Question

Find out the percentage composition of each element present in sulphuric acid

Answer 1

Mass of the components/ Total mass of compounds×100

S-atom is present

further molar mass mass of H2So4=2+32+64=98

Hence, percentage of sulphur32/ 98×100=32.65%

Hope it helps u❤️❤️❤️

Answer 2

Sulphuric acid (H₂SO₄)

 

Solution :

$\bf Molar \:\:mass\:\: of\:\: H_{2} SO_{4} = 2 \times \:\: Atomic \:\: mass\:\: of\:\: H + 1 \times Atomic\:\: mass\:\: of\:\: S + 4 \times Atomic\:\: mass \:\:of\:\: O$

 

Molar mass of H₂SO₄ = 2 × 1 + 1 × 32 + 4 × 16

Molar mass of H₂SO₄ = 98 u

 

Now,

$\bf Percentage\:\: of\:\:Hydrogen\:\: (H) = \frac{Mass\:\:of \:\:Hydrogen}{Molecular\:\:mass\:\:of\:\:H_2SO_4}\times 100$

$\bf Percentage \:\:of\:\: Hydrogen\:\:(H) = \frac{2\times 1u}{98u}\times 100$

Percentage of Hydrogen (H) = $\boxed{\boxed{\bf 2.04 \% }}$

 

Now,

$\bf Percentage \:\:of \:\:Sulphur\:\: (S) = \frac{Mass\:\:of \:\:Sulphur}{Molecular\:\:mass\:\:of\:\:H_2SO_4}\times 100$

$\bf Percentage \:\:of \:\:Sulphur\:\: (S) = \frac{32u}{98u}\times 100$

Percentage of Sulphur (S) = $\boxed{\boxed{\bf 32.65 \% }}$

   

Now,

$\bf Percentage\:\: of \:\:Oxygen \:\:(O) = \frac{Mass\:\:of \:\:Oxygen}{Molecular\:\:mass\:\:of\:\:H_2SO_4}\times 100$

$\bf Percentage \:\:of\:\: Oxygen \:\:(O) =\frac{4\times 16u}{98u}\times 100$

Percentage of Oxygen (O) = $\boxed{\boxed{\bf 65.30 \% }}$