A `0.109 g` sample of a pure gaseous compound occupies `112 mL` at `100^(@) C` and 750 torr. What is the molecular mass of the compound ?
Strategy : We first use the ideal gas law , `pV = nRT` , to find the number of moles of the gas . Then knowing the mass of that number of moles of the gas , we calculate the molar mass which is numerically equal to the molecular mass . Alternatively , we use Eq.(5.21) directly to get the molar mass.

Explanation

`p = 750 "torr" xx (1 "atm")/( "760 torr") = 0.987 "atm"`
`V = 0.112 L`
`T = 100^(@) C + 273^(@) = 373 K`
`n = (pV)/(RT) = ((0.987 atm) (0.112 L))/((0.0821 ("L atm")/("mol" K)) (373 K))`
`= 0.00361 "mol"`
Now , `n = ("Mass"(m))/("Molar mass" (M))`
`:. "Molar mass "(M) = ("Mass")/("Number of moles") = (0.109 g)/(0.00361 "mol")`
` = 30.2 g "mol"^(-1)`
Hence , molecular mass is `30.2 u "molecule"^(-1)`.
Alternatively,
`M("molar mass") = (mRT)/(pV)`.
`= ((0.109 g) (0.0821 L "atm" K^(-1) "mol"^(-1)) (373 K))/((0.987 "atm") (0.112 L))`
`= 30.2 g "mol"^(-1)`