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CLASSIFYING CHEMICAL REACTIONS Review of Writing and Balancing Pg. 58-64 Review: When counting elements, don’t forget to look at both the subscript and the coefficient. For example: P2O5 = has 2 phosphorus atoms and 5 oxygen atoms 2P2O5 = has 4 phosphorus atoms and 10 oxygen atoms Because there are 2 molecules (indicated by the coefficient) and 2 atoms in each molecule (indicated by the subscript) – So you multiply!! Remember: Never change a subscript to balance an equation! O2(g) + H2(g) H2O(l) Is unbalanced – but you can’t change it the following way! O2(g) + H2(g) H2O2(l) X Make sure the coefficients are the lowest whole-number ratio : 2O2(g) + 4H2(g) 4H2O(l) This is a balanced formula but these are not the lowest numbers you could use: O2(g) + 2H2(g) 2H2O(l) Balancing Chemical Equations 1) Write the chemical formulas for the reactants and products including the states Cu(s) + AgNO3(aq) Ag(s) + Cu(NO3)2(aq) 2) Balance the element (atom or ion) present in the greatest number by multiplying by the lowest coefficient possible (NO3)2(aq) = 2 present (lowest coefficient possible to balance = 2) Cu(s) + 2AgNO3(aq) Ag(s) + Cu(NO3)2(aq) 3) Repeat step 2 for the rest of the elements Now we have 2 Ag, so balance the other side Cu(s) + 2AgNO3(aq) 2Ag(s) + Cu(NO3)2(aq) 4) Count elements on each side of the final equation to ensure they balance: 1 Cu(s) = 1 Cu(aq) ; 2Ag = 2Ag(s) ; 2 NO3 = (NO3)2(aq) What do you remember? You learned about five reaction types, can you match them up Composition (Formation) CH4(g) + O2(g) CO2(g) + H2O(g) Decomposition Mg(s) + O2(g) MgO(s) Combustion Cu(s) + AgNO3(aq) Ag(s) + Cu(NO3)2(aq) Single Replacement CaCl2(aq) + Na2CO3(aq) CaCO3(s) + NaCl(aq) Double Replacement H2O(l) O2(g) + H2(g) Classifying Chemical Reactions: Composition (Formation) 2Mg(s) + O2(g) 2MgO(s) element + element compound Predict and balance the following: Li(s) + Cl2(g) Na(s) + F2(g) Ba(s) + N2(g) Classifying Chemical Reactions: Decomposition 2H2O(l) O2(g) + 2H2(g) compound element + element Predict and balance the following: NaCl(s) Sr3P2(s) Cs2O(s) Classifying Chemical Reactions: Combustion CH4(g) + 2O2(g) CO2(g) + 2H2O(g) fuel + oxygen carbon dioxide + water Predict and balance the following: C3H8(g) + O2(g) C2H6(g) + O2(g) C4H10(g) + O2(g) Classifying Chemical Reactions: Single Replacement 2AgNO3(aq) + Cu(s) Cu(NO3)2(aq) + 2Ag(s) compound + element compound + element 2NaI(aq) + Cl2 (g) I2 (s) + 2NaCl(aq) Predict and balance the following: NaBr(aq) + O2(g) CuCl2(aq) + Al(s) Li2CO3(aq) + K(s) Classifying Chemical Reactions: Double Replacement 2AgNO3(aq) + CuCl2(aq) Cu(NO3)2(aq) + 2AgCl(aq) compound + compound compound + compound CaCI2(aq) + Na2CO3 (aq) 2NaCl (aq) + CaCO3(s) Predict and balance the following: NaBr(aq) + MgO(aq) CuCl2(aq) + AlF3(aq) Li2CO3(aq) + K2O(aq) Using the solubility table: Chemical Reaction Equations Section 7.1 pg. 278-285 Reaction Assumptions Reactions are spontaneous – reactions will occur when all the reactants are mixed together Reactions are fast – the reaction must occur within a reasonable time (see pg. 280) Reactions are quantitative – one that is more than 99% complete; in other words, at least one reactant is completely used up Reactions are stoichiometric – means that there is a simple whole-number ratio of chemical amounts of reactants and products (the coefficients for a balanced equation do not change) Net Ionic Equations A chemical reaction equation that includes only reacting entities (molecules, atoms and/or ions) and omits any that do not change Writing Net Ionic Equations: 1) Write a complete balanced chemical equation 2) Dissociate all high-solubility ionic compounds, and ionize all strong acids to show the complete ionic equation 3) Cancel identical entities that appear on both the reactant and product sides 4) Write the net ionic equation, reducing coefficients if neccessary When cancelling spectator ions, they must be identical in every way: chemical amount, form (atom, ion, molecule) and state of matter Practice Write the net ionic equation for the reaction of aqueous barium chloride and aqueous sodium sulfate. (Refer to the solubility table) 1) BaCl2(aq) + Na2SO4(aq) BaSO4(s) + 2NaCl(aq) 2) Ba2+(aq) + 2Cl-(aq) +2Na+(aq) + SO42-(aq) BaSO4(s) + 2Na+(aq) + 2Cl-(aq) (Complete ionic equation) 3) Ba2+(aq) + 2Cl-(aq) +2Na+(aq) + SO42-(aq) BaSO4(s) + 2Na+(aq) + 2Cl-(aq) 4) Ba2+(aq)) + SO42-(aq) BaSO4(s) (Net ionic equation) Ions that are present but do not take part in (change during) a reaction are called spectator ions (like spectators at a sports game: they are present but do not take part in the game) Practice Write the net ionic equation for the reaction of zinc metal and aqueous copper (II) sulfate (Refer to the solubility table) 1) Zn(s) + CuSO4(aq) Cu(s) + ZnSO4(aq) 2) Zn(s) + Cu2+(aq) + SO42-(aq) Cu(s) + Zn2+(aq) + SO42-(aq) (Complete ionic equation) 3) Zn(s) + Cu2+(aq) + SO42-(aq) Cu(s) + Zn2+(aq) + SO42-(aq) 4) Zn(s) + Cu2+(aq) Cu(s) + Zn2+(aq) (Net ionic equation) For entities involving strong acids, H+(aq) is used as a matter of convenience over H3O+(aq) Practice Write the net ionic equation for the reaction of hydrochloric acid and barium hydroxide 1) 2HCl(aq) + Ba(OH)2(aq) BaCl2(aq) + 2HOH(l) aka: 2H2O(l) 2) 2H+(aq) + 2Cl-(aq) + Ba2+(aq) + 2OH-(aq) Ba2+(aq) + 2Cl-(aq) + 2H2O(l) (Complete ionic equation) 3) 2H+(aq) + 2Cl-(aq) + Ba2+(aq) + 2OH-(aq) Ba2+(aq) + 2Cl-(aq) + 2H2O(l) 4) H+(aq)) + OH-(aq) H2O(l) (Net ionic equation) – coefficients reduced to 1 Limiting and Excess Reagents Cu(s) + AgNO3(aq) What is in the container when the reaction is finished? Cu(NO3)2(aq) + Ag(s) (Cu2+(aq) = blue) But how do you know the reaction is done?? There is still copper left! (a) Copper wire and a beaker with aqueous silver nitrate solution (b) A few moments after the wire is immersed (c) The beaker contents after 24 h Limiting and Excess Reagents When no further changes appear to be occurring, we assume that all of the AgNO3(aq) that was initially present has now been completely reacted. A limiting reagent is the reactant whose entities are completely consumed in a reaction, meaning the reaction stops. In order to make sure this happens, more of the other reactant must be present than is required An excess reagent is the reactant whose entities are present in surplus amounts, so that some remain after the reaction ends.. In our reaction: much more copper was used than needed (evidenced by the unreacted copper) so we assume the reaction ended when no more silver ions were left, so silver nitrate was the limiting reagent. Homework: WB pg 59 Q 10 WB pg 71 Q 42 WB pg 91 Q 1-3 * Classifying & Balancing Equations WS Stoichiometry Section 7.2 pg. 286-293 Stoichiometry A method of problem-solving using known quantities of one entity in a chemical reaction to find out unknown quantities of another entity in the same reaction Always requires a balanced chemical equation Series of steps for measurement calculations Always involves a mole ratio to “switch substances” Writing chemical reactions, net ionic equations and dissociation equations is essential Converting grams to moles (and vice versa) is also essential Gravimetric Stoichiometry Gravimetric Stoichiometry – method used to calculate the masses of reactants or products in a chemical reaction Gravimetric = mass measurement Using gravimetric stoichiometry, we can apply our knowledge of balanced chemical reactions and mass to mol conversions to: Predict the mass of product we will get from a reaction Estimate the amount of reactant we need for a reaction to produce a certain mass of product Gravimetric Stoichiometry Steps: Step 1: Write a balanced chemical reaction equation List the measured mass (given), the unknown quantity (required) and conversion factors (M = molar mass) Step 2: Convert the mass of the measured substance to moles Step 3: Calculate the moles of the unknown substance using the mole ratio required given Step 4: Convert from moles of the required substance to its mass. Stoichiometry Calculations (Measured quantity) solids/liquids m n mole ratio solids/liquids m (Required quantity) n How many grams of oxygen are required to completely burn 10.0 g of propane (C3H8(g))? The balanced chemical equation is: C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g) The number of moles of propane reacting is: (MASS to MOLES) 10.0 g x 1 mol = 0.227 mol 44.11 g The number of moles of oxygen produced is: (MOLE RATIO) 0.227 mol x 5 mol = 1.13 mol 1 mol The mass of oxygen produced is: (MOLES TO MASS) 1.1335 mol x 32.00 g = 36.3 g 1 mol Practice #2 20.0 g of butane (C4H10(g)) are completely burned in a lighter. How many grams of CO2(g) are produced? Start with a balanced chemical equation: 2 C4H10(g) + 13 O2(g) 8 CO2(g) + 10 H2O (g) m = 20.0g m=? M = 58.14 g/mol M = 44.01 g/mol Practice #2 (Team Unit Analysis) 2 C4H10(g) + 13 O2(g) 8 CO2(g) + 10 H2O (g) m = 20.0g m=? M = 58.14 g/mol M = 44.01 g/mol 2) Mass to moles, mole ratio, moles to mass 20.0 g x 1 mol x . 8 mol CO2 x 58.14 g 2 mol C4H10 44.01 g = 60.6 g 1 mol Practice #3 (Unit Analysis) What mass of iron (III) oxide is required to produce 100.0 g of iron? Fe2O3(s) + 3 CO(g) 2 Fe(s) + 3 CO2(g) m=? m = 100.0g M = 159.70g/mol M = 55.85 g/mol m Fe2O3(s): 100.0 g x 1 mol x 1 mol x 159.70 g 55.85 g 2 mol 1 mol = 143.0 g Fe2O3 Practice Remember to keep the unrounded values in your calculator for further calculation until the final answer is reported. WB pg 77-80 ALL Pg. 290 #8-14 (Answers pg. 785) Percent Yield for Reactions We can use stoichiometry to test experimental designs, technological skills, purity of chemicals, etc. We evaluate these by calculating a percent yield. This is the ratio of the actual (experimental) quantity of product obtained and the theoretical (predicted) quantity of product obtained from a stoichiometry calculation Percent yield = actual yield predicted yield x 100 Some forms of experimental uncertainties: All measurements (limitations of equipment) Purity of chemical used (80-99.9% purity) Washing a precipitate (some mass is lost through filter paper) Estimation of reaction completion (qualitative judgements i.e. color) Percent Yield Example #1 Example: In a chemical analysis, 3.00 g of silver nitrate in solution was reacted with excess sodium chromate to produced 2.81 g of filtered, dried precipitate. Predicted value: 2AgNO3(aq) + Na2CrO7(aq) Ag2CrO7(s) +2NaNO3(aq) m = 3.00 g M = 169.88g/mol 3.00g x 1 mol x 169.88g 1 mol x 2 mol m= M = 331.74 g/mol 331. 74 g = 2. 93 g 1 mol Percent yield = actual yield x 100% predicted yield = 2.81g x 100% 2.93g = 95.9% Testing the Stoichiometric Method Stoichiometry is used to predict the mass of precipitate actually produced in a reaction. Filtration is used to separate the mass of precipitate actually produced in a reaction Testing the Stoichiometric Method What mass of lead is produced by the reaction of 2.13 g of zinc with an excess of lead(II) nitrate solution? Design: A known mass of zinc is place in a beaker with an excess of lead(II) nitrate solution. The lead is produced in the reaction is separated by filtration and dried. The mass of the lead is determined Prediction: Zn(s) + Pb(NO3)2(aq) Zn(NO3)2(aq) + Pb(s) m = 2.13 g M = 65.41 g/mol 2.13 g x 1 mol x 1 65.41 g 1 m=? M = 207.2 g/mol x 207.2 g = 1 mol 6.75 g Testing the Stoichiometric Method Prediction: 6.75 g of lead will be produced (Stoichiometric calculation) Evidence: In the beaker, crystals of a shiny black solid were produced, and all the zinc disappeared. Mass of filter paper = 0.92 g Mass of dried filter paper plus lead = 7.60g Analysis: mass of lead product = 7.60g – 0.92g = 6.68g According to the evidence collected, 6.68 g of lead precipitated. Evaluation: % difference = (experimental – predicted) x 100% predicted 6.68 – 6.75 x 100% 6.75 = 1% Given such a small difference, which can readily be accounted for by normal sources of error, the prediction is clearly verified, and so the stoichiometric method for predicting the mass of lead formed in this experiment is judged to be acceptable. Testing the Stoichiometric Method Prediction: 6.75 g of lead will be produced (Stoichiometric calculation) Analysis: mass of lead product = 7.60g – 0.92g = 6.68g According to the evidence collected, 6.68 g of lead precipitated. Evaluation: Could you also calculate the % yield ? Percent yield = actual yield predicted yield x 100% = 6.68 g x 100% 6.75g = 99.0% Homework Pg 293 Q 1,2 & 6-10 WB pg 85-87 Q 3, 5, 6, 9, 10, 11 Gas Stoichiometry Section 7.3 pg.294-298 Molar Volume STP = 22.4L/mol SATP = 24.8 L/mol Molar volume is the same for all gases at the same temperature and pressure (remember, all gases have the same physical properties) At STP, molar volume = 22.4 L/mol (101.325 kPa and 0oC) At SATP, molar volume = 24.8 L/mol (100 kPa and 25oC) This can be used as a conversion factor just like molar mass! At STP, one mole of gas has a volume of 22.4 L, which is approximately the volume of 11 “empty” 2 L pop bottles. Comparing Kelvin and Celsius Scales To convert degrees Celsius to Kelvin , you add 273. K = oC + 273 To convert Kelvin to degrees Celsius, you subtract 273. oC = K - 273 Examples: What is 254 K in oC ? -19oC What is -34oC in K ? 239K Gas Stoichiometry Many chemical reactions involve gases as a reactant or a product Gas Stoichiometry – the procedure for calculating the volume of gases as products or reactants Gases also have a molar volume (L/mol) rather than concentration. This is the conversion factor used to convert (litres of gas) to (moles of gas) The Ideal Gas Law (PV = nRT) may also be required to: A) find the number of moles of reactant B) Find the V, P, or T of the product Example #1 If 300g of propane burns in a gas barbecue, what volume of oxygen measured at SATP is required for the reaction? Remember: 24.8L/mol for SATP C3H8(g) + m = 300g 44.11g/mol 300 g x 1 mol 44.11 g 5O2(g) 3CO2(g) + 4H2O(g) V=? 24.8L/mol x 5 mol 1 mol x 24.8 L = 843 L O2(g) 1 mol **Remember – molar volume is the conversion factor for gases just like molar mass is the conversion factor in gravimetric stoichiometry Example #2 Hydrogen gas is produced when sodium metal is added to water. What mass of sodium is necessary to produce 20.0L of hydrogen at STP? Remember: 22.4L/mol for STP 2Na(s) + 2H2O (l) 2NaOH(aq) + H2(g) m=? 22.99g/mol 20.0L x 1 mol 22.4 L V = 20.0L 22.4L/mol x 2mol 1 mol x 22.99g = 41.1 g Na(s) 1 mol **Remember – molar volume is the conversion factor for gases just like molar mass is the conversion factor in gravimetric stoichiometry Example #3 If the conditions are not STP or SATP, the molar volume cannot be used! You must use the ideal gas law to find the gas values using moles determined from stoichiometry What volume of ammonia at 450kPa and 80oC can be obtained from the complete reaction of 7.5kg of hydrogen with nitrogen? 2N2(g) + 3H2(g) m = 7500g M = 2.02 g/mol 2NH3(g) m=? P = 450kPA T = 353.13K 7500 g x 1 mol x 2 = 2475.2475 mol NH3(g) 2.02 g 3 PV = nRT V = nRT P = (2475.2475 mol)(8.314kpa•L/mol•K)(353.15K) (450kPa) = 16150.10L 1.6 x 104 L of NH3(g) 1. Gas Stoichiometry Summary Write a balanced chemical equation and list the measurements, unknown quantity symbol, and conversion factors for the measured and required substances. 2. Convert the measured quantity to a chemical amount using the appropriate conversion factor 3. Calculate the chemical amount of the required substance using the mole ratio from the balanced chemical equation. 4. Convert the calculated chemical amount to the final quantity requested using the appropriate conversion factor. Stoichiometry Calculations (Measured quantity) solids/liquids gases gases solids/liquids (Required quantity) m V, T, P V,T,P m n mole ratio n Homework Pg. 299 #1-6 WB pg 85-87 Q 2, 4 & 8 Solution Stoichiometry Section 7.4 pg. 300-303 Solution Stoichiometry The majority of work in research and industry involves solutions. Recall that solutions are easy to handle and are usually easier to control in reactions. The major difference compared to gravimetric and gas stoichiometry is that we use molar concentration (mol/L) as a conversion factor rather than molar mass or molar volume Solution stoichiometry – the procedure for calculating the molar concentration or volume of solution products or reactants Example #1 Ammonia and phosphoric acid solutions are used to produce ammonium hydrogen phosphate fertilizer. What volume of 14.8mol/L NH3(aq) is needed to react with 1.00kL of 12.9mol/L of H3PO4(aq)? 2NH3(aq) + V=? 14.8mol/L 1.00kL H3PO4(aq) (NH4)2HPO4(aq) V = 1.00kL 12.9 mol/L x 12.9 mol 1L x 2mol x 1 L = 1.74 kL 1mol 14.8 mol = 1.74 x 103 L Example #2 In an experiment, a 10.00 mL sample of sulfuric acid solution reacts completely with 15.9 mL of 0.150 mol/L potassium hydroxide. Calculate the amount concentration of the sulfuric acid. H2SO4(aq) + 2KOH(aq) V = 10.00mL c=? 2H2O(l) + K2SO4(aq) V = 15.9 mL 0.150 mol/L 15.9mL x 0.150 mol x 1mol 1L 2 mol x 1 = 0.119 mol/L 10.0 mL Gravimetric, Gas and Solution Stoichiometry Summary 1. Write a balanced chemical equation and list the quantities and conversion factors for the given substance and the one to be calculated. 2. Convert the given measurement to its chemical amount using the appropriate conversion factor 3. Calculate the amount of the other substance using the mole ratio from the balanced chemical equation. 4. Convert the calculated chemical amount to the final quantity requested using the appropriate conversion factor. Remember to use the Ideal Gas Law for all gases not at STP or SATP Stoichiometry Calculations (Measured quantity) solids/liquids gases solutions solutions gases solids/liquids (Required quantity) m V, T, P c, V c, V V,T,P m n mole ratio n Guided Practice WB pg 85-87 Q 1, 7, 12-13 Introduction to Chemical Analysis Using Stoichiometry Section 8.1-8.2 (Pg. 314-319) Chemical Analysis Analysis of an unknown chemical sample can include both qualitative and quantitative analysis Qualitative – the identification of a specific substance present Quantitative – the determination of the quantity of a substance present Three methods commonly used to do this: Colorimetry – analysis by color, which used light emitted, absorbed or transmitted by a chemical (Section 8.1) Gravimetric Analysis – stoichiometric calculations from a measured mass of a reagent (Section 8.2) Titration Analysis – uses stoichiometric calculations from a measured solution volume of a reagent (Section 8.4) Colorimetry Ion Solution colour Group 1, 2, 17 Colourless Cr2+(aq) Blue Cr3+(aq) Green Co2+(aq) Pink Cu+(aq) Green Cu2+(aq) Blue Fe2+(aq) Pale-green Fe3+(aq) Yellow-brown Mn2+(aq) Pale pink Ni2+(aq) Green CrO4 2- (aq) Which solution is which? potassium dichromate sodium chloride sodium chromate Yellow Cr2O72-(aq) Orange MnO4-(aq) Purple potassium permanganate nickel (II) nitrate copper (II) sulfate 6 2 4 1 3 5 Colorimetry Aqueous ions can sometimes be identifies qualitatively by eye, (like the previous example) but for a more precise identification, technology must be used Spectrophotometer – is a device that measures the quantity of light absorbed at any desired wavelength when a light beam is passed through a solution. Can measure the concentration of any desired coloured ion, even in a solution that has several different mixed colours. It can be adjusted to only “see” one precise wavelength (colour) selected Colorimetry We can also use flame test to detect the presence of several metal ions. The technology has become more sophisticated now, instead we use atomic absorption spectrophotometers, which analyze the light absorbed by samples vaporized in a flame Gravimetric Analysis Lab technicians sometimes perform the same chemical analysis on hundreds of samples every day. For example, in a medical laboratory, blood and urine samples are routinely analyzed for specific chemicals such as cholesterol and sugar. In many industrial and commercial laboratories, technicians read the required quantity of a chemical from a graph that has been prepared in advance. This saves the time and trouble of doing a separate stoichiometric calculation for each analysis performed. By completing the Analysis of the following investigation report, you will be illustrating this practice. (see pg. 317) Gravimetric Analysis Purpose: The purpose of this lab exercise is to use a graph of a precipitation reaction’s stoichiometric relationship to determine the mass of lead(II) nitrate present in a sample solution. Problem: What mass of lead(II) nitrate is in 20.0 mL of a solution? Design: Samples of two different lead(II) nitrate solutions are used. Each sample is reacted with an excess quantity of a potassium iodide solution, producing lead(II) iodide, which has a low solubility and settles to the bottom of the beaker (Figure 1). After the contents of the beaker are filtered and dried, the mass of lead(II) iodide is determined. Use the Evidence pg. 317 Analysis: The analysis is completed by graphing Table 1 and then reading the mass of lead(II) nitrate present in each solution from the graph. Gravimetric Analysis Analysis: Read the mass of lead(II) nitrate present in each solution from the graph. According to the evidence collected and the graph: The mass of lead(II) nitrate in solution 1 = 3.15 g solution 2 = 5.40 g. Precipitation Completeness 1. Precisely measure a sample volume of the solution containing the limiting reagent. 2. Add (while stirring) an approximately equal volume of the excess reagent solution. 3. Allow the precipitate that forms to settle, until the top layer of solution is clear. 4. With a medicine dropper, add a few more drops of excess reagent solution. Allow the drops to run down the side of the container, and watch for any cloudiness that may appear when the drops mix with the clear surface layer 5. If any new cloudiness is visible, the reaction of the limiting reagent sample is not yet complete. Repeat steps 2 to 4 of this procedure as many times as necessary, until no new precipitate forms during the test in step 4. 6. When no new cloudiness is visible (the test does not form any Colorimetry Analysis Pg. 316 #1-4 Gravimetric Analysis Pg. 319 #1, 2, 4 Section 8.3 pg. 320-324 Limiting and Excess Reagents In any chemical reaction, it is easy to run out of one or another reactant – which has an impact on the amount of products that can result from a reaction To solve these problems you must identify which of the reactants is going to run out first. This is the “limiting reagent” The other is the “excess reagent” Example: LEGO – If you had three yellow blocks and four red blocks, you could only make three yellow/red combinations – because there is not enough yellow blocks to make four. The yellow block is the limiting reagent and the red block is the excess reagent + = Identifying Limiting and Excess Reagents WHY DO WE CARE?? It is often desirable to know how much excess reagent is required to ensure that a reaction goes to completion. The general rule is to assume that a reasonable quantity of excess reagent is to use 10% more than the quantity required (Not the case in commercial chemistry) When you know the quantity of more than one reagent, you may need to know which one will limit the reaction. 1) You want to test the stoichiometric method using the reaction of 2.00 g of copper(II) sulfate in solution with an excess of sodium hydroxide in solution. What would be a reasonable mass of sodium hydroxide to use? To answer this question, you need to calculate the minimum mass required and then add 10%. CuSO4(aq) + 2 NaOH(aq) Cu(OH)2(s) + Na2SO4(aq) 2.00g m=? 159.62g/mol 40.00g/mol 2.00 g x 1 mol x 2 159.62 g 1 1.00 g x 1.10%= 1.10g x 40.0g 1 mol = 1.00 g 10% = 0.10g Practice pg. 321 #2 3) If 10.0 mol of methane and 10.0 mol of oxygen react, which is the limiting reagent. CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) 10.0 mol 10.0 mol n CH4(g): 10.0 mol x 2 1 = 20.o mol n O2(g): 10.0 mol x 1 = 5.00 mol 2 20.0 mol of oxygen would be required to react with 10.0 mol of methane 5.00 mol of methane would be required to react with 10.0 mol of oxygen Since 20.0 mol of oxygen is required to react with 10.0 mol of methane, but only 10.0 mol is available, oxygen is the limiting reagent. How much methane would be left? 5.0 mol – 5.00 mol = 5.00 mol Practice pg. 324 #3b-d 2) If 10.0g of copper is placed in solution of 20.0g of silver nitrate, which reagent will be the limiting reagent? All reactants must be converted to moles, then using the mole ratio, determine which reactant will run out first. Cu(s) + 2 AgNO3(aq) 2 Ag(s) + Cu(NO3)2(aq) 10.0g 63.55 g/mol 20.0g 169.88g/mol n Cu(s): 10.0g x 1 mol = 0.157 mol x 2 = 0.315 mol 63.55 g 1 n AgNO3: 20.0g x 1 mol = 0.118 mol x 1 169.88 g = 0.0589 mol 2 You must test one of the values using the mole ratio. Assume one chemical is completely used up and see if enough of the second chemical is present. That much silver nitrate is not available so copper is not the limiting reagent More copper than that is available so silver nitrate is the limiting reagent 3) From the previous example, where 10.0 g of copper reacts with 20.0 g of silver nitrate, what mass of copper will be in excess? (leftover when the reaction is complete) ER Cu(s) LR + 2 AgNO3(aq) 2 Ag(s) + Cu(NO3)2(aq) 0.157 mol 0.118 mol n Cu(s): 0.118 mol x 1 = 0.0589 mol x 63.55 g = 2 1 mol 3.74 g of copper will be required in this reaction 10 g mass – 3.74 gof= silver 6.3 g ofwill copper be leftover 4) What be will produced? n Ag(s): 0.118 mol x 2 = 0.118 mol x 107.87 g = 2 1 mol Practice pg. 324 #4 12.7 g of silver will be produced in this reaction Putting it all together… In an experiment, 26.8g of iron (III) chloride in solution is combined with 21.5g of sodium hydroxide. Which reactant is in excess, and by how much? What mass of precipitate will be obtained? FeCl3(aq) + 3NaOH(aq) Fe(OH)3(s) + 3NaCl(aq) 26.8g 21.5g m=? 162.20g/mol 40.00g/mol 106.88g/mol nFeCl3 = 26.8g x 1 mol = 0.165 mol x 3 = 0.496 mol 162.20g 1 nNaOH = 21.5 g x 1 mol = 0.538 mol 40.0g There is more NaOH than this so FeCl3 is the LR 0.538mol – 0.496 mol = 0.042 mol x 40.00 g = 1.7 g NaOH (excess) 1 mol 0.165 mol x 1 x 106.88 g = 17.7 g of Fe(OH)3(s) 1 mol Limiting and Excess Reagents Summary Identify the limiting reagent by choosing either reagent amount, and use the mole ratio to compare the required amount with the amount actually present. The quantity in excess is the difference between the amount of excess reagent present and the amount required for complete reaction. A reasonable reagent excess to use to ensure complete reaction is 10%. Homework Pg. 327 #4, 6-8 Titration Analysis Section 8.4 – pg. 328-332 Experimental Design for Analysis Experimental designs discussed so far have been QUALitative (flame test, solution colour, litmus test, conductivity, solubility) QUANtitative experimental designs allow measures amounts to be obtained. The experimental design chosen for analysis depends on equipment, time available, and degree of accuracy required. Experimental Design for Analysis There are four types of analytical experimental designs: 1. Crystallization – the solvent is vaporized from a solution, with or without heating, leaving a solid behind which is measured for mass. 2. Filtration – a low solubility solid, produced from a single or double replacement reaction, is separated with a filter then dried so mass can be measured (filtrate = solution that goes through) 3. Gas Collection – a gas, formed as a product of a reaction, is collected and volume, pressure and temperature are measured. 4. Titration – a solution in a burette (called the “titrant”) is progressively added to a measured volume of another solution (called the “sample”) in an Erlenmeyer flask. The volume of the titrant at the endpoint is measured. Chemical Analysis by Titration Titration – is a common experimental design used to determine the amount concentration of substances in solution. The solution of known concentration may be either the titrant or the sample; it makes no difference to the analysis Titration breakdown: Carefully adding a solution (titrant) from a burette into a measured, fixed volume of another solution (sample) in an Erlenmeyer flask until the reaction is judged to be complete Chemical Analysis by Titration Burette – precisely marked glass cylinder with a stopcock at one end. Allows precise, accurate measurement and control of the volume of reacting solution. When doing a titration, there will be a point at which the reaction is complete; when chemically equivalent amounts of reactants have combined. This is called the equivalence point: Equivalence point – the point during a titration at which the exact theoretical chemical amount of titrant has been added to the sample. (QUANTITATIVE) To measure this equivalence point experimentally, we look for a sudden change in an observable property, such as color, pH, or conductivity. This is called the endpoint. (QUALITATIVE) Chemical Analysis by Titration An initial reading of the burette is made before any titrant is added to the sample. Then the titrant is added until the reaction is complete; when a final drop of titrant permanently changes the colour of the sample. The final burette reading is then taken. The difference between the readings is the volume of titrant added. Near the endpoint, continuous gentle swirling of the solution is important Chemical Analysis by Titration A titration should involve several trials, to improve reliability of the answer. A typical requirement is to repeat titrations until three trials result in volumes within a range of 0.2mL. These three results are then averaged before carrying out the solution stoichiometry calculation; disregard any trial volumes that do not fall in the range. Titration Tips Any property of a solution such as colour, conductivity, or pH that changes abruptly can be used as an endpoint. However, some changes may not be very sharp or may be difficult to measure accurately. This may introduce error into the experiment. Any difference between the titrant volumes at the empirical (observed) endpoint and the theoretical equivalence point is known as the titration error. How to read a burette: **Initial readings will always be lower (1 mL) than final readings (13.2 mL) because 0 is at the top of the burette. You will see this in the example problem next. Sample Problem Determine the concentration of hydrochloric acid in a commercial solution. A 1.59g mass of sodium carbonate, Na2CO3(s), was dissolved to make 100.0mL of solution. Samples (10.00mL) of this standard solution were then taken and titrated with the hydrochloric acid solution. Trial 1 2 3 4 Final burette reading (mL) 13.3 26.0 38.8 13.4 Volume of HCl(aq) added 13.1 12.7 12.8 12.8 Indicator colour Red The titration evidence collected is below. Methyl orange Initial burette reading (mL) 0.2 13.3 26.0 0.6 indicator was used. Orange Orange Orange TIP: In titration analysis, the first trial is typically done very quickly. It is just for practice, to learn what the endpoint looks like and to learn the approximate volume of titrant needed to get to the endpoint. Greater care is taken with subsequent trials Sample Problem First: Determine the concentration of the 100.0 mL sodium carbonate solution. 1.59 g Na2CO3(s) x 1 mol 105.99g x 1 0.100L = 0.150 mol/L Second: Write a balanced chemical equation: 2HCl(aq) + Na2CO3(aq) H2CO3(aq) + 2NaCl(aq) *12.8 mL c=? 10.0 mL c = 0.150 mol/L c HCl (aq): 10.0 mL x 0.150 mol x 2 x 1 L 1 12.8mL * The volume of HCl(aq) used is an average of trials 2, 3, and 4. = 0.234 mol/L Using titration analysis, the concentration of the commercial hydrochloric acid solution is 0.234 mol/L Titration Analysis Summary Titration is the technique of carefully controlling the addition of a volume of solution (the titrant) from a burette into a measured fixed volume of a sample solution until the reaction is complete. The concentration of one reactant must be accurately known. The equivalence point is the point at which the exact theoretical (stoichiometric) reacting amount of titrant has been added to the sample. The endpoint is the point during the titration at which the sudden change of an observable property indicates that the reaction is complete. Several trials must be completed. When at least three trials result in values that are all within a range of 0.2 mL, those values are averaged. The average value is used for the stoichiometry calculation. Homework Pg. 331 #1-4 – go through answers as a class Volumetric Stoichiometry – Titration Worksheet homework Acid–Base Titration Curves and Indicators Section 8.5 (The last one!) Pg. 333-339 Titration Curves A titration curve is a graph of the pH (vertical axis) versus the amount of the reagent progressively added to the original sample. As the equivalence point is approached, there is a rapid change in the pH. When a titration is done to create a pH curve, the addition of titrant is not stopped at the endpoint, but is continued until a large excess has been added. So what is happening? The initial addition of the titrant (in the burette) to the acid does not produce large changes. This relatively flat region of the pH curve is where a buffering action occurs. As the titration proceeds, and base is added, some of the acid is reacted with the added base, but anywhere before the equivalence point some excess acid will remain, so the pH stays relatively low. • Very near the equivalence point, a small excess of acid becomes a small excess of base with the addition of a few more drops, so the pH abruptly changes. • The equivalence point is the centre of this change, where the curve is the most vertical. What reaction is happening? HCl(aq) + NaOH(aq) HOH(l) + NaCl(aq) Net-ionic equation? Do you remember how? H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) H2O(l) + Na+(aq) + Cl-(aq) H+(aq) + OH-(aq) H2O(l) (net ionic equation) For convenience, hydrogen ions are written in their simplest (Arrhenius) form Why was the equivalence point 7? Remember water has a neutral pH of 7, and the spectator ions are neutral, so a strong monoprotic acid-strong monoprotic base titration must have a pH of 7 at the equivalence point. Equivalence points It is important to note, that the equivalence point pH is 7 ONLY for strong acid-strong base reactions. • For every other acid-base reaction, the equivalence point solution will contain ions or molecules that are not spectators – so titration curves must be done empirically to determine the equivalence point . Gener al Rule Strong Acid to Weak Base: pH at equivalence point is always lower than 7 Strong Base to Weak Acid: pH at equivalence point is always higher than 7 Strong Acid (HCl)– Weak Diprotic Base (Na2CO3) The equivalence point is still below 7 but do you notice anything else? If you observe the curve closely, you see that there are two places where the curve steepens as the titration proceeds. This happens because the base is diprotic, meaning that it will react with two hydrogen ions; so the hydrogen ions attach to the carbonate one at a time. We use the second reaction equivalence point, because we want the pH value when the reaction is complete. Why does the curve start at the top? Because now a strong acid is being added to the weak base, not vice versa Strong Acid (HCl)– Weak Diprotic Base (Na2CO3) CO32-(aq) + H+(aq) HCO3-(aq) – first equivalence point HCO3-(aq)+ H+(aq) H2CO3(aq) – second equivalence point Net reaction: CO32-(aq) + 2H+(aq) H2CO3(aq) You will want to choose an indicator that changes color at the second equivalence point Strong Acid (HCl)– Weak Diprotic Base (Na2CO3) Why is the equivalence point less than 7??? Balanced reaction: Na2CO3(aq) + 2HCl(aq) 2NaCl(aq) + H2CO3(aq) Net ionic equation: 2Na+ (aq) + CO32-(aq) + 2H+(aq) + 2Cl-(aq) So at the equivalence point, there is neutral water, neutral spectator ions, and 2Na+ (aq) + 2Cl-(aq) + H2CO3(aq) some weak acid H2CO3(aq). What do you think keeps the pH below 7? Why do we care about titration curves? Acid base reaction pH curves provide a wealth of information: Initial pH levels Equivalence point volume of titrant Number of reaction steps Equivalence point pH for indicator selection; so the endpoint observed for the indicator chosen will closely match the equivalence point of the reaction • Thymol blue is an unsuitable indicator for this titration because it changes colour before the equivalence point (pH 7). • Alizarin yellow is also unsuitable because it changes colour after the equivalence point. Simulation • Bromothymol blue is suitable because its endpoint pH of 6.8 (assume the middle of its pH range) closely matches the reaction equivalence point pH of 7, and the colour change is completely on the vertical portion of the pH curve. Final Tips • It is also critical that the amount of indicator used be extremely small. • Some of the titrant volume is used to react with the indicator to make it change color. But if the amount of indicator is small, the volume of titrant used this way will be very small, and the accuracy of the titration will not be affected. Summary An indicator for an acid–base titration analysis must be chosen to have an endpoint (change of colour) at very nearly the same pH as the pH at the equivalence point of the reaction solution. The pH of the solution at the equivalence point for a strong monoprotic acid–strong monoprotic base reaction will be 7. The pH of the solution at the equivalence point for any other acid–base reaction must be determined Homework Pg. 336 #1-4 Pg. 339 #1-3, 5-7, 9 Unit 4 Review Exam Next Week – chapter 7-8