Chemistry, 13.07.2019 18:30 tanabugg

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.

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Answer from: ryan23allen

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 = $\frac{mass}{\text{molar mass of C}}$

= $\frac{27.2 g}{12 g/mol}$

= 2.26 mol

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

= $\frac{72.8 g}{16 g/mol}$

= 4.55 mol

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

C = $\frac{ 2.26}{ 2.26}$

= 1

O = $\frac{4.55}{ 2.26}$

= 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 = $\frac{mass}{\text{molar mass of C}}$

= $\frac{42.9 g}{12 g/mol}$

= 3.57 mol

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

= $\frac{57.1 g}{16 g/mol}$

= 3.57 mol

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

C = $\frac{3.57}{3.57}$

= 1

O = $\frac{3.57}{ 3.57}$

= 1

Hence, empirical formula of the second compound is CO.

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

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Consider 100.0-g samples of two different compounds consisting only of carbon and oxygen. one compou...

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