# 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?

##### 1 Answer
Sep 6, 2017

Combustion analysis!

$157 m g \cdot \frac{{10}^{-} 3 g}{m g} \cdot \frac{C {O}_{2}}{44.01 g} \approx 3.57 \cdot {10}^{-} 3 m o l$

$42.8 m g \cdot \frac{{10}^{-} 3 g}{m g} \cdot \frac{{H}_{2} O}{18.02 g} \cdot \frac{2 H}{{H}_{2} O} \approx 4.75 \cdot {10}^{-} 3 m o l$

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

${C}_{1.33} H$

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

${C}_{4} {H}_{3}$