Question

calculate the percentage composition of nitrogen in urea. urea formula is NH2CONH4. N=14
H=1
C=12
O=16

Answer 1

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%

Answer 2

Answer: Percentage of nitrogen in urea is 46.66 %.

Explanation:

The chemical formula for urea is $NH_2CONH_2$

To calculate the percentage composition of an element in a compound, we use the equation:  

$\%\text{ composition}=\frac{\text{Mass of nitrogen element}}{\text{Mass of compound}}\times 100$

In 1 mole of urea, 2 moles of nitrogen atoms are there.

Mass of urea = $[(2\times 14)+16+12+(4\times 1))]=60g/mol$

Mass of nitrogen element = $[2\times 14]=28g/mol$

Putting values in above equation, we get:  

$\%\text{ composition of nitrogen}=\frac{28}{60}\times 100=46.66\%$

Hence, percentage of nitrogen in urea is 46.66 %.