1. 6.7 Empirical Formula

2. Formulas Percent composition allows you to calculate the simple ratio of the atoms in a compound. Empirical Formula: formula of a compound that expresses lowest whole number ratios of atoms Molecular Formula: Actual formula of a compound showing the number of atoms present

3. Formulas Examples: MolecularEmpirical C 8 H 18  C 6 H 12 O 6 

4. Formulas Examples: MolecularEmpirical C 8 H 18  C 4 H 9 C 6 H 12 O 6 

5. Formulas Examples: MolecularEmpirical C 8 H 18  C 4 H 9 C 6 H 12 O 6  CH 2 O

6. Formulas H 2 O 2. Empirical or molecular? MOLECULAR!!!! It can be reduced to HO HO = Empirical formula

7. Example An oxide of aluminum is formed by the reaction of 4.151 g of aluminum with 3.692 g of oxygen. Calculate the empirical formula. 1.Determine the number of grams of each element Al: 4.151gand O: 3.692g

8. Example 2. Convert mass to moles n Al = m/MM =4.151/26.98 = 0.1539 mol Al n O = m/MM =3.692/16.00 = 0.2308 mol O

9. Example 3. Find ratio by dividing each element by the smallest amount 4. Multiply by a factor to get whole numbers 0.1539 moles Al 0.1539 = 1.000 mol Al 0.2308 moles O 0.1539 = 1.500 mol O O = 1.500 x 2 = 3 Al = 1.000 x 2 = 2 therefore, Al 2 O 3

10. Example 1.4.550 g of Co reacts with 5.475 g Cl to form a binary compound. Determine the empirical formula. 2.2.000 g of iron metal is heated in air. It reacts with oxygen to achieve a final mass of 2.573 g. Determine the empirical formula.

11. Example 3.The most common form of nylon (Nylon-6) is 63.38% carbon, 12.38% nitrogen, 9.80% hydrogen and 14.14% oxygen. Calculate the empirical formula for Nylon-6.

12. Exit Ticket A sample of lead arsenate, an insecticide used against the potato beetle, contains 1.3813 g lead, 0.00672g of hydrogen, 0.4995 g of arsenic, and 0.4267 g of oxygen. Calculate the empirical formula for lead arsenate.

13. Homework Read section 6.7 Questions p. 292 #1 p. 293 #2-9