Answer:
The correct option is (2) 35.0%
Explanation:
If we want to calculate the percent composition by mass of nitrogen in NH4NO3 we first need to calculate the mass of nitrogen in one mole of NH4NO3.
Because of the molecular form of NH4NO3 (it has 2 ''N'') we know that one mole of NH4NO3 contains 2 moles of N.
The molar mass of N (nitrogen) is [tex]14.0067\frac{g}{mole}[/tex]
This means that 1 mole of N contains 14.0067 grams of N.
Because we have 2 moles of N ⇒ [tex](2moles).(14.0067\frac{g}{mole})=28.0134g[/tex]
We know that in 1 mole of NH4NO3 we have 2 moles of N(28.0134 g of N).
To calculate the percent composition by mass of nitrogen we need to divide the mass of nitrogen in one mole of NH4NO3 by the mass of one mole of NH4NO3 and then multiply it by 100.
We know that in 1 mole of NH4NO3 there are 80.0 g of NH4NO3 (Because of the molar mass data) ⇒
[tex](\frac{28.0134g}{80.0g}).100=35.016[/tex] % ≅ 35.0%
We find that the percent composition by mass of nitrogen in NH4NO3 is 35.0%
We conclude that the correct option is (2) 35.0%
