how many grams of nitrogen are present in 46.34 grams of ammonium nitrate?

Answer 1

ammonium nitrate is NH4NO3

moleculear weight is 80 g / mole
nitrogen weight per mole is 28 g / mole

% N by weight in a mole = 100 * 28 /80 = 35%

46.34 g amonium nitrate is 46.34*0.35= 16.22 g N

Answer 2

molar mass=mass/# of moles

ammonium nitrate has a formula of NH4NO3 and has a molar mass of 14+4+14+(16x3)=80g/mol

therefore # of moles of NH4NO3 = 46.34g/80g/mol
=0.57925 mol.
N has a molar mass of 14g/mol

from the equation of formation :
N2+ 2H2+ 3/2 O2------> NH4NO3

THE MOLE RATIO OF N TO NH4NO3 IS 1:1
therefore:
0.57925 mol of N reacts to produce 0.57925 mol of NH4NO3

hence;
mass of N =molar mass x # of moles
= 14g/mol x 0.57925 mol
=8.11g