1. Determining Chemical Formulas Read pp. 107-120 Extension Questions p. 120 #1, 5-7

2. Law of Constant Composition “A compound contains elements in certain fixed proportions (ratios) and in no other combinations, regardless of how the compound is prepared or where it is found in nature.”

3. How do we determine chemical formulas? 1.Mass Spectrometer Finds the molar mass of a compound A small sample is vapourized into a gas phase and then ionized (becomes charged) into ions. The detector sorts and separates the ions according to their mass and charge. Results portrayed in a bar graph that tells you the composition of the compound. DO NOT COPY!

4. How do we determine chemical formulas? 2. Combustion Analyzer Finds the % of C, H, O and maybe N in a compound Sample of compound is burned using oxygen gas at 980 Celsius. Products are now CO 2 and H 2 O  masses are weighed to determine % composition. DO NOT COPY!

5. Percent Composition The percentage (by mass) of EACH element in a compound.

6. Empirical Formula A formula that gives you the lowest ratio of atoms in a compound. However, it does not tell you the exact number of each type of atom.

7. Example 1 What is the empirical formula of a compound with 39.99% C, 6.727% H and 53.28% O? 1. Find masses in 100 g sample m C = 39.99 x 100 gm H = 6.727 x 100 g 100 100 = 39.99 g = 6.727 g m O = 53.28 x 100 g 100 = 53.28 g

8. 2. Find moles for each element n C = 39.99 gn H = 6.727 gn O = 53.28 g 12.01 g/mol 1.01 g/mol 16.00 g/mol = 3.33 mol = 6.66 mol = 3.33 mol 3. State the ratio, then divide by lowest ratio n C : n H : n O 3.33 : 6.66 : 3.33 1 : 2 : 1 The empirical formula is CH 2 O

9. Example 2 The percentage composition of a compound is 69.9% iron and 30.1% oxygen. What is the empirical formula of the compound? 1. Find mass (in 100 g sample) m Fe = 69.9 x 100 gm O = 30.1 x 100 g 100 100 = 69.9 g = 30.1 g 2. Find moles for each element n Fe = 69.9 gn O = 30.1 g 55.85 g/mol 16.00 g/mol = 1.25 mol = 1.88 mol

10. 3. State ratio then divide by lowest ratio n Fe : n O 1.25 : 1.88 1 : 1.5 (you need WHOLE #s, so multiply both #s by 2 in order to get whole #s) 2 : 3 Therefore, the empirical formula is Fe 2 O 3.

11. Molecular Formula A formula that tells you the exact number of atoms in one molecule of the compound. It may be equal to its empirical formula or a multiple of that formula. To find this, divide the molar mass of the known sample by the empirical mass.

12. Example 3 From Example 1, the sample was known to have a molecular mass of 180.18 g/mol. What is the molecular formula of this compound? Empirical formula: CH 2 O (from Example 1 slide) 1. Find molar mass of EF M = 1(12.01 g/mol) + 2(1.01 g/mol) + 1(16.00 g/mol) = 30.03 g/mol 2. Find ratio of M actual : M EF then divide by lowest # 180.18 g/mol : 30.03 g/mol 6 : 1 (so the molar mass of the sample is 6 times greater than the molar mass of the EF sample)

13. 3. Calculate MF using ratio MF = 6 (EF) = 6(CH 2 O) = C 6 H 12 O 6 The molecular formula of the compound is C 6 H 12 O 6.

14. Example 4 (when EF is not given) A combustion analyzer found that the % composition of a compound was 32.0% C, 6.70% H, 42.6% O, and 18.7% N. The mass spectrometer found the molar mass to be 75.8 g/mol. What is the MF of the compound? 1. Calculate mass of each element in 100 g sample m C = 32.0 x 100 gm H = 6.70 x 100 gm O = 42.6 x 100 g 100 100 100 m N = 18.7 x 100 g 100

15. 2. Find moles of each element n C = 32.0 gn H = 6.70 gn O = 42.6 g 12.01 g/mol 1.01 g/mol 16.00 g/mol = 2.66 mol = 6.63 mol = 2.66 mol n N = 18.7 g 14.01 g/mol = 1.33 mol 3. State ratio (and divide by lowest #) n C : n H : n O : n N 2.66 : 6.63 : 2.66 : 1.33 2 : 5 : 2 : 1 Empirical formula is C 2 H 5 O 2 N

16. 4. Find molar mass of EF M = 2(12.01 g/mol) + 5(1.01 g/mol) + 2(16.00 g/mol) + 1(14.01 g/mol) M = 75.00 g/mol 5. Find ratio of M actual : M EF and divide by lowest # 75.8 g/mol : 75.00 g/mol 1 : 1 6. Calculate MF using ratio MF = 1(EF) = 1(C 2 H 5 O 2 N) = C 2 H 5 O 2 N The molecular formula is C 2 H 5 O 2 N.