Question

Calculate the percentage composition of each element in urea CON2H4 (C=12 N=14 O=16 and H=1)

Answer 1

$\Large{\underline{\underline{\bf{Given:}}}}$

• Compound CH₄N₂O (Urea)

$\Large{\underline{\underline{\bf{To\:find:}}}}$

• Percentage compisition of Carbon, Hydrogen, Nitrogen and Oxygen in the compound.

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ Molar mass of CH₄N₂O = 12+4+28+16 = 60

Percentage composition of carbon:

→ Molar mass of Carbon in CH₄N₂O =12

→ % Composition of C = molar mass of C/molar mass of CH₄N₂O×100

   % Composition of C = 12/60×100

   $\boxed{\bold{Percentage\:of\:carbon\:=\:20\%}}$

Percentage composition of hydrogen:

→ Molar mass of Hydrogen in CH₄N₂O = 4

→ % Composition of H = molar mass of H/molar mass of CH₄N₂O×100

  % Composition of H = 4/60×100

  $\boxed{\bold{Percentage\:of\:hydrogen\:=\:6.67\%}}$

Percentage composition of nitrogen:

→ Molar mass of Nitrogen in CH₄N₂O = 28

→ % Composition of N = molar mass of N/ molar mass of CH₄N₂O×100

   % Composition of N = 28/60×100

   $\boxed{\bold{Percentage\:composition\:of\:nitrogen\:=\:46.67\%}}$

Percentage composition of oxygen:

→ Molar mass of Oxygen in CH₄N₂O = 16

→ % Composition of O = molar mass of O/ molar mass of CH₄N₂O×100

  % Composition of O = 16/60×100

  $\boxed{\bold{Percentage\:composition\:of\:Oxygen\:=\:26.67\%}}$

Answer 2

Hope it helps you

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