Question

What is the molecular formula of a compound with the following percentages: 75.69% C, 8.80% H, and 15.51% O?

Answer 1

C `C_7 H_9O`

Explanation:

I might be mistaken, but here's what I did.
Assume you have 100 g of the compund. Then you take these percentages as normal weight, meaning, you can use formula for mole number. You count it for every atom and then compare the results.

So:
`n = (m)/"M"`
For C it's 6,3, for H 8,8 and for O it's 0,96

6,3:8,8:0,96 /*100
630:880:96 /:96
7:9:1