a Example of organic compound contain 0.624 g carbon 0.065 hydrogen 0. 364g carbon 0.364g nitrogen and oxygen if the relative molecular mass of the compound is 194 what is the molecular formula c = 12 h = 1 n = 14 o=16
To determine the molecular formula, we need to calculate the empirical formula first.
First, calculate the number of moles of each element based on the given masses:
- Carbon (C): 0.624 g / 12 g/mol = 0.052 moles
- Hydrogen (H): 0.065 g / 1 g/mol = 0.065 moles
- Nitrogen (N): 0.364 g / 14 g/mol = 0.026 moles
- Oxygen (O): (194 g/mol - (0.624 g + 0.065 g + 0.364 g)) / 16 g/mol = 7.9475 moles
Next, divide each number of moles by the lowest number of moles to get the mole ratios:
- Carbon: 0.052 moles / 0.026 moles = 2
- Hydrogen: 0.065 moles / 0.026 moles = 2.5
- Nitrogen: 0.026 moles / 0.026 moles = 1
- Oxygen: 7.9475 moles / 0.026 moles = 305.288
Since we cannot have fractional subscripts in a molecular formula, we need to round the ratios to the nearest whole number:
- Carbon: 2 (unchanged)
- Hydrogen: 2.5 ≈ 3
- Nitrogen: 1 (unchanged)
- Oxygen: 305.288 ≈ 305
Therefore, the empirical formula is C2H3NO305.
To find the molecular formula, we need to determine the ratio between the empirical formula mass and the relative molecular mass:
Empirical formula mass = (2 * 12.01 g/mol) + (3 * 1.01 g/mol) + (1 * 14.01 g/mol) + (305 * 16.00 g/mol)
Empirical formula mass = 798.35 g/mol
Molecular formula ratio = Relative molecular mass / Empirical formula mass
Molecular formula ratio = 194 g/mol / 798.35 g/mol
Molecular formula ratio ≈ 0.243
Finally, multiply the empirical formula by the molecular formula ratio to get the molecular formula:
C2H3NO305 * 0.243 = C0.486H0.729N0.243O74.215 ≈ C½H¾NO74
Therefore, the molecular formula of the organic compound is C½H¾NO74 or simplified as CH2NO37.