What is the empirical formula of the unknown compound

An unknown compound contains only C, H, and O. Combustion of 3.10 g of this compound produced 7.57 g of CO2 and 3.10 g of H2O. What is the empirical formula of the unknown compound? CHO

Combustion Equation: ?CHO + ?O2 ===> ?CO2 + ?H2O with unknown coefficients


First use the molar mass of CO2 to calculate the number of moles of CO2 = .172 mol

7.57g/(44.01g/mol) = .172 mol CO2

do the same process for H2O - you get .172 mol H2O


Then use the molar mass of C to get the number of grams of C produced:

.172*12.01 = 2.0657g C

same process for H: .172*(2 H atoms per mol of H2O)*1.008 = .346752 g H

these values must be the same on the right side of the equation

so subtract them from the mass of the original sample to get the amount of O in that sample

3.10 - 2.0657 - .346752 = .687548 g O

now convert that to moles:

.687548g/(16.0g/mol) = .04297 mol O

now we want to compare the ratios of O to C and to H

.172 mol C / .04297 mol O = 4 so there are 4 moles C to each mol of O in the original compound

similarly,

(2*.172 mol H)/ .04297 mol O = 8 so there are 8 moles H to each mol of O in the original compound

Thus we have C4H8O or some multiple such as C8H16O2

but the empirical formula is just C4H8O (since all it is asking for is the simplest ratios of each atom)

http://www.chemteam.info/Mole/Combustion…

Look at example #3

CO3H2 (carbonic acid)