Chapter 3 part 2
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Chapter 3 part 2

Course Number: CHEM 1010, Spring 2011

College/University: UOIT

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Chapter 3 Formulas, Equations and Moles Lecture 6 Agenda: (CHEM 1010) Practice Quiz (Answer will be included in inked version) Problem 2.113, 2116 Limiting Reagents Problem Concentrations of Reactants in Solution: Molarity Diluting Concentrated Solutions Solution Stoichiometry Titration Percent Composition and Empirical Formulas Determining Empirical Formulas: Elemental Analysis Practice quiz The poisonous...

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3 Chapter Formulas, Equations and Moles Lecture 6 Agenda: (CHEM 1010) Practice Quiz (Answer will be included in inked version) Problem 2.113, 2116 Limiting Reagents Problem Concentrations of Reactants in Solution: Molarity Diluting Concentrated Solutions Solution Stoichiometry Titration Percent Composition and Empirical Formulas Determining Empirical Formulas: Elemental Analysis Practice quiz The poisonous gas hydrogen cyanide (HCN) is produced by the high-temperature reaction of ammonia (NH3) with methane (CH4). Hydrogen is also produced in this reaction. (a)Write a balanced equation for the reaction that occurs. HCN produced (PRODUCT) from rxn NH3 with CH4 (REACTANTS) PS H2 is also produced (PRODUCT) NH3 + CH4 HCN + H2 N balance and C balance need to balance the Hs NH3 + CH4 HCN + 3H2 Practice quiz The poisonous gas hydrogen cyanide (HCN) is produced by the hightemperature reaction of ammonia(NH3) with methane (CH4). Hydrogen is also produced in this reaction. (a) NH3 + CH4 HCN + 3H2 (b) Suppose 500.0g of methane is mixed with 200.0 g of ammonia. MW(CH4) = 12.01 + 4.04 = 16.05 g/mol Mol(CH4) = 500.0 g / 16.05 g/mol = 31.15264 mol MW(NH3) = 14.01 + 3.03 = 17.04 g/mol Mol(NH3) = 200.0 g / 17.04 g/mol =11.73709 mol Limiting reagent is NH3 Practice quiz Problem 2.113 5th edition -The mass percent of an element in a compound is the mass of the element (total mass of the elements atoms in the molecule) divided by the mass of the compound (total mass of all atoms in the molecules) times 100%. What is the mass percent of each element in acetaminophen? C8H9NO2 Problem 2.116 5th edition Zinc has atomic mass A=65.39 amu and has five naturally occurring isotopes: 64 Zn 63.929 amu 48.63% 66 Zn x amu 27.90% 67 Zn 66.927 amu 4.10% 68 Zn 67.925 amu 18.75% 70 Zn 69.925 amu 0.62% What is the isotopic mass of 66Zn? Av atomic mass = (amu64Zn*%64Zn) +(amu66Zn*%66Zn) + . You know Av atomic mass and all the other numbers except for x. Solve for x. Do You Understand Limiting Reagents? In one process, 124 g of Al are reacted with 601 g of Fe 2O3 2Al + Fe2O3 Al2O3 + 2Fe Calculate the mass of Al2O3 formed. g Al mol Al g Fe2O3 mol Fe2O3 needed OR mol Al needed mol Fe2O3 124 g Al x 1 mol Al 27.0 g Al x g Fe2O3 needed 1 mol Fe2O3 2 mol Al Start with 124 g Al g Al needed 160. g Fe2O3 = x 1 mol Fe2O3 367 g Fe2O3 need 367 g Fe2O3 Have more Fe2O3 (601 g) so Al is limiting reagent. Do You Understand Limiting Reagents? Use limiting reagent (Al) to calculate amount of product that can be formed. g Al mol Al mol Al2O3 2Al + Fe2O3 124 g Al x 1 mol Al 27.0 g Al x 1 mol Al2O3 2 mol Al g Al2O3 Al2O3 + 2Fe 102. g Al2O3 = x 1 mol Al2O3 234 g Al2O3 Limiting Reagents Challenge Hydrogen and chlorine react to yield hydrogen chloride. RXN How many grams of HCl are formed from reaction of 3.56g of H2 with 8.94 g of Cl2? Which reactant is limiting? (problem 3.71 5th) Solutions A solution is a homogenous mixture of two or more substances. Solution Composition The is(are) the substance(s) present in the smaller amount(s). The amount. is the substance present in the larger Solution Solvent Solute Soft drink (l) H2O Sugar, CO2 Air (g) N2 O2, Ar, CH4 Soft Solder (s) Pb Sn Dissolution vs reaction Ni(s) + HCl(aq) NiCl2(aq) + H2(g) dry NiCl2(s) Dissolution is a physical change - you can get back the original solute by evaporating the solvent. If you cant, the substance didnt dissolve, it reacted. Aqueous Solution and Concentration Solutions in which water is the dissolving medium are called . The term concentration is used to indicate the amount of solute dissolved in a given quantity of solvent or solution. Concentration expressions The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. Most widely used way of quantifying concentration in chemistry is (M or mol/L). The molarity of a solution is defined as the number of moles of solute in a liter volume of solution: mol solute M = molarity = L solution Concentration Units (Molarity) Example: 1 Molar (1M) HCl contains 1 mole per liter of solution. 1M HCl, 1M NaOH, 1M NaCl, each contain 1 mole of solute in 1 liter of solution. 2M HCl contains 2 mole of HCl in 1 liter of solution. A 1M solution could also contain 0.5 mole in 0.5 liters of solution. Solution Stoichiometry Challenge What mass of KI is required to make 500. mL of a 2.80 M KI solution? volume KI 500. mL x M KI 1L 1000 mL moles KI x 2.80 mol KI 1 L soln MWKI x grams KI 166 g KI 1 mol KI = 232 g KI Solution Stoichiometry How many grams of solute would you use to prepare 250.00 mL of 0.600M ethyl alcohol (C2H6O). Problem 3.81a 5th Dilution Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution. Dilution Add Solvent Moles of solute before dilution (ni) = Moles of solute after dilution (nf) Calculating Dilutions The ability to prepare a solution of a specific concentration from a stock is very important in chemistry. Note that the number of moles of solute does not change on dilution so: ni = nf and since n = M x V: Mi Vi = M f Vf You can also choose M1V1=M2V2 OR C1V1=C2V2 **Choose the combo that will work for you.** Calculating Dilutions Calculate the volume of a 6.2 M NaOH solution needed to prepare 500.0 mL of a 1.5 M solution. Mi = 6.2 mol/L Mf = 1.5 mol/L Vi = ? Vf = 500.0 mL Vi = MfVf = (1.5mol/L)(500.0mL) = 120 mL Mi (6.2 mol/L) Mi Vi = M f Vf Calculating Dilutions If 15.0 mL of a 12.0 M HCl solution were diluted to a volume of 100.0 mL, what would the concentration of the new solution be? Mi = mol/L Mf = mL Vf = mol/L Vi = mL Mf = MiVi = ( Vf mol/L)( ( mL) = mol/L mL) Mi Vi = M f Vf Concentration Challenge The sterile saline solution used to rinse contact lenses can be made by dissolving 400mg of NaCl in sterile water and diluting to 100mL. What is the molarity of the solution? Solution Stoichiometry aA + bB Volume of Solution of A Molarity of A Moles of A cC + dD Moles of B Mole Ratio Between A & B (Coefficients) Volume of Solution of B Molarity of B Solution Stoichiometry What volume of 0.250 M H2SO4 is needed to react with 50.0 mL of 0.100 M NaOH? H2SO4(aq) + 2NaOH(aq) Volume of Solution of H2SO4 Molarity of H2SO4 Moles of H2SO4 Na2SO4(aq) + 2H2O(l) Moles of NaOH Mole Ratio Between H2SO4 and NaOH Volume of Solution of NaOH Molarity of NaOH Which solution do we know all the info about NaOH or H2SO4? Solution Stoichiometry Na2SO4(aq) + 2H2O(l) H2SO4(aq) + 2NaOH(aq) Moles of NaOH available: 50.0 mL NaOH x 0.100 mol 1L x 1L 1000 mL = 0.00500 mol NaOH Volume of H2SO4 needed: 0.00500 mol NaOH x 1 mol H2SO4 2 mol NaOH / 0.250 M = 0.0100 L 10.0 mL solution (0.250 M H2SO4) Solution Stoichiometry Challenge The concentration of cholesterol (C27H46O) in normal blood is approximately 0.005M. How many grams of cholesterol are in 750 mL of blood? Problem 3.17 Titrations In a titration, a solution of accurately known concentration is added gradually to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Slowly add known to unknown UNTIL the indicator changes color. Special Equipment Used in Titrations Buret: Volumetric Flask: Titrations Important terms in titrations: : When the quantity of added titrant is the exact amount necessary for a stoichiometric reaction. The equivalence point is what is theoretically measured in a titration (actually measure the endpoint). : Marked by a sudden change in a physical property of the solution: A colour change. A change in the adsorption of light. A change in the voltage or current between a pair of electrodes. The endpoint occurs at or past the equivalence point. Titrations Important terms in titrations: Indicator: A compound that changes its physical property (e.g. colour) near the equivalence point. Change caused by the disappearance of the analyte or the appearance of excess titrant. Titration error: The difference between the endpoint and the equivalence point. : A titration performed without analyte present to detect the size of the titration error. Special Equipment Used in Titrations Buret: A precision volume measuring device with a capacity of 50mL. Volumes can be read to two decimal places ie. 10.45 mL. Volumetric Flask: Allows for the preparation of solutions of accurate concentration (ie. 0.25 M). Flask is filled with solvent to the line to make the desired concentration. Review of Acids and Bases Acid will donate a proton (H+ ion) when dissolved in water. Base will donate a hydroxyl group (OH- ion) when dissolved in water. Mi Vi = M f Vf Equation can become MAVA=MBVB for Acid and Base Example Titration 1 If 30.0 mL of NaOH neutralizes 45.2 mL of 0.100 M HCl solution calculate the molarity of the base. HCl + NaOH NaCl + H2O 1 mol 1 mol nA nB and so nA = nB MAVA = MBVB MB = MAVA = (0.100 mol/L)(45.2mL) = 0.151 mol/L VB (30.0 mL) Example Titration 2 In an acid-base titration, 34.5 mL of NaOH solution required 27.5 mL of 0.0500 M H2SO4 solution to reach a neutral end point. Calculate the molarity of the base. H2SO4 + 2 NaOH Na2SO4 + 2 H2O 1 mol 2 mol nA nB and nB = 2 nA or MB V B = Empirical Formula and Percent Composition Composition: Expressed by identifying the elements present and giving the mass percent of each. Formula: It tells only the ratios of the atoms in a compound. Formula: It tells the actual numbers of atoms in a compound. It can be either the empirical formula or a multiple of it. molecular mass multiple = empirical formula mass Percent Composition Percent composition of an element in a compound = n x molar mass of element x 100% molar mass of compound n is the number of moles of the element in 1 mole of the compound 2 x (12.01 g) x 100% = 52.14% 46.07 g 6 x (1.008 g) %H = x 100% = 13.13% 46.07 g 1 x (16.00 g) %O = x 100% = 34.73% 46.07 g %C = C2H6O 52.14% + 13.13% + 34.73% = 100.0% Percent Composition Molecular formula for ethyl chloride: C2H5Cl Molar mass = g/mol Percent Composition: % C = 2( ) x 100 % H = 5() x 100 = % Cl = x 100 = + + = % % % = 100.0 % Empirical Formula and Percent Composition A colorless liquid has a composition of 84.1% carbon and 15.9% hydrogen by mass. Determine the empirical formula. Also, assuming the molar mass of this compound is 114.2 g/mol, determine the molecular formula of this compound. Mass percents Molar masses Moles Mole ratios Subscripts Relative mole ratios Empirical Formula and Percent Composition Assume you have 100g of the sample: Mole of carbon: 84.1 g C x 1 mol C 12.0 g C = 7.01 mol C Mole of hydrogen: 15.9 g H x 1 mol H 1.0 g H = 15.9 mol H You could assume 1 g of sample but the numbers are a little harder to work with. Empirical Formula and Percent Composition Empirical formula: C7.01H15.9 C7.01H15.9 smallest value for the ratio C1H2.27 7.01 = C1H2.27 7.01 C1x4H2.27x4 = C4H9 need whole numbers Molecular formula: multiple = 114.2 57.0 =2 C4x2H9x2 = C8H18 Empirical formula mass = (4*12g/mol) + (9*1g/mol) = 57g/mol Empirical Formula and Percent Composition Challenge What is the empirical formula of an ingredient of Bufferin tablets that has the percentage composition C14.25%, O 56.93% Mg28.83% by mass?
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