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 $\ce{P_4}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.