Question

What is the molecular mass of a compound if 560mL have a mass of 1.10g at STP?

Answer 1

The molar mass (molecular mass in grams) is `"45 g"`.

Explanation:

Note: This is going to be a long answer.

First determine the number of moles.

`"STP=273.15 K and 100 kPa"`

Use the ideal gas law.

`PV=nRT`,
where `P` is pressure in kiloPascals `("kPa")`, `V` is volume in liters `("L")`, `n` is moles, `R` is the gas constant, and `T` is temperature in Kelvins `("K")`.

Given/Known

`P="100 kPa"`

`T="273.15 K"`

`V=560cancel"mL"xx(1"L")/(1000cancel"mL")="0.56 L"`

`R="8.3144598 L kPa K"^(-1) "mol"^(-1)"`

Unknown

moles, n

Equation

`PV=nRT`

Solution
Rearrange the equation to isolate `n` and solve.

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

`n=((100cancel"kPa")(0.56cancel"L"))/((8.3144598cancel "L" cancel"kPa"cancel" K"^(-1) "mol"^(-1)")` `=` `"0.024658 mol"`

`n"=0.024658 mol"` (keeping a couple of guard digits)

Determine the molecular (molar) mass.

Given/Known

`n"=0.024658 mol"`

`"m"="mass"="1.10 g"`

Unknown

Molar mass: `"MM"`

Equation

`n=("m")/("MM")`

Solution
Rearrange the equation to isolate `"MM"` and solve.

`"MM"xx("m")/(n)`

`"MM"xx(1.10"g")/(0.024658 "mol")=`

`"MM"="45 g/mol"` (rounded to two significant figures due to 0.56 L)

Answer 2

`"45 g/mol"`

Explanation:

The idea here is that you need to use the fact that at STP conditions, one mole of any ideal gas occupies exactly `"22.7 L"` - this is known as the molar volume of a gas at STP.

This means that if you know the volume of the gas, you can determine how many moles the sample contains by using the known molar volume at STP

`560color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(1000color(red)(cancel(color(black)("mL")))) * "1 mole"/(22.7color(red)(cancel(color(black)("L")))) = "0.02467 moles"`

The molar mass of the compound, which tells you what the exact mass of one mole of the gas is, can be determined by using the mass of the sample.

`M_"M" = m/n`

`M_"M" = "1.10 g"/"0.02467 moles" = "44.59 g/mol"`

Rounded to two sig figs, the nswer will be

`M_"M" = color(green)("45 g/mol")`