# How many grams of helium are contained in a #"2.0-L"# balloon at #30.0^@"C"# and at #"735 mmHg"# ?

##### 1 Answer

#### Explanation:

The idea here is that you can use the **ideal gas law equation** to find the number of *moles* of helium present in your sample and the **molar mass** of the gas to convert that to *grams*.

You could also rewrite the ideal gas law equation in terms of the mass of helium present in the sample to get the answer directly.

So, you know that the ideal gas law equation looks like this

#color(blue)(ul(color(black)(PV = nRT)))#

Here

#P# is the pressure of the gas#V# is the volume it occupies#n# is the number of moles of gas present in the sample#R# is theuniversal gas constant, equal to#0.0821("atm L")/("mol K")# #T# is theabsolute temperatureof the gas

You also know that the mass of the gas, let's say

#color(blue)(ul(color(black)(m = n/M_M)))#

This means that you can rewrite the ideal gas law equation as

#PV = m/M_M * RT#

Rearrange to solve for

#m = overbrace((PV)/(RT))^(color(blue)("the number of moles")) * M_M#

Finally, plug in your values to find the mass of the sample--do not forget to convert the temperature from degrees Celsius to Kelvin and the pressure from mmHg to atmospheres!

#m = ( 735/760 color(red)(cancel(color(black)("atm"))) * 2.0color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (30.0 + 273.15)color(red)(cancel(color(black)("K")))) * "4.003 g" color(red)(cancel(color(black)("mol"^(-1))))#

#color(darkgreen)(ul(color(black)(m = "0.31 g")))#

The answer is rounded to two **sig figs**, the number of sig figs you have for the volume of the gas.