A metal forms two oxide . The higher oxide contains 80% metal . 0.72 grams of the lower oxide gave 0.8 g of the higher oxide when oxidized . Show that the data illustrates the law of multiple proportions.
A metal forms two oxide . The higher oxide contains 80% metal . 0.72 grams of the lower oxide gave 0.8 g of the higher oxide when oxidized . Show that the data illustrates the law of multiple proportions.
Percentage of metal in higher oxide = 80%
Percentage of oxygen = (100-80)% = 20%
Thus, 4 parts of metal combines with 1 part of oxygen.
Now, 0.72 g of lower oxide on oxidation gives 0.8 g of higher oxide.
Let us assume that the mass percent of metal in 0.8 g of higher oxide is same as that of 0.72 g of lower oxide.
Then,
Mass of metal in higher oxide = (80/100) x 0.8 g = 0.64 g
Mass of oxygen in higher oxide = (0.8 - 0.64) g = 0.16 g
Now,
Mass of oxygen in lower oxide = (0.72-0.64) g = 0.08 g
Ratio of oxygen which combines with a fixed mass of metal is 0.16 : 0.08 = 2 :1 is a simple whole number ratio.
Hence, law of multiple proportions is illustrated.
Percentage of metal in higher oxide = 80%
Percentage of oxygen = (100-80)% = 20%
Thus, 4 parts of metal combines with 1 part of oxygen.
Now, 0.72 g of lower oxide on oxidation gives 0.8 g of higher oxide.
Let us assume that the mass percent of metal in 0.8 g of higher oxide is same as that of 0.72 g of lower oxide.
Then,
Mass of metal in higher oxide = (80/100) x 0.8 g = 0.64 g
Mass of oxygen in higher oxide = (0.8 - 0.64) g = 0.16 g
Now,
Mass of oxygen in lower oxide = (0.72-0.64) g = 0.08 g
Ratio of oxygen which combines with a fixed mass of metal is 0.16 : 0.08 = 2 :1 is a simple whole number ratio.
Hence, law of multiple proportions is illustrated.
