1. GRAB A CALCULATOR AND GET OUT A PIECE OF PAPER FOR NOTES. Empirical Formulas

2. Empirical Formula An empirical formula is a formula that shows the smallest whole number ratio of the elements in a compound.

3. Empirical Formula Review   Percent composition shows the percentage of the compound made up of each element Mass of Element _______________ X 100 Mass of Compound  CO 2 Fe 2 O 3  We use the percent composition to calculate the empirical formula

4. 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 C1 mol C=4.050 mol C 12.01 g C Hydrogen  8.16 g H1 mol H=8.10 mol H 1.008 g H Oxygen  43.20 g O1 mol O=2.700 mol O 15.999 g O

5. 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

6. 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  3.00 mol x 2 = 6 Oxygen  1.00 mol x 2 = 2 So, the empirical formula is C 3 H 6 O 2

7. 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.