calculate the percentage composition of the elements present in urea[CO(NH₂)₂]

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Solution

$M o l e c u l a r m a s s o f u r e a = {(N H_{2})}_{2} C O = 2 \times 14 + 4 \times 1 + 1 \times 12 + 1 \times 16 = 28 + 4 + 12 + 16 = 60 g % o f n i t r o g e n = \frac{28}{60} \times 100 = 46.67 % % o f h y d r o g e n = \frac{4}{60} \times 100 = 6.67 % % o f c a r b o n = \frac{12}{60} \times 100 = 20 % % o f o x y g e n = \frac{16}{60} = 26.67 %$

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