Question

Calculate the number of oxygen atoms in 3.2 g of $H_2{\mathrm{SO}}_4$? (3 marks)

Answer 1

Given:
Mass of $H_2{\mathrm{SO}}_4$ = 3.2 g
No.of. oxygen atoms = ?
Solution:
Molar mass of $H_2{\mathrm{SO}}_4$
= $(2\times 1)+32+(4\times 16)$= 98 g/mol (0.5 mark)
No. of moles = $\frac{\mathrm{Mass}\mathrm{in}\mathrm{grams}}{\mathrm{Molar}\mathrm{mass}}$= $\frac{3.2}{98}$= 0.032 moles (0.5 mark)
No. of molecules of $H_2{\mathrm{SO}}_4$ present in 0.032 moles
= No. of moles × Avogadro's number = $0.032\times 6.023\times {10}^{23}$= $0.193\times {10}^{23}$ (1 mark)
One molecule of $H_2{\mathrm{SO}}_4$ contains 4 atoms of oxygen.
No. of oxygen atoms present in $0.193\times {10}^{23}$ molecules of $H_2{\mathrm{SO}}_4$
= $0.193\times {10}^{23}\times 4$= $7.72\times {10}^{22}$ (1 mark)
Hence, 3.2 g of $H_2{\mathrm{SO}}_4$ contains $7.72\times {10}^{22}$ atoms of oxygen.