7.3 – NOTES Molecular Formulas

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7.3 – NOTES Molecular Formulas

E. Molecular formula Since more than 1 compound could have the same empirical formula, in order to determine the molecular formula we need more information, such as the molar mass. Molecular formula: the actual whole number ratio of atoms For example, benzene has the molecular formula C6H6, but the empirical formula is CH.

The process is the same as empirical formula with one additional step. Divide the molar mass given by the mass of the empirical formula. Multiply this factor by the numbers in the empirical formula.

Examples: 1. 82.7% of a compound is carbon and 17.3% is hydrogen. If the molar mass is 58.12 g, what is the molecular formula?   C – 82.7 g (1 mol/ 12.0 g) = 6.8916 / 6.8916 = 1 x 2 = 2 H – 17.3 g (1 mol/ 1.01 g) = 17.128/6.8916 = 2.5 x 2 = 5 C2H5 = mass of 29.1 58.12/ 29.1 = 2 C2H5 x 2 = C4H10

2. Caffeine has the composition: 49. 5% carbon, 5. 19% hydrogen, 28 2. Caffeine has the composition: 49.5% carbon, 5.19% hydrogen, 28.9% nitrogen and 16.5% oxygen. If the molar mass of caffeine is 194 g, what is the molecular formula?   C – 49.5 g (1 mol/ 12.0 g) = 4.125/ 1.0312 = 4 H – 5.19 g (1 mol/ 1.01 g) = 5.1386/ 1.0312 = 5 N – 28.9 g (1 mol/ 14.0 g) = 2.0642/ 1.0312 = 2 O – 16.5 g (1 mol/ 16.0 g) = 1.0312 / 1.0312 = 1 C4H5N2O = 97.1 194/ 97.1 = 2 C4H5N2O x 2 = C8H10N4O2