1. Calculate the number of moles of the named chemical in each of the following.

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2 1. Calculate the number of moles of the named chemical in each of the following. a. 1.32g of ammonium sulfate, (NH4)2SO4 moles = mass/gfm gfm (NH4)2SO4 = 132 g moles = 1.32/132 = 0.01 moles b. 200cm 3 of 3 mol l -1 sodium hydroxide, NaOH moles = CxV = 3 x 200/1000 = 0.6 moles cm 3 of a solution of magnesium chloride, MgCl2, is known to contain 0.5 moles of chloride ions. What mass of magnesium does the solution contain? Ratio of Mg: Cl = 1: 2. If compound contains 0.5 moles Cl it must contain 0.5/2 = 0.25 moles Mg Mass Mg = n x gfm = 0.25 x 24.3 = g 3. In an experiment to prepare aspirin, 10 g of 2 hydroxybenzoic acid, HOC6H4COOH, was treated with excess ethanoyl chloride. The equation for the reaction is CH3COCl + HOC6H4COOH CH3COOC6H4COOH + HCl Calculate the mass of aspirin formed in this reaction. gfm HOC6H4COOH = 138 g moles of HOC6H4COOH = mass/gfm = 10/138 = mol As g there is a one to one ratio moles of aspirin made = mol gfm of aspirin = 180 g mass of aspirin = n x gfm = x 180 = g

3 4. Calculate the percentage by mass of nitrogen in urea CO(NH2)2 %N = (mass of N / gfm of compound) x 100 gfm CO(NH2)2 = 60 g %N = (28/60) x 100 = 46.7% 5. Titanium is manufactured by heating titanium(iv) chloride with sodium. TiCl4 + 4Na Ti + 4NaCl If this reaction is only 75% efficient, calculate the mass of sodium is required to produce 100 kg of titanium? Moles of titanium produced = mass/gfm = /47.9 = mol {for 75%} Moles of titanium produced for100% efficiency = 100/75 x = mol Mole ratio of Ti: 4Na = 1:4 Moles of Na required = x 4 = mol Mass of Na required = n x gfm = x 23 = g = kg g of magnesium was used to make hydrated magnesium sulfate crystals, MgSO4.7H2O The mass of crystals produced was 9.84 g. Calculate the percentage yield. Moles of Mg = mass/gfm = 1.20/24.3 = Mole ratio of Mg:MgSO4.7H2O = 1:1 gfm of MgSO4.7H2O = (remember to include the 7H2O) Theoretical mass of MgSO4.7H2O produced = n x gfm = x = g % yield = (actual yield/theoretical yield) x 100 = (9.84/12.17) x 100 = 80.9% 7. Zinc reacts with dilute hydrochloric acid according to the equation. Zn + 2HCl ZnCl2 + H g of zinc is mixed with 250 cm 3 of 2 mol l -1 hydrochloric acid. Calculate which reactant is in excess.

4 Zn + 2HCl 1 mol 2 mol Moles of Zn present = mass/gfm = 1.625/65.4 = mol Moles of HCl required to react with zinc = x 2 = mol Moles of HCl present = C x V = 2 x 250/1000 = 0.5 mol HCl is in excess cm 3 of hydrogen peroxide, H2O2, was titrated with mol l -1 acidified potassium permanganate. The end - point was reached when 16.7 cm 3 of the permanganate solution had been added. 5H2O2 + 2MnO H + 2Mn O2 + 8H2O Calculate the concentration of the hydrogen peroxide in moles per litre. Moles of permanganate = C x V = x 16.7/100 = mol Mole ratio of 2MnO4 - : 5H2O2 = 2:5 Moles of H2O2 = x 5/2 = mol Concentration of H2O2 = n/v = /0.025 = mol l Seaweeds are a rich source of iodine in the form of iodide ions. The mass of iodine in seaweed can be found using the procedure outlined below g of seaweed is dried in an oven and ground into a fine powder. Hydrogen peroxide solution is then added to oxidise the iodide ions to iodine molecules. Using starch solution as an indicator, the iodine solution is then titrated with sodium thiosulfate solution to find the mass of iodine in the sample. The balanced equation for the reaction is shown. 2Na2S2O3(aq) + I2(aq) 2NaI(aq) + Na2S4O6(aq) In this analysis of seaweed, 14.9 cm 3 of mol l 1 sodium thiosulfate solution was required to reach the end-point. Calculate the mass of iodine present in one gram of the seaweed sample. Moles of thiosulfate = C x V = x 14.9/1000 = mol

5 Mole ratio of 2Na2S2O3: I2 = 2:1 Moles of I2 = /2 = mol gfm I2 = 254 g Mass of I2 = n x gfm = x 254 = (in g of seaweed) Mass of I2 in 1 g of seaweed = /50 = g 1. Titration with solutions of potassium bromate (KBrO3) can be used to determine the concentration of arsenic (III) ions. The balanced equation is: 3H3AsO3 + BrO3 - Br - + 3H3AsO4 What is the concentration of As(III) in a solution if cm 3 of mol l -1 KBrO3 is needed to titrate cm 3 of the As(III) solution? Moles of bromate = C x V = x 22.35/1000 = mol Mole ratio = 3H3AsO3 : BrO3 - = 3 : 1 Moles of As(III) = = x 3 = mol Concentration of As(III) = n/v = /0.050 = mol l Alcohol(ethanol) levels in blood can be determined by a redox titration with potassium dichromate according to the balanced equation: C2H5OH(aq) + 2Cr2O7 2- (aq) + 16H + (aq) 2CO2(g) + 4Cr 3+ (aq) + 11H2O(l) a. What is the blood alcohol level in mol l -1 if 8.76 ml of mol l -1 K2Cr2O7 is required for titration of a cm 3 sample of blood? Moles of dichromate = C x V = x 8.76/1000 = mol Mole ratio C2H5OH : 2Cr2O7 2- = 1:2 Moles of C2H5OH = = /2 = mol

6 Concentration of C2H5OH = n/v = / = mol l -1 b. Suggest why an indicator is not required for this titration. As the dichromate ion is orange and the chromium(iii) ion is green, the reaction is self indicating. 3. To determine the concentration of chloride ions in seawater it is titrated with silver(i) nitrate solution. 25 cm 3 of raw seawater was diluted to 250 cm 3 in a volumetric flask. A 25 cm 3 sample of the diluted seawater was pipetted into a conical flask and a few drops of potassium chromate(vi) indicator solution was added. On titration with mol l -1 silver nitrate solution, 13.8 cm 3 was required to react with the chloride ions in the diluted sample. The equation for this reaction is Ag + (aq) + Cl - (aq) Ag + Cl - (s) a. Silver(I) nitrate is a primary standard. What is meant by primary standard? A chemical with high purity and gram formula mass which is stable in air. It must be soluble and stable when dissolved. b. What type of reaction is this? Precipitation reaction. c. Suggest the name of a chemical which could be used in a control experiment. A soluble chloride. Sodium chloride would be ideal. d. Calculate the concentration of chloride ions in the undiluted seawater. Moles of silver(i) nitrate = C x V = 0.10 x 13.8/1000 = mol Mole ratio Ag + : Cl - = 1:1 Moles of Cl - = mol (25 cm 3 of diluted seawater) Moles of Cl - in 250 cm 3 flask = x 10 = mol

7 Moles of Cl - in 25 cm 3 of undiluted seawater = mol Concentration of Cl - = n/v = /0.025 = mol l Brass is an alloy consisting mainly of copper and zinc. To determine the percentage of copper in a sample of brass, 2.63 g of the brass was dissolved in concentrated nitric acid and the solution diluted to 250 cm 3 in a standard flask. Excess potassium iodide solution was added to 25.0 cm 3 of this solution, iodine being produced according to the equation: 2Cu 2+ (aq) + 4I (aq) 2CuI(s) + I2(aq) The iodine formed was titrated with 0.10 mol l 1 sodium thiosulfate solution, Na2S2O3(aq), the volume required for complete reaction being 24.8 cm 3. I2(aq) + 2S2O3 2 (aq) colourless 2I (aq) + S4O6 2 (aq) colourless a. How could the end-point for the titration be made more obvious? Add starch as an indicator. b. Explain why the potassium iodide solution could be measured out in a measuring cylinder instead of a pipette. The potassium iodide is in excess which means the volume added need not be highly accurate and so the accuracy provided by a measuring cylinder is sufficient. b. How many moles of sodium thiosulfate were required in the titration? Moles thiosulfate = C x V = 0.10 x 24.8/1000 = mol c. Calculate the percentage by mass of copper in the sample of brass. Mole ratio 2S2O3 2 : I2 = 2 : 1 Moles of I2 produced = /2 = mol Mole ratio I2 : 2Cu 2+ = 1 :2

8 Moles of Cu 2+ produced = x 2 = mol (25 cm 3 sample) Moles of Cu 2+ in 250 cm 3 flask = x 10 = mol Mass of copper = n x gfm = x 63.5 = g % copper in the brass = (1.5748/2.63) x 100 = 59.9% 5. In an experiment to determine the percentage by mass of ammonium sulfate, (NH4)2SO4, in a fertiliser, 3.80 g of the fertiliser was dissolved in water and made up to 250 cm 3 in a standard flask.. To 25.0 cm 3 portions of this solution, an excess of methanal was added. 4NH4 + (aq) + 6HCHO(aq) C6H12N4(aq) + 4H + (aq) + 6H2O(l) The H + (aq) ions produced were titrated with mol l -1 sodium hydroxide solution. The average volume required to neutralise these H + (aq) ions was 28.0 cm 3. a. Why was the methanol added in excess to the fertiliser solution? To make sure all the ammonium ions reacted. b. Calculate the number of moles of hydrogen ions produced by the methanol in the 25 cm 3 of fertiliser solution. Moles of hydroxide ions = C x V = x 28/1000 = mol Mole ratio OH - : H + = 1 : 1 Moles of H + = mol c. Calculate then number of moles of ammonium ions in the 250 cm 3 standard flask. Mole ratio 4H + : 4NH4 + = 1 : 1 Moles of ammonium ions in 25 cm 3 sample = mol Moles of ammonium ions in 250 cm 3 standard flask = x 10 = mol d. Calculate the number of moles of ammonium sulfate in the 250 cm 3 standard flask. Mole ratio NH4 + ammonium : (NH4)2SO4 ammonium sulfate = 2 : 1 Moles of ammonium sulfate = 0.028/2 = mol

9 e. Calculate the percentage of ammonium sulfate in the fertiliser. Mass of ammonium sulfate = n x gfm gfm ammonium sulfate = 132 g Mass = x 132 = g % ammonium sulfate = (1.848/3.8) x 100 = 48.6% 6. Some bleaches use hypochlorite ions (OCl - ) as the bleaching agent. The concentration of the hypochlorite can be found by adding a diluted sample of the bleach to an excess of ethanoic acid and potassium iodide. This releases iodine, the concentration of which can be determined using a standard thiosulfate solution with starch indicator. OCl - (aq) + 2H + (aq) + 2I - (aq) 2S2O3 2- (aq) + I2(aq) I2(aq) + Cl - (aq) + H2O(l) 2I - (aq) + S4O6 2- (aq) 25.0 cm 3 of bleach was diluted to 250 cm 3 in a standard flask cm 3 samples of diluted bleach were pipetted into conical flasks containing an excess of ethanoic acid and potassium iodide. Each sample was titrated against sodium thiosulfate solution (concentration mol l -1 ). The results are shown in the table below. Titration Burette reading Initial/cm Final/cm a. Why was the titration carried out three times? To increase the reliability of the result. It is not correct to say increase accuracy. b. Why was ethanoic acid added to the bleach/iodide mixture? To provide the hydrogen ions needed for the reaction. c. Calculate the concentration of hypochlorite ions in the undiluted bleach. Average titre (from titration 2 and 3) = ( )/2 = cm 3 Moles of thiosulfate = C x V = x 13.50/1000 = mol

10 Mole ratio 2S2O3 2- : I2 = 2 : 1 Moles of iodine produced = /2 = mol Mole ratio I2 : OCl - = 1 : 1 Moles of OCl - = mol (in 10 cm 3 of diluted bleach) Moles of OCl - in 250 cm 3 standard flask = = x 250/10 = mol 1. Before Concentration 1947, silver of OClcoins - undiluted were made bleach from (25 an cm 3 alloy ) = n/v of = silver,copper /0.025 and = 0.51 nickel. mol lto -1 determine the metal composition, a coin weighing g was dissolved in nitric. acid and the resulting solution diluted to 1000 cm 3 in a standard flask. A 100 cm 3 portion was treated in the following way. Hydrochloric acid (0.20 mol l 1 ) was added to this solution until precipitation of silver(i) chloride was complete. The precipitate was recovered by filtration. It was washed and dried and found to weigh 0.60 g. a. (i) Calculate the percentage by mass of silver in the coin. GFM AgCl = 143.4g Mass of Ag in precipitate = 107.9/ = g {in 100 cm 3 of sample} Mass of Ag in coin = 1000/ = 4.51 g Percentage of Ag in coin = 4.51/ = 45.1% (ii) How could you tell when precipitation was complete? Add some AgNO3 to filtrate. There should be no more precipitate. b. The filtrate was treated to reduce the copper(ii) ions to copper(i) ions. Ammonium thiocyanate solution was added to precipitate the copper as copper(i) thiocyanate: Cu + (aq) + CNS (aq) CuCNS(s) After filtration, drying and weighing, the precipitate was found to weigh 0.31 g. Calculate the percentage by mass of copper in the coin. GFM CuCNS = g Mass of Cu = 63.5/ = {in 100 cm 3 of sample} Mass of Cu in coin = 1000/ = 1.62 g

11 Percentage of Cu in coin = 16.2% 2. Crystals of hydrated sodium carbonate left exposed to the atmosphere gradually lose some of their water of crystallisation. The formula of the crystals may be given by Na2CO3. xh2o, where x has a numerical value between 0 and g of the crystals was dissolved in water and made up to 250 cm 3 of solution in a standard flask. To determine the value of x in the formula, 25 cm 3 of the sodium carbonate solution was titrated with 1.0 mol l 1 hydrochloric acid cm 3 of the acid was required for neutralisation. {Hint: Carbonate ions react with hydrochloric acid in a 1:2 ratio} a. Calculate the mass of sodium carbonate (Na2CO3) in 16.0 g of the crystals. Moles of HCl = 15/ = Moles of Na2CO3 {in 25 cm 3 of solution} = ½ x = Moles of Na2CO3 {in 250 cm 3 of solution} = x 10 = 0075 Mass of in 16 g of crystals = n x gfm = x 106 = 7.95 g b. Find the value of x in the formula Na2CO3.xH2O. Mass of water = = 8.05 g Moles of water = mass/gfm = 8.05/18 = Moles of Na2CO3 = 7.95/106 = So Na2CO3 : H2O Moles : Divide by smallest 0.075/0.075 : 0.447/ : 5.96 {this is rounded to 6}

12 Formula is therefore Na2CO3. 6H2O 3. A 5.02 g sample of a silver alloy was analysed as follows. The sample was completely dissolved in excess dilute nitric acid and then treated with an excess of sodium chloride solution, causing a white precipitate of silver(i) chloride. The precipitate had a mass of 2.37 g. Calculate the percentage mass of silver in the alloy Formula of silver(i) chloride is AgCl GFM = g Moles of silver(i) chloride = mass/gfm = 2.37/143.5 = mol Mass of silver(1:1 ratio of Ag:AgCl) = n x gfm = x 108 = 1.78 g Percentage of silver in alloy = (1.78/5.02) x 100 = 35.5% Alternatively mass of silver = (108/143.5) x 2.37 = 1.78 g then calculate percentage. 1. An experiment was carried out to determine the % manganese in a sample of stainless steel. A series of standard permanganate solutions were prepared from a stock solution of mol l -1 solution of permanganate and used to produce the calibration graph shown below.

13 a. One of the standard solutions used to prepare the calibration graph had a concentration of 1 x 10-4 mol l -1. Describe, giving appropriate volumes and apparatus, how this solution could be prepared from the mol l -1 solution of permanganate. As 1x10-4 (0.0001) is ten times as dilute as 0.001, the original solution must be diluted with deionised water by a factor of 10. Accuracy is very important here so pipettes, burettes and standard flasks should be used. not measuring cylinders.l Pipette 10 cm 3 of mol l -1 into a 100 cm 3 standard flask {could also use 25 cm 3 into 250 cm 3 etc}. Fill the flask to 1 cm below the graduation line with deionised water. Use a dropper to add more deionised water up to the graduation mark. Stopper the flask and invert it to ensure thorough mixing. b. The results of the experiment are shown below. Mass of steel used = 0.19 g Absorbance of permanganate solution = 0.25 Total volume of permanganate solution = 100 cm 3 Use the graph and the results to calculate the percentage, by mass, of manganese in the sample of stainless steel. From the graph, 0.25 absorbance gives a concentration of 1.4 x 10-4 mol l -1 MnO4 -. Moles of MnO4- = C x V = 1.4 x 10-4 x 0.1 = 1.4 x 10-5 Mass of Mn = n x gfm = 1.4 x 10-5 x 54.9 = g Percentage Mn = ( /0.19) x 100 = = 0.41% 2. Margaret analysed a sample of 100 cm 3 of contaminated water for its copper content. Suitable chemicals were added to the water to produce a blue coloured copper

14 compound, and the intensity of the colour was measured using a colorimeter. A series of 5 solutions with known copper concentrations were prepared in a similar way and the colorimeter reading produced by each was recorded as shown in the table below.. Margaret analysed a sample of 100 cm 3 of contaminated water for its copper content. Suitable chemicals were added to the water to produce a blue coloured copper compound, and the intensity of the colour was measured using a colorimeter. A series of 5 solutions with known copper concentrations were prepared in a similar way and the colorimeter reading produced by each was recorded as shown in the table below. Concentration of copper/ milligrams per litre (mg l -1 ) Absorbance a. The solutions appeared blue because they were absorbing part of the visible region of the electromagnetic spectrum. Which colours of the spectrum were being absorbed? Blue and green. b. Suggest the most appropriate coloured filter for use in the colorimeter and explain your choice. A yellow filter as this is the colour most absorbed by the blue coloured solution. c. Draw a calibration graph of the colorimeter results Absorbance

15 0.1 0 d. The colorimeter reading for the contaminated water sample was Calculate the number of moles of copper in the water sample. Concentration Cu /mg l -1 Red line on graph (which is too thick but had to be so you can see it) gives a concentration of 132 mg l -1 ( your value may not identical but it should be close to this) Mass of copper in 100 cm 3 = 0.1 x 132 = 13.3 mg Moles of copper = mass/gfm = /63.5 = 2.08 x 10-4 mol 3. The diagram shows white light passing through a cyan coloured solution. a. The steps shown below outline how the concentration of the cyan solution could be determined by colorimetry. Put the steps in the correct order. Order is b > d > a > e > c b. Explain why the absorbance graph shown below is NOT that of the cyan coloured solution.

16 The graph shows that the solution has maximum absorbance at 510 nm (green light) and low absorbance at around 625nm (red light). This means the solution is transmitting red and so will appear red in colour.

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19 A list of learning outcomes for the topic is shown below. When the topic is complete you should review each learning outcome. Your teacher will collect your completed notes, mark them, and then decide if any revision work is necessary. Need Help Revise Understand State that chemical reaction can occur at different rates. Give examples of both fast and slow reactions. State that decreasing the particle size of a solid will increase its surface area. State that increasing the surface area of a solid will increase reaction rate. State that increasing the concentration of a reactant will increase reaction rate. State that increasing the temperature of a reaction will increase reaction rate.

20 State that a catalyst is a chemical which increases the rate of a reaction but remains unchanged at the end of the reaction {E.g. If a catalyst weighs 1g at the start of the reaction, 1g will be present at the end of the reaction} State that the collision theory says that a reaction will take place if particles collide with sufficient energy. State that the ACTIVATION ENERGY is the minimum energy required for a reaction to occur. Be able to draw a labelled diagram of the apparatus used to generate, collect measure a gas. and Be able to relate the rate of a reaction to the slope of the line in a reaction progress graph. Be able to calculate the reaction rate from the gradient of a line in a reaction progress graph and give the correct rate units. Be able to determine the time a reaction is complete from a reaction progress graph. state that a catalyst saves money by lowering the energy needed for a reaction State that the finishing volume/mass/concentration in a reaction progress is related to the initial quantity of reactants used. graph Be able to draw lines on a reaction progress graph which identify changes in the initial reaction conditions. State that in all chemical reactions the reaction rate slows down from the start of the reaction to the end of the reaction. State that, in general, reactions slow down due to the reactants getting used up. I have discussed the learning outcomes with my teacher. My work has been marked by my teacher. Teacher Comments.

21 Date. Pupil Signature. Teacher Signature.

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