[artwill872]

Combustion analysis of a 13.42g sample of estradiol (which contains only carbon, hydrogen, and oxygen) produced 39.01 g CO₂ and 10.65 g H₂O. The molar mass of estradiol is 272.38 g/mol.

Find the molecular formula for estradiol.

I keep getting C₉H₁₂O. I know the formula for estradiol is just double all of these values. Does this have something to do with oxygen needing to be diatomic in the molecular formula?

Anyway, I start out by finding moles of C and moles of H.. then getting each of them into grams. Once I get that, I figure out what difference I need to make up between their totals and the original estradiol sample in grams. That difference should be the amount of oxygen I need in the estradiol formula. So I take the grams of O, get it into moles and make a pseudoformula.

Then I do the standard procedure of dividing each subscript by the smallest and rounding.

I've done this problem three times since yesterday. I thought I might have just messed up with calculations somewhere but that doesn't seem to be the case. What am I overlooking?

[sjb]

Combustion analysis of a 13.42g sample of estradiol (which contains only carbon, hydrogen, and oxygen) produced 39.01 g CO₂ and 10.65 g H₂O. The molar mass of estradiol is 272.38 g/mol.

Find the molecular formula for estradiol.

I keep getting C₉H₁₂O. I know the formula for estradiol is just double all of these values. Does this have something to do with oxygen needing to be diatomic in the molecular formula?

Anyway, I start out by finding moles of C and moles of H.. then getting each of them into grams. Once I get that, I figure out what difference I need to make up between their totals and the original estradiol sample in grams. That difference should be the amount of oxygen I need in the estradiol formula. So I take the grams of O, get it into moles and make a pseudoformula.

Then I do the standard procedure of dividing each subscript by the smallest and rounding.

I've done this problem three times since yesterday. I thought I might have just messed up with calculations somewhere but that doesn't seem to be the case. What am I overlooking?

All you can calculate from combustion analysis is empirical formulae (in lowest terms). What would the molecular weight of the compound be, if it had formula  C₉H₁₂O? Nothing to do with diatomic oxygen, how could you calculate the formula for e.g. ethanol in this case?

[artwill872]

Ok.. so as expected, the molar mass of my empirical formula is exactly half that of the molar mass provided for estradiol.

Now I see that however many times the empirical formula's molar mass goes into the given molar mass of the compound, that number needs to be multiplied to the subscripts in the empirical formula by to get the molecular formula.

Thanks.