Percent Composition Percent = part / whole Example: MgO

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Presentation transcript:

Percent Composition Percent = part / whole 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

Empirical Formulas Gives the lowest whole-number ratio of elements and compounds in a formula 4 Steps: 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

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 = 1.003 (can round to 1.0) = 1.0 4) Ratio = 1:1 Formula = OH

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 0.338 0.336 1.35 0.336 = 1.00 Hg = 1.00 S = 4.02 O 4) Ratio = 1:1:4 Formula = HgSO4

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 = 1.5 O = 1.00 Fe 4) Ratio = 1 : 1.5 5) Mult ratio by 2 = 2 : 3 Formula = Fe2O3

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

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

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