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

2. 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?

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

4. Writing Empirical Formulas
3. Divide all mole amounts by the smallest mole amount 75 g Hg ( 1 mole ) = mol Hg = g Hg g Cl ( 1 mole) = mol Cl = 1.9  2 35 g Cl 0.375

5. Writing Empirical Formulas
4. The answer from step 3 now will be the subscripts for each element 75 g Hg ( 1 mole ) = mol Hg = g Hg g Cl ( 1 mole) = mol Cl = 1.9  2 35 g Cl HgCl2

6. Mole Rounding Rules If the mole amount ends in this decimal… .25 x 4
If you multiply you must multiply to all numbers

7. Example with Rounding 36.84% N and 63.16% O 36.84 g N (1 mole) =
63.16 g O (1 mole) =

8. Example with Rounding
36.84% N and 63.16% O g N (1 mole) = 2.63 mol N 14 g N g O (1 mole) = 3.95 mol O 16 g O

9. Example with Rounding
36.84% N and 63.16% O g N (1 mole) = 2.63 mol N = 1 14 g N g O (1 mole) = 3.95 mol O = g O 2.63

10. Example with Rounding 36.84% N and 63.16% O
36.84 g N (1 mole) = mol N = 1 x2 =2
14 g N
63.16 g O (1 mole) = 3.95 mol O = x 2= 3
16 g O
N2O3

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

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

13. 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