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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

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

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 BH 3

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

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

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

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 12.01g7.13 7.13 14.36g x 1mol = 14.21 mol H CH 2 1.01g________________________________ 12.01 + (2x 1.01)= 14.03g/mol CH 2 42.08/ 14.03= 2.99  3 CH 2 x 3= C 3 H 6 Empirical formula

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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