# Chemistry

When 1 gram of sex hormone (contains C, H, and O) was burned in a combustion analysis, 1.5 grams of CO2 and 0.405 grams of H2O were obtained. The molar mass was found to be 352g/mol. What is the molecular formula for the sex hormone?

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1. Convert 1.5 g CO2 to g C.
Convert 0.405 g H2O to g H.
Determine O by
1.00 g sample- gC - gH = g O.

Convert g each to moles.
moles C = g/atomic mass C.
moles H = g/atomic mass H.
moles O = g/atomic mass O.

Now find the ratio of the atoms to each other with the smallest number being 1.00. The easy way to do that is to divide the smallest number by itself; therefore, that number will be 1.00 and the other numbers will follow. Round to whole numbers (BUT don't round too much). If one of the numbers is further away than about 0.2 or so from a whole number, I would multiply the first set of numbers by 2, then 3, then 4, then 5, etc until all are reasonably close to whole numbers. That gives you the empirical formula.
To find the molecular formula, divide 352/mass empirical formula and round that answer to a whole number. That will be n in the following:
(CxHyOz)n where x,y,z are the subscripts in the empirical formula. Post your work if you need help through this. Be glad to help.

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2. okay,this is what i have so far:
so is the chemical formula:
O2 + CxHyOz --> CO2 + H2O ?

CO2: 0.41 g of C
1.09 g of O2
H2O: 0.045 g of H2
0.36 g of O

Empirical Formula: CH2O -> n = 0.033 mol

C: 0.033 mol x 12g/mol = 0.396 g of C
H2: 0.033 mol x 2g/mol = 0.066 g of H2
O: 0.033 mol x 16g/mol = 0.528 g of O

is this right?

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3. .41 (I used 0.409 since I'm allowed three places) is correct.
.045 g H is correct; not H2.
Then 1 - 0.409 - 0.045 = 0.546 g Oxygen.

Then 0.409/12 = ? moles C
0.045/1 =? moles H
0.546/16 = ? moles O.
Then continue.

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