posted by gloria on .
In 1934, Edward A. Doisy of Washington Univeristy extracted 1361 kg og hog ovaries to isolate a few milligrams of a potent female hormone. Doisy burned 5.00 mg of this precious sample in oxygen and found that 14.54 mg of CO2 and 3.97 mg of H2O were generated.
(a) determine the empirical formula of the hormone
(b) the molecular weight of the hormone was later determined to be 272 g/mol. determine the molecular formula of the hormone.
Convert 14.54 mg CO2 to mols C.
0.01454/44.01 = ?
Convert 3.97 g H2O to mols H.
0.00397*2/18.015 = ?
Find the whole number ratio for C and H and that will be the empirical formula.
For molecular formula,
(272/(mass empirical formula) = x
round x to a whole number and molecular formula is
I know how to find the empirical formula if you find the moles of each element. I am unsure about what to do when given compounds. How do I find how many moles there are of carbond and hydrogen?
14.54 mg = 0.01454 g CO2.
mols CO2 = (0.01454/molar mass CO2).
There is 1 mol C in 1 mol CO2; therefore, the mol C mols CO2.
mols H2O = grams/molar mass H2O.
Then 2x that for mols H.
Okay thank you, and what about oxygen? Do i do anything with oxygen?
To be honest with you I missed that part of the problem when I read it. What I've told you is right but we need to rearrange some of it.
sample + oxygen = CO2 + H2O
I guess we are to assume that if we convert to carbon and hydrogen that the rest of the sample was oxygen. Does your problem say that we are to do that. Anyway, I'll assume that.
So we convert CO2 to grams C.
0.01454 x (atomic mass C/molar mass CO2) = 0.01454 x (12.01/44.01) = grams C.
Same for the H2O.
grams H = 0.00397 x (2*atomic mass H)/molar mass H2O = 0.00397 x (2/18.015) = ? grams H.
Then 0.00500 - g C - g H = g O.
You can use mols C you already have.
You can use mols H you already have.
Add mols O = grams O/16 = ?
Then find the ratio for the empirical formula and continue for the molecular formula. Sorry I didn't read the problem very well. I'm glad you brought it to my attention.
I got the right answer by following DrBob222's step
for empirical formula C=3.97mg; H=0.44mg; and O=5mg-3.97mg-0.44mg=0.59mgO Then 0.59mg/16g=0.036mmol. Then you good to go!
Thank you DrBob222:D