Question

6. A metal forms two oxides, A and B. 1 g of A on reduction gave 0.881 g of

metal, while 1 g of B yielded 0.788 g of metal. Show that these facts are in
agreement with the law of multiple proportions.
010​

Answer 1

Here's what I get.

Step-by-step explanation:

The Law of Multiple Proportions states that when two elements A and B combine to form two or more compounds, the masses of B that combine with a given mass of A are in the ratios of small whole numbers.

That is, if one compound has a ratio r₁ and the other has a ratio r₂, the ratio of the ratios r is in small whole numbers.

1. Oxide A

Mass of O = 1 - 0.881 = 0.119 g

`r_(1) = \frac{\text{mass of O}}{\text{mass of M}} = ( 0.119)/(0.881) = 0.1351`

2. Oxide B

Mass of O = 1 - 0.788 = 0.212 g

`r_(2) = ( 0.212)/(0.788) = 0.2690`

3. Ratio of the ratios

`r = (r_(1))/(r_(2)) = ( 0.1351)/(0.2690) = (1 )/(1.992) \approx (1)/(2)\\\\\text{The relative amounts of O per gram of M in the two oxides are in the ratio $\large \boxed{\mathbf{(1)/(2)}}$}`