Aim # 7: How can we determine the empirical and molecular formulas of a compound? H.W. # 7 Study pp. 174-182 Ans. ques. p. 189 # 33, 34 p. 190 # 37,40,42 H.W. # 7a Study pp. 183-185 Ans. ques. p. 190 # 44,45,46,49,50

I Determining empirical formulas A. Problem: Find the empirical formula of a compound, a sample of which consists of 11.2g of iron and 21.3g of chlorine. 1. Determine the number of moles of each element in the sample. 11.2g Fe x 1 mole Fe = 0.200 mol Fe 55.9g Fe 21.3g Cl x 1 mole Cl = 0.600 mol Cl 35.5g Cl

# of mole Fe atoms = 0.200 = 1 # of moles Cl atoms 0.600 3 2. From the number of moles, determine the smallest whole-number ratio of the atoms of each element in the compound. # of mole Fe atoms = 0.200 = 1 # of moles Cl atoms 0.600 3 3. Use the numbers from the whole-number ratio as the subscripts in the empirical formula. 1 atom Fe : 3 atoms Cl FeCl3 Remember, the metal comes first!

1. Assume that you have a 100-g sample of the compound. B. Problem: Determine the empirical formula of the compound with a percentage composition of 27.3% carbon and 72.7% oxygen. 1. Assume that you have a 100-g sample of the compound. 27.3% of 100g = 27.3g C 72.7% of 100g = 72.7g O The problem may now be solved as before. 2. 27.3g C x 1 mol C = 2.28 mol C 12.0g C 72.7g O x 1 mol O = 4.54 mol O 16.0g O

3. # of moles C atoms = 2.28 = 1 ≈ 1 # of moles O atoms 4.54 1.99 2 1 atom C : 2 atoms O CO2 Problem: Find the empirical formula of the compound that consists of 75.0% carbon and 25.0% hydrogen.

75. 0 g C x 1 mol C = 6. 25 mol C 12. 0g C 25. 0 g H x 1 mol H = 25 75.0 g C x 1 mol C = 6.25 mol C 12.0g C 25.0 g H x 1 mol H = 25.0 mol H 1.00g H # moles C = 6.25 = 1 # moles H 25.0 4 Formula: CH4

II To determine the molecular formula of a compound, Derive the empirical formula. Calculate gram-molecular mass gram-formula mass of empirical formula Multiply the subscripts of the empirical formula by the number calculated in step 2, above.

the gram-formula mass of CH2 is 14g. e.g. If the empirical formula of the compound is CH2 and the gram molecular mass is 56g, the gram-formula mass of CH2 is 14g. The ratio is 56g = 4 = 1 14g 1 The molecular formula is 4(CH2) C4H8

Problem: A compound of phosphorous and oxygen contains 56. 4% oxygen Problem: A compound of phosphorous and oxygen contains 56.4% oxygen. If the gram-molecular mass is 284g, what is the molecular formula of the compound? Ans. A 100-g sample of the compound contains 56.4g O and 43.6g P. 43.6g P x 1 mol P = 1.41 mol P 31.0g P 56.4g O x 1 mol O = 3.53 mol O 16.0g O mol P = 1.41 = 1 = 2 3.53 2.5 5 empirical formula: P2O5 gram-formula mass: 2(31.0g) + 5(16.0g) = 142.0g

Therefore, the molecular formula is 2(P2O5) = P4O10 284g = 2 142g 1 Therefore, the molecular formula is 2(P2O5) = P4O10 Problem: Ethylene glycol, the substance used in automobile antifreeze, is composed of 38.7% C, 9.7% H, and 51.6% O by mass. Its molar mass is 62.1 g/mol. a) What is the empirical formula of ethylene glycol? b) What is its molecular formula?

Problem Determine the empirical and molecular formulas of nicotine, a component of tobacco: 74.1% C, 8.6% H, and 17.3% N by mass. Nicotine’s molar mass is 160 ± 5g.