Final answer:
The molecular formula of the compound with a molar mass of 219 g/mol and percent composition of 32.88% C, 4.14% H, 19.18% N, and 43.80% O is C6H9N3O6.
Step-by-step explanation:
To determine the molecular formula of a compound with a known percent composition and molar mass, you must first assume a sample size of 100 g so that the percent composition can directly translate to grams. With given percentages, we have 32.88 g of carbon (C), 4.14 g of hydrogen (H), 19.18 g of nitrogen (N), and 43.80 g of oxygen (O). The molar mass of each element is: C (12.01 g/mol), H (1.01 g/mol), N (14.01 g/mol), O (16.00 g/mol). By dividing the grams by the corresponding molar mass, we find the number of moles of each element in our 100 g sample.
- For carbon: 32.88 g ÷ 12.01 g/mol = 2.738 moles of C
- For hydrogen: 4.14 g ÷ 1.01 g/mol = 4.099 moles of H
- For nitrogen: 19.18 g ÷ 14.01 g/mol = 1.369 moles of N
- For oxygen: 43.80 g ÷ 16.00 g/mol = 2.738 moles of O
We then divide each by the smallest number of moles to get a ratio:
For carbon, hydrogen, nitrogen, and oxygen that ratio (to the nearest whole number) becomes 2:3:1:2. This reflects the empirical formula of the compound, which is C2H3NO2. To find the molecular formula, we compare the empirical formula mass to the molar mass given (219 g/mol). The empirical formula mass of C2H3NO2 is approximately 30.03 g/mol for C, 3.03 g/mol for H, 14.01 g/mol for N, and 32.00 g/mol for O, totaling 79.07 g/mol.
The ratio of the molar mass to the empirical formula mass is 219 g/mol ÷ 79.07 g/mol = 2.77. Since we cannot have a fraction of atoms in a molecule, we round to the nearest whole number, which is 3. Therefore, the molecular formula is three times the empirical formula: C6H9N3O6. This is the molecular formula of the compound.