Unit 6 The Mole Concept

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1 Chemistry Form 3 Page 62 Ms. R. Buttigieg Unit 6 The Mole Concept See Chemistry for You Chapter 28 pg See GCSE Chemistry Chapter 5 pg Relative atomic mass. The relative atomic mass () of an atom is the number of times an atom is heavier than one twelfth of a carbon-12 atom. The average mass number (taking the isotopes in consideration) is approximately equal to the relative atomic mass. The two are not exactly equal as in the average mass number, the mass of the electrons and the binding energy is not considered. Mass number = mass of protons and neutrons. = average mass number considering also the mass of electrons and binding energy. E.g. For Chlorine there are 2 isotopes 75% of them are 35 Cl and 25% are 35 Cl. Average Mass Number = (75 x 35) + (25 x 37) = We are used to collective terms to describe a number of objects, e.g. a dozen eggs etc. In chemistry the term mole (abbreviation mol) is used in the same way. A mole is that amount of matter which contains 6 x particles ( ) This number is called Avogadro s constant (L). The mass of one mole of atoms: We can know the number of moles that are present in a given mass of a particular element (the can be obtained from the periodic table) by using a simple equation. If the equation is re-arranged we can find the mass for a given number of moles. Moles of Atoms = Mass OR mass = Moles of Atoms x How many moles are there in 3g of carbon? What is the mass of 7 moles of carbon? Moles = 3 = 0.25 moles 12 Mass = 7 x 12 = 84 g of C = 12 Practice on mole( s)/ mass and mole( s)/ atoms interconversion. Work out Chemistry for You pg. 352 nos 1-5 (mass to moles); pg. 353 nos 6-10 (moles to mass)

2 Chemistry Form 3 Page 63 Ms. R. Buttigieg 6.2 Relative molecular/formula mass. The relative molecular mass of any molecule is worked out by adding together the relative atomic masses of all the atoms in the molecule. Relative formula mass is used for ions. Write if the relative formula or molecular masses (R.F.M. or R.M.M.) of the following can be found and calculate them. a) NH 3 b) CO c) O 2 d) CuSO 4 e) CaCO 3 f) N 2 g) SO 2 _ Moles and mass of Molecules A similar equation as those used in the previous page is used Moles of Molecules/Ions = Mass OR mass = Moles of Molecules/Ions x R.F.M R.F.M How many moles are there in 4g of carbon dioxide (CO 2 )? Step one Add the.s = 44 of C = 12 of O = 16 Step two moles = 4 44 = 0.09 moles or 1 / 11 moles 4 What is the mass of 0.6 moles of carbon dioxide (CO 2 )? Step one Add the.s = 44 Step two mass = 0.6 x 44 = 26.4 g Work out Chemistry for You Pg. 353 numbers 11-15

3 Chemistry Form 3 Page 64 Ms. R. Buttigieg 6.3 Applications of the mole concept. Experimental determination of chemical formulae of compounds. To work out the formula, one must know what masses of the elements combine. This must be found by experiment. For example Magnesium and Oxygen combine like this: Magnesium + Oxygen Magnesium Oxide crucible Method: 1. A clean crucible and lid are weighed. 2. A coil of clean magnesium ribbon is added, and the new mass found. tripod 3. Then the crucible is heated for some time. 4. Then the Bunsen burner is removed from under the crucible, and the lid is carefully raised a little, so that oxygen enters and the magnesium reacts, but without losing the magnesium oxide powder. 5. When the magnesium no longer continues to flare up, the lid is removed and the crucible heated. 6. When the crucible cools, the crucible, lid and contents are weighed. 7. The crucible is heated again, left to cool and weighed, until two consecutive weighing (one after the other) are the same. a. This is done to ensure that all the magnesium has reacted. Mass of Crucible and lid a 14.63g Mass of crucible, lid and magnesium b 14.87g Mass of crucible, lid and magnesium oxide c 15.03g Mass of magnesium b a 0.24g Mass of oxygen that combined with magnesium c - b 0.16g Magnesium From these numbers, the formula can be found. Magnesium Oxygen Masses reacting Number of moles Ratio of moles 1 1 Formula MgO Chemistry 4U Work out Pg. 358 no. 1-5 See pg. 359 This is called the empirical formula of the compound and is shows the simplest ratio of the atoms present.

4 Chemistry Form 3 Page 65 Ms. R. Buttigieg 2 nd Part - Experimental determination of chemical formulae of compounds. By reduction using hydrogen to prepare Copper Oxide Concentrated hydrochloric acid Copper (II) oxide Heat Hydrogen gas Heat Anhydrous calcium chloride to dry the hydrogen Black copper (II) oxide contains copper and oxygen combined. By the removal of oxygen in a stream of hydrogen the masses of copper and oxygen which were combined can be determined. 1. A boat (container) is weighed. 2. It is then weighed again with pure dry copper (II) oxide in it and placed in a hard glass tube. 3. Hydrogen is produced as shown, passed through calcium chloride and over the copper oxide. 4. This will burn in exposure to a flame till all the oxygen is expelled. 5. Eventually a reddish brown powder of copper is left. 6. When cooled the boat is weighed again and the mass of copper and oxygen determined. Copper (II) oxide + hydrogen Copper + Water Mass of boat Mass of boat + copper oxide Mass of boat + copper Mass of oxygen Mass of copper 4.32 g 5.61 g 5.35 g 0.26 g 1.03 g Copper Oxygen Masses reacting Number of moles 1.03 = = Ratio of moles 1 1 Formula CuO

5 Chemistry Form 3 Page 66 Ms. R. Buttigieg HW: Work out the following: Write down the number of moles of atoms in: 1. 60g of Carbon 2. 3g of aluminium 3. 40g of iron (III) oxide, Fe 2 O g of calcium carbonate, CaCO g of hydrogen molecules, H 2 What are the masses of the following? moles of water, H 2 O 7. 2 moles of ethanol, C 2 H 5 OH moles of ammonium nitrate, NH 4 NO moles of lead(ii) nitrate, Pb(NO 3 ) 2. Determination of water of Crystallization An example is barium chloride BaCl 2.xH 2 O. 1. A clean crucible is weighed with the lid grams of barium chloride crystals are added and everything is weighed. 3. The crucible with lead is then heated on the pipe triangle on a tripod till the water of crystallization is driven off. 4. It is then allowed to cool in a dessicator (to exclude moisture) and weighed. 5. It is then heated again, cooled and reweighed. 6. This is done till a constant mass is obtained which shows that all water has been removed. An example of values that can be obtained. Mass of Crucible and lid a 4.40 g Mass of crucible, lid and barium chloride crystals b 6.45 g Mass of crucible, lid and barium chloride anhydrous c 5.35 g Mass of water removed b - c 1.1 g Mass of anhydrous barium chloride c - a 0.95 g xh 2 O = BaCl 2 (b-c) (c-a) And inserting the relative molecular masses 18x = (b-c) 208 (c-a) and x = (b-c) x 208 (c-a) 18 and x = (1.1) x 208 (0.95) 18 Therefore x = 13

6 Chemistry Form 3 Page 67 Ms. R. Buttigieg 6.4 Percentage Composition by Mass a. Calculating the percentage composition by mass of an element in a compound. Example 2: What is the percentage of Carbon in C3H6? RAM of C = 12, RAM of H = 1. The RFM is (3 x 12) + (6 x 1), = = 42. The percentage of Carbon in the compound = 12 x 3 x 100% = 85.7% is Carbon 42 Example 3: What is the percentage of Calcium in CaCO3? RAM of H = 1, RAM of C = 12, RAM of O = 16, RAM of Ca = 40, The RFM = Percentage of Calcium = Work out Chemistry for You. Pg. 360 numbers 1-5

7 Chemistry Form 3 Page 68 Ms. R. Buttigieg b. Calculating percentage composition by mass of water of crystallization in a hydrated salt. Example 1: Calculate the percentage of water of crystallization in Na 2 CO H 2 O. RAM of H = 1, RAM of C = 12, RAM of O = 16, RAM of Na = 23 The RFM is (2 x 23) (3 x 16) + (10 x 18) = 286 g. The percentage of Water in the compound = 18 x 10 x 100% = 63% is Water 286 Example 2: Calculate the percentage of water of crystallization in CuSO 4. 5 H 2 O. RAM of H = 1, RAM of Cu = 64, RAM of O = 16, RAM of S = 32 The RFM = Percentage of Water = Example 3: Calculate the percentage of water of crystallization in CuSO H 2 O. The RFM = Percentage of Water = Example 4: Calculate the percentage of water of crystallization in NaCl. 7 H 2 O. RAM of H = 1, RAM of Cl = 35.5, RAM of O = 16, RAM of Na = 23 The RFM = Percentage of Water = Example 5: Calculate the percentage of water of crystallization in NaCl. 4 H 2 O. The RFM = Percentage of Water =

8 Chemistry Form 3 Page 69 Ms. R. Buttigieg c. Determining empirical formula of a compound from the percentage composition by mass. Example 1: Calculate the formula of a compound containing hydrogen and sulphur. It contains 6 % hydrogen and 64 % sulphur. RAM of H = 1, RAM of S = 32 Number of moles = Percentage Moles of H = 6 = 6 Moles of S = 94 = The ratio of the moles present H : S 6 : 2.9 Divide by the smallest possible number = 3 2 : 1 So the empirical formula is H 2 S Example 2: Calculate the formula of a compound containing iron, sulphur and oxygen. It contains 28 % iron and 24 % sulphur. RAM of Fe = 56, RAM of O = 16, RAM of S = 32 a. Percentage of Oxygen in the compound = 100 ( ) = 48 % b. Number of moles = Percentage Moles of Iron = 28 = 0.5 Moles of Sulphur = 34 = 0.75 Moles of Oxygen = 48 = Therefore ratio of moles Iron Sulphur Oxygen A number which divides into all of them is 0.25 So we get Fe 2 S 3 O 12 - this is the empirical formula Work out Chemistry for You. Pg. 361 numbers 6-8

9 Chemistry Form 3 Page 70 Ms. R. Buttigieg Example : Calculate the simplest formula of a compound which has the composition: Carbon = %, oxygen = 27.6%, hydrogen = 10.3 % RAM of C = 12, RAM of H = 1, RAM of O = 16 Moles = Ratio of Moles Divide by The empirical formula is d. Deriving the formula of a hydrated compound, given its percentage composition by mass Example 1: Calculate the simplest formula of a compound which has the composition: Magnesium = 9.8 %, sulphur = 13 %, oxygen = 26%, water of crystallization = 51.2 % RAM of Mg = 24, RAM of S = 32, RAM of H2O = 18, RAM of O = 16 Moles = Ratio of Moles Divide by The empirical formula is

10 Chemistry Form 3 Page 71 Ms. R. Buttigieg 6.5 Mole/Mass relationships in chemical equations Law of Conservation of Mass The total mass of reactants is equal to the total mass of products. This applies for all chemical reactions and is known as the Law of Conservation of Mass. 2 Ca (s) + O 2 (g) 2 CaO (s) 2 moles + 1 mole 2 moles 2 x x 2 2 x (40+16) =80g =32g 112g This was understood by the Greeks but was first explained in this way again by Antoine Lavoisier in This can be used to calculate masses of products formed from given masses of reactants before they are carried out. Using the mole ratios from a balanced equation, to calculate the mass of a reactant/ product required/ or produced from a given mass of reactant/ product. Example 1: The following equation shows the complete combustion of methane in oxygen. What mass of carbon dioxide will be formed from 4g of methane? CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (l) Moles reacting: Masses reacting: moles of CH 4 give moles of CO 2 g of CH 4 give g of CO 2 4 g of CH 4 give g of CO 2 Example 2: The shows the reaction of sodium carbonate with hydrochloric acid. What mass of sodium chloride is produced when 10.6g of sodium carbonate is reacted with excess HCl? Na 2 CO 3 (s) + 2 HCl (aq) 2NaCl (aq) + CO 2 (g) + H 2 O (l) Moles reacting: Masses reacting: moles of Na 2 CO 3 give moles of NaCl g of Na 2 CO 3 give g of NaCl 10.6 g of CH 4 give g of NaCl

11 Chemistry Form 3 Page 72 Ms. R. Buttigieg Deriving a balanced chemical equation by converting the masses of reactants and/ or products into mole ratios. 8 g of hydrogen react with excess oxygen to give 72 g of water. Calculate the moles involved and hence balance the equation for this reaction. Hydrogen + Oxygen Water According to the Law of Conservation of Mass: Mass of reactants = Mass of reactants 8 + = 72 Therefore Mass of Oxygen reacting = 72 8 = g Moles reacting: Hydrogen = 1 g = 1 mole Oxygen = 16 g = 1 mole Water = 18 g = 1 moles 8 g = 8 moles g = 72 g = Therefore mole H 2 : mole O 2 : mole H 2 O So the balanced equation (and the most simplified is) = Work out Chemistry for You pg. 363 numbers 1-2 Work out GCSE Chemistry pg. 77 numbers 1 3 Question: ( of Mg = 24; of O = 16) 4.0g of Magnesium metal reacted with oxygen according to the following equation. 2 Mg (s) + O 2 (g) 2 MgO (s) i) Calculate the number of moles of atoms in 4.0g of magnesium. ii) Calculate the mass of magnesium oxide that is formed.

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