1. Calculating Empirical Formula Using percentage or mass to find the Empirical Formula

2. Empirical Formula Empirical Formula is the lowest whole number ratio of elements in a compound. Molecular Formula the actual ratio of elements in a compound The two can be the same: H 2 O. Or Different: –CH 2 empirical formula –C 2 H 4 molecular formula –C 3 H 6 molecular formula

3. Calculating Empirical Formula Just find the lowest whole number ratio –C 6 H 12 O 6 becomes CH 2 O –CH 4 N is the lowest ratio It is not just the ratio of atoms, it is also the ratio of moles of atoms –In 1 molecule of CO 2 there is 1 atom of C and 2 atoms of O –In 1 mole of CO 2 there is 1 mole of carbon and 2 moles of oxygen

4. Combustion Analysis Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this –C is determined from the mass of CO 2 produced –H is determined from the mass of H 2 O produced –O is determined by difference after the C and H have been determined

5. Calculating Empirical Formula If you have mass, skip the next 2 steps If you have %, pretend that you have a 100 gram sample of the compound. That is, change the % to grams. Convert the grams to moles for each element. Write the number of moles as subscripts in a chemical formula. Divide each number by the least number. Multiply the results to get rid of any fractions.

6. Example Calculate the empirical formula of a compound composed of 38.7 % C, 16.2 % H, and 45.1 %N. Assume 100 g so: 38.7 g C x 1mol C = 3.22 mole C 12.0 gC 16.2 g H x 1mol H = 16.2 mole H 1.0 gH 45.1 g N x 1mol N = 3.22 mole N 14.0 gN

7. Determining Mole Ratio 3.22 mole C 16.2 mole H 3.22 mole N C 3.22 H 16.2 N 3.22 If we divide all of these by the smallest one It will give us the empirical formula

8. Divide by smallest to get Empirical Formula The ratio is 3.22 mol C = 1 mol C 3.22 molN 1 mol N The ratio is 16.2 mol H = 5 mol H 3.22 molN = 1 mol N C 1 H 5 N 1 is the Empirical Formula

9. A compound is 43.64 % P and 56.36 % O. What is the empirical formula? 43.6 g P x 1mol P = 1.4 mole P 31.0 gP 56.4 g O x 1mol O = 3.5 mole O 16 gO P 1.4 O 3.5

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

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

12. Caffeine is 49.5% C, 5.15% H, 28.9% N and 16.5% O. What is its empirical formula?

13. 49.5 g C 5.15 g H 28.9 g N 16.5 g O = 4.1 mol = 5.15 mol = 2.06 mol = 1.03 mol Since they are close to whole numbers we will use this formula We divide by lowest (1mol O) and ratio doesn’t change

14. C 4.12 H 5.15 N 2.06 O 1 empirical mass = 97g OR C 4 H 5 N 2 O 1

15. Empirical to molecular Since the empirical formula is the lowest ratio, the actual molecule could weigh more (by whole number multiples Divide the actual molar mass by the mass of one mole of the empirical formula. Caffeine has a molar mass of 194 g. what is its molecular formula?

16. Find x if 194 g 97 g = 2 C4H5N2O1C4H5N2O1 C 8 H 10 N 4 O 2. 2 X

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. Example- Data to moles 71.65 Cl 24.27C 4.07 H. = 2.0mol = 4.0mol

19. Moles to Empirical Formula C 2 H 4 Cl 2 Its molar mass is known (from gas density) is known to be 98.96 g. What is its molecular formula? We divide by lowest (2mol ) C 1 H 2 Cl 1 Lowest ratio = Empirical Formula would give an empirical wt of 48.5g/mol

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

21. Molecular Formula 2 X C 1 H 2 Cl 1 = C 2 H 4 Cl 2