Chemical analysis of vitamin D3 yields the following composition: 84.31 C, 4.16 O, and 11.53 H with a molar mass of 384 g/mol. What is the empirical and molecular formulas for vitamin D3?
To find the empirical formula, we need to determine the simplest whole number ratio of the elements present in the compound.
Step 1: Convert the percentages to grams.
Let's assume we have 100 grams of vitamin D3.
Given:
Carbon (C) = 84.31%
Oxygen (O) = 4.16%
Hydrogen (H) = 11.53%
Therefore:
Carbon (C) = 84.31 g
Oxygen (O) = 4.16 g
Hydrogen (H) = 11.53 g
Step 2: Convert grams to moles.
Using the molar mass, determine the number of moles for each element:
Carbon (C):
Molar mass of C = 12.01 g/mol
Number of moles of C = 84.31 g / 12.01 g/mol ≈ 7.02 mol
Oxygen (O):
Molar mass of O = 16.00 g/mol
Number of moles of O = 4.16 g / 16.00 g/mol ≈ 0.26 mol
Hydrogen (H):
Molar mass of H = 1.01 g/mol
Number of moles of H = 11.53 g / 1.01 g/mol ≈ 11.42 mol
Step 3: Determine the empirical formula.
Divide the moles of each element by the smallest number of moles obtained in Step 2 (in this case, oxygen), and round to the nearest whole number ratio:
Carbon (C): 7.02 mol / 0.26 mol ≈ 27
Oxygen (O): 0.26 mol / 0.26 mol = 1
Hydrogen (H): 11.42 mol / 0.26 mol ≈ 44
Therefore, the empirical formula of vitamin D3 is C27H44O.
To determine the molecular formula, we need to know the molar mass of the compound. Given that the molar mass of vitamin D3 is 384 g/mol, we can calculate the ratio between the empirical formula mass and the molar mass:
Empirical formula mass:
(27 x molar mass of C) + (44 x molar mass of H) + (1 x molar mass of O)
= (27 x 12.01 g/mol) + (44 x 1.01 g/mol) + (16.00 g/mol)
≈ 380.78 g/mol
Ratio:
384 g/mol (molar mass) / 380.78 g/mol (empirical formula mass) ≈ 1
Therefore, the molecular formula of vitamin D3 is C27H44O.