(a) Combustion analysis of toluene, a common organic
solvent, giv...

Question

(a) Combustion analysis of toluene, a common organic
solvent, gives 5.86 mg of CO2 and 1.37 mg of H2O. If the
compound contains only carbon and hydrogen, what is its
empirical formula?

Accepted Answer

Hi everyone today, we have a problem asking us to determine the empirical formula of a hydrocarbon. If combustion analysis revealed the following composition, 23.093g of carbon dioxide and 4.701g of water. So our first step here is going to be to calculate the molar mass of carbon dioxide. So we have carbon Which has a molar mass of 12 and there's only one of them, So that equals 12. And then we have oxygen, which has a molar mass of 16, and we have two of them, Which equals 32. So 32 plus equals 44. And now we want to start out with our .09, 3 grams of carbon dioxide. and we are going to divide by the Molar mass. So 40 for grams of carbon dioxide Over one Under mole of carbon dioxide. And then we're going to multiply by the multiple ratio to get moles of carbon. So we're going to have one mole of carbon over one mole of carbon dioxide. So our grams of carbon dioxide are going to cancel out and our moles of carbon dioxide are going to cancel out And that equals 0. moles of carbon. Now we need to do the same thing with ox with water. So we have hydrogen with a molar mass of one and we have two of them. So that equals two. And then we have oxygen with a molar mass of 16 And we have one of them. So that equals one and equal 16 And then two plus equals 18. And now we want to start out with our four .701g of water. And we're going to divide by our molar mass so 18 g of water. And that will give us moles of water. And then we're gonna multiply by our multiple ratio. So we have two moles of hydrogen For every one mole of water. So our grams of water is going to cancel out and our moles of water are going to cancel out. And that is going to give us 0. moles of hydrogen. And now we need to divide the two by each other. So 0.52 divided by 0.52 is obviously one. So we have a 1-1 ratio which gives us an empirical formula of C. H. And that is our final answer. Thank you for watching. Bye.

(a) Combustion analysis of toluene, a common organic solvent, gives 5.86 mg of CO2 and 1.37 mg of H2O. If the compound contains only carbon and hydrogen, what is its empirical formula?

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