General Chemistry question - emperical formula!!! please help me?
Lysine, an essential amino acid in the human body, contains C, H, O and N. In one experiment, the complete combustion of 2.175 g of lysine gave 3.94 g of CO2, and 1.89 g of H2O. In a separate experiment the same mass of lysine sample gave 0.506 g of NH3. What is the emperical forumla of lysine
Update:With explanation and detailed steps please :)
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using molar masses, find the mass of C,H,& N in the products :
3.94 g of CO2 @ 12.011 g C / 44.01 g CO2 = 1.075 g carbon
1.89 g of H2O @ 2.016 g H2 / 18.01 g H2O = 0.211 g Hydrogen
0.506 g NH3 @ 14.01 g N / 17.031 g NH3 = 0.416 g nitrogen
they total:
1.075 g C & 0.211 g H & 0.416 g N = 1.704 g CHN
find grams of Oxygen:
2.175 g total - 1.704 g CHN = 0.471 g oxygen
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finmd moles , using molar mass:
1.075 g C @ 12.01 g/mol = 0.0895 mol C
0.211 g H @ 1.008 g/mol = 0.208 mol H
0.416 g N @ 14.01 g/mol = 0.0297 mol N
0.471 g O @ 16.00 g/mol = 0.0295 mol O
get the mole ratio, by dividing by the smallest # of moles:
0.0895 mol C / 0.0295 = 3 mol C
0.208 mol H / 0.0295 = 7 mol H
0.0297 mol N / 0.0295 = 1 mol N
0.0295 mol O / 0.0295 = 1 mol O
your answer is
C3H7N1O1