Question

Exhaled air contains \(74.5 \% \mathrm{~N}_{2}, 15.7 \% \mathrm{O}_{2}, 3.6 \% \mathrm{CO}_{2},\) and \(6.2 \% \mathrm{H}_{2} \mathrm{O}\) (mole percent). (a) Calculate the molar mass of exhaled air. (b) Calculate the density of exhaled air at \(37^{\circ} \mathrm{C}\) and \(757 \mathrm{~mm} \mathrm{Hg}\) and compare the value you obtained with that of ordinary air \((\mathrm{MM}=29.0 \mathrm{~g} / \mathrm{mol})\) under the same conditions.

Short answer

The molar mass of exhaled air is 29.65 g/mol. The density of exhaled air at 37°C and 757 mmHg is 1.184 g/L, while the density of ordinary air under the same conditions is 1.158 g/L. This shows that exhaled air is slightly denser than ordinary air under the same conditions.

Step-by-step solution

Calculate the molar mass of each component

To find the molar mass of exhaled air, we first find the molar mass of each component: \(\mathrm{N}_2\): 28.02 g/mol \(\mathrm{O}_2\): 32.00 g/mol \(\mathrm{CO}_2\): 44.01 g/mol \(\mathrm{H}_2\mathrm{O}\): 18.02 g/mol

Find the weighted average of the molar masses

We will now use these values and the mole percentages to find the weighted average of the molar masses: Molar Mass of Exhaled Air = \((74.5 \% \times 28.02 \frac{g}{mol})+(15.7 \% \times 32.00 \frac{g}{mol})+(3.6 \% \times 44.01 \frac{g}{mol})+(6.2 \% \times 18.02 \frac{g}{mol})\)

Calculate the molar mass of exhaled air

Plugging in the values, we get: Molar Mass of Exhaled Air = \(0.745 \times 28.02 +0.157 \times 32.00 + 0.036 \times 44.01 + 0.062 \times 18.02 = 29.65 \frac{g}{mol}\) (b) Calculate the density of exhaled air at \(37^{\circ} \mathrm{C}\) and \(757 \mathrm{~mm} \mathrm{Hg}\):

Convert temperature and pressure to proper units

First, we need to convert the given temperature and pressure to Kelvin and atm: Temperature: \(37^\circ \mathrm{C} + 273.15 = 310.15 \mathrm{K}\) Pressure: \(757 \mathrm{~mm} \mathrm{Hg} \times \frac{1 \mathrm{~atm}}{760 \mathrm{~mm} \mathrm{Hg}} = 0.996 \mathrm{~atm}\)

Apply the ideal gas law to find the density

Using the ideal gas law, we can write the equation to find the density of the exhaled air: \(\rho = \frac{P \times M}{R \times T} \) Where \(\rho\) is the density, \(P\) is the pressure, \(M\) is the molar mass, \(R\) is the ideal gas constant (0.0821 \(\frac{L \cdot atm}{mol \cdot K}\)), and \(T\) is the temperature.

Calculate the density of exhaled air

Plugging in the values, we get: Density of Exhaled Air = \(\frac{0.996 \mathrm{~atm} \times 29.65 \frac{g}{mol}}{0.0821 \frac{L \cdot atm}{mol \cdot K} \times 310.15 \mathrm{K}} = 1.184 \frac{g}{L}\)

Compare the density of exhaled air with ordinary air

To find the density of ordinary air at the same conditions, we can use the given molar mass of 29.0 \(\frac{g}{mol}\) and apply the same calculations: Density of Ordinary Air = \(\frac{0.996 \mathrm{~atm} \times 29.0 \frac{g}{mol}}{0.0821 \frac{L \cdot atm}{mol \cdot K} \times 310.15 \mathrm{K}} = 1.158 \frac{g}{L}\) Comparing the two densities, we can see that exhaled air is slightly denser than ordinary air under the same conditions.