Chapter 4. The Mole Concept


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1 Chapter 4. The Mole Concept Introduction If you were to take one volume (Eg. 1cm 3 ) of every element, weigh them and rank them according to their weights you would discover that they followed the periodic table with H first, He second across the table. Atoms come in various weights The difference between the mass of individual atoms is due to addition of particles (+ and n) Because we can not individually measure atoms, we must relate their mass to a standard. We have arbitrarily chosen Carbon  12 and assigned it the amu of amu Avogadro s Hypothesis GayLussac discovered that when two gaseous elements formed a new compound, the ratios of the volumes of gases used were simple whole number ratios. Avogadro explained the phenomenon by stating Equal volumes of gases (at the same temperature) and pressure contain equal numbers of particles. Weighing equal volumes of same T and P  see p.97 Figure 42 1
2 Mole  the amount of a substance that contains 6.02x10 23 particles (Avogadro s number  N) The molar mass of an element or compound   the number of grams necessary to have one mole of atoms or molecules of that compound Eg. The molar mass of H is 1.00g atoms H 2 is 2.0 molecules C is N is (average) Do example 45 to 411on overhead figure HWK Due tomorrow p.104 #49 The Molar Volume of a Gas At standard Temperature and Pressure (STP) one mole of gas occupies 22.4L (1 Litre = 1,000mL) Standard Temperature = 0 degrees C (273 degrees K) degrees K = degrees C
3 Standard Pressure = 1 atmosphere (~ sea level) or Pa (pascals) Example Given: Find: 12 litres of CO STP How many moles? Find: How much mass? Find: How many particles? New Memory Aid figure 3
4 Empirical Formulae obtained from experimental data  represents the smallest whole number ratio of atoms in the compound  ignores average weight of isotopes Eg. Experimentally one unit of carbon 1 mole C 12.0 g C combines with two units of oxygen 2 moles O 32.0 g O to give one unit of carbon dioxide 1 mole CO g CO 2 Therefore: One mole CO 2 has a mass of 44g One molecule CO 2 has a mass of 44g CO 2 x 1 mole = 7.31x1023 g mole CO x10 23 molecules molecule 4
5 Given: Experimental data Problem: Determine the Empirical Formula of the compound Rules To Find Empirical Formula 1. Determine the mass of each element in a sample of the compound 2. The mass of each element is divided by its molar mass ( ) to determine the number of moles of each element in the sample of the compound 3. The number of moles of each element is divided by the smallest number From Heath (p.120) 1. Given: 4.20g N, 12.0g O Find: Empirical Formula i) Moles of N 4.20g N x 1 mole N = moles 14.01g ii) Moles of O 12.0g O x 1 mole O = moles iii) N O / / = N O 2.5 *must be whole number ratios iv) 2(NO 2.5 ) = N 2 O 5 5
6 2. Given: 4.80g C; 0.40g H Find: Empirical Formula i) Moles of C 4.80g x 1 mole = 0.4 moles C g C ii) Moles of H 0.40g x 1 mole = 0.4 moles H 1.00g H iii) C 0.4 H 0.4 /0.4 /0.4 = CH 4. Given: 0.5g H; 9.40g O molecular mass  34g Find: Empirical Formula i) Moles of H 0.59g x 1 mole = 0.59 moles 1.00g H ii) Moles of O 9.40g x 1 mole = 0.59 moles 16.00g O of moles to give the ratio of atoms in the compound. Example 4.12, 4.13 Molecular Formulae  may or may not be the same as EF. If not  always some multiple of empirical formula 6
7  to determine the molecular formula of a compound, you must also be given the molecular compound Given: A 10.00g sample of a compound composed of H g O The molecular mass of the compound is 34g. Find: Molecular formula Solution: i) find empirical formula ii) find the empirical formula weight iii) divide the molecular mass by the empirical weight to give the multiple by which you determine molecular formula iv) Multiply empirical formula by multiple to give molecular formula So 1) Empirical Formula i) moles of each element 0.59g H x 1 mole H = 0.58 mol H 1.01g H ii) EF 9.40g O x 1 mol O = 0.59 mol O 16.0g O Therefore the ratio is 1:1 so empirical formula is HO 2) Empirical Formula Weight H = 1x1g + 17g O = 1x16g + 7
8 3) Molecular Mass/EF Weight 34g (given) = 2 multiple 17g 4) Multiply EF to give Molecular Formula (HO) 2 = H 2 O 2 (hydrogen peroxide) 410 Percent Composition Given: A compound of known formula Calculate: The percent composition of the component elements Example: H 2 O i) H  2 mol H = 2 (1.01) 2.02 O  1 mol O = 1 (16.00) ii) H: 2.02 x 100% = 11.2% O: x 100% = 88.8% Given: Two compounds with known masses of their components Problem: Are they the same? Example  Sample A contains 28g Fe & 8g O Total wt. Is 36g Sample B contains 112g Fe & 48g O Total wt is 160g Solution: Check for Fe % content and compare A  28g Fe x 100% = 78% 36g Total NOT THE SAME! 8
9 B  112g Fe x 100% = 70% 160g Total Given: A  a sample that is 78% Fe (rest O) B  a sample that is 70% Fe (rest O) Problem: Find the formulas! 1) Find percentage oxygen A  100%  78% = 22% B  100%  70% = 30% *Assume 2) Find grams of Fe in 100g of A 78% = 78g Fe 78g 100g A Find grams of O in 100g of A 22% = 22g O 22g 100g A 3) Find moles of Fe 78g Fe x 1 mol = 1.4 mole 55.8g Fe 4) Find moles of O 22g O x 1 mol = 1.4 moles 16.0g O Therefore, the ratio is 1.4:1.4 or FeO 9
10 411 Molarity Molarity is a method of measuring the concentration of a solution whereby the molar number is the number of moles of solute per liter of solution. Therefore: Molarity (M) = moles of solute Liters of solution Consider the NaCl solution: In class exercise: Choose a data point and calculate the molarity of this solution. Example 1 How many moles of HCl are in 1.45 L of a 2.25 M solution? 10
11 Example 2 What volume of a 0.40 M solution of NaOH(aq) is required to obtain 64 g of NaOH? Example 3 What is the molarity of a solution formed by dissolving 8.50 g of silver nitrate, Ag NO3, in water and making the final volume up to ml? 11
12 Exercises Review and Practice p120 Chapter Review 12
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