1. Molecular Formulas

2. 2 A molecular formula is equal or a multiple of its empirical formula has a molar mass that is the product of the empirical formula mass multiplied by a small integer molar mass = a small integer empirical mass is obtained by multiplying the subscripts in the empirical formula by the same small integer Relating Molecular and Empirical Formulas

3. 3 Some Compounds with Empirical Formula CH 2 O

4. Calculating a Molecular Formula from an Empirical Formula 4

5. 5 Determine the molecular formula of a compound that has a molar mass of 78.11 g and an empirical formula of CH. STEP 1 Calculate the empirical formula mass. Empirical formula mass of CH = 13.02 g STEP 2 Divide the molar mass by the empirical formula mass to obtain a small integer. 78.11 g = 5.999 ~ 6 13.02 g Finding the Molecular Formula

6. 6 STEP 3 Multiply the empirical formula by the small integer to obtain the molecular formula. Multiply each subscript in C 1 H 1 by 6. Molecular formula = C 1x6 H 1x6 = C 6 H 6 Finding the Molecular Formula (continued)

7. 7 A compound has a molar mass of 176.1g and an empirical formula of C 3 H 4 O 3. What is its molecular formula? 1) C 3 H 4 O 3 2) C 6 H 8 O 6 3) C 9 H 12 O 9 Learning Check

8. 8 STEP 1 Calculate the empirical formula mass. C 3 H 4 O 3 = 88.06 g/EF STEP 2 Divide the molar mass by the empirical formula mass to obtain a small integer. 176.1 g (molar mass) = 2 88.06 g (empirical formula mass) STEP 3 Multiply the empirical formula by the small integer to obtain the molecular formula. molecular formula = 2 x empirical formula C 3x2 H 4x2 O 3x2 = C 6 H 8 O 6 (2) Solution

9. 9 A compound contains 24.27% C, 4.07% H, and 71.65% Cl. The molar mass is about 99 g. What are the empirical and molecular formulas? Molecular Formula

10. 10 STEP 1 Calculate the empirical formula mass. 24.27 g C x 1 mol C = 2.021 mol of C 12.01 g C 4.07 g H x 1 mol H = 4.04 mol of H 1.008 g H 71.65 g Cl x 1 mol Cl = 2.021 mol of Cl 35.45 g Cl Solution

11. 11 Solution (continued) 2.021 mol C =1 mol of C 2.021 4.04 mol H =2 mol of H 2.021 2.02 mol Cl=1 mol of Cl 2.021 Empirical formula = C 1 H 2 Cl 1 = CH 2 Cl Empirical formula mass (EM) CH 2 Cl = 49.48 g

12. 12 Solution (continued) STEP 2 Divide the molar mass by the empirical formula mass to obtain a small integer. Molar mass = 99 g = 2 Empirical formula mass 49.48 g STEP 3 Multiply the empirical formula by the small integer to obtain the molecular formula. 2 x (CH 2 Cl) C 1x2 H 2x2 Cl 1x2 = C 2 H 4 Cl 2

13. 13 A compound is 27.4% S, 12.0% N, and 60.6 % Cl. If the compound has a molar mass of 351 g, what is the molecular formula? Learning Check

14. 14 STEP 1 Calculate the empirical formula mass. In 100 g, there are 27.4 g S, 12.0 g N, and 60.6 g Cl. 27.4 g S x 1 mol S = 0.854 mol of S 32.07 g S 12.0 g N x 1 mol N = 0.857 mol of N 14.01 g N 60.6 g Cl x 1mol Cl = 1.71 mol of Cl 35.45 g Cl Solution

15. 15 STEP 2 Divide the molar mass by the empirical formula mass to obtain a small integer. 0.854 mol S = 1.00 mol of S 0.854 0.857 mol N = 1.00 mol of N 0.854 1.71 mol Cl = 2.00 mol of Cl 0.854 empirical formula = SNCl 2 empirical formula mass = 116.98 g Solution (continued)

16. STEP 3 Multiply the empirical formula by the small integer to obtain the molecular formula. Molar mass = 351 g = 3 Empirical formula mass 116.98 g Molecular formula = (SNCl 2 ) 3 = S 3 N 3 Cl 6 Solution (continued)

17. Review Topics We have covered the following: – Stoichiometry – Limiting Reactant – Percent Composition – Empirical formula – Molecular Formula

18. 1. Write a balanced chemical equation. 2. Identify known & unknown. 3. Line up conversion factors. – Mole ratio - moles  moles – Molar mass -moles  grams – Molar volume -moles  liters gas Core step in all stoichiometry problems!! Mole ratio - moles  moles 4. Check answer.

19. How many grams of silver will be formed from 12.0 g copper in a silver nitrate solution? How many grams of KClO 3 are required to produce 9.00 L of O 2 at STP?

20. Reacting Amounts In a table setting, there is 1 plate, 1 fork, 1 knife, and 1 spoon. How many table settings are possible from 5 plates, 6 forks, 4 spoons, and 7 knives? What is the limiting item?

21. B. Calculations Involving a Limiting Reactant

22. 22

23. If you start with 35.0 grams of Lead (II) Nitrate and 18.0 grams of sodium iodide, how many grams of sodium nitrate can be formed?

24. Theoretical Yield –The maximum amount of a given product that can be formed when the limiting reactant is completely consumed. The actual yield (amount produced) of a reaction is usually less than the maximum expected (theoretical yield). Percent Yield –The actual amount of a given product as the percentage of the theoretical yield.

25. Empirical Formula Determination 1.Given the percent of each element, base calculation on 100 grams of compound. Determine moles of each element in 100 grams of compound. 2.Divide each value of moles by the smallest of the values. (This gives one element the smallest number of moles necessary) 3.Multiply each number by an integer to obtain all whole numbers.