1. Empirical and Molecular Formulas Topic #20

2. Empirical and Molecular Formulas Empirical --The lowest whole number ratio of elements in a compound. Molecular -- the actual ratio of elements in a compound Sometimes the two can be the same. CH 2 empirical formula C 2 H 4 molecular formula C 3 H 6 molecular formula H 2 O both

3. Calculating Empirical Just find the lowest whole number ratio C 6 H 12 O 6 CH 4 N

4. Determine the empirical formulas for each of the following molecular formulas. 1. C 8 H 18 …..______________ 2. H 2 O 2 ……______________ 3. HgCl 2 …..______________ 4. C 3 H 8 …….______________ 5. Na 2 C 2 O 4..._____________ 6. H 2 O......________________ 7. C 4 H 8 ….._______________ 8. C 4 H 6 …..________________ 9. C 7 H 12 …._______________

5. You can calculate empirical formulas if you know the percent composition………….

6. Calculating Empirical Pretend that you have a 100 gram sample of the compound. That is, change the % to grams. Convert the grams to mols for each element. Write the number of mols as a subscript in a chemical formula. Divide each number by the least number. Multiply the result to get rid of any fractions.

7. Example Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N. Assume 100 g so 38.67 g C 1mol C = 3.220 mole C 12.01 gC 16.22 g H 1mol H = 16.09 mole H 1.01 gH 45.11 g N 1mol N = 3.219 mole N 14.01 gN

8. 3.220 mole C 16.09 mole H 3.219 mole N C 3.22 H 16.09 N 3.219 If we divide all of these by the smallest one It will give us the empirical formula

9. Example The ratio is 3.220 mol C = 1 mol C 3.219 molN The ratio is 16.09 mol H = 5 mol H 3.219 molN C 1 H 5 N 1 is the empirical formula A compound is 43.64 % P and 56.36 % O. What is the empirical formula?

10. 43.6 g P 1mol P = 1.4 mole P 30.97 g P 56.36 g O 1mol O = 3.5 mole O 16 gO P 1.4 O 3.5

11. Divide both by the lowest one The ratio is 3.5 mol O = 2.5 mol O 1.4 mol P P 1.4 O 3.5 P 1 O 2.5

12. Multiply the result to get rid of any fractions. P 1 O 2.5 2 X = P 2 O 5

13. Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula? Empirical Mass=

14. Empirical to molecular Since the empirical formula is the lowest ratio the actual molecule would weigh more. By a whole number multiple. Divide the actual molar mass by the mass of one mole of the empirical formula.

15. Caffeine has a molar mass of 194 g. what is its molecular formula? Find x if 194 g 97 g = 2 C4H5N2O1C4H5N2O1 C 8 H 10 N 4 O 2. 2 X

16. Example A compound is known to be composed of 71.65 % Cl, 24.27% C and 4.07% H. Its molar mass is known (from gas density) is known to be 98.96 g. What is its molecular formula?

17. Example A compound is known to be composed of 71.65 % Cl, 24.27% C and 4.07% H. Its molar mass is known (from gas density) is known to be 98.96 g. What is its molecular formula?

18. Its molar mass is known (from gas density) is known to be 98.96 g. What is its molecular formula? would give an empirical wt of 48.5g/mol = 2 =

19. 1. What’s the empirical formula of a molecule containing 65.5% carbon, 5.5% hydrogen, and 29.0% oxygen? 2.If the molar mass of the compound in problem 1 is 110 grams/mole, what’s the molecular formula? 3.What’s the empirical formula of a molecule containing 18.7% lithium, 16.3% carbon, and 65.0% oxygen? 4.If the molar mass of the compound in problem 3 is 73.8 grams/mole, what’s the molecular formula?