1. 6.7 & 6.9 Empirical & Molecular Formulae pp. 289 – 293, 296 - 300

2. Empirical & Molecular Formulae The chemical word equation for the reaction between aluminum and bromine liquid is as follows: aluminum + bromine liquid → aluminum bromide Its balanced chemical equation is: 2Al(s) + 3Br 2 (l) → Al 2 Br 6 (s)

3. Empirical & Molecular Formulae At first glance, the formula for aluminum bromide in the equation, A l2 Br 6 appears to be a strange one. Aluminum’s valence is 3 and bromine’s valence is 1 and using the cross-over rule, the formula would end up as being AlBr 3. However, through chemical analysis, the molecular formula is Al 2 Br 6.

4. The molecular formula is the exact formula of the compound. The empirical formula is the simplest whole number ratio of the elements’ atoms in a compound. – The empirical formula of aluminum bromide is AlBr 3. In many cases, the empirical formula is the molecular formula of a compound. – Water, H 2 O, is such a case.

5. The elements’ respective subscripts in the formula show a mole ratio relationship. The formula for water, H 2 O, shows us that on a macroscopic level, there are 2 moles of hydrogen for every 1 mole of oxygen atoms. To calculate an empirical or a molecular formula for a compound, the given masses for each element must be converted into their respective whole number mole values.

6. Sample Calculation What is the empirical formula of a compound which is found by analysis to contain 2.2% hydrogen, 26.7% carbon and 71.1% oxygen? If we assume a 100g sample, G: m H = 2.2g m C = 26.7g m O = 71.1g R: Empirical Formula CHO = ?

7. A: n = m ÷ M S: n H = 2.2 g ÷ 1.01 g/mol = 2.18 mol n C = 26.7 g ÷ 12.01 g/mol = 2.22 mol n O = 71.1 g ÷ 16.00 g/mol = 4.44 mol Mole Ratio:HCO:: 2.18 mol2.22 mol4.44 mol: : Dividing each by the lowest mole value, 2.18 1 mol1.02 mol2.04 mol: :

8. If the values are within 0.1 of a whole number, the numbers can be rounded off to the nearest whole number 1 mol 2 mol: : Therefore, the empirical formula is HCO 2

9. Example # 2 A hydrocarbon, upon analysis, shows the following mole to mole ratio relationship between carbon and hydrogen Mole Ratio:CH: 0.166 mol0.444 mol: Dividing each by the lowest mole value, 0.166 1 mol2.67 mol:

10. 2.67 cannot be rounded, so try multiplying each number by 2 3 mol8.02 mol: 2 mol5.3 mol: 5.3 cannot be rounded, so try multiplying each number from the previous page by 3 Round each to the nearest whole number 3 mol 8 mol : Therefore, the empirical formula is C 3 H 8

11. Molecular Formulae The molecular formula of a compound gives you the actual composition of a molecule. Its calculation is not much different from that of the empirical formula. Once you calculate the empirical formula you need to do one further calculation: x = M compound ÷ M empirical formula You will then multiply each subscript in the empirical formula by x

12. Sample Calculation The empirical formula of a compound is determined to be CH. Its molecular molar mass is 104.16 g/mol. Determine its molecular formula. G: M CH = 13.02 g/mol M compound = 104.16 g/mol R: Molecular formula of compound

13. A: x = M compound ÷ M empirical formula S: x = 104.16 g/mol ÷ 13.02 g/mol = 8 P: Therefore the molecular formula is C 8 H 8

14. Homework Read pp. 289 – 292 – Answer # 1 – 11 p. 293 Read pp. 304 – 305 – Answer # 1 – 9 p. 300