AP chem question CH3 please help?

Question

Please show how to do and give answer to :
Combustion analysis of 1.00g of the male sex hormone, testosterone, yields 2.90 g of CO2 and .875 g of H20. What are the mass percents of carbon, hydrogen, and oxygenin testosterone?

Answer 1

*Notice* my calculations are simplified and done in my head, redo them for your homework

Simplified:
C burns with O2 to make CO2 (C + O2 -> CO2)
H burns with O2 to make H2O (2H + 0.5 O2 -> H2O)

Molecular weights of:
C = 12 g/mol
O = 16 g/mol
H = 1 g/mol
CO2 = 44 g/mol
H2O = 18 g/mol

So, 2.90 g of CO2 is 2.90 g / 44 g/mol = .0659 mol CO2 = .0659 mol C
0.875 g of H2O is 0.875 g / 18 g/mol = .0486 mol H2O = .0243 mol H (2 H's are used for each mol of H2O)

You don't know the amount of oxygen in the sample, since O2 was available from the atmosphere during the burning, so you need to find the weights of C and H to determine the amount of O.

.0659 mol C = .0659 mol * 12 g/mol = .791 g C
.0243 mol H = .0243 mol * 1 g/mol = 0.0243 g H
1.000 g Testosterone - .791 g C - 0.024 g H = 0.185 g O

So Testosterone is:
.791 / 1.000 * 100% = 79.1 % C
.0243 / 1.000 * 100% = 2.4% H
.185 / 1.000 * 100% = 18.5 % O

Checking the math: 79.1 + 2.4 + 18.5 = 100%

Like I said, this is approximate, but it should show you how to do it, and your result should be close to mine.