Question

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?

Answer 1

Answer:

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