The weight of a bottle filled with $250 \; mL$ air is $ 152.827 \;g $
The bottle filled with butane has a weight of $ 153.131 \; g $
The measurements were made at $ 21 ^o C $ and $ 770 \; mm \; Hg$
Calculate the molar mass of butane!
$\rho_{air}=1,293\;\frac{g}{L}$ s.t.p.
First calculate the mass $m_1$ of the air:
Let $ V_o $ (L) be the volume that air would occupy s.t.p.
Ideal Gas Law:
$\frac{V_o}{V}=\frac{P\cdot T_o}{P_o\cdot T}$
$\frac{V_o}{0,25}$ $=$ $\frac{770\cdot 273}{760\cdot 294}$
$V= 0,235 \;L$
Mass of the air:
$m_1=1,293\cdot 0,235$ $= $ $0,304\; g$
Then calculate the mass $m_2$ of butane:
Masse of the bottle = $152,827-0,304= 152,523\; g$
Mass of the butane:
$m_2=153,131-152,523=0,608\; g$
Then calculate the number of moles $n_2$ of butane:
Ideal Gas Law:
$P\cdot V=n_2RT$
with $P=\frac{770}{760}\; atm$
$n_2=\frac{770\cdot 0,250}{760\cdot 0,082\cdot 294}=0,01049\; mol$
Finally calculate the molar mass $M_2$ of butane:
$M_2=\frac{m_2}{n_2}$ $ =$ $\frac{0,608}{0,01049}=58,0 \frac{g}{mol}$
