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? [2 marks]
b. What is its empirical formula? [3 marks]
To find the mass of alkaloid combusted, we can use the Law of Conservation of Mass. The total mass of products (CO2, H2O, and NO2) must be equal to the mass of the alkaloid.
a. Mass of alkaloid combusted = Mass of products
= Mass of CO2 + Mass of H2O + Mass of NO2
= 210.6 g + 152.1 g + 110.0 g
= 472.7 g
Therefore, the mass of alkaloid combusted is 472.7 g.
b. To find the empirical formula, we need to determine the ratio of carbon, nitrogen, and hydrogen in the alkaloid.
First, let's calculate the moles of each element in the products:
- Moles of carbon (C) in CO2 = mass of CO2 / molar mass of CO2
= 210.6 g / (12.01 g/mol + 2 * 16.00 g/mol)
= 7.864 mol
- Moles of hydrogen (H) in H2O = mass of H2O / molar mass of H2O
= 152.1 g / (2 * 1.01 g/mol + 16.00 g/mol)
= 8.949 mol
- Moles of nitrogen (N) in NO2 = mass of NO2 / molar mass of NO2
= 110.0 g / (14.01 g/mol + 2 * 16.00 g/mol)
= 2.081 mol
Next, we need to find the simplest whole-number ratio of moles. We divide each mole value by the smallest mole value (in this case, 2.081 mol) and round to the nearest whole number:
- C: 7.864 mol / 2.081 mol ≈ 3.77 ≈ 4
- H: 8.949 mol / 2.081 mol ≈ 4.30 ≈ 4
- N: 2.081 mol / 2.081 mol = 1
Therefore, the empirical formula of the alkaloid is C4H4N.