Question

A sample of `O_2` gas is stored at 30.0 °C and 755 torr. If the volume was 125mL, how much did the oxygen weigh?

Answer 1

`"0.160 g"`

Explanation:

Your strategy here will be to

• use the ideal gas law equation to determine how many moles of oxygen gas you have in that sample

• use oxygen gas' molar mass to find how many grams would contain that many moles

The ideal gas law equation looks like this

`color(blue)(PV = nRT)" "`, where

`P` - the pressure of the gas
`V` - the volume it occupies
`n` - the number of moles of gas present in the sample
`R` - the universal gas constant, usually given as `0.0821 ("atm" * "L")/("mol" * "K")`
`T` - the temperature of the gas

Now, before plugging in your values, you need to make sure that the units given to you match those used in the expression of the universal gas constant.

More specifically, you need to have a pressure in atm, a temperature in Kelvin, and a volume in liters, so don't forget to keep track of the unit conversions needed to get you to these units.

So, rearrange the ideal gas law equation to solve for `n`

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

This means that you have

`n = (755/760color(red)(cancel(color(black)("atm"))) * 125 * 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 + 30.0)color(red)(cancel(color(black)("K")))) = "0.004989 moles O"_2`

As you know, a substance's molar mass tells you what the mass of one mole of that substance is. In this case, oxygen has a molar mass of `"31.9988 g/mol"`, which means that one mole of oxygen as will have a mass of `"31.9988 g"`.

This means that you sample will have a mass of

`0.004989 color(red)(cancel(color(black)("moles"))) * "31.9988 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)("0.160 g O"_2)`