Question

calculate the percentage of nitrogen in ammonium nitrate (NH4 NO3) (N=14,H=1,O=16)​

Answer 1

Answer:

Molecular weight = 14 + 1×4+14 +16×3 =14+4+14+48 = 80

% of N = (14+14)/ 80 × 100 = 28/80×100 =35%

Explanation:

Hope it helped

Answer 2

Answer: 35%

Explanation:

$%N\ in\ NH_{4}NO_{3}=(Mass\ of\ nitrogen\ present\ in\ compound/Total\ amount\ of\ individual\ elements)\times 100\\\\Molecular\ weight\ of\ NH_{4}NO_{3}=2\times 14+4+16\times 3=28+4+48=80\\\\%N=(2\times 14/80)\times 100=(28/80)\times 100=35%$

The Percentage element in a compound is equal to the amount of that particular element in the compound divided by the total amount of individual elements present in the compound multiplied by 100.

Amount of N in compound Ammonium Nitrate=2×14=28

Molecular weight of compound=80

$%N=28/80\times 100=35%$

%N= (28/80)×100 = 35%

35% of Nitrogen is present in Ammonium nitrate.