What is the molecular weight of a gas if a 15.0 g sample has a pressure of 836 mm Hg at 25°C in a 2.00 L flask? (R= 0.0821 L atm/ mol K)

1 Answer
Apr 9, 2018

The molecular weight of the gas is #"167 u"#.

The molar mass is #"167 g/mol"#.

Explanation:

Use the equation for the ideal gas law to determine moles of the gas. Then divide the mass of gas by the moles to get molar mass.

#PV=nRT#,

where:

#P# is pressure, #V# is volume, #n# is moles, #R# is the gas constant, and #T# is the temperature in Kelvins.

Since the given value for #R# is in #"L atm"/"mol K"#, you need to convert the pressure in #"mmHg"# to #"atm"#.

#"1 atm"##=##"760.0 mmHg"#

#836color(red)cancel(color(black)("mmHg"))xx("1 atm")/(760.0color(red)cancel(color(black)("mmHg")))="1.10 atm"#

The given temperature is #25^@"C"#, so you need to convert it to Kelvins.

#25^@"C + 273.15"="298 K"#

Organize the data:

Known/Given

#P="1.10 atm"#

#V="2.00 L"#

#R="0.0821 L atm K"^(-1) "mol"^(-1)"#

#T="298 K"#

Unknown

#n#

Solution

Rearrange the equation to isolate #n#. Plug in the known values and solve.

#n=(PV)/(RT)#

#n=((1.10color(red)cancel(color(black)("atm")))xx(2.00color(red)cancel(color(black)("L"))))/((0.0821 color(red)cancel(color(black)("L")) color(red)cancel(color(black)("atm")) color(red)cancel(color(black)("K"))^(-1) "mol"^(-1))xx(298color(red)cancel(color(black)("K"))))="0.0899 mol"#

Molar mass of gas

#M=("15.0 g")/("0.0899 mol")="167 g/mol"# (rounded to three significant figures)

Molecular weight is numerically equal to the molar mass.

#"M"_r="167 u"#