urea is a very important nitrogenous fertilizer.Its formlais CON2H4.Calculate the percentage of nitrogen in urea.(C=12,O=1,N=14 and H=1)
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urea=> NH2-CO-NH2
total mass=60
mass of nitrogen = 14, 2 molecules hence m=28
N%= mass of nitrogen X 100
mass of urea
= 28 X 100
60
N%= 46.6%
total mass=60
mass of nitrogen = 14, 2 molecules hence m=28
N%= mass of nitrogen X 100
mass of urea
= 28 X 100
60
N%= 46.6%
Answered by
3
Elemental composition of CON2H4: ... C, Carbon, 12.0107, 1, 19.9994. O, Oxygen ... N, Nitrogen, 14.0067, 2, 46.6460.
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