I don't think the problem is stated correctly. Where it says 0.287 g N2 gas it should say 0.287 g N
am = atomic mass
mm = molar mass
grams C = g CO2 x (am C/mmCO2) =2.72 x (12/44) = 0.742 g C
grams H = g H2O x (2*am H/mm H2O) = 1.29 x (2/18) = 0.143 g H
grams N = 0.287 g N2 but the problem should say 0.287 g N
grams O = 1.50 g total - g C - g H - g N =
grams O = 1.50 -0.742 - 0.143 - 0.287 = 0.328 grams O.
Convert to mols.
mols C = g/molar mass = 0.742/12 = 0.0618
mols H = 0.143/1 = 0.143
mols N = 0.287/14 = 0.0205
mols O = 0.328/16 = 0.0205
Now you want to find the whole number ratio of the elements to each other with no number being smaller than 1.0. The easy to do that is to divide the smallest number by itself and treat the other numbers the same.
mols C = 0.0618........... 0.0618/.0205 = 3.1 and round to 3.0
mols H = 0.143/1 = 0.143..... 0.143/0.0205 = 6.98 and round to 7
mols N = 0.287/14 = 0.0205 ............0.0205/0.0205 = 1.0
mols O = 0.328/16 = 0.0205 ........... 0.0205/0.0205 = 1.0
So the empirical formula is C3H7ON with an empirical mass = 73
Second part of the problem tells you the molecular formula is 146. You know that empirical mass x a whole number = molar mass
So you see the whole number is 2 and the molecular formula is
(C3H7ON)2 or C6H14O2N2
Lysine is a compound composed of carbon, hydrogen, nitrogen and oxygen. When 1.50 g of
lysine is burned, 2.72 g of carbon dioxide, 1.29 g of water and 0.287 g of nitrogen gas are
produced. What is the empirical formula of lysine? If the molar mass of lysine is 146.19
g/mol, what is the molecular formula?
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