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UNIT 4

UNIT 4. Percent Composition. Calculating Mass % (or Weight %). Once you have the correct chemical formula for a compound, you can calculate the percent by mass of any element or ion in the compound: mass % of an element in a compound = 100 x mass of the element in the compound

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UNIT 4

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  1. UNIT 4 Percent Composition

  2. Calculating Mass % (or Weight %) Once you have thecorrect chemical formula for a compound, you can calculate the percent by mass of any element or ion in the compound: mass % of an element in a compound = 100 x mass of the element in the compound total mass of the compound Example: Calculate the mass % of Mg in Mg(OH)2 (molar mass of Mg is 24.305 g, molar mass of Mg(OH)2 is 58.3197 g) mass % Mg = 100 x 24.305 g = 41.675 % 58.3197 g Mass % is also called weight %.

  3. Calculating Mass % (or Weight %) Example: Calculate the mass % of O and H in Mg(OH)2 (molar mass of O is 16.00 g, molar mass of H is 1.008 g, and the molar mass of Mg(OH)2 is 58.3197 g) mass % O = 100 x 2 (16.00) g = 54.87 % 58.3197 g mass % H = 100 x2 (1.008) g = 3.456 % 58.3197 g The total mass of an element in a compound is the number of moles of atoms of the element multiplied by the molar mass of the element.

  4. Calculating % Composition The percent composition of a compound is the mass % of every element in the compound. The sum of the mass %’s of the elements in the compound must be 100%. Example: Calculate the percent composition of Mg(OH)2: mass % Mg = 100 x 24.305 g = 41.675 % 58.3197 g mass % O = 100 x 32.00 g = 54.87 % 58.3197 g mass % H = 100 x 2.016 g = 3.457 % 58.3197 g _________ 100.00%

  5. Calculating the Empirical Formula of a Compound from its % Composition • If you can determine the % composition from the formula of a compound, you can also work the other way and determine theempirical formulaof a compoundfrom its % composition. • To calculate themolecular formulaas well, you need to know themolar massof the compound.

  6. Calculating the Empirical Formula of a Compound from its % Composition Example: A compound contains 30.4% nitrogen and 69.6% oxygen. What is its empirical formula? In 100.0 g of the unknown compound, there would be 30.4 g of N and 69.6 g of O. Since the numbers shown in empirical formulas refer to moles, we now convert the masses to moles: 30.4 g N x 1 mol N =2.17 mol N 69.6 g O x 1 mol O = 4.35 mol O 14.007 g 15.9994 g This gives us N2.17O4.35 as our formula. It must be reduced to a ratio of small whole numbers. N2.17O4.35 N1O2.00 NO2 2.172.17 This is the empirical formula…NOT the molecular formula!!!

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