Calculate the empirical and molecular formula of the molecule giving rise to the molecular ion peak at 148 m/z. The percentage composition by weight is 64.80 % carbon, 13.62 % hydrogen, and 21.58 % oxygen

Step 1, pick out the important information from the question: Carbon: Hydrogen: Oxygen % by weight: 64.8: 13.62: 21.58 Step 2, from the periodic table, find out the relative atomic masses of each element: Carbon: Hydrogen: Oxygen % by weight: 64.8: 13.62: 21.58 Mass: 12: 1: 16 Step 3, Divide the percentage by weight by the mass of the element to give an atomic ratio: Carbon: Hydrogen: Oxygen % by weight: 64.8: 13.62: 21.58 Mass: 12: 1: 16 Atomic ratio: 5.40: 13.62: 1.35 Step 4, divide the atomic ratios for each element by the smallest atomic ratio. In this case, each ratio will be divided by 1.35. This is to calculate the empirical formula. Carbon: Hydrogen: Oxygen % by weight 64.8: 13.62: 21.58 Mass: 12: 1: 16 Atomic ratio: 5.40: 13.62: 1.35
Empirical number: 4.00: 10.08: 1.00 Empirical formula: C4H10O Step 5, referring back to the question, we are also asked to calculate the molecular formula. We know from the question that the molecular mass of the compound is 148. Therefore if we work out the mass of the empirical formula, the molecular formula should become apparent. Mass of Empirical formula: (12x4)carbon+ (1x10)hydrogen +(16x1)oxygen=74 Molecular mass 148 (from question) / 74 = 2 This means that the empirical formula must be multiplied by two to obtain the molecular formula. Empirical formula: C4H10O Molecular formula C8H16O2