Question

The composition of Air in weight percent is:

• A. $\text{N}_2 - 77\%, \text{O}_2 - 23\%$
• B. $\text{N}_2 - 21\%, \text{O}_2 - 79\%$
• C. $\text{N}_2 - 79\%, \text{O}_2 - 21\%$
• D. $\text{N}_2 - 23\%, \text{O}_2 - 77\%$

Hint:

Remember these two critical constants for chemical engineering exams: Air is $21:79$ by volume/mole, but switches to $23:77$ by mass/weight because oxygen molecules are heavier than nitrogen molecules.

Answer 1

The Correct Option is A

Concept: Atmospheric air is standardly approximated as a binary mixture of nitrogen ($\text{N}_2$) and oxygen ($\text{O}_2$). By volume (or mole percent), air is widely known to be composed of approximately:
• Nitrogen ($\text{N}_2$): $79\text{ mol \%}$ (or $\text{vol \%}$)
• Oxygen ($\text{O}_2$): $21\text{ mol \%}$ (or $\text{vol \%}$) To find the percentage composition by weight (mass percent), we must factor in the respective molar masses of the gases:
• Molecular weight of Nitrogen ($M_{\text{N}_2}$) = $28\text{ g/mol}$
• Molecular weight of Oxygen ($M_{\text{O}_2}$) = $32\text{ g/mol}$

Step 1: Select a basis and calculate individual component masses.
Let us assume a basis of $100\text{ moles}$ of air. Based on standard volume fractions, this yields: \[ n_{\text{N}_2} = 79\text{ moles} \] \[ n_{\text{O}_2} = 21\text{ moles} \] Now, compute the absolute mass contribution of each component: \[ \text{Mass of }\text{N}_2 = n_{\text{N}_2} \times M_{\text{N}_2} = 79 \times 28 = 2212\text{ g} \] \[ \text{Mass of }\text{O}_2 = n_{\text{O}_2} \times M_{\text{O}_2} = 21 \times 32 = 672\text{ g} \]

Step 2: Calculate total combined mass.
\[ \text{Total Mass} = \text{Mass of }\text{N}_2 + \text{Mass of }\text{O}_2 \] \[ \text{Total Mass} = 2212 + 672 = 2884\text{ g} \] *(Note: This implies the average molecular weight of dry air is $\frac{2884}{100} = 28.84\text{ g/mol}$).*

Step 3: Evaluate weight percentages.
Calculate the weight fraction of nitrogen: \[ \text{Weight \% of }\text{N}_2 = \left( \frac{\text{Mass of }\text{N}_2}{\text{Total Mass}} \right) \times 100 = \left( \frac{2212}{2884} \right) \times 100 \approx 76.699\% \approx 77\% \] Calculate the weight fraction of oxygen: \[ \text{Weight \% of }\text{O}_2 = \left( \frac{\text{Mass of }\text{O}_2}{\text{Total Mass}} \right) \times 100 = \left( \frac{672}{2884} \right) \times 100 \approx 23.301\% \approx 23\% \] Thus, by weight, air contains approximately $77\%\text{ }\text{N}_2$ and $23\%\text{ }\text{O}_2$.