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Empirical and Molecular Formulas Chem 11 Empirical formula: shows the lowest whole number ratio of elements in a compound; also called the simplest formula Molecular formula: shows the actual number of atoms that make up a molecule Ex: Glucose Molecular formula Empirical formula C6H12O6 CH2O Analysis of an unknown sample to determine the percent composition of elements can be used to find the empirical formula. However, there can be many compounds that have the same empirical formula: Ex: Benzene, C6H6 and acetylene, C2H2 are very different substances but both have the empirical formula CH. How to find the empirical formula of a compound from the percent composition: Example 1: Calculate the empirical formula of a compound that is 25.9 % nitrogen and 74.1 % oxygen. Solution 1: 1. Assume 100 g of sample and determine the mass of each element. Mass of nitrogen = 25.9g Mass of oxygen = 74.1g 2. Find the number of moles of each element. nN = 25.9 g = 1.85mol N 14.0067 g / mol nO = 74.1g = 4.63mol O 15.9994 g / mol 3. Write the ratio of moles of each element, then reduce this ratio by dividing through by the lowest number of moles. If necessary, multiply each term by a factor to convert decimals to whole numbers. Ratio (divide by smallest number of moles) N : O 1.85 4.63 : 1.85 1.85 1 : 2.5 (we don’t use fractions in chemical formulas, so we need to multiply to get whole numbers) 2:5 Therefore our empirical formula would be N 2 O5 Try: Write the empirical formula for a substance that is 38.7% C, 16.3% H and 45.1 % N. Again assume 100g sample to find mass of each element present: 38.7g C, 16.3g H, 45.1g N Find moles of each (in order as listed in question): nC = 38.7 g = 3.222molC 12.0107 g / mol nH = 16.3g = 16.17molH 1.00794 g / mol nN = 45.1g = 3.220molN 14.0067 g / mol Write ratios of each and divide by smallest number of moles. Multiply, if necessary, to get whole numbers throughout: 3.222 16.17 3.220 : : 3.220 3.220 3.220 Therefore our empirical formula would be CH 5 N 1: 5 :1 How to determine the molecular (actual) formula, given the empirical formula and the molar mass: Example 1: A substance has the empirical formula CH2O, and a molar mass of 180 g/mol. What is the molecular formula for this compound? Solution 1: 1. Using the empirical formula, calculate the mass of one mole of the compound (this is molar mass). 12.0107g/mol + (2 x 1.00794g/mol) + 15.9994g/mol = 30.02598g/mol 2. Divide the given molar mass by the calculated molar mass of the empirical compound. Then multiply this factor by the empirical formula to obtain the molecular formula. 180 g / mol =6 30.02598 g / mol Try: So 6 times the E.F. is the M.F which gives us C 6 H 12 O6 . Empirical formula: C2HCl What is the molecular formula? molar mass = 181.5 g/mol Molar mass = (2 x 12.0107g/mol) + 1.00794g/mol + 35.453g/mol = 60.48234g/mol 181.5 g / mol =3 60.48234 g / mol Therefore our M.F. is C 6 H 3Cl 3