What mass in "mg" of "O"_2 gas is found in "469.9 mL" of it contained at "755 mm Hg" and 60^@ "C"?

May 13, 2017

The Creative Chemist spots the volume $V$ of gas, as well as its pressure $P$ and temperature $T$ from a mile away and immediately jumps to the conclusion that the major assumption is the gas being ideal.

He thus pulls out the ideal gas law from out of nowhere (presumably some textbook from underneath his dusty study desk), and carefully examines the units.

$P V = n R T$

He is unfazed by the volume and pressure units of $\text{mL}$ and $\text{mm Hg}$, and brushes off the temperature units of $\text{^@ "C}$ as something he sees all the time.

Looking only for the mass of the gas, he filters out the rest and realizes that the mols are:

$n = \frac{P V}{R T}$

and thus that the mass is merely the molar mass $M$ multiplied by the whole thing:

$n M = m = \frac{P M V}{R T}$

He is admittedly too lazy to use the universal gas constant in any other units than $\text{L"cdot"atm/mol"cdot"K}$ for these types of calculations, and performs rapid unit conversion, morphing the units of everything else to match $R$:

$\textcolor{b l u e}{{m}_{{O}_{2} \left(g\right)}} = \left(\left(755 \cancel{\text{mm Hg" xx cancel"1 atm"/(760 cancel"torr"))("31.998 g/"cancel"mol")(469.9 cancel"mL" xx cancel"1 L"/(1000 cancel"mL")))/((0.082057 cancel"L"cdotcancel"atm""/"cancel"mol"cdotcancel"K")(60 + 273.15 cancel"K}}\right)\right)$

$=$ $\textcolor{b l u e}{{\text{0.546 g O}}_{2} \left(g\right)}$

He is feeling nice enough to give the mass in $\text{mg}$...

$=$ $\text{546 mg}$