Combustion analysis of 45.62 mg of toluene, a commonly used solve...

Question

Combustion analysis of 45.62 mg of toluene, a commonly used solvent, gives 35.67 mg of H2O and 152.5 mg of CO2. What is the empirical formula of toluene?

Accepted Answer

Hello. Everyone in this question, we're asked to identify the empirical formula of a compound only having carbon and hydrogen atoms. And the reaction produced 12.25 g of carbon dioxide And 6.246 g of water. We need to first find the most of carbon and hydrogen. We have 12.25 grams of carbon dioxide. And in one mall of carbon dioxide we have the molar mass And it's gonna be 12.01 g for carbon Plus 16 for oxygen and there's two oxygen. This give us 44 .01g and in one more of C. 02 we have one more of carbon. MS give us 0.28 malls of carbon. Then we also have 6.246 g of H 20. And in one mole of H 20. You have the molar mass. So we have 1.01 g for hydrogen and there's two plus for oxygen. And this will give us 18.02 g and one more of water. We have two moles of hydrogen. And this will give us 0.7 powers of hydrogen. And now to find the smallest ratio we need to divide by the smallest number of moles. But for carbon we have 0.28 Divided by 0.28. This will give us one and for hydrogen 0.7 Divided by 0.28. And this will give us 2.5. So we have seats H 2.5. But since this is not a whole number we need to multiply the whole formula by two. And this gives us C2 age five. And it's going to be be thanks for watching my video and I hope it was helpful.

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