Question

A sample of a compound containing only carbon and hydrogen is burned, producing 157 mg CO2 and 42.8 mg H2O. What is the empirical formula?

Answer 1

Combustion analysis!

`157mg * (10^-3g)/(mg) * (CO_2)/(44.01g) approx 3.57*10^-3mol`

`42.8mg * (10^-3g)/(mg) * (H_2O)/(18.02g)* (2H)/(H_2O) approx 4.75*10^-3mol`

If you divide the largest molar mass by the smallest...

`C_1.33H`

Annoying... multiply by three so integers are the subscripts:

`C_4H_3`