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### Workbook: Quiz 1 2015 #5 Find empirical formula

Posted: Mon Oct 03, 2016 12:34 pm
6.40 g of a compound was burned in the air and produced 8.80 g CO2 and 7.20 g H2O. Find the empirical formula of the compound.

The answer says it is CH4O, but I did all the steps and checked multiple times but I keep getting CH2O.
What did I do wrong? ### Re: Workbook: Quiz 1 2015 #5 Find empirical formula  [ENDORSED]

Posted: Mon Oct 03, 2016 1:11 pm
Most likely, you did not multiply the moles of H2O by 2 to get moles of H.

Starting from the beginning, the first step is find how many moles of C and H there are in the compound. To do so we divide the masses of CO2 and H2O by their respective molar masses:

(8.80 g CO2)/(12.01g/mol + 2 (16 g/mol O) ) = .19995 mol CO2

Since there is 1 mol C for every 1 mol of CO2, there are .19995 mol C.

(7.20 g H2O)/ (2(1.0079 g/mol H) +16 g/mol O)=.39978 mol H2O

Since there are 2 mol of H for every 1 mol of H2O, there are .79956 mol H.

Now we must calculate the grams of C and H in the compound and subtract it from the total mass of the compound to find the mass of O.

.19995 mol C x 12.01 g/mol = 2.401 g C
.79956 mol H x 1.0079 g/mol= .8059 g H

6.40 g compound - (2.401 g C + .8059 g H ) =3.193 g O

3.193 g O / 16 g/mol O = .19957 mol O

Now we have the mol of C H and O. If we use this to find the empirical formula, we should get CH4O.