Determining an empirical formula by combustion analysis

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Amy Luu 2G
Posts: 105
Joined: Wed Sep 18, 2019 12:19 am

Determining an empirical formula by combustion analysis

Postby Amy Luu 2G » Wed Oct 09, 2019 10:35 pm

I'm confused on how to solve Self Test M.4A:
When 0.528 g of sucrose (a compound of carbon, hydrogen, and
oxygen) is burned, 0.306 g of water and 0.815 g of carbon dioxide are formed.
Deduce the empirical formula of sucrose.

I started by finding the mass of H, C and got 0.0342g H, 0.222g C and then used that to solve for the mass of O which i got 0.272g O. I also found the moles of H, C and O and divided them by the smallest number which was 0.0170 and got a ratio of 1.994:1.088:1 (H:C:O). However, the correct answer is C12H22O11. Im not sure how they got this answer

Posts: 164
Joined: Fri Aug 30, 2019 12:16 am

Re: Determining an empirical formula by combustion analysis

Postby sarahsalama2E » Wed Oct 09, 2019 11:47 pm

So first, given the values of .306 g of water and .815 g of CO2, you would need to convert these values into moles--making sure that you are using mol:mol ratio for the specific element (either C or H) when doing these calculations.
Ex: .815 g CO2 x 1 mol/44.01 g CO2 x 1 mol c/1 mol co2--> .01855 mol C--> .222 g C

Repeat this calculation for Hydrogen (H) in the water and obtain the moles and grams for H

add up the grams for Carbon and hydrogen that are obtained in these previous steps and subtract from .528 g Sucrose. Because we know sucrose contains c,h, and o we can do this and we can also do this because of the law of conservation of mass. Doing so we get the resultant grams of oxygen and we can then find the moles of O.
Like any other Empirical Formula problem, we can then divide all the moles (C,H,and O) by the smallest number (which in this case is the mol of o) to get the ratio of the respective elements. we do have to multiply by 11 however to make sure we get whole numbers. The answer is C12H22O11, after multiplying 1.089 mol . C, 2 mol H, and 1 mol of O by 11.

Deepika Reddy 1A
Posts: 125
Joined: Thu Jul 11, 2019 12:15 am

Re: Determining an empirical formula by combustion analysis

Postby Deepika Reddy 1A » Wed Oct 09, 2019 11:48 pm

You have to use the mass percentage composition of hydrogen, oxygen, and carbon. So you already found the mass of hydrogen, carbon, and oxygen. Now, just divide those by the overall mass of sucrose which is 0.528 to get the mass percentage composition of each. Then you can use that to find the empirical formula.

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