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Percent Composition

Percent Composition. Example: MgO Find molar mass of whole compound: MgO = 24.3 + 16.00 = 40.3 grams % Mg (by mass) = mass of Mg = 24.3 g x 100 = 60.3% molar mass of MgO 40.3 g

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Percent Composition

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  1. Percent Composition Example: MgO Find molar mass of whole compound: MgO = 24.3 + 16.00 = 40.3 grams % Mg (by mass) = mass of Mg = 24.3 g x 100 = 60.3% molar mass of MgO 40.3 g % O (by mass) = mass of O = 16.0 g x 100 = 39.7% molar mass of MgO 40.3 g Percent = part / whole

  2. Empirical Formulas • 4Steps: • Change the percent to grams • Convert grams to moles • Divide each of the moles by the smallest number to find the ratio • Round ratio to a whole number. If the ratio is not a whole number (ex. 1.5) multiply each element by 2 to get a whole number Gives the lowest whole-number ratio of elements and compounds in a formula

  3. Examples A compound contains 94.1% Oxygen and 5.9% Hydrogen. What is its empirical formula? • 94.1% = 94.1g O • 5.9% = 5.9g H 5.9g H 1 mol H 1 mol O • 94.1g O = 5.9 mol H = 5.88 mol O 16 g O 1 g H 5.9 5.88 • 5.88 • 5.88 = 1.003 (can round to 1.0) = 1.0 4) Ratio = 1:1 Formula = OH

  4. Examples A compound contains 67.6% Mercury and 10.8% Sulfur and 21.6% Oxygen. What is its empirical formula? • 67.6% = 67.6g Hg • 10.8% = 10.8g S • 21.6% = 21.6g O 67.6g Hg = 0.336 mol Hg 10.8g S = 0.338 mol S 21.6g O = 1.35 mol O • 0.336 • 0.336 0.338 0.336 1.35 0.336 = 1.00 Hg = 1.00 S = 4.02 O 4)Ratio = 1:1:4 Formula = HgSO4

  5. Examples What is the empirical formula for a compound containing 70.0% Fe and 30.0% O? • 70.0% = 70.0g Fe • 30.0% = 30.0g O 70.0g Fe = 1.25 mol Fe 30.0g O = 1.875 mol O 1.875 1.25 • 1.25 • 1.25 = 1.5 O = 1.00 Fe 4) Ratio = 1 : 1.5 5) Mult ratio by 2 = 2 : 3 Formula = Fe2O3

  6. Molecular Formulas • Same as empirical formula, or a simple whole-number multiple of it • Steps: • Calculate the mass in grams of the empirical formula provided • Divide the molar mass by the mass of the empirical formula • Multiply this whole number ratio by the empirical formula

  7. Examples Calculate the molecular formula of the compound whose molar mass is 60.0g and empirical formula is CH4N. 1) CH4N = 12 + 4(1) + 14 = 30.0 g • 60.0g • 30.0g = 2.0 3) 2.0 (CH4N) = C2H8N2

  8. Examples What is the molecular formula of ethylene glycol (CH3O), used in antifreeze. It has a molar mass of 62 g/mol. 1) CH3O = 31.0 g 3) 2.0 (CH3O) = C2H6O2 • 62.0g • 31.0g = 2.0 Find the molecular formula of C3H2Cl, which is mothballs. Its molar mass is 147 g/mol. 1) C3H2Cl = 73.0 g 3) 2.0 (C3H2Cl) = C6H4Cl2 • 147.0g • 73.0g = 2.0

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