34.05 mL of phosphorus vapour weighs 0.0625 g at 546 °C and 0.1 bar pressure. What is the molar mass of phosphorus?

I tried to solve the above question and got 1249.8g. The book's answer sheet lists the answer to be 1247.7 g.

I don't understand how the heck that happened? I'm pretty sure a mole of phosphorous doesn't weigh more than 1.2 kg.


To start your book didn't use significant figures properly. There is no way that the weight can have 3 significant figures in the mass and then conjure 6 significant figures in the molecular weight. To make matters worse the pressure is only given to 1 significant figure. YUCK!

Use PV = nRT to determine the moles of gas present.

$\mathrm{n =}\dfrac{\mathrm{PV}}{\mathrm{RT}} = \dfrac{0.1 \text{ bar} \times 0.03405 \text{ L}}{\text{0.08314 L bar K}^{−1} \text{mol}^{−1} \times 819\text{ K}} = 5.000 \times 10^{-5}\text{ mol}$

$\text{MW} = \dfrac{\text{m}}{\text{mol}} = \dfrac{ 0.0625\text{ g}}{5.000 \times 10^{-5}\text{ mol}} = 1,249.8\text{ g/mol}$

This is of course wrong. It would seem that the pressure should have been 1.00 bar. That would have given an answer of 125 g/mol. Using 31.0 g/mol as the atomic weight of phosphorous would reveal that the phosphorous vapor was $\ce{P_4}$.

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