How many grams of nitrogen are in 48.5g ammonium nitrate?

Answer 1

Ammonium Nitrate = NH4NO3

or

N2H4O3

the total mass of a ammonium nitrate molecule is as follows: 2*14u + 4*1u + 3*16u = 28u + 4u + 48u = 80u

now, the total mass of Nitrogen in one ammonium nitrate is 2*14u = 28u.

Then, we divide 28 (the mass of Nitrogen) by 80 (total mass)= 28/80 = 0.35, which is the ratio for Nitrogen mass divided by total mass.

then, we get the 48.5 grams (total mass) and multiply it by this ratio (0.35): 48.5 * 0.35 = 16.975 grams of Nitrogen in 48.5 grams of Ammonium Nitrate.

Answer 2

In ammonium nitrate (NH4NO3), there is one nitrogen atom in the formula. The molar mass of nitrogen is approximately 14 g/mol. Therefore, in 48.5 g of ammonium nitrate, there are approximately 14 g of nitrogen.