How do I set up this problem?
Etaomidate, a hypnotic compound, has the percent composition: 68.83%C; 6.60%H; 11.47%N; and 13.10%O. its molecular weight is about 240 g/mol.
a) Determine the empirical formula.
b) Determine the molecular formula.

Question

How do I set up this problem?
Etaomidate, a hypnotic compound, has the percent composition: 68.83%C; 6.60%H; 11.47%N; and 13.10%O. its molecular weight is about 240 g/mol. 
a) Determine the empirical formula.
b) Determine the molecular formula.

jfraser · Answer

turn those percentages into relative masses. This is easy if you assume you have 100g of this unknown substance, that way 68.83% turns into 68.83g with no math. Do the same for all the other element components. Then turn each mass into moles by dividing by the respective molar masses. Then divide by the smallest to get a small whole number ratio

jfraser · Answer

I'll do carbon for you

jfraser · Answer

$$68.83\%C * \frac{100g}{100\%} = 68.83g C$$$$68.83g C * \frac{1mol \space C}{12g \space C} = 5.74mol \space C$$

candyme · Answer

Oh ok. I think I was confused by the 240. Where does that play in?

jfraser · Answer

once you find an empirical formula, it may or may not be the actual formula. The molecular weight of 240 tells us something like this:

if the empirical formula is (and i'm completely making this up) C3H10NO, then that has an empirical weight of 80g. The ACTUAL molecule has a weight of 240g, three times heavier.

This tells us that the molecular (actual) formula is THREE TIMES HEAVIER than the empirical formula, in order to make the masses match.

candyme · Answer

Ok. Thank you so much for all your help!

jfraser · Answer

YW. let me know what you get, and i'll see if it agrees

candyme · Answer

I got:
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