A gaseous hydrocarbon collected over water at a temperature of 21° C and a barometric pressure of 753 torr occupied a volume of 48.1 mL. The hydrocarbon in this volume weighs 0.1133 g. What is the molecular mass of the hydrocarbon?
1 Answer
Explanation:
Before focusing on anything else, use the vapor pressure of water at
As you know, gases collected over water will also contain water vapor. Simply put, the volume of gas collected over water will contain molecules of hydrocarbon and molecules of water.
This means that you can use Dalton's Law of partial pressures to determine the exact pressure of the hydrocarbon. At
http://www.endmemo.com/chem/vaporpressurewater.php
This means that you can write
#P_"mixture" = P_"water" + P_"hydrocarbon"#
#P_"hydrocarbon" = "753 torr" - "18.59 torr" = "734.41 torr"#
Your next step will be to use the ideal gas law equation to determine how many moles of this gaseous hydrocarbon you have in this sample.
#color(blue)(PV = nRT)" "# , where
Plug in your values and solve for
#PV = nRT implies n = (PV)/(RT)#
#n = (734.41/760color(red)(cancel(color(black)("atm"))) * 48.1 * 10^(-3)color(red)(cancel(color(black)("L"))))/(0.0821( color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 21.0)color(red)(cancel(color(black)("K"))))#
#n = "0.0019247 moles"#
So, your hydrocarbon sample has a mass of
#1 color(red)(cancel(color(black)("mole"))) * "0.1133 g"/(0.0019247color(red)(cancel(color(black)("moles")))) = "58.87 g"#
Rounded to three sig figs, the answer will be
#M_M = color(green)("58.9 g/mol")#