Question

Consider a gaseous binary compound with a molar mass of. When$\mathbf{1}\mathbf{.39}\mathit{g}$ of this compound is completely burned in excess oxygen, $\mathbf{1}\mathbf{.21}\mathit{g}$of water is formed. Determine the formula of the compound.

Short answer

The formula for the compound is$S{i}_2{H}_6\text{\hspace{0.17em}\hspace{0.17em}or\hspace{0.17em}\hspace{0.17em}}{N}_4{H}_6$

Step-by-step solution

Hydrogen atoms in compound

Mass of$H$ in $H_{2}O$ is$0.136g$

This mass is the same as the mass of$H$ in a gaseous compound

$\frac{0.136g}{1.36g}=0.0978$

This is the molar mass percent of hydrogen atoms in compound

So, there are 6 hydrogen atoms in the compound.

Accessory element in a compound

$62.09g/\text{mol\hspace{0.17em}\hspace{0.17em}}-1.01g/\text{mol\hspace{0.17em}\hspace{0.17em}}=56.03g/\text{mol}$

So the gaseous compound have more molar mass

Conclusion

Therefore, the formula for the compound is$S{i}_2{H}_6\text{\hspace{0.17em}\hspace{0.17em}or\hspace{0.17em}\hspace{0.17em}}{N}_4{H}_6$