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Formulas and Percent Composition

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Percent Composition The percent composition is the percentage by mass of each element in a compound This helps distinguish compounds made up of the same elements –FeO or Fe 2 O 3 In Iron (III) Oxide, Iron is 69.94% Fe and 30.06% O In Iron (II) Oxide, Iron is 77.73% Fe and 22.27% O

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Percent Composition Calculate the percent composition of copper (I) sulfide. –1. Write the molecular formula Cu 2 S –2. Find the molar mass of each element in the compound (2 mol Cu)(63.55 g/mol Cu) = g Cu (1 mol S)(32.07 g/mol S) = g S Molar mass of Cu 2 S = g/mol –3. Calculate the percent by mass of each element (127.1g Cu)/(159.2g Cu 2 S) x 100 = 79.85% Cu (32.07g S)/(159.2g Cu 2 S) x 100 = 20.15% S –4. Check to be sure it adds to 100%

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Percent Composition Calculate the percent composition of aluminum nitrate. Al = 12.67% N = 19.73% O = 67.60%

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Empirical Formulas The Empirical Formula is the simplest whole number ratio among the atoms in a compound –Ammonium Nitrite –NH 4 NO 2 –What is the simplest whole number ratio of each element in the compound? N 2 H 4 O 2 becomes NH 2 O –This is the empirical formula

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Percent Composition and Empirical Formula One can determine the Empirical Formula from percent composition –Chemical analysis of a liquid shows that it is 60.0% C, 13.4% H, and 26.6% O by mass. What is the empirical formula for the liquid? 1. Assume you have g of the substance –Then you have 60.0g of C 13.4g of H and 26.6g of O

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Percent Composition and Empirical Formula 2. Convert from mass to moles –(60.0g C)(1 mol C/12.011g C) = 5.00 mol C –(13.4g H)(1 mol H/1.01g H) = 13.3 mol H –(26.6g O)(1 mol O/16.00g O) = 1.66 mol O 3. Divide all moles by the smallest number to find the ratio –5.00 mol C/1.66 = 3.01 mol C –13.3 mol H/1.66 = 8.01 mol H –1.66 mol O/1.66 = 1.00 mol O 4. This is your Empirical Formula –C3H8O–C3H8O

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Empirical Formulas Benzoic Acid contains 68.8% Carbon, 4.95% Hydrogen, and 26.2% Oxygen. Find the empirical formula. –(68.8gC)(1molC/12.01gC) = 5.73 mol C –(4.95gH)(1molH/1.01gH) = 4.90 mol H –(26.2gO)(1molO/16.00gO) = 1.64 mol O Divide by 1.64 (smallest number) –5.73molC/1.64 = 3.5 mol C –4.90molH/1.64 = 3.0 mol H –1.64molO/1.64 = 1.0 mol O

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Empirical Formulas Can you have C 3.5 H 3 O? No! You must multiply all by 2 so you can have whole numbers 3.5 x 2, 3 x 2 and 1 x 2 Your final empirical formula is C 7 H 6 O 2

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Molecular Formulas Molecular formulas are multiples of empirical formulas –CH 2 O C 2 H 4 O 2 is acetic acid (2 x Empirical Formula) C 6 H 12 O 6 is Glucose (6 x Empirical Formula) –You can determine the molecular formula from the empirical formula and the mass of the compound

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Molecular Formulas The empirical formula for a compound is P 2 O 5. Its experimental molar mass is 284 g/mol. Determine the molecular formula for the compound. –Determine the molar mass of the empirical formula g/mol –Find the ratio of the compound molar mass to the empirical formula molar mass (284 g/mol)/( g/mol) = 2.00 –Multiply the empirical formula by this number 2(P 2 O 5 ) = P 4 O 10 –Verify the results with a check of the molar mass of the new formula

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