Question

Consider 100.0-g samples of two different compounds consisting only of carbon and oxygen. one compound contains 27.2 g of carbon and the other has 42.9 g of carbon. how can these data support the law of multiple proportions if 42.9 is not a multiple of 27.2? show that these data support the law of multiple proportions.

Answer 1

Step-by-step explanation:

It is known that a mole of carbon atom contains a mass of 12 g and a mole of oxygen atom contains a mass of 16 g.

As it is given that first compound has 27.2 g C. So, amount of oxygen present in this compound will be as follows.

(100 - 27.2) g

= 72.8 g of O

Hence, calculate the moles of C and O as follows.

No. of moles of C =
`\frac{mass}{\text{molar mass of C}}`

=
`(27.2 g)/(12 g/mol)`

= 2.26 mol

No. of moles of O =
`\frac{mass}{\text{molar mass of O}}`

=
`(72.8 g)/(16 g/mol)`

= 4.55 mol

Therefore, mole ratios of both C and O will be calculated as follows.

C =
`( 2.26)/( 2.26)`

= 1

O =
`(4.55)/( 2.26)`

= 2

Hence, empirical formula of the first compound is
`CO_(2)`.

Since, it is also given that mass of carbin in another compound is 42.9 g. So, mass of oxygen present will be as follows.

(100 - 42.9) g

= 57.1 g

Hence, calculate the moles of C and O as follows.

No. of moles of C =
`\frac{mass}{\text{molar mass of C}}`

=
`(42.9 g)/(12 g/mol)`

= 3.57 mol

No. of moles of O =
`\frac{mass}{\text{molar mass of O}}`

=
`(57.1 g)/(16 g/mol)`

= 3.57 mol

Therefore, mole ratios of both C and O will be calculated as follows.

C =
`(3.57)/(3.57)`

= 1

O =
`(3.57)/( 3.57)`

= 1

Hence, empirical formula of the second compound is CO.

Therefore, carbon and oxygen combine to form 2 different compounds in small whole numbers.