Question 10.5.2: Use Graham’s law of effusion to calculate molar mass. A samp......
Use Graham’s law of effusion to calculate molar mass.
A sample of ethane, C_{2}H_{6}, effuses through a small hole at a rate of 3.6 × 10^{-6} mol/h. An unknown gas, under the same conditions, effuses at a rate of 1.3 × 10^{-6} mol/hr. Calculate the molar mass of the unknown gas.
You are asked to calculate the molar mass of an unknown gas.
You are given the effusion rate and identity of a gas and the effusion rate of the unknown gas measured under the same conditions.
Use Equation 10.14 and the molar mass of ethane to calculate the molar mass of the unknown gas.
\frac{rate_{1}}{rate_{2}} = \sqrt{\frac{M_{2}}{M_{1}} }
\frac{3.6 \times 10^{-6} \text{ mol/h}}{1.3 \times 10^{-6} \text{ mol/h}} =\sqrt{\frac{M_{2}}{30.07 \text{g/mol}} }
M_{2} = 230 g/mol
Is your answer reasonable? The unknown gas effuses at a slower rate than ethane, so its molar mass should be greater than that of ethane.