1. Empirical Formulas

2. How would a scientist determine the identity of a substance that they made?

3. Percent Composition Mass of element in compound X 100%
Total mass of compound
Repeat for each element in compound
Will add up to 100

4. Empirical Formula
The symbols for the elements combined in the compound, with subscripts showing the smallest whole-number ratio of the different atoms in the compound.
Ionic = formula unit
Covalent = reduced molecular formula

5. If you know the percent composition of a compound, then you can determine its empirical formula.
We do this by first assuming we have a g sample so that the percentage values can be converted to grams.
We then convert the grams to moles (to account for atoms having different molar masses)
To correct for our assumption of g we then divide each number of moles by the smallest number.

6. Mass composition (in 100.0g) = Composition in moles =
What is the empirical formula if a compound contains 32.4% Na, 22.6% S, & 45.0% O?
Mass composition (in 100.0g) =
32.4 g Na, 22.6 g S, 45.0 g O
Composition in moles =
32.4 g Na x (1 mol Na / g Na) = 1.41 mol Na
22.6 g S x (1 mol S / g S) = mol S
45.0 g O x (1 mol O / g O) = 2.81 mol O
Smallest whole-number mole ratio
1.41 mol Na / .705 = mol Na
.705 mol S / .705 = 1.00 mol S
2.81 mol O / .705 = 3.99 mol O

7. Percent
Composition
Mass
Composition
Composition
in moles
Smallest
whole-number
mole ratio

8. A compound’s empirical formula can also be used to determine its molecular formula.
x (empirical formula) = molecular formula
x = whole-number factor

9. So… CH  C6H6 Molecular formula mass = 78.11 amu
Determine the molecular formula of the compound with an empirical formula of CH and a mass of amu.
Molecular formula mass = amu
Empirical formula mass =
C + H = amu amu = amu
x (empirical formula) = molecular formula
x (13.02 amu) = amu
x = amu / amu = 5.999
So… CH  C6H6

10. In the green workbook: Pg 67 #1 b Pg 77 #1 a & b Pg 79 #1 a & b