# 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

*number of mole* of gas

Plug in your values and solve for **match** those used in the expression of the universal gas constant

#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 **molar mass**, which tells you what the mass of **one mole** of a substance is, will be

#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")#