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1.25 g of cupric oxide prepared from copper nitrate on heating in hydrogen gave 1 g of copper. 0.6248 g of cupric oxide prepared from copper carbonate by the same method have 0.5 of copper. Show that these results illustrate the law of definite proportions

1. From Cupric Nitrate

Weight of Copper = 1 g

Weight of Cupric oxide = 1.25 g

Weight of Oxygen = 0.25 g

% of Copper = (Weight of Copper / Weight of Cupric oxide) x 100

?% of Copper = 1/1.25 x 100

                      = 0.8 x 100 = 80 %

lll^ly % of Oxygen = 0.25/1.25 x 100

                          = 20%

2. From Copper Carbonate:

Weight of Copper = 0.5 g

Weight of Cupric oxide = 0.6248 g

Weight of Oxygen = 0.6248 - 0.5 = 0.1248 g

% of Copper = (Weight of Copper / Weight of Cupric oxide) x 100

?% of Copper = 0.5/ 0.6248 x 100

                      = 80 %

lll^ly % of Oxygen = 0.1248/ 0.6248 x 100

                          = 20 %

Since the ratio by weights of Copper and Oxygen in the two compounds remain the same, the law of definite proportions is illustrated.