Empirical Formulas Simplest Chemical formula for a compound

Writing Empirical Formulas 1. Drop the % sign and add g (for grams) to

Writing Empirical Formulas 2. Change all gram amounts to moles using the substances molar

Writing Empirical Formulas 3. Divide all mole amounts by the smallest mole amount 75

Writing Empirical Formulas 4. The answer from step 3 now will be the subscripts

Mole Rounding Rules If the mole amount ends in this decimal… • . 25

Example with Rounding 36. 84% N and 63. 16% O 36. 84 g N

Molecular Formula Based on the actual number of atoms of each type in the

Calculating Molecular Formula mass = # Empirical Formula mass The number now needs to

Molecular Formula example A substance has a molecule formula mass of 42 amu and

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Empirical Formulas Simplest Chemical formula for a compound

Writing Empirical Formulas 1. Drop the % sign and add g (for grams) to the element 75% Hg and 25% Cl 75 g Hg and 25 g Cl Why can we do this?

Writing Empirical Formulas 2. Change all gram amounts to moles using the substances molar mass (look on periodic table) 75 g Hg ( 1 mole ) = 0. 375 mol Hg 200 g Hg 25 g Cl ( 1 mole) = 0. 714 mol Cl 35 g Cl

Writing Empirical Formulas 3. Divide all mole amounts by the smallest mole amount 75 g Hg ( 1 mole ) = 0. 375 mol Hg = 1 200 g Hg 0. 375 25 g Cl ( 1 mole) = 0. 714 mol Cl = 1. 9 2 35 g Cl 0. 375

Writing Empirical Formulas 4. The answer from step 3 now will be the subscripts for each element 75 g Hg ( 1 mole ) = 0. 375 mol Hg = 1 200 g Hg 0. 375 25 g Cl ( 1 mole) = 0. 714 mol Cl = 1. 9 2 35 g Cl 0. 375 Hg. Cl 2

Mole Rounding Rules If the mole amount ends in this decimal… • . 25 x 4 • . 33 x 3 • . 5 x 2 • If you multiply you must multiply to all numbers

Example with Rounding 36. 84% N and 63. 16% O 36. 84 g N (1 mole) = 63. 16 g O (1 mole) =

Example with Rounding 36. 84% N and 63. 16% O 36. 84 g N (1 mole) = 2. 63 mol N 14 g N 63. 16 g O (1 mole) = 3. 95 mol O 16 g O

Example with Rounding 36. 84% N and 63. 16% O 36. 84 g N (1 mole) = 2. 63 mol N = 1 14 g N 2. 63 63. 16 g O (1 mole) = 3. 95 mol O = 1. 5 16 g O 2. 63

Example with Rounding 36. 84% N and 63. 16% O 36. 84 g N (1 mole) = 2. 63 mol N = 1 x 2 =2 14 g N 2. 63 63. 16 g O (1 mole) = 3. 95 mol O = 1. 5 x 2= 3 16 g O 2. 63 N 2 O 3

Molecular Formula Based on the actual number of atoms of each type in the compound Empirical formula C 3 H 8 Molecular Formula C 6 H 16 C 12 H 48

Calculating Molecular Formula mass = # Empirical Formula mass The number now needs to be multiplied to the subscripts in the empirical formula

Molecular Formula example A substance has a molecule formula mass of 42 amu and the empirical formula for a compound is CH 2. What is the molecule formula for this compound Molecular Formula Mass = 42 amu = 3 Empirical Formula Mass 14 amu CH 2 C 3 H 6