Question #67709

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Question #67709

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Answer 1 · 2017-09-08T22:47:12

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

$6.78 g \cdot \frac{C O_{2}}{44.0 g} \approx 0.154 m o l$ $6.78g * (CO_2)/(44.0g) approx 0.154mol$
$1.94 g \cdot \frac{H_{2} O}{18.0 g} \cdot \frac{2 H}{H_{2} O} \approx 0.216 m o l$ $1.94g * (H_2O)/(18.0g) * (2H)/(H_2O) approx 0.216mol$
$0.432 g \cdot \frac{N_{2}}{28.0 g} \cdot \frac{2 N}{N_{2}} \approx 0.0309 m o l$ $0.432g * (N_2)/(28.0g) * (2N)/(N_2) approx 0.0309mol$

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

$C_{5} H_{7} N$ $C_5H_7N$

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Question #67709

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