# Calculating mass of krypton from mole fraction

There are two gases in a container: krypton and carbon dioxide. If the mass of the gases is 35 grams, and total pressure of the container is 0.708 atm, and the pressure of krypton is 0.250 atm. What is the mass of the krypton?

I've found the mole fraction of krypton is 0.3531. First, I think if I multiply mole fraction of krypton and mass of the gases, I'd get the mass of krypton. But my teacher said if the answer is false. So what is the correct answer?

• I think your method is correct. Jan 19, 2014 at 8:07

Let $m_{\ce{CO2}}$ be the mass of carbon dioxide and $m_{\ce{Kr}}$ be the mass of krypton in the vessel. These two are related by:

$$m_{\ce{CO2}}+m_{\ce{Kr}}=35\ \text{g}\ \ \ \ \ (1)$$

If we can get a second equation in both variables, you can solved the system for each variable.

You were able to determine the mole fractions $\chi$:

$$\chi_{\ce{Kr}}=\frac{n_{\ce{Kr}}}{n_T}=\frac{P_{\ce{Kr}}}{P_T}=\frac{0.250\ \text{atm}}{0.708 \ \text{atm}}= 0.353\ \ \ \ \ (2)$$

$$\chi_{\ce{CO2}}=1-\chi_{\ce{Kr}}=0.647 \ \ \ \ \ (3)$$

Let $n_{\ce{CO2}}$ be the number of moles of carbon dioxide and $n_{\ce{Kr}}$ be the the number of moles of krypton in the vessel (its a little easier to work in moles). We can generate two relationships between the numbers of moles using 1) the masses and 2) the mole fractions.

From the masses, since the mass of a sample of substance is equal to the number of moles times the molar mass (or formula weight).

$$m_x=n_x\cdot FW_x\ \ \ \ \ (4)$$

$$m_\ce{Kr}=n_\ce{Kr}\cdot FW_\ce{Kr}=n_\ce{Kr}\cdot (83.798\ \text{g/mol})\ \ \ \ \ (5)$$ $$m_{\ce{CO2}}=n_{\ce{CO2}}\cdot FW_{\ce{CO2}}=n_{\ce{CO2}}\cdot (44.01\ \text{g/mol})\ \ \ \ \ (6)$$ $$n_{\ce{CO2}}\cdot (44.01\ \text{g/mol})+n_\ce{Kr}\cdot (83.798\ \text{g/mol})=35\ \text{g}\ \ \ \ \ (7)$$

From mole fractions

$$\dfrac{\chi_{\ce{CO2}}}{\chi_{\ce{Kr}}}=\dfrac{\dfrac{n_{\ce{CO2}}}{n_T}}{\dfrac{n_{\ce{Kr}}}{n_T}}=\dfrac{n_{\ce{CO2}}}{n_{\ce{Kr}}}=\dfrac{0.647}{0.353}=1.832\ \ \ \ \ (8)$$ $$n_{\ce{CO2}}=1.832\cdot n_{\ce{Kr}} \ \ \ \ \ (9)$$

Substitute equation (9) into equation (7) and solve for $n_{\ce{Kr}}$ and convert to $m_{\ce{Kr}}$.