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

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

Jul 6, 2017

Here's how you can do that.

#### Explanation:

The idea here is that the molar mass of the gas represents the mass of exactly $1$ 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 $n$

$P V = n R T \implies n = \frac{P V}{R T}$

Now, you know that $\text{206 ng}$ of gas contain $n$ moles of gas. To convert the mass from nanograms to grams, use the fact that

$\text{1 g} = {10}^{9}$ $\text{ng}$

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) $\text{g}$

This means that you will have

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

$= \frac{2.06 \cdot {10}^{- 7}}{\frac{P V}{R T}} \textcolor{w h i t e}{.} \text{g" * 1/"1 mole}$

$= \frac{2.06 \cdot {10}^{- 7}}{\frac{P V}{R T}} \textcolor{w h i t e}{.} {\text{g mol}}^{- 1}$

To find the value of $n$, simply plug in your values into the ideal gas law equation. Do not forget that you need to do some conversions and that you have $R = 0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$.

• 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 $\approx {\text{63.0 g mol}}^{- 1}$.