Question

'A sample contains both CO2 and Ne in unknown quantities. If the sample contains a combined total of 1.75 mol and has a total mass of 65.3 g, what are the percentages of CO2 and Ne in the sample by mole?

Answer 1

%`CO_2` by mole = 71.9%

%`Ne` by mole = 28.1%

Explanation:

Molecular weight of `CO_2` is 12.011+2(15.999)=44.01 gm/mole

Molecular weight of `Ne` is 20.18 gm/mole.

Let the number of moles of `CO_2` be `n_(CO_2)`.

Let the number of moles of `Ne` be `n_(Ne)`.

From the problem statement

`n_(CO_2)+n_(Ne)=1.75`

and

`44.01n_(CO_2)+20.18n_(Ne)=65.3`

Solve by substitution

`44.01n_(CO_2)+20.18(1.75-n_(CO_2))=65.3`

`n_(CO_2)=(65.3-20.18*1.75)/(44.01-20.18)=1.258` moles of `CO_2`

`n_(Ne)=1.75-1.258=0.492` moles of `Ne`.

%`CO_2` by mole = `1.258/1.75=71.9%`

%`Ne` by mole = `0.492/1.75=28.1%`