Question #67709

Sep 8, 2017

The mass of the compound is irrelevant in this problem, probably meant to confuse us.

$6.78 g \cdot \frac{C {O}_{2}}{44.0 g} \approx 0.154 m o l$
$1.94 g \cdot \frac{{H}_{2} O}{18.0 g} \cdot \frac{2 H}{{H}_{2} O} \approx 0.216 m o l$
$0.432 g \cdot \frac{{N}_{2}}{28.0 g} \cdot \frac{2 N}{{N}_{2}} \approx 0.0309 m o l$

Divide every number that's larger than the smallest amount of moles and we get...

${C}_{5} {H}_{7} N$