Urea is a very important nitrogenous fertilizer. Its formula is $C O N_{2} H_{4}$ . Calculate the percentage of carbon in urea.

(C = 12, O = 16, N = 14 and H = 1)

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Solution

Urea has the formula $N H_{2} C O N H_{2}$ , i.e. $C O N_{2} H_{4}$
Molecular mass of urea = 60 g/mol

Atomic mass of N = 14 & number of atoms = 2 i.e. 28 g/mol of urea

Thus the % = $\frac{28}{60}$ $\times 100$ = 46.66 %

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Urea is a very important nitrogenous fertilizer. Its formula is CON2H4. Calculate the percentage of carbon in urea. (C = 12, O = 16, N = 14 and H = 1)

Urea has the formula NH2CONH2 , i.e. CON2H4 Molecular mass of urea = 60 g/mol Atomic mass of N = 14 & number of atoms = 2 i.e. 28 g/mol of urea Thus the % = 2860 × 100 = 46.66 %

Urea is a very important nitrogenous fertilizer. Its formula is $C O N_{2} H_{4}$ . Calculate the percentage of carbon in urea.

(C = 12, O = 16, N = 14 and H = 1)

Urea has the formula $N H_{2} C O N H_{2}$ , i.e. $C O N_{2} H_{4}$
Molecular mass of urea = 60 g/mol

Atomic mass of N = 14 & number of atoms = 2 i.e. 28 g/mol of urea

Thus the % = $\frac{28}{60}$ $\times 100$ = 46.66 %