6.488 g of lead combine directly with 1.002g of oxygen to form lead peroxide (PbO_(2)). Lead peroxide is also produced by heating lead nitrate and it was found that the percentage of oxygen present in lead peroxides 13.38 percent. Use these data to illustrate the law of constant composition.

6.488 g of lead combine directly with 1.002g of oxygen to form lead pe

1.	6.488 g of lead combine directly with 1.002g of oxygen to form lead peroxide (PbO_(2)). Lead peroxide is also produced by heating lead nitrate and it was found that the percentage of oxygen present in lead peroxides 13.38 percent. Use these data to illustrate the law of constant composition.
Answer» Solution :STEP 1. To calculate the PERCENTAGE of oxygen in first experiment. Mass of peroxide formed `=6.488+1.002=7.490g`. 7.490 g of lead peroxide contain 1.002 g of oxygen `THEREFORE 100G` of lead peroxide will contain oxygen `=(1.002)/(7.490)xx100=13.38g`, i.e., oxygen present = `13.38%` Step. To compare the percentage of oxygen in both the experiments. Percentage of oxygen in `PbO_(2)` in the first experiment = 13.38 Step 2. To compare the percentage of oxygen in both he experiments. Percentage of oxygen in `PbO_(2)` in the second experiment = 13.38 Since the percentage composition of oxygen in both the SAMPLES of `PbO_(2)` is identical, the above data illustrate the law of constant composition.