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A sample of $\pu{1.42 g}$ of helium and an unweighted quantity of oxygen gas are mixed in a flask at room temperature. The partial pressure of helium in the flask is $\pu{42.5 torr}$, and partial pressure of oxygen gas is $\pu{158 torr}$. What is the mass of the oxygen in the container?

As volume is not stated, I assume it's constant, and temperature is a constant too. Therefore, from the ideal gas law

$$\frac{p}{n} = \frac{RT}{V}$$

For helium and oxygen gas, the above formula is applicable, and the mass of oxygen can be determined:

\begin{align} \frac{\pu{0.5592 atm}}{1.42/4.0026} &= \frac{\pu{0.20789 atm}}{m(\ce{O2})/31.99} \\ \to m(\ce{O2}) &= \pu{4.22 g} \end{align}

This is my attempt on finding the mass. However, I believe it's not right. Why is that the case?

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  • $\begingroup$ Why did you think 4.2 grams was incorrect? You have a case of equationitis using equations instead of thought patterns. Start over taking one step at a time expressing the chemical-math principle involved, ensure correctness, then proceed to the next step, etc. Dump the calculator and use a slide rule or do longhand arithmetic become intimate with the calculations. $\endgroup$
    – jimchmst
    Mar 11 at 20:06

1 Answer 1

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I don't see any flaws in your argumentation. I agree that $R, T, V = \mathrm{const}$, hence

$$\frac{p(\ce{He})}{n(\ce{He})} = \frac{p(\ce{O2})}{n(\ce{O2})}\tag{1}$$

$$m(\ce{O2}) = \frac{p(\ce{O2}) \cdot M(\ce{O2})}{p(\ce{He}) \cdot M(\ce{He}) } \cdot m(\ce{He})\tag{2}$$

Just one minor thing: you incorrectly converted partial pressure of helium: $\pu{42.5 Torr}$ is $\pu{0.056 atm}$, not $\pu{0.559 atm}$, therefore your resulting mass should be 10 times higher.

In fact, there was no need to convert $\pu{Torr}$ to $\pu{atm}$ in the first place. Don't do unnecessary steps if you can avoid them. At the same time don't drop any units and always remember to include them in your final calculation.

$$m(\ce{O2}) = \frac{(\pu{158 Torr})(\pu{31.99 g mol-1})}{(\pu{42.5 Torr})(\pu{4.00 g mol-1})}(\pu{1.42 g}) = \pu{42.22 g}\tag{3}$$

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