A sample of a plant alkaloid, containing ONLY carbon, nitrogen and hydrogen, is combusted in air to
produce 210.6 g CO2, 152.1 g H2O and 110.0g of NO2.
a. What mass of alkaloid was combusted?
b. What is its empirical formula?
To solve this problem, we need to follow these steps:
a. Calculate the moles of CO2, H2O, and NO2 produced.
b. Use the stoichiometry to find the moles of carbon, hydrogen, and nitrogen present in the alkaloid.
c. Calculate the molar mass of the empirical formula.
d. Calculate the mass of the alkaloid.
Step a:
We need to convert the masses of CO2, H2O, and NO2 to moles using their respective molar masses.
Molar mass of CO2:
C = 12.01 g/mol
O = 16.00 g/mol (x2)
Molar mass of CO2 = 12.01 + 16.00 + 16.00 = 44.01 g/mol
Moles of CO2 = Mass of CO2 / Molar mass of CO2 = 210.6 g / 44.01 g/mol = 4.777 mol
Molar mass of H2O:
H = 1.01 g/mol (x2)
O = 16.00 g/mol
Molar mass of H2O = 1.01 + 1.01 + 16.00 = 18.02 g/mol
Moles of H2O = Mass of H2O / Molar mass of H2O = 152.1 g / 18.02 g/mol = 8.436 mol
Molar mass of NO2:
N = 14.01 g/mol
O = 16.00 g/mol (x2)
Molar mass of NO2 = 14.01 + 16.00 + 16.00 = 46.01 g/mol
Moles of NO2 = Mass of NO2 / Molar mass of NO2 = 110.0 g / 46.01 g/mol = 2.391 mol
Step b:
Now we use the stoichiometry to find the moles of carbon, hydrogen, and nitrogen present in the alkaloid.
From the balanced combustion reaction, we can see that each mole of CO2 corresponds to 1 mole of carbon in the alkaloid.
Moles of carbon = Moles of CO2 = 4.777 mol
From the balanced combustion reaction, we can see that each mole of H2O corresponds to 2 moles of hydrogen in the alkaloid.
Moles of hydrogen = 2 * Moles of H2O = 2 * 8.436 mol = 16.872 mol
From the balanced combustion reaction, we can see that each mole of NO2 corresponds to 2 moles of nitrogen in the alkaloid.
Moles of nitrogen = 2 * Moles of NO2 = 2 * 2.391 mol = 4.782 mol
Step c:
The molar mass of the empirical formula can be calculated using the atomic masses.
Molar mass of carbon = 12.01 g/mol
Molar mass of hydrogen = 1.01 g/mol
Molar mass of nitrogen = 14.01 g/mol
Molar mass of the empirical formula = (Molar mass of carbon * moles of carbon) + (Molar mass of hydrogen * moles of hydrogen) + (Molar mass of nitrogen * moles of nitrogen)
= (12.01 g/mol * 4.777 mol) + (1.01 g/mol * 16.872 mol) + (14.01 g/mol * 4.782 mol)
= 57.507 g + 17.050 g + 67.188 g
= 141.745 g/mol
Step d:
Finally, we can calculate the mass of the alkaloid using the molar mass of the empirical formula.
Mass of the alkaloid = Molar mass of the empirical formula * Moles of the empirical formula
= 141.745 g/mol * 1 mol
= 141.745 g
Therefore,
a. The mass of the alkaloid that was combusted is 141.745 g.
b. The empirical formula of the alkaloid is C4H17N4.