1. A sample of ethane C2H6 diffuses at a rate of 3.6 x 10^-6 mol/hr. An unknown gas under the same conditions diffuses at a rate of 1.3 x 10^-6 mol/hr. Calculate the molar mass of... 1. A sample of ethane C2H6 diffuses at a rate of 3.6 x 10^-6 mol/hr. An unknown gas under the same conditions diffuses at a rate of 1.3 x 10^-6 mol/hr. Calculate the molar mass of the unknown gas. 2. Which gas in each of the following pairs diffuses more rapidly, and what are the relative rates of diffusion? a. Kr and O2 b. N2 and acetylene (C2H2) Read more

Steps to Solve

1. Calculate the molar mass of ethane ($C_2H_6$)

The molar mass of ethane ($C_2H_6$) is calculated by summing the atomic masses of its constituent atoms: $2 \times \text{Atomic mass of Carbon} + 6 \times \text{Atomic mass of Hydrogen} = (2 \times 12.01) + (6 \times 1.008) = 24.02 + 6.048 = 30.068 \text{ g/mol}$. Rounding to a reasonable number of significant figures gives $30.07 \text{ g/mol}$.

2. Apply Graham's Law of Diffusion

Graham's Law states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. This can be expressed as:

$$ \frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{M_2}{M_1}} $$

where $\text{Rate}_1$ and $\text{Rate}_2$ are the rates of diffusion of gas 1 and gas 2, respectively, and $M_1$ and $M_2$ are their respective molar masses.

3. Rearrange Graham's Law to solve for the unknown molar mass

Let gas 1 be ethane and gas 2 be the unknown gas. We have: $\text{Rate}_1 = 3.6 \times 10^{-6} \text{ mol/hr}$, $M_1 = 30.07 \text{ g/mol}$, and $\text{Rate}_2 = 1.3 \times 10^{-6} \text{ mol/hr}$. We want to find $M_2$.

$$ \frac{3.6 \times 10^{-6}}{1.3 \times 10^{-6}} = \sqrt{\frac{M_2}{30.07}} $$

$$ \left(\frac{3.6}{1.3}\right)^2 = \frac{M_2}{30.07} $$

$$ M_2 = 30.07 \times \left(\frac{3.6}{1.3}\right)^2 $$

4. Calculate the molar mass of the unknown gas

$$ M_2 = 30.07 \times \left(\frac{3.6}{1.3}\right)^2 \approx 30.07 \times (2.769)^2 \approx 30.07 \times 7.667 \approx 230.5 \text{ g/mol} $$

5. Determine which gas diffuses more rapidly in pair a: Kr and $O_2$

Krypton (Kr) has an atomic mass of approximately 83.8 g/mol, while oxygen ($O_2$) has a molar mass of approximately 32.00 g/mol ($2 \times 16.00$). Since lighter gases diffuse more rapidly, $O_2$ will diffuse more rapidly than Kr.

6. Calculate the relative rates of diffusion for Kr and $O_2$

$$ \frac{\text{Rate}{O_2}}{\text{Rate}{Kr}} = \sqrt{\frac{M_{Kr}}{M_{O_2}}} = \sqrt{\frac{83.8}{32.00}} \approx \sqrt{2.619} \approx 1.62 $$ Oxygen diffuses approximately 1.62 times faster than Krypton.

7. Determine which gas diffuses more rapidly in pair b: $N_2$ and acetylene ($C_2H_2$)

Nitrogen ($N_2$) has a molar mass of approximately 28.02 g/mol ($2 \times 14.01$). Acetylene ($C_2H_2$) has a molar mass of approximately 26.04 g/mol ($2 \times 12.01 + 2 \times 1.008$). Since acetylene is slightly lighter than nitrogen, acetylene will diffuse more rapidly.

8. Calculate the relative rates of diffusion for $N_2$ and $C_2H_2$

$$ \frac{\text{Rate}{C_2H_2}}{\text{Rate}{N_2}} = \sqrt{\frac{M_{N_2}}{M_{C_2H_2}}} = \sqrt{\frac{28.02}{26.04}} \approx \sqrt{1.076} \approx 1.037 $$ Acetylene diffuses approximately 1.037 times faster than nitrogen.