Molecular weight of vanillin = 152 g/mol
Futher molecular formula of vanillin is C8H8O3
Atomic weight of oxygen = 16 g/mol
Thus, 152 g of vanillin contains 16 g of oxygen
∴ 0.045 g (45 mg) of vanillin contains [tex] \frac{16X0.045}{152} [/tex] = 0.00473 g
Also, number of moles of vanillin in 0.045 g sample = [tex] \frac{weight}{molecular.weight} = \frac{0.045}{152} = 2.96X10^{-4} [/tex]
Now, 1 mole = 6.023 X 10^23 molecules
∴ 2.96 X 10^-4 mole = 1.78 X 10^20 molecules
From molecular formula, it can be seen that 1 molecule of vanallin contain 3 atoms of oxygen
∴1.78 X 10^20 molecules contain 3 X 1.78 X 10^20 = 5.34 X 10^20 oxygen atoms
