# Finding molar mass of uranium fluoride using Graham’s law of diffusion

### Problem

A sample of uranium fluoride is found to effuses at the rate of $$\pu{17.7 mg/h}.$$ Under comparable conditions, gaseous iodine effuses at the rate of $$\pu{15.0 mg/h}.$$ What is the molar mass of the uranium fluoride?

$$\pu{354 g mol^-1}$$

### My approach

Perhaps I’m still having issues with significant figures, but I am getting the wrong answer. I know to use Graham’s law of diffusion, which leaves me with:

$$\frac{\pu{17.7 mg h^-1}}{\pu{15.0 mg h^-1}} = \sqrt\frac{\pu{253.8 g mol^-1}}{x}$$

I get $$x = \pu{181.70 g mol^-1}.$$ I know that the molecular mass of uranium fluoride should be less than iodine, so the answer makes no sense to me.

What am I doing wrong?

• Check the molar mass of uranium hexafluoride.
– Ed V
Apr 5 at 2:12
• Also check the exact wording of the problem, to see if the flow rates are correctly associated to the two gases.
– Ed V
Apr 5 at 2:35
• I agree with both the above - UF6 is 352 g/mol, and you get that if you take 15/17.7 rather than the other way around. Either you misread or they miswrote. Smaller is faster; KE = 1/2 mv^2 is the same on average for all gas particles moving through a pore at a given temperature. Apr 5 at 3:42

Graham's law of diffusion deals with the molar effusion rate rather than mass effusion rate.

This can be confirmed by the description of Graham's Law on Wikipedia.

An excerpt from the aforementioned page,

$$\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$$

where:
$$r_1$$ is the rate of effusion for the first gas (volume or amount per unit time)
$$r_2$$ is the rate of effusion for the second gas
$$M_1$$ is the molar mass of gas 1
$$M_2$$ is the molar mass of gas 2

Now the mathematical part,

\begin{align} r_{\ce{UF6}} &= \pu{17.7 mg/h} = \frac{\pu{17.7 mmol/h}}{M_{\ce{UF6}}} \\ r_{\ce{I2}} &= \pu{15 mg/h} = \frac{\pu{15 mmol/h}}{M_{\ce{I2}}} \end{align}

By Graham's Law,

\begin{align} \frac{r_{\ce{UF6}}}{r_{\ce{I2}}} &= \sqrt{\frac{M_{\ce{I2}}}{M_{\ce{UF6}}}} \\ \frac{17.7}{M_{\ce{UF6}}} \cdot \frac{M_{\ce{I2}}}{15} &= \sqrt{\frac{M_{\ce{I2}}}{M_{\ce{UF6}}}} \\ \frac{17.7}{15} &= \sqrt{\frac{M_{\ce{I2}}}{M_{\ce{UF6}}}} \cdot \frac{M_{\ce{UF6}}}{M_{\ce{I2}}} \\ \frac{17.7}{15} &= \sqrt{\frac{M_{\ce{UF6}}}{M_{\ce{I2}}}}\\ M_{\ce{UF6}} &= \left(\frac{17.7}{15}\right)^2 M_{\ce{I2}}\\ &= \pu{353.39 g mol^-1} \end{align}