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?
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
To find the empirical formula of lysine, we need to determine the ratio of the atoms present in the compound.
1. Calculate the number of moles of carbon dioxide (CO2) produced:
- Carbon mass in CO2 = 12.01 g/mol
- Oxygen mass in CO2 = 2 × 16.00 g/mol
- Moles of CO2 = mass of CO2 / molar mass of CO2
Moles of CO2 = 2.72 g / (12.01 g/mol + 2 × 16.00 g/mol) = 0.0802 mol
2. Calculate the number of moles of water (H2O) produced:
- Hydrogen mass in H2O = 2 × 1.01 g/mol
- Oxygen mass in H2O = 16.00 g/mol
- Moles of H2O = mass of H2O / molar mass of H2O
Moles of H2O = 1.29 g / (2 × 1.01 g/mol + 16.00 g/mol) = 0.0718 mol
3. Calculate the number of moles of nitrogen gas (N2) produced:
- Moles of N2 = mass of N2 / molar mass of N2
Moles of N2 = 0.287 g / (2 × 14.01 g/mol) = 0.0102 mol
4. Calculate the ratio of moles of carbon, hydrogen, and nitrogen to the smallest number of moles (0.0102 mol in this case):
- Carbon ratio = 0.0802 mol / 0.0102 mol ≈ 7.9
- Hydrogen ratio = 0.0718 mol / 0.0102 mol ≈ 7.0
- Nitrogen ratio = 0.0102 mol / 0.0102 mol = 1.0
5. Round the ratios to the nearest whole number to get the empirical formula:
- Carbon ratio ≈ 8
- Hydrogen ratio ≈ 7
- Nitrogen ratio = 1
Therefore, the empirical formula of lysine is C8H7N.
To find the molecular formula, we need to know the molar mass of the compound. Given that the molar mass of lysine is 146.19 g/mol, we can calculate the number of empirical formula units in one molecular formula unit:
6. Calculate the empirical formula mass:
- Carbon mass = 12.01 g/mol × 8 = 96.08 g/mol
- Hydrogen mass = 1.01 g/mol × 7 = 7.07 g/mol
- Nitrogen mass = 14.01 g/mol × 1 = 14.01 g/mol
- Empirical formula mass = sum of the masses of carbon, hydrogen, and nitrogen
Empirical formula mass = 96.08 g/mol + 7.07 g/mol + 14.01 g/mol = 117.16 g/mol
7. Calculate the ratio of the molar mass to the empirical formula mass:
- Ratio = molar mass / empirical formula mass
Ratio = 146.19 g/mol / 117.16 g/mol = 1.25
8. Round the ratio to the nearest whole number to determine the number of empirical formula units in the molecular formula:
- Ratio ≈ 1
Therefore, the molecular formula of lysine is the same as its empirical formula, C8H7N.
To determine the empirical formula of lysine, we need to calculate the ratio of the elements present.
1. Start by converting the given masses of carbon dioxide (CO2), water (H2O), and nitrogen gas (N2) to moles.
- The molar mass of CO2 is 44.01 g/mol, so 2.72 g of CO2 is equal to (2.72 g / 44.01 g/mol) = 0.062 mol of CO2.
- The molar mass of H2O is 18.02 g/mol, so 1.29 g of H2O is equal to (1.29 g / 18.02 g/mol) = 0.072 mol of H2O.
- The molar mass of N2 is 28.02 g/mol, so 0.287 g of N2 is equal to (0.287 g / 28.02 g/mol) = 0.010 mol of N2.
2. Next, determine the mole ratio of the elements present by dividing each number of moles by the smallest number of moles. In this case, the smallest number of moles is 0.010 mol of N2.
- The mole ratio of CO2 is (0.062 mol CO2 / 0.010 mol N2) ≈ 6.
- The mole ratio of H2O is (0.072 mol H2O / 0.010 mol N2) ≈ 7.
- The mole ratio of N2 is (0.010 mol N2 / 0.010 mol N2) = 1.
3. Finally, determine the empirical formula using the mole ratios obtained. The empirical formula represents the simplest whole-number ratio of atoms in a compound.
- The empirical formula of lysine is N1C6H7O2 since each element is present in the ratio of 1:6:7:2.
To find the molecular formula of lysine, we need to know its molar mass.
1. Calculate the molar mass of the empirical formula by summing up the molar masses of the individual elements.
- The molar mass of N is 14.01 g/mol, so (1 x 14.01 g/mol) = 14.01 g/mol.
- The molar mass of C is 12.01 g/mol, so (6 x 12.01 g/mol) = 72.06 g/mol.
- The molar mass of H is 1.01 g/mol, so (7 x 1.01 g/mol) = 7.07 g/mol.
- The molar mass of O is 16.00 g/mol, so (2 x 16.00 g/mol) = 32.00 g/mol.
- The molar mass of the empirical formula N1C6H7O2 is (14.01 + 72.06 + 7.07 + 32.00) = 125.14 g/mol.
2. Calculate the ratio of the molar mass of the molecular formula to the molar mass of the empirical formula.
- The molar mass of lysine is given as 146.19 g/mol.
- The ratio is (146.19 g/mol / 125.14 g/mol) ≈ 1.17.
3. Finally, write the molecular formula of lysine by multiplying the subscripts in the empirical formula by the ratio obtained.
- The molecular formula of lysine is approximately N1C6H8O2, obtained by multiplying 1:6:7:2 by 1.17.
Therefore, the empirical formula of lysine is N1C6H7O2, and the approximate molecular formula is N1C6H8O2.