The following problem is from Principles of General Chemistry, Silberberg, 1st edition:

One way to utilize naturally occurring uranium ($0.72\%$ $\ce{^235U}$ and $99.27\%$ $\ce{^238U}$) as a nuclear fuel is to enrich it (increase its $\ce{^235U}$ content) by allowing gaseous $\ce{UF6}$ to effuse through a porous membrane. From the relative rates of effusion of $\ce{^235UF6}$ and $\ce{^238UF6}$, find the number of steps needed to produce uranium that is $\ce{3.0 mol}\ \%$ $\ce{^235U}$, the enriched fuel used in many nuclear reactors.

Elsewhere in the text the following ratio is given:

$$\frac{\text{Rate}(\ce{^235UF6})}{\text{Rate}(\ce{^238UF6})} = \sqrt\frac{M(\ce{^238UF6})}{M(\ce{^235UF6})} = 1.0043$$

I've found two different ways to solve this that give different answers.

  1. Solution 1 based on similar problem in LibreTexts Chemistry:

$$\begin{align} 1.0043^n &= \frac{0.030}{0.0072} \\[5pt] n &= 333 \end{align}$$

  1. Solution 2 based on similar problem in Principles of Modern Chemistry by Oxtoby, Gillis, Butler:

$$\begin{align} 1.0043^n &= \left(\frac{3.0}{97.0}\right) \left(\frac{99.27}{0.72}\right) \\[5pt] n &= 338 \end{align}$$

Which of these approaches is correct? 2 makes more sense to me but 1 is closer to the answer given in the book (332).


The second approach looks more correct to me. The first approach is using raw percentages without converting to a ratio between isotopes, and the ratio rates is going to be directly related to the ratio of isotopes, not raw percentages.

From the comments, ed: Well, $1.0043$ is the ratio between rates, so each stage will cause the ratio between isotopes to change by this factor. I simple set up: $$\frac{0.72}{99.27} \times 1.0043^{n} = \frac{3}{97}$$ which is the same equation as method 2.

$\frac{0.72}{99.27}$ is the initial ratio (before any enrichment). You need to increase it successively by $1.0043$ until you get to $\frac{3}{97}$, the desired mol%.

  • 3
    $\begingroup$ Can you expand on this, perhaps with numerical examples that justify your answer? As it reads now, it's a comment and is primarily opinion-based with no quantitative information illustrating/justifying your reasoning. $\endgroup$ – Todd Minehardt Oct 17 '16 at 20:21
  • $\begingroup$ Right that's where solution 2 came from. Should have been more explicit. $\endgroup$ – abjske Oct 17 '16 at 21:34

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