### Empirical Formulas

```Empirical Formulas
GRAB A CALCULATOR AND GET OUT A PIECE
OF PAPER FOR NOTES.
Empirical Formula
 An empirical formula is a formula that shows the
smallest whole number ratio of the elements in a
compound.
Empirical Formula
 Review 
 Percent composition shows the percentage of the compound
Mass of Element
_______________
Mass of Compound


CO2
X
100
Fe2O3
We use the percent composition to calculate the empirical
formula
Empirical Formula
 Example –
 Methyl acetate is made up of 48.64% carbon, 8.16 % hydrogen,
and 43.20% oxygen. Find the empirical formula methyl
acetate.
 Step 1  find the moles of each element assuming you have a
total mass of 100 g
Carbon  48.64 g C
1 mol C
=
4.050 mol C
12.01 g C
Hydrogen  8.16 g H
1 mol H
=
8.10 mol H
1.008 g H
Oxygen  43.20 g O
1 mol O
=
2.700 mol O
15.999 g O
Empirical Formula
 Example –
 Methyl acetate is made up of 48.64% carbon, 8.16 % hydrogen,
and 43.20% oxygen. Find the empirical formula methyl
acetate.
 Step 2  Divide all moles by the smallest number
Carbon 
4.050 mol C = 1.500 mol
2.70 mol
Hydrogen  8.10 mol H = 3.00 mol
2.70 mol
Oxygen  2.700 mol O = 1.00 mol
2.70 mol
Empirical Formula
 Example –
 Methyl acetate is made up of 48.64% carbon, 8.16 % hydrogen,
and 43.20% oxygen. Find the empirical formula methyl
acetate.
 Step 3  Multiply each by the same number to get a whole
number
Carbon 
1.500 mol x 2 = 3
Hydrogen 
Oxygen 
3.00 mol x 2 = 6
1.00 mol x 2 = 2
So, the empirical formula is
C3H6O2
 http://youtu.be/AFqwtY7m2PI
 Calculate the percent composition of Calcium Chloride
assuming you have a 100 g sample.
 Determine the emperical formula for a compound with
60.98% arsenic and 39.02% sulfur.
 Determine the emperical formula for a compound that
contains 74.03% C, 8.70% H, and 17.27 % N.
```