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Stoichiometric Calculations

Taking N2 a...

Question

Taking $N_{2}$ and $O_{2}$ as main components of the air ( $79$ % $N_{2}, 21$ % $O_{2}$ by volume) what is the molecular mass of air? How has it been arrived at?

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Solution

For a mixture of gases, the average molecular mass is taken.

Average molecular mass $= \sum x_{i} M_{i} = x_{N_{2}} M_{N_{2}} + x_{o_{2}} M_{o_{2}}$ where $x_{N_{2}}$ and $x_{o_{2}}$ are mole fractions of $N_{2}$ and $O_{2}$ and $M_{N_{2}}$ and $M_{O_{2}}$ are their molecular masses.

As equal volumes contain equal number of moles, therefore, their molar ratio is same as ratio of their volumes.

$∴ x_{N_{2}} = 0.79, x_{O_{2}} = 0.21.$ Also $M_{N_{2}} = 28 u, M_{O_{2}} = 32 u$

$∴$ Average molecular mass
$= 0.79 \times 28 + 0.21 \times 32 u$
$= 22.12 + 6.72 u = (28.84 u)$

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