## Q. 12.13

Calculating the Molar Mass of a Gas Using the Ideal Gas Law in Modified Form

A 3.30 g sample of chlorine gas (Cl$_2$) occupies a volume of 1.20 L at 741 mm Hg and 33°C. Based on this data, calculate the molar mass of chlorine gas.

## Verified Solution

Rearranging the modified ideal gas equation

$PV = \Bigl( \frac{m}{M} \Bigr) RT$

to isolate M on the left side of the equation gives

$M = \frac{mRT}{PV}$

This equation can be used to directly calculate molar mass since all quantities on the right side of the equation are known.

$m = 3.30 \textrm{ g}, \quad R= 62.36 \frac{\textrm{mm Hg} \cdot \textrm{L}}{\textrm{mole} \cdot \textrm{K}} , \quad T = 33° \textrm{C} + 273 = 306$ K

P = 741 mm Hg,       V = 1.20 L

Substitution of these known values in the equation gives

$M = \frac{(3.30 \textrm{ g}) \Bigl( 62.36 \frac{\cancel{\textrm{mm Hg}} \cdot \cancel{\textrm{L}}}{\textrm{mole} \cdot \cancel{\textrm{K}}} \Bigr) (306 \cancel{\textrm{K}})}{(741 \cancel{\textrm{mm Hg}}) (1.20 \cancel{\textrm{L}})}$

All units cancel except for grams per mole, the units of molar mass.

Doing the arithmetic, we obtain a value of 70.8 for the molar mass of Cl$_2$.

$M = \frac{3.30 \times 62.36 \times 306}{741 \times 1.20} \frac{\textrm{g}}{\textrm{mole}} = 70.817732 \frac{\textrm{g}}{\textrm{mole}}$     (calculator answer)
= $\textbf{70.8} \frac{\textbf{g}}{\textbf{mole}}$          (correct answer)

Answer Double Check: Is the answer reasonable? Yes. The calculated molar mass can be compared to that obtained using atomic masses from the periodic table. The periodic table molar mass of chlorine (Cl$_2$) is 70.90 g/mole. The answer obtained, using the ideal gas law and allowing for experimental error, is consistent with the periodic table value. 