LAW OF RECIPROCAL PROPORTION:

If two elements B and C react with the same mass of a third element (A), the ratio in which they do so will be the same or simple multiple if B and C reacts with each other.

The above law is the basis of law of equivalent masses, which will be described latter in details.

For example: 

Ratio of masses of carbon and sulphur which combine with fixed mass (32 parts) of oxygen is

                                                12:32 or 3:8                                                                … (1)

In CS₂ ratio of masses of carbon and sulphur is 12: 64 or 3:16                                 … (2)

The two ratios (1) and (2) are related to each other by 3/8:3/18 or 2:1

ILLUSTRATIVE EXAMPLE (11): One part of an element A combines with two parts of B (another element). Six parts of element C combine with four parts of element B. If A and C combines together, the ratio of their masses will be governed by:

                        (A) Law of definite proportions                      (B) law of multiple proportions

                        (C) Law of reciprocal proportions                  (D) law of conservation of mass

SOLUTION:  

From this when A combined with C the ratio is A/C = 1/3 Hence (C) is correct.    

ILLUSTRATIVE EXAMPLE (12): Hydrogen combines with chlorine to form HCl. It also combines with sodium to form NaH. If sodium and chlorine also combine with each other, they will do so in the ratio of their masses as          

                        (A) 23: 35.5                                                     (B) 35.5: 23

                        (C) 1: 1                                                            (D) 23: 1

SOLUTION:  1 atom of hydrogen combined with 1 atom of chlorine and 1 atom of sodium. So when sodium reacts with chlorine the ratio is 1: 1 by atom. While 23: 35.5 by mass. Hence (A) is correct.

ILLUSTRATIVE EXAMPLE (13): Copper sulphide contains 66.6% Cu, copper oxide contains 79.9% copper and sulphur trioxide contains 40% Sulphur. Show that these data illustrates law of reciprocal proportions.

SOLUTION:  

In copper sulphides Cu: S mass ratio is 66.6: 33.4

In sulphur trioxide Oxygen: Sulphur (O: S) mass ratio is 60:40

Now in CuS 33.4 parts of sulphur combines with Cu = 66.6 parts

40.0 parts of sulphur combines with cu=66.68*40/33.4=79.8 parts

Now the ratio of masses of Cu and O which combines with same mass (40 parts) of sulphur separately is 79.8:60                                                                                       … (1)

                                                Cu: O ratio by mass in CuO is 79.9: 20.1        … 2)

                                                Ratio 1: Ratio 2 = 1:3

Which is simple whole number ratio; hence law of reciprocal proportion is proved.

TRY YOURSELF:

Exercise 9. Carbon combines with hydrogen to form three compounds A, B and C. The  percentage of hydrogen in A, B and C are 25, 14.3 and 7.7 respectively. Which law of chemical combination is illustrated?

Exercise 10. 61.8g of A combines with 80 g of B. 30.9g of A combines with 106.5g of C, B and C combine to form compound CB₂. Atomic weight of C and B are respectively 35.5 and 6.6. Show that the law of reciprocal proportion is obeyed.

Exercise 11. Carbon dioxide contains 27.27% carbon, carbon disulphide contains 15.97% carbon and sulphur dioxide contains 50% sulphur. Show that these figures illustrate the law of reciprocal proportions.

`Exercise 12. Aluminium oxide contains 52.9% aluminium and carbon dioxide contains 27.27% carbon. Assuming the validity of the law of reciprocal proportions, calculate the percentage of aluminium in aluminium carbide.