to do the same with water. We want to get the mass of hydrogen. So we have 0.3 2, 5 g of water And we're going to multiply that by one mole of water Over waters molar mass, which is 18.02. And then we're going to multiply it by two moles of hydrogen because we have two moles of hydrogen in H 20. And that's over one mole of water. And then we're going to multiply by hydrogen smaller mass, which is 1. And that's over one mole of hydrogen. So our grams of water are canceling out our moles of water are canceling out our moles of hydrogen are canceling out leaving us with g of hydrogen and that is 0.036g of hydrogen. Next we want to do the same for nitrogen. So that is 0.04 g of nitrogen. And we're going to multiply that by one mole of into, divided by It's more mass. So 28.02 g of into. And then we're going to multiply that by two moles of nitrogen over one mole of into. And we're gonna multiply it by its smaller mass, which is 14.01 g over one mole. So we have our grams of into canceling out our moles of, into canceling out our moles of nitrogen, canceling out leaving us with grams of nitrogen. And that is 0.042, one g of nitrogen. And now to find our mass of oxygen, we are just going to take our sample mass, our zero for 87 g. And we're going to subtract the grams that we just found for our other elements. So minus zero 0. -0. minus 0.04- And that equals 0. grams of oxygen. Now our next step is to convert these two moles. So we're going to start out with our carbon here, We have 0.36 g of carbon. We're going to multiply it by one mole Divided by its smaller mass, which is 12 and that equals 0. moles of carbon. Next we have our 0. g of hydrogen. We're going to multiply that by one mole divided by its smaller mass which is 1.01 And that gives us zero moles of hydrogen. And now for nitrogen we have 0.04- grams of nitrogen. And we're going to multiply that by one mole Divided by its smaller mass which is 14.01 g And that equals 0.003 moles of nitrogen. And next we have 0.04, g of oxygen times one mole Divided by its smaller mass which is 16. And that equals 0. moles of oxygen. Now we need to divide all of these by the smallest which is 0.003. So of course with oxygen that will be one nitrogen, that will be one hydrogen that will equal and carbon That will equal 10. So our empirical formula is C. 10 H. 12 N. O. And now to find the molecular mass, we are going to add up our masses here. So we have 12 . times Plus 1.01 times Because we have 12 hydrogen and then plus 14.1 for R. One nitrogen Plus 16. For our one oxygen. And that equals 162.24. And now if we take our original Mueller mass, which is 324 And divide it by 162.24, That equals two. So we need to multiply everything in our empirical formula by two, Which gives us c. 20 age into 02. And that is our final molecular mass. Thank you for watching. Bye.

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Key Concepts

Combustion Analysis

Empirical Formula

Molecular Formula