A metal forms two oxides, the higher oxide contains 80% metal, 0.72g of the lower oxide gave 0.8g of higher oxide on heating. Show that the data illustrates the law of multiple proportions?
A metal forms two oxides, the higher oxide contains 80% metal, 0.72g of the lower oxide gave 0.8g of higher oxide on heating. Show that the data illustrates the law of multiple proportions?
ANSWER:- In higher oxide
Metal = 80%
Oxygen = (100-80)% = 20%
Therefore, we can say that 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. It can be assumed that the mass percent of metal in 0.8 g is same as that of 0.72 g of lower oxide. So,
Mass of metal in higher oxide = (80/100) x 0.8 g = 0.64 g
If in 0.8 g of higher oxide 0.64 g is metal then mass of oxygen present in higher oxide will be (0.8 - 0.64) g = 0.16 g
Since, lower oxide contains the same mass of metal as that of higher oxide, we need to calculate the mass of oxygen in lower oxide.
Mass of oxygen in lower oxide = (0.72-0.64) g = 0.08 g
According to Law to multiple proportions if two elements combine with each other to form two different compounds then the ratio of masses of that element which combines with the other element whose mass is fixed in both the compounds, will be in small whole numbers. Now, in the given problem the mass of metal in both the oxide is fixed so, for the data to illustrate the law of multiple proportion the ratio of mass of oxygen in both the oxides should be in whole numbers.
Now, mass of oxygen in higher oxide : mass of oxygen in lower oxide = 0.16 : 0.08 = 2 : 1. Therefore, it can be said that the given data depicts law of multiple proportion.
Hope it makes you clear.
In higher oxide
Metal = 80%
Oxygen = (100-80)% = 20%
Therefore, we can say that 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. It can be assumed that the mass percent of metal in 0.8 g is same as that of 0.72 g of lower oxide. So,
Mass of metal in higher oxide = (80/100) x 0.8 g = 0.64 g
If in 0.8 g of higher oxide 0.64 g is metal then mass of oxygen present in higher oxide will be (0.8 - 0.64) g = 0.16 g
Since, lower oxide contains the same mass of metal as that of higher oxide, we need to calculate the mass of oxygen in lower oxide.
Mass of oxygen in lower oxide = (0.72-0.64) g = 0.08 g
According to Law to multiple proportions if two elements combine with each other to form two different compounds then the ratio of masses of that element which combines with the other element whose mass is fixed in both the compounds, will be in small whole numbers. Now, in the given problem the mass of metal in both the oxide is fixed so, for the data to illustrate the law of multiple proportion the ratio of mass of oxygen in both the oxides should be in whole numbers.
Now, mass of oxygen in higher oxide : mass of oxygen in lower oxide = 0.16 : 0.08 = 2 : 1. Therefore, it can be said that the given data depicts law of multiple proportion.
Hope it makes you clear.
