What is the empirical formula of the compound?

Question

Combustion analysis of 1.500 g of an unknown compound containing only carbon and hydrogen produced 4.554 g of CO2 and 2.322 g of H2O. What is the empirical formula of the compound?

Answer 1

1.5 g CxHyOz + 5.376 g O2 -> 4.554 g CO2 + 2.322 g H2O
CxHyOz + .168 mol O2 -> .1035 mol CO2 + .129 mol H2O
multiply everything by 77.5 to get nice whole numbers for the molar coefficients
CxHyOz + 13 O2 -> 8 CO2 + 10 H2O
Balance the C and H.
C8H20Oz + 13 O2 -> 8 CO2 + 10 H2O
Balance the O.
C8H20 + 13 O2 -> 8 CO2 + 10 H2O
Because you can't stick 20 hydrogens on a skeleton of 8 carbons, the molecular formula is likely to be C4H10, and since both 4 and 10 are divisible by 2, the empirical formula is C2H5.

Check: would burning 1.5 g of C4H10 (butane) yield those products?
1.5 g C4H10 = .0259 mol
.0259 mol C4H10 + .168 mol O2 -> .1035 mol CO2 + .129 mol H2O
multiply by 77.5 again
2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O

That's pretty much the balanced equation for this reaction.

Answer 2

1) First, the possible reaction of the unknown compound:

CxHyOz + heat -------> CO2 + H2O

and molecular masses of each product:

M(CO2)=44 g/mol
M(H2O) = 18 g/mol

and mass of the compound:

mCxHyOz = 1.500 g

2) We know the masses of product. We have to calculate the number of moles:

number of moles CO2 = 4.554 g / 44 g/mol = 0.1035 moles of CO2
number of moles H2O = 2.322 g / 18 g/mol = 0.129 moles of H2O

3) We have to know what is the ratio between moles of CO2 respect to water or viceversa. It is better to get integer numbers so:

moles H2O/moles CO2 = 0.129/0.1035 =1.25

4) This result tell us that each mol of the unknown compound yields 1.25 moles of water for each mol of CO2

5) If we take round numbers, we can say that relation between water and CO2 is almost 1:1, so the empirical formula of the unknown compound is:

CH2O3 (because is as if the compound was entirely formed by one mol of water and one mol of CO2)


That´s it!

Good luck!