Find the w/w percentage nitrogen in an ammonium nitrate solution

Question

MCQ Question:

What is the w/w percentage of nitrogen (N) in an ammonium nitrate solution with a density of 1.05 g/cm3 and a molar concentration of 2.1 mol dm-3?

(a) 2.8 %

(b) 11.2 %

(c) 16.0 %

(d) 5.6 %

(e) 4.2 %

My attempt:

1 dm^3 of ammonium nitrate contains 2.1 mol (from 2.1 mol dm^-3)

1 dm^3 also contains 1050 g of ammonium nitrate (from 1.05 g/cm^3)

Mass of N in NH4NO3 = 28/80 = 7/20

Using Mr = mass/moles = 1050/2.1 = 500 g/mol of NH4NO3

Therefore Mr of N = 500 x (7/20) = 175 g/mol

I'm not sure what to do after this. The answer from the mark scheme is d) 5.6%.

Is there another method I should follow?

Answer 1

I don't understand your calculation. For example, on the 2nd line, you state that $\pu{1 dm^3}$ contains $1050$ g ammonium nitrate. No ! It contains $1050$ g of a solution. The ratio $7/20$ is exact, and I call it later $0.35$. But I don't understand why you mean with "Mr", and the next line.

I have another approach.

$\pu{2.1 mol}$ $\ce{NH4NO3}$ weighs : $\pu{2.1 mol· 80 g/mol = 168 g}$.

The mass of $\ce{N}$ in $168$ g $\ce{NH4NO3}$ is $0.35·168 g = 58.8$ g

The weight proportion on $\ce{N}$ in the whole solution (weighing $1050$ g) is $58.8/1050 = 0.056 = 5.6$%