9 volumes of a gaseous mixture contain an organic compound and oxygen sufficient for its combustion. Burning the compound yielded 4 volumes of $C O_{2}$ , 6 volumes of $H_{2} O$ and 2 volumes of $N_{2}$ , find the molecular formula of the compound?

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Solution

In the case of gases, an equal amount of moles of occupying equal volumes at a given temperature and pressure. Let the volume of the organic compound be x. Therefore, the volume of a will be 9-a.

$a C_{x} H_{y} N_{z} + (9 - a) O_{2} \to 4 C O_{2} + 6 H_{2} O + 2 N_{2}$

1 mol of C combines with 1 mol of $O_{2}$ . Hence, 4 mol of $C O_{2}$ contains 4 moles of $O_{2}$ .

Similarly, since 1 atom of O reacts with 2 atoms of H, it can be known that for every 1 vol. of $O_{2}$ , there are two volumes of $H_{2}$ hence two volumes of $H_{2} O$ . Therefore, 6 vol of $H_{2} O$ would be produced by 3 vol of Oxygen.
$6 H_{2} + 3 O_{2} \to 6 H_{2} O$

4 + 3 = 7 vol of Oxygen was present in the mixture.

This gives us:
9 - a = 7
a = 2

$2 C_{x} H_{y} N_{z} + 7 O_{2} = 4 C O_{2} + 6 H_{2} O + 2 N_{2}$
$C_{x} H_{y} N_{z} + 72 O_{2} = 2 C O_{2} + 3 H_{2} O + N_{2}$

1 mol of C gives 1 mol of C.
x = 2

Similarly,
y = 6
z = 2
Compound: $C_{2} H_{6} N_{2}$

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9 volumes of a gaseous mixture contain an organic compound and oxygen sufficient for its combustion. Burning the compound yielded 4 volumes of CO2, 6 volumes of H2O and 2 volumes of N2, find the molecular formula of the compound?

9 volumes of a gaseous mixture contain an organic compound and oxygen sufficient for its combustion. Burning the compound yielded 4 volumes of $C O_{2}$ , 6 volumes of $H_{2} O$ and 2 volumes of $N_{2}$ , find the molecular formula of the compound?