# How do you calculate the molar mass of a gas at 347 torr and 77 degrees Celsius if 206 ng occupies 0.206 #mu#L?

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I know that I need to use the ideal gas equation, PV=nRT but unsure how to set up the problem.

I know that I need to use the ideal gas equation, PV=nRT but unsure how to set up the problem.

##### 1 Answer

Here's how you can do that.

#### Explanation:

The idea here is that the **molar mass** of the gas represents the mass of exactly **mole** of gas.

In order to calculate the molar mass of the gas, you need to divide the total mass of the sample of gas expressed in *grams* by the number of moles of gas it contains.

To find the number of moles of gas present in the sample, use the **ideal gas law equation**. I assume that you're familiar with how the ideal gas law equation works, so I'll leave the actual calculations for you as practice.

Rearrange the equation to solve for

#PV = nRT implies n = (PV)/(RT)#

Now, you know that **moles** of gas. To convert the mass from *nanograms* to *grams*, use the fact that

#"1 g" = 10^9# #"ng"#

You will end up with

#206 color(red)(cancel(color(black)("ng"))) * "1 g"/(10^9color(red)(cancel(color(black)("ng")))) = 2.06 * 10^(-7)# #"g"#

This means that you will have

#M_"M gas" = (2.06 * 10^(-7)color(white)(.)"g")/(n color(white)(.)"moles")#

# = (2.06 * 10^(-7))/((PV)/(RT))color(white)(.)"g" * 1/"1 mole"#

# = (2.06 * 10^(-7))/((PV)/(RT)) color(white)(.)"g mol"^(-1)#

To find the value of **Do not** forget that you need to do some conversions and that you have

#P:" " "torr " -> " atm " ["1 atm = 760 torr"]# #T:" " ""^@"C " -> " K " ["K" = ""^@"C" + 273.15]# #V: " " mu"L " -> " L " ["1 L" = 10^(6)color(white)(.)mu"L"]#

I think that the molar mass of the gas should come out to be