Consider a compound that is 67.57% C, 9.92% H, and 22.50% O by mass. Assume that we have a 100 g sample of this compound. Also consider that the molecular formula mass of this compound is 284.4 amu. What are the subscripts in the actual molecular formula for this compound?

Question

Chemistry

Frank Richardson

28 August, 12:22

Consider a compound that is 67.57% C, 9.92% H, and 22.50% O by mass. Assume that we have a 100 g sample of this compound. Also consider that the molecular formula mass of this compound is 284.4 amu. What are the subscripts in the actual molecular formula for this compound?

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Answers (1)

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Lesly Adams · Answer 1

The subscripts in the molecular formula are 16 for C, 28 for H and 4 for O

Explanation:

Let's work with the mass of compound and the percentage composition.

100 g of compound has 67.57 g of C, 9.92 g of H and 22.5 g of O

To find out in the molecular weight we must make a rule of three:

100 g of compound has 67.57 g of C, 9.92 g of H and 22.5 g of O

Then 284.4 g of compound would have:

(284.4. 67.57) / 100 = 192 g of C

(284.4. 9.92) / 100 = 28 g of H

(284.4. 22.5) / 100 = 64 g of O

We have to divide each mass, to determine the moles of each element

192 g / 12 g/mol = 16 mol C

28 g / 1 g/mol = 28 mol H

64 g / 16 g/mol = 4 mol O

The subscripts in the molecular formula are 16 for C, 28 for H and 4 for O