# A gas sample weighing 3.78 grams occupies a volume of 2.28 L at STP. What is the molecular mass of the sample?

##### 1 Answer

#### Explanation:

The trick here is to realize that if you know the *volume* of a gas at **STP**, you can use the fact that **mole** of any ideal gas occupies **moles** of gas you have in your sample.

#color(red)(ul(color(black)("Under STP conditions: " "1 mole of an ideal gas = 22.7 L")))#

In your case, you know that your sample of gas occupies * currently* defined as a pressure of

This means that your sample will contain

#2.28 color(red)(cancel(color(black)("L"))) * overbrace("1 mole gas"/(22.7color(red)(cancel(color(black)("L")))))^(color(blue)("molar volume of a gas at STP")) = "0.10044 moles gas"#

Now, the **molar mass** of the gas is the mass of exactly **mole** of the gas. In your case, you know that you get **for every** **moles**, which means that you have

#1 color(red)(cancel(color(black)("mole"))) * "3.78 g"/(0.10044color(red)(cancel(color(black)("moles")))) = "37.6 g"#

Since this is the mass of

#color(darkgreen)(ul(color(black)("molar mass = 37.6 g mol"^(-1))))#

The answer is rounded to three **sig figs**.

**SIDE NOTE** *A lot of textbooks and online sources still use the old definition of STP conditions, i.e. a pressure of*

*and a temperature of*

*Under these conditions,* **mole** of any ideal gas occupies*If this is the value of the molar volume of a gas at STP given to you, simply redo the calculations using* *instaed of*