1. Empirical & Molecular Formulas Chapter 3

2. Mathematical Methods STEP 1 Obtain in the laboratory, the number of grams OR the weight percentage of each element in the compound. This can be found by: breaking down the existing compound into its elements building the compound from the elements

3. STEP 2 Determine the number of moles of each element in the compound. Determine the number of moles of each element in the compound. Use dimensional analysis with the atomic mass of the element.

4. STEP 3 Find the simplest whole number ratio of the moles of each element. Since each mole contains the same number of atoms, the simplest whole number ratio of the moles is also the simplest whole number ratio of the atoms of each element. To do this, take all the mole values and divide them by the SMALLEST one The answers are the subscripts in the empirical formula

5. .05 Rule After step #3, if the value is within.05 of a whole number (+ 0.05 or - 0.05), then the value may be rounded to that whole number. The values used in these problems are obtained by experimentation. The 0.05 rule allows for experimental error.

6. IF the application of the.05 Rule does not produce whole numbers, then ALL of the results of step 3 must be multiplied by the same smallest integer that WILL produce values that can be rounded to whole numbers by the.05 Rule.

7. EXAMPLE: A compound is found to contain 72.3% Fe and 27.7% O by weight. Calculate the empirical formula. STEP 1 In 100 g of compound there would be 72.3 g Fe and 27.7 g O

8. 72.3g Fe 1 mole Fe X —————— 55.8 g Fe = 1.296 mole Fe 27.7g O 1 mole O X —————— 16.0 g O = 1.731 mole O 1.296 mole 1.731 mole 1.296 mole = 1.000 =1.336 X 3 = 3.00 = 3 X 3 = 4.01 = 4 Fe 3 O 4

9. Example of a hydrated compound CaSO 4  7 H 2 O Compounds with molecules of water held in their crystal structure Very common in nature Water can be removed by heating, leaving behind what is called the anhydrous compound (CaSO 4 for the above example)

10. Naming -- the following is tacked on the name obtained from the ions H 2 O monohydrate2 H 2 O dihydrate 3 H 2 O trihydrate4 H 2 O tetrahydrate 5 H 2 O pentahydrate6 H 2 O hexahydrate 7 H 2 O heptahydrate 8 H 2 O octahydrate 9 H 2 O nonahydrate 10 H 2 O dekahydrate CaSO 4  7 H 2 O -- named as calcium sulfate heptahydrate

11. Finding the empirical formula of a hydrate Find the empirical formula of a hydrate of CaSO 4 hydrate that is 28.5% H 2 O To solve this problem, find the simplest mole ratio between the anhydrous part of the compound (CaSO 4 ) and the water (H 2 O) H 2 O =28.5 % CaSO 4 =71.5 % (100% - 28.5%)

12. 1 mole CaSO 4 71.5 g CaSO 4 X ------------------- =.5250 mole CaSO 4 136.2 g CaSO 4. (molar mass of CaSO 4 ) 1 mole H 2 O 28.5 g H 2 O X ----------------- = 1.583 mole H 2 O 18.0 g H 2 O.5250 ------- = 1.00 = 1.5250 1.583 ------- = 3.01 = 3.5250 CaSO 4  3 H 2 O When you are finding formulas of hydrates they ALWAYS come out even!

13. Molecular Formula A molecular formula tells the actual number of atoms of each element in a molecule. It is a multiple of the empirical formula.

14. Molecular Formula Problem: The compound with a molar mass of 171.0 g/mole that contains 14.0% carbon, 41.5% chlorine, and 44.4% fluorine is a chlorofluorocarbon (CFC) that was once used as a refrigerant, but is now on the list of chemicals known to be ozone depleting. What is the molecular formula of this compound?

15. Empirical formula =CClF 2 Molar Mass of Empirical Formula: C: 1(12.0) = 12.0 Cl: 1(35.5) = 35.5 F: 2(19.0) = 38.0 85.5 (CClF 2 ) 2 =C 2 Cl 2 F 4

16. Combustion Analysis A 3.489g sample of a compound containing C, H, and O yields 7.832 g of CO 2, and 1.922g of water upon combustion. What is the simplest formula of the compound?

17. Since it’s combustion, you know C x H y O z + O 2  CO 2 + H 2 O All the C goes into the CO 2 All the H goes into the H 2 O (our job is to find x, y, and z!)

18. First find the amounts of C, H, and O Using the MM of CO 2 : Using the MM of H 2 O:

19. X 2 ≈ 5 X 2 ≈ 6 X 2 = 2 C5H6O2C5H6O2