In order to solve this problem, you will need to use the ideal gas law:
#PV=nRT#,
where:
#P# is pressure, #V# is volume, #n# is moles, #R# is the gas constant, and #T# is the temperature in Kelvins.
The gas laws use the Kelvin temperature scale, so the temperature in Celsius must be converted to Kelvins by adding 273.15..
Determine the molar mass of butane, #"C"_4"H"_10"# by multiplying the atomic weight for each element by its subscript. Add the molar masses of each element, and record the answer in g/mol. (You can also look up the molar mass.)
#MM_"CH4H10"=(4xx"12.011 g/mol C") + (10xx"1.008 g/mol H")="58.124 g/mol"#
Known:
#P = "0.963 atm"#
#V = "20 L"#
#R ="0.08206 L atm /K mol"#
#T = "27"^@"C + 273.15 = 300 K"#
#MM_"butane" = 58.124 "g/mol"#
Unknown:
mass of butane
Moles Butane
Rearrange the equation to isolate #n#. Plug in the known values and solve.
#n = (PV)/(RT)#
#n =(0.963 color(red)cancel(color(black)("atm"))xx20 color(red)cancel(color(black)("L")))/(0.08206 color(red)cancel(color(black)("L")) color(red)cancel(color(black)("atm")) color(red)cancel(color(black)("K"))^(-1) "mol"^(-1)xx300color(red)cancel(color(black)("K")))= "0.8 mol C"_4"H"_10"# (rounded to one significant figure because 20 L has only one significant figure)
Determine the mass of butane in the cylinder.
#0.8 color(red)cancel(color(black)("mol C"_4"H"_10))xx(58.124 "g C"_4"H"_10)/(1color(red)cancel(color(black)("mol C"_4"H"_10)))="50 g C"_4"H"_10"# (rounded to one significant figure)
Convert mass in grams to kilograms.
#50color(red)cancel(color(black)("g C"_4"H"_10))xx(1"kg")/(1000color(red)cancel(color(black)("g")))="0.05 kg C"_4"H"_10"#
The mass of butane in the cylinder is #"50 g"#, or #"0.05 kg"#.
