A sample of mass 2.00 g from a pungent compound was extracted from skunk spray. Upon analysis, the sample was found to produce 3.39 g of CO₂ and 1.63 - of H₂O. In addition to C and H, the compound also contains sulphur. The molar mass of the compound was found to be 88.17 g/mol. What is the molecular formula? 

C_xH_yS_z + O₂ ---> ACO₂ + BH₂O

Molar mass CO₂: 44.01 g/mol

Molar mass H₂O: 18.02 g/mol

We need to determine the molar ratio so we can balance our equation

(3.99 g)/(44.01 g/mol)=0.091 mol

(1.63 g)/(18.02 g/mol)=0.091 mol

(2.00 g)/(88.17 g/mol)=0.023 mol

(0.091/0.023)=4

So we have a 1:4 ratio of our unknown to product

C_xH_yS_z + 6O₂ ---> 4CO₂ + 4H₂O (we don't need to worry about O, but I am including it for completeness)

So we have 4 C and 8 H to balance

4(12)+8(1.01)=56.08 g/mol (this is the mass of just C and H)

We subtract this mass from the molar mass given to determine how many S are in the formula

88.17-56.08=32.09 (this is the molar mass of sulphur)

So the formula is C₄H₈S