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II. Formula Calculations

Ch. 3 – The Mole. II. Formula Calculations. A. Percentage Composition. the percentage by mass of each element in a compound. 127.10 g Cu 159.17 g Cu 2 S. 32.07 g S 159.17 g Cu 2 S. A. 1 st Way % Composition. Find the % composition of Cu 2 S.  100 =. %Cu =. 79.85% Cu. %S =.

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II. Formula Calculations

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  1. Ch. 3 – The Mole II. Formula Calculations

  2. A. Percentage Composition • the percentage by mass of each element in a compound

  3. 127.10 g Cu 159.17 g Cu2S 32.07 g S 159.17 g Cu2S A. 1st Way % Composition • Find the % composition of Cu2S.  100 = %Cu = 79.85% Cu %S =  100 = 20.15% S

  4. A. Percentage Composition • How many grams of copper are in a 38.0-gram sample of Cu2S? Cu2S is 79.85% Cu (38.0 g Cu2S)(0.7985) = 30.3 g Cu

  5. 28 g 36 g 8.0 g 36 g A. 2nd Way % Composition • Find the percentage composition of a sample that is 28 g Fe and 8.0 g O.  100 = 77.78% Fe %Fe =  100 = 22.22% O %O =

  6. 36.04 g 147.02 g A. Hydrate % Composition • Find the mass percentage of water in calcium chloride dihydrate, CaCl2•2H2O? 24.51% H2O %H2O =  100 =

  7. CH3 B. 2 Types of formulas Empirical Formula: • Smallest whole number ratio of atoms in a compound C2H6 reduce subscripts

  8. C2H6 Molecular Formula: • “True Formula” - the actual number of atoms in a compound CH3 empirical formula ? molecular formula

  9. B. Empirical Formula 1. Find mass (or %) of each element. 2. Find moles of each element. 3. Divide moles by the smallest # to find subscripts. 4. When necessary, multiply subscripts by 2, 3, or 4 to get whole #’s.

  10. 1.85 mol 1.85 mol B. Empirical Formula • Find the empirical formula for a sample of 25.9% N and 74.1% O. 25.9 g 1 mol 14.01 g = 1.85 mol N = 1 N 74.1 g 1 mol 16.00 g = 4.63 mol O = 2.5 O

  11. N2O5 B. Empirical Formula N1O2.5 Need to make the subscripts whole numbers  multiply by 2

  12. C. Molecular Formula 1. Find the empirical formula. 2. Find the empirical formula mass. 3. Divide the molecular mass (given) by the empirical mass (step 2). 4. Multiply each subscript by the answer from step 3.

  13. C. Molecular Formula • The molecular mass of the compound is 216 g/mol (this must be given!) • Empirical Formula: N2O5 • Empirical Form. Mass: 108.02 g/mol • 216/108.02 = 2 • (N2O5)2 = N4O10

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