Determining molecular formula

[preet]

What is the molecular formula of a compound which contains 92.3% carbon and 7.7% hydrogen by mass, if 0.39g of its vapour occupies an equal volume of .01g of hydrogen gas under identical conditions?

I don't understand how to do this question. I can figure out the empirical formula... but I don't know how to get the molecular formula. Any help would be appreciated.

[chem_tr]

This is not very hard, I think. Let's take a look.

I'll start from the second part. An equal volume means equal mole amount at identical circumstances, as you know. You'll just find how many moles are there in 0.01 g of H2 gas, and make it equal to the gas you're asking about.

If 0.39 grams of this gas is equal to the mole amount you'll find, then one mole of this gas is the molecular weight of the gas.

About molecular formulae, try dividing percentage results into their corresponding atomic weights, for example, if a substance gives 20% of carbon, then you'll divide 0.20 to 12. What you'll do next is dividing the numbers into the smallest one to obtain a 1 in the series, and expressing them in terms of relative magnitudes of the smallest one.

[preet]

I don't really understand what you mean by an equal volume means equal mole amount at identical circumstances...I don't think we've learned that. And in the last part, isn't that to get the empirical formula?

[so-crates]

[ ignore this, didn't read the entire post]

[chem_tr]

Considering your question involves ideal gases, one mole of gas is known to occupy 22.4 liters in 1 atm ordinary pressure (in 0 °C=273,15 K). So you can directly calculate how many moles are there in a given volume of the gas, by using the known formula, P \times V=n \times R \times T

The last part is used to obtain empirical formula, you are right, and molecular formula can be derived from empirical one, because you will know the molecular formula from the volume calculations, and from there you can multiply the empirical formula with a constant number to obtain molecular formula, I think.