Question

Air contains $23\%$ of oxygen and $77\%$of nitrogen by weight. What is the percentage of oxygen by volume is?

Answer 1

Step 1: Given data

Percentage of oxygen$=23\%$

Percentage of nitrogen$=77\%$

Step 2: Calculating the percentage of oxygen

Let the mass of air$=100\mathrm{g}$

Therefore, the mass of oxygen in $100\mathrm{g}$ of air$=23\mathrm{g}$

and mass of nitrogen in $100\mathrm{g}$ of air$=77\mathrm{g}$

Also, the molecular mass of oxygen$=32\mathrm{g}$

and the molecular mass of nitrogen$=28\mathrm{g}$

Now, the number of moles of oxygen$=\frac{\mathrm{given}\mathrm{mass}}{\mathrm{molecular}\mathrm{mass}}$$\Rightarrow \frac{22\mathrm{g}}{32\mathrm{g}}$

and number of moles of nitrogen$=\frac{\mathrm{given}\mathrm{mass}}{\mathrm{molecular}\mathrm{mass}}$$\Rightarrow \frac{77\mathrm{g}}{28\mathrm{g}}$

As one mole of gas requires $22.4\mathrm{L}$ of volume at Standard Temperature and Pressure (STP) conditions, So

The volume occupied by moles of oxygen$=\frac{23\mathrm{g}}{32\mathrm{g}}\times 22.4\mathrm{L}\Rightarrow 16.1\mathrm{L}$

The volume occupied moles of nitrogen$=\frac{77\mathrm{g}}{28\mathrm{g}}\times 22.4\mathrm{L}\Rightarrow 61.6\mathrm{L}$

Therefore, the total volume of air$=$volume occupied by moles of oxygen$+$volume occupied by moles of nitrogen

$=16.6\mathrm{L}+61.6\mathrm{L}\Rightarrow 77.7\mathrm{L}$

So percentage by volume of oxygen$=\frac{\mathrm{volume}\mathrm{occupied}\mathrm{by}\mathrm{moles}\mathrm{of}\mathrm{oxygen}}{\mathrm{total}\mathrm{volume}\mathrm{of}\mathrm{air}}\times 100$

$=\frac{16.1}{77.7}\times 100\Rightarrow 20.7\%$

Therefore, the percentage of oxygen by volume is $20.7\%$