Question

Lead forms three oxides A, B, and C the quantity of oxygen in each of the oxides A, B and C is 7.143 % ,10.345 % and13.133 % respectively. show that the law of proportions is obeyed

Answer 1

Answer: -

Mass of lead = 207

Mass of oxygen= 16

For the first oxide A

Oxide = 7.143

Lead = 100 - 7.143 = 88.57

Ratio of atoms = $\frac{88.57}{207}$ : $\frac{7.143}{16}$

= 0.43 : 0.44

= 1 : 1

For the first oxide B

Oxide = 10.345

Lead = 100 - 10.345 = 89.655

Ratio of atoms = $\frac{89.655}{207}$ : $\frac{710.345}{16}$

= 0.43 : 0.646

= 2:3

For the first oxide C

Oxide = 13.133

Lead = 100 - 13.133 = 86.867

Ratio of atoms = $\frac{86.867}{207}$ : $\frac{13.133}{16}$

= 0.41 : 0.82

= 1 : 2

Law of multiple proportions, statement that when two elements combine with each other to form more than one compound, the weights of one element that combine with a fixed weight of the other are in a ratio of small whole numbers.

From above we see the valencies of lead are 2+, 3+ and 4+. Hence they are whole numbers and the law of multiple proportions is observed.