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Warm-Up: To be turned in

Warm-Up: To be turned in. 3.6 mol NaNO 3 = ______ g 228.50 g MgCl 2 = ______ mol. Empirical & Molecular Formulas. Empirical formula. Formula that shows the smallest whole-number ratio of elements in a compound Does not always indicate the actual # of atoms in the compound.

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Warm-Up: To be turned in

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  1. Warm-Up: To be turned in 3.6 mol NaNO3 = ______ g 228.50 g MgCl2 = ______ mol

  2. Empirical & Molecular Formulas

  3. Empirical formula • Formula that shows the smallest whole-number ratio of elements in a compound • Does not always indicate the actual # of atoms in the compound

  4. Determining Empirical Formulas • If given % composition: • Assume you have 100.0 g total • Percentages become the # of g for each element • Calculate # of mol for each atom • Divide all mol values by smallest mol value • Round to nearest whole number • If given grams: • Skip steps 1 and 2

  5. Example A compound contains 78.1% B and 21.9% H. 78.1g x ___1__ = 7.22 mol B 10.81g B 21.9g x ___1__ = 21.7 mol H 1.01g H 7.22 mol B : 21.7 mol H = 1 mol B: 3.01 mol H 7.22 mol 7.22 mol BH3

  6. Practice Determine the empirical formula for a compound containing the following: 40.0% C, 6.7% H, 53.3% O

  7. Molecular Formula • Actual formula of a compound • Contains the same ratio of atoms as the empirical formula • Must know the molar mass of the compound

  8. Determining Molecular Formulas • Determine the empirical formula • Calculate molar mass of empirical formula • Divide actual molar mass by empirical molar mass • Determines the multiplier • Multiply empirical formula by multiplier

  9. Example 85.64% C and 14.36% H, mol mass= 42.08 g/mol 85.64g x _1mol_ = 7.13 mol C 7.13mol: 14.21mol 12.01g 7.13 7.13 14.36g x 1mol = 14.21 mol H CH2 1.01g________________________________ 12.01 + (2x 1.01)= 14.03g/mol CH2 42.08/ 14.03= 2.99  3 CH2 x 3= C3H6 Empirical formula

  10. Practice 2 Determine the empirical and molecular formula for a compound that contains the following: 5.9% H, 94.1% O, molar mass= 34.02g/mol

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