Simplest Chemical formula for a compound Empirical Formulas Simplest Chemical formula for a compound

Writing Empirical Formulas 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 HgCl2

Mole Rounding Rules If the mole amount ends in this decimal… .25 x 4 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 x2 =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 N2O3

Molecular Formula Based on the actual number of atoms of each type in the compound Empirical formula C3H8 Molecular Formula C6H16 C12H48

Calculating Molecular Formula 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 CH2. What is the molecule formula for this compound Molecular Formula Mass = 42 amu = 3 Empirical Formula Mass 14 amu CH2  C3H6