Question

Both FeO and Fe2O3 contain only iron and oxygen. The mass ratio of oxygen to iron for each compound is given in the following table:

Compound mass O : mass Fe
FeO 0.2865
Fe2O3 0.4297
Show that these data are consistent with the law of multiple proportions.

Answer 1

- For FeO:

`(O)/(Fe)=(16)/(55.8)=0.2865`

- For Fe₂O₃:

`(O)/(Fe)=(16*3)/(55.8*2)=0.43`

Step-by-step explanation:

Hello,

The law of multiple proportions states that: "if two elements form more than one compound between them, then the ratios of the masses of the second element which combine with a fixed mass of the first element will always be ratios of small whole numbers". In this manner, both FeO and Fe₂O₃ are in such way that they have the small whole numbers in their structures, thus, to compute the required ratios we perform the following mathematical relationships:

- For FeO:

`(O)/(Fe)=(16)/(55.8)=0.2865`

- For Fe₂O₃:

`(O)/(Fe)=(16*3)/(55.8*2)=0.43`

Wherein such values are consistent with the given data.

Best regards.

Answer 2

Step-by-step explanation:

Mass of O / Mass of Fe = .2865/1 = 112 x .2865/112 = 32.088/112

Mass of O / Mass of Fe = .4297 /1 = 112 x .4297 / 112 = 48 / 112

Ratio of mass of O That reacts with constant mass of Fe of 112 g is as follows

32.088 : 48 = 32/16 : 48/16

= 2 : 3

Hence given data are consistent with the law of multiple proportions.