Ask your homework questions to teachers and professors, meet other students, and be entered to win $600 or an Xbox Series X 🎉Join our Discord!

Like

Report

Numerade Educator

Like

Report

Problem 42 Hard Difficulty

An unknown compound has the formula $\mathrm{C}_{x} \mathrm{H}_{1} \mathrm{O}_{2}$. You burn $0.1523 \mathrm{g}$ of the compound and isolate $0.3718 \mathrm{g}$ of $\mathrm{CO}_{2}$ and $0.1522 \mathrm{g}$ of $\mathrm{H}_{2} \mathrm{O}$. What is the empirical formula of the compound? If the molar mass is $72.1 \mathrm{g} / \mathrm{mol}$, what is the molecular formula? (See Exercise 4.9.)

Answer

Discussion

You must be signed in to discuss.
Top Chemistry 102 Educators
Stephanie C.

University of Central Florida

Lizabeth T.

Numerade Educator

Stephen P.

Drexel University

Allea C.

University of Maryland - University College

Chemistry 102 Bootcamp

Lectures

Join Bootcamp

Video Transcript

this is gonna be a fairly lengthy problem. In order to determine the molecular formula of the compound, we'll start with a massive carbon dioxide convert. The massive carbon dioxide, two moles of carbon dioxide using the molar mass carbon dioxide than moles. Carbon dioxide, two moles, carbon molds, carbon, two grams carbon. Now we know the mass of carbon that was present in the compound. Knowing originally the massive carbon dioxide produced, we can calculate the mass of hydrogen that was present in the compound by figuring out how the massive hydrogen present in the given mass of water that was produced will go from grams of water. Two moles of water using Mueller, massive water moles of water to most of hydrogen. Recognizing there's two moles of hydrogen for every one mole of water and then moles of hydrogen, two grams of hydrogen. Now that we know the grams of hydrogen and the grams of carbon that was present in the compound, the only other element is oxygen. So if we take the total mass of the compound minus the mass that is carbon and hydrogen, what's left over should be the massive his oxygen. Okay, now we know the masses of carbon, hydrogen and oxygen present in the compound. We need to know the moles of carbon, hydrogen and oxygen president compound to get the empirical formula and then ultimately to get the molecular formula. So let's take thes masses and calculate the moles of carbon and the moles of oxygen and the moles of hydrogen. We can do that by starting again with the, um, grams carbon dioxide. And instead of going to Gramps, Carbon will just stop it. Moles carbon do the same thing with water as we did before. But instead of going two grams of hydrogen, we will stop at moles of hydrogen. And then for the last one, we know the grams of oxygen from our subtraction calculation so we can convert the grams of oxygen we just calculated into moles of oxygen by dividing by the molar mass of a mole of oxygen atoms. Okay, now we have moles of our of carbon, hydrogen and oxygen would like whole numbers. So we're gonna divide by the smallest, and we'll get one mole of oxygen for every eight moles of hydrogen for every formals of carbon for the empirical formula. Would BC for a Jato. And that empirical formula has a molar mass of 72.1 grams per mole, simply adding up the formals carbon, eight miles hydrogen and one mole of oxygen. And it turns out that that is the empirical formula. Moller, Mass. Which is equal to the given molecular formula. Mueller, Mass. So the empirical formula is the molecular formula. C four h h o.

Weber State University
Top Chemistry 102 Educators
Stephanie C.

University of Central Florida

Lizabeth T.

Numerade Educator

Stephen P.

Drexel University

Allea C.

University of Maryland - University College

Chemistry 102 Bootcamp

Lectures

Join Bootcamp