Calculate the percentage composition of nitrogen in urea ( $H_{2} N C O N H_{2}$ ).

46.6%

N

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43.6%

N

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44.6%

N

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42.6%

N

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Solution

The correct option is C 46.6% $N$
Molecular weight of $H_{2} {N C O N H}_{2} = 4 + 28 + 12 + 16 = 60 g$
$60 g$ of urea contains $28 g$ of $N$
$∴$ % of $N = \frac{28}{60} \times 100 = 46.6$ %

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Calculate the percentage composition of nitrogen in urea (H2NCONH2).

The correct option is C 46.6% NMolecular weight of H2NCONH2=4+28+12+16=60g60g of urea contains 28g of N∴ % of N=2860×100=46.6%

Calculate the percentage composition of nitrogen in urea ( $H_{2} N C O N H_{2}$ ).

The correct option is C 46.6% $N$
Molecular weight of $H_{2} {N C O N H}_{2} = 4 + 28 + 12 + 16 = 60 g$
$60 g$ of urea contains $28 g$ of $N$
$∴$ % of $N = \frac{28}{60} \times 100 = 46.6$ %