Question

Vanillin, c8h8o3 (m = 152 g/mol), is the molecule responsible for the vanilla flavor in food. how many oxygen atoms are present in a 45.0 mg sample of vanillin?

Answer 1

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 $\frac{16X0.045}{152}$ = 0.00473 g

Also, number of moles of vanillin in 0.045 g sample = $\frac{weight}{molecular.weight} = \frac{0.045}{152} = 2.96X10^{-4}$
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