Question

At STP, 0.106 g of a gas sample occupies a volume of 62.2 mL. What is the molar mass of this gas? Keep 3 sig figs for your final answer.

Answer 1

Is it three sig figs or four?

Answer 2

Start with the ideal gas equation; we need number of moles n so we solve for n
\[pV = nRT \]

then our equation becomes this
\[\frac{ pV }{ RT } = n \]

\[STP = 1 atm, 298K, 25C = Standard temperature and pressure \]

plug everything in
\[\frac{ pV }{ RT } = n \frac{ 1atm*0.062L }{ (0.08L, atm, mol, * 298) } = 2.54x10^{-3} mol\]

we know that 0.00254 moles contains .106 grams but how many grams are in 1 mole.
we can do this by setting up a proportion.

\[\frac{ mol }{ x,grams } = \frac{ 0.00254,moles }{ 0.106 grams }\]

cross multiply

\[0.00254 moles, x = 1.06 grams \]

1 mol contains 44.732 grams

\[\frac{ 0.106grams }{ 0.00254mol } = 44.732 grams \]