The Concepts and Overall Exam 2
The Concepts and Overall Exam 2 CHEM 1307
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This 27 page Study Guide was uploaded by Kyle A. Headen on Monday October 17, 2016. The Study Guide belongs to CHEM 1307 at Texas Tech University taught by Whittlesey in Fall 2016. Since its upload, it has received 21 views. For similar materials see Experimental Principles of Chemistry 1 in Chemistry at Texas Tech University.
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CHEM 1307 – Fall 2016 Topics to Study for Exam 2 Chapter 3: Molecules, Compounds, and Chemical Equations Sections 3.9, 3.10, and 3.11 (NOT 3.12) Be able to calculate the mass or the mass percent composition of an element in a compound when given the formula. (Section 3.9) Be able to determine the empirical formula of a compound when given the masses or the mass percents of its elements, or the data from combustion analysis. (Section 3.10) Be able to determine the molecular formula of a compound when given the empirical formula and the molecular mass (molar mass). (Section 3.10) Be able to balance a chemical equation. (Section 3.11) Chapter 4: Chemical Quantities and Aqueous Reactions All Sections Reaction stoichiometry calculations: be able to calculate the amount needed (for a reactant) or the amount produced (for a product) for any reactant or product in a chemical equation when given the amount of one of the species. (Section 4.2) Be able to identify the limiting reactant in a chemical equation when given the amounts of all reactants, and to use that knowledge to calculate the amounts of the other reactants that will be consumed, the amounts of the other reactants that will be left over, and the amounts of products that will be produced. (Section 4.3) Be able to calculate the theoretical yield of a product. (Section 4.3) Be able to calculate the percent yield of a product when given the actual yield. (Section 4.3) Memorize and be able to use the formula for the molarity of a solution. (Section 4.4) Memorize and be able to use the formula for dilution of a solution. (Section 4.4) Be able to use the formula for the molarity of a solution to calculate the number of moles of a species in a chemical equation when given the volume of the solution and its molarity. (Section 4.4) Be able to identify whether or not a compound is an electrolyte. (Section 4.5) Be able to predict whether or not an ionic compound is soluble. (Table 4.1, Section 4.5) Be able to predict whether or not a combination of ions will form an insoluble product in a precipitation reaction. (Sections 4.5 and 4.6) Know the definitions of molecular equation, complete ionic equation, and net ionic equation. (Section 4.7) Know the definition of a spectator ion, and be able to identify them in a reaction. (Section 4.7) CHEM 1307 – Fall 2016 Topics to Study for Exam 2 Be able to write a net ionic equation for any reaction involving ions (precipitation reactions, acid-base reactions, and oxidation-reduction reactions) (Sections 4.7, 4.8, and 4.9). Know the Arrhenius definitions of acids and bases. (Section 4.8) Know the definitions of hydronium ion, diprotic acid, and polyprotic acid. (Section 4.8) Be able to identify an acid-base reaction, and know what products are formed. (Section 4.8) Know the acids and bases in Table 4.2, including which ones are strong (black print) and which are weak (red print). (Section 4.8) Be able to identify whether a binary acid (H Ex or an oxyacid (H EO x isystrong or weak. (class notes) Understand the concept of titration, and the definitions of equivalence point and indicator. (Section 4.8) Be able to identify a gas evolution reaction. (Section 4.8) Be able to assign charges (oxidation states) to all elements in a substance. (Section 4.9) Be able to identify an oxidation-reduction reaction. (Section 4.9) Be able to identify the oxidizing agent and the reducing agent in an oxidation-reduction reaction, as well as which species are oxidized and which are reduced. (Section 4.9) Be able to identify a combustion reaction. (Section 4.9) Chapter 5: Gases All Sections Know the common units of pressure (bar, Pa, atm, mmHg, inHg, and torr) and memorize 5 the following conversions: 1 mmHg = 1 torr 1 atm = 760 torr 1 bar = 10 Pa (Section 5.2) Know Boyle’s Law, Charles’s Law, and Avogadro’s Law. (Section 5.3) Know the Ideal Gas Law and how to use it in calculations. (Section 5.4) Know the values for Standard Temperature and Pressure (STP). (Section 5.5) Memorize the volume of one mole of gas at STP (22.4 L). (Section 5.5) Be able to identify which value of R to use in calculations, based upon the units. Be able to use density, mass, and molar mass in calculations using the Ideal Gas Law. (Section 5.5) Be able to use partial pressures and mole fractions in calculations using the Ideal Gas Law. Be able to account for the partial pressure of water in experiments where a gas is collected over water. (Section 5.6) Be able to use the Ideal Gas Law to calculate the number of moles of a species in a chemical equation. (Section 5.7) Know the assumptions underlying the Kinetic Molecular Theory of Gases. (Section 5.8) CHEM 1307 – Fall 2016 Topics to Study for Exam 2 Memorize the formula for the root mean square speed of a gas molecule and be able to use it in calculations. (Equation 5.26, Section 5.8) Have a qualitative understanding of the relationship between mass and velocity for a gas molecule at a given temperature (i.e. the larger the mass the lower the velocity). Memorize the formula for Graham’s Law of Effusion for a gas and be able to use it in calculations. (Equation 5.27, Section 5.9) Have a qualitative understanding of the relationship between mass and rate of effusion for a gas at a given temperature (i.e. the larger the mass the lower the rate of effusion). Have a qualitative understanding of the differences between a real gas and an ideal gas. Know how the properties of a real gas differ from those of an ideal gas, and under which conditions they most resemble each other and which conditions they differ the most. (Section 5.10) Chapter3 Formulasand things to remember: Writinga Formulafroma Name: Forbinary nonmetalcompounds: Thenumerical prefixes will tell you howmanyof each element is present. Ifthere is noprefix, assumethat there is oneof that element. Forionic compounds: You mustknowthecharges on all oftheions. Write downtheions with their charges, anddetermine thesmallest numberofeach ion that is needed so that thetotalnumberofpositivecharges is equal to thetotalnumberofnegative charges. Foracids: Figureout thecharge on theanion. Write downthenumberofH+ ions that are needed to balance thenumberofnegative charges on the anion. Mass Percent: Knowingthemass percent of an element in acompoundis usefulfordetermining theformulaofthe compound. Themass percent ofan element can be calculated fromtheformula, as follows: 1) Find thetotalmass ofeverything in theformula(themolarmass). 2) Foreach element, dividethe totalmass ofthat element in the compoundbythemolarmass of thecompound. This gives the massfraction foreach element. 3) Multiply each massfraction by 100toget themasspercent foreach element. Example: C20H24N2O2 (Quinine,usedinthe treatmentof malaria) C: (20)(12.011)=240.22 H: (24)(1.00794)=24.19056 N: (2)(14.0067)=28.0134 O: (2)(15.9994)=31.9988 Total molar mass forC20H24N2O2is324.42gmol-1 %C: 240.22/324.42x100=74.05% %H: 24.19056/324.42x100=7.46% %N: 28.0134/324.42x100=8.63% %O: 31.9988/324.42x100=9.86% Empirical Formula: Theformulaofan unknowncompoundcanbedetermined in thefollowingway: I. Measurethemass ofa sampleofthe unknowncompound. II. Break downthe compoundusingchemical methodsand determine themass (ormass percent) of each element present. III. Convert themassof each element into moles (numbersofatoms)ofthat element. IV. Determine thesimplest wholenumberrelationship between the elements in thecompound. This wholenumberrelationship is called the"empirical formula”. We usethemolarmasses to convert whatwemeasure in thelaboratory (massesofthe elements in the compound)towhatwe wouldlike to know(thenumberof atomsof each element in the compound). In manyproblems youare given the masspercents of theelements present and youmust determinethe formula. Problem: A puresampleof anew chemical compoundwasanalyzed and wasfoundto havethefollowing percentages, by weight, oftheelements C, H, and Cl: C 43.01%; H 2.58; Cl 54.41%. What is the empirical formulaofthis compound? Solution: Themass percentages aretrue forany size sampleofthis compound,andthe formulathat weare trying to find is trueforany sizesample, so wemay chooseany size that makesthe arithmetic easier. Since percentages are"parts perhundred,"wewill choose100g ofcompoundforthecalculation. Stepsto findthe EmpiricalFormula: Step 1: Calculate thenumberofgrams (mass)ofeach element present. Step 2: Convert themassesto moles (numberofatoms)of each element. Step 3: Adjustthenumbersby multiplying bya commonfactorsothat wehavea whole-number relationship between them. Step 1 Calculate themass ofeach element present. A 100g samplehas 43.01gofcarbon. 100g x 43.01/100 = 43.01g A 100g samplehas 2.58gof hydrogen. A 100g samplehas 54.41gofchlorine. Step 2 Calculate themoles of each element present. 43.01gC x 1 molC/12.01115C= 3.581molC 2.58g Hx 1mol H/1.00797 = 2.560molH 54.41gCl x1 molCl/35.453=1.535molCl Step 3 Adjustthenumbersofmoles to get awhole-numberrelationship. Therelationship between theelements will bemaintained if you multiply ordivideall ofthem by the samenumber. First, divideall the numbersby thesmallest one. 3.581/1.535=2.333C 2.560/1.535=1.668H 1.535/1.535=1Cl Forevery 1 Clthere are very close to 2(1/3)Cand 1 (2/3)H. We can get wholenumbers forall of theelements if wemultiply all of them by3. (3)(2.333)= 6.999C = 7 C (3)(1.668)= 5.004H = 5 H (3)(1)=3.000Cl= 3 Cl Theempirical formulaforthecompoundis C7H5Cl3,which is thesimplest whole-numberrelationship between theatoms ofeach element. In somesituationsyoumay be given themasses oftheelements present and youmust determinethe formula. A 13.0100gsampleofa new chemical compoundwasanalyzed and wasfoundto havethefollowing masses oftheelements C and H: C 4.1100g H 0.6895g Therest ofthe compoundwasmadeup ofoxygen. What is theempirical formulaofthis compound? Step 1: We already havethemasses of each element present except foroxygen, soweadd themasses of thecarbon andhydrogen andsubtract that sumfrom thetotalmass to findthe oxygen. Step 2: We then convert thosemasses to moles (numberof atoms)ofeach element. Step 3: Oncewedetermine howmany moles ofeach element are present, weadjustthosenumbers so that wehaveawhole-numberrelationship between them. Step 1 Calculate themass ofeach element present. 4.1100gC + 0.6895gH = 4.7995gC +H 13.0100gtotal- 4.7995gC +H =8.2105gO Step 2 Calculate themoles of each element present. 4.1100gC x1 molC/12.01115 = 0.342182molC 0.6895gHx 1molH/1.00797 =0.6840481molH 8.2105gOx 1 molO/15.9995 =0.5131722molO Step 3 Adjustthenumbersofmoles to get awhole-numberrelationship. First,divide all thenumbersby the smallest one. 0.342182/0.342182 =1 C 0.6840481/0.342182 =1.9999H 0.5131722/0.342182 =1.4999O Forevery oneC there are 2 Hand 1 (1/2)O. We can get wholenumbers forall of theelements if wemultiply all of them by2. (2)(1) = 2 C (2)(2) = 4 H (2)(1.5) = 3 O Theempirical formulaforthecompoundis C2H4O3. MolecularFormula: Theempirical formulais thesimplest whole-numberrelationship between the atomsof thethedifferent elements in thecompound However, themolecules ofthecompoundmay bemadeup ofmany moreatoms,still in thesame proportionto each other. Forexample, thechemically different compoundsC2H4,C3H6,C4H8,andC5H10allhavetheempirical formulaCH2 (all havetwohydrogensforevery carbon). Theformulaofany of thesemolecules could bewritten as a multipleof theempirical formula,i. e. C2H4 could bewritten (CH2)2,andC3H6wouldbe(CH2)3,andin general any compoundwith this empirical formulacould bewritten as (CH2)n,wheren is an integer. Ifthe molecular massis known(fromexperiment) andthe empirical formulais known,thevalueofn (andhence themolecular formula)can beobtained by dividing the molecular mass bythe massof the empirical formula. Thenumbern is thenumberof empirical formula"units"it takes to makeup a molecule. Stepsfor MolecularFormula: (1) Find themolarmass fortheempirical formula. (2) Divide themolecular massby themass forthe empirical formulato get n. (3) Multiply theempirical formulaby n to get the molecularformula. Example: Theempirical formulaof thecompoundis C2H4O3andits molecular massis 228.158. Whatis the molecular formula? Step 1 Themolar massfortheempirical formulaC2H4O3is: (2)(12.01115)+(4)(1.00797)+(3)(15.9995)=76.05268molarmass Step 2 Themolar massforthemolecular formuladivided bythe molarmassfortheempirical formulais: 228.158/76.05268=2.9999995=3 Step 3 Themolecular formulais (C2H4O3)3,whichis C6H12O9. Chapter 4: Things to Remember and steps to follow Stoichiometry The balanced equation for a reaction will tell you the relative numbers of the reactants and products, but in the lab we can't count out molecules to do a reaction, we can only weigh out the masses of the reactants and products. The following method can be used to calculate the amounts (in grams) of reactants needed and the amounts (in grams) of products that could be formed. Calculating Quantities in a Reaction Equation- NH3 + O2 N2 + H2O Balanced Equation- 4 NH3 + 3 O2 2 N2 + 6 H2O The first step is to balance the equation. Put in the known quantity and indicate what you are trying to find. (grams, moles, or molar mass). 80grams to 4NH3 Calculate the molar mass of NH3. Molar mass of NH3 = (1) *(14.0067) + (3) *(1.00794) = 17.0305 g mol-1 Now calculate the number of moles of NH3 in 80.0 g of NH3. 80.0 g NH3 x 1 mol NH3/ (17.0305 g NH3) = 4.6975 mol NH3 Calculate the number of moles of O2 that can react with 4.6975 moles of NH3. It’s important to remember the coefficients in front of the elements of the balanced equations should be used to find moles. First you take the number of O2 and divide it by the number in front of NH3. In other words, just multiply 3mols over 4 mols by 4.6975 mols of NH3. 4.6975 mol NH3 x 3 mol O2/4 mol NH3 = 3.5231 mol O2 To find the mass of O2 you will multiply the mol by the molar mass of O2. Molar mass of O2 = (2)(15.9994) = 31.9988 g mol-1 3.5231 mol O2 x 31.9988 g O2/1 mol O2 = 112.7350 g O2 Rounding off to the correct number of significant figures, the answer is 113 g O2. You will repeat the previous steps to finding mols and mass for the other two the exact same way as shown above. Make sure you put the NH3 coefficient as the denominator. When calculating moles for others. Limiting Reagent In many reactions, the reactants may not all be present in the exactly right amounts (an extra amount of one of the reactants may be available). The reactant that is used up first is called the limiting reagent. The limiting reagent controls the amounts of the products that can be formed. Products can only be formed until the limiting reagent is used up, at which point the reaction stops. Steps to Determine Limiting Reagents 1) Balance the chemical equation for the reaction. 2) Pick one of the reactants (it doesn't matter which one) and, using the balanced equation, calculate how much of each of the other reactants would be needed to use it all up. 3) Compare the amounts of the other reactants that are needed to the amounts that you actually have. If you have enough of all the other reactants, then the reactant that you picked will be used up first, and therefore it is the limiting reagent. 4) If you do not have enough of one of the other reactants, then it will be used up first and is therefore the limiting reagent. If more than one of the other reactants is insufficient, then you will have to perform the calculation again using one of the reactants that is in short supply. 5) Once you have determined which reactant is the limiting reagent, use that amount to calculate the amounts of products formed or the amounts of other reactants that are needed. Example: Suppose that you wish to burn 50.00 g of C3H8 (propane) in an enclosed space that contains 100.0 g of oxygen. Which reactant, the propane or the oxygen, will be used up first? How much carbon dioxide will be formed? The balanced equation for the reaction is C3H8 + 5 O2 3 CO2 + 4 H2O Using the diagram for calculating quantities in a reaction, we'll pick one of the reactants (let’s say propane) and calculate how much of the other reactant (oxygen) would be needed to completely use it up. Molar mass of C3H8 = (3)(12.0107) + (8)(1.00794) = 44.09562 g mol-1 50.00 g C3H8 x 1 mol C3H8/44.09562 g C3H8= 1.133899 mol C3H8 Next we calculate the number of moles of O2 needed. C3H8 + 5 O2 3 CO2 + 4 H2O 1.133899 mol C3H8 x 5 mol O2/1 mol C3H8 = 5.669495 mol O2 Now calculate the mass of O2needed. Molar mass of O2 = (2)(15.9994) = 31.9988 g mol-1 5.669495 mol O2 x 31.9988 g O2/ 1 mol O2 = 181.4170 g O2 The Other way to solve this problem is to solve the mass the Oxygen. 181.42 g of oxygen is needed to completely use up all of the propane, however only 100.0 g of oxygen is available. The oxygen will be used up first, and therefore it is the limiting reagent even though there is more oxygen by mass (100.0 g) than propane (50.00 g). Since the amount of oxygen determines how far the reaction will go, we must use this amount to calculate how much carbon dioxide is formed or how much propane is consumed. First calculate the number of moles of O2that we have. 100.0 g O2 x 1 mol O2/ 31.9988 g O2=3.125117 mol O2 Calculate the number of moles of CO2produced. C3H8 + 5 O2 3 CO2 + 4 H2O 3.125117 mol O2 x (3 mol CO2)/(5 mol O2)= 1.875070 mol CO2 Now calculate the mass of CO2produced. Molar mass of CO2 = (1)(12.0107) + (2)(15.9994) = 44.0095 g mol-1 1.875070 mol CO2 x 44.0095 g CO2/1 mol CO2 = 82.5209 g CO2 Round off to the correct number of significant figures (from 100.0 g O2) Answer: 82.52 g CO2 Percent Yield For any reaction, if you know the balanced equation and the amount of the limiting reagent present, then the amount of product that could be formed can be calculated as shown previously. This is known as the theoretical yield. However, the amount of product that is actually obtained is almost always less than the theoretical maximum amount. Some of the product is lost in the course of manipulating the material and sometimes there is incomplete reaction or other minor products are formed. It is often useful to know what fraction of the product was obtained compared to the theoretical maximum amount that could have been obtained. Percent yield = actual yield/ theoretical yield x 100 In the limiting reagent problem shown previously, in which propane combined with oxygen, the theoretical yield of carbon dioxide was 82.52 g. Suppose only 63.45 g of CO2 was obtained. The percent yield of CO2 would be: % yield = actual yield/ theoretical yieldx 100 = 63.45 g/82.52 g x 100 % yield = 76.89 % Concentration A solution is a homogeneous mixture of two or more substances. The term “solution” usually refers to a liquid mixture, but it is possible to have gaseous solutions and even “solid solutions”. One of the components of the solution is usually present in a much greater amount than the other components. This is called the solvent. Each minor component is called a solute. The amount of solute per unit of solution is called the concentration of that component, and it can be expressed in several ways. Molarity Concentrations can be expressed in terms of moles per unit volume. Concentration= moles/volume Molarity= moles of solute/liters of solution Molarity is abbreviated as “M”. The concentration of a chemical species is often symbolized by writing the formula for the species in square brackets. Example: [Cu2+] = 0.01 M means that the concentration of Cu2+ ions in the solution is 0.01 moles per liter. What is the molarity of a solution that contains 0.750 moles of NaOH in 2.50 L of solution? Molarity= moles of solute/ liters of solution Molarity= 0.750 mol NaOH/2.50 L solution = 0.300 M How many grams of Na2SO4are needed to make 2.50 L of a 0.500 M Na2SO4solution? Molarity = moles of solute/liters of solution We know the volume of the solution and the molarity, so we can calculate the number of moles of Na2SO4needed and then convert the moles to grams. Moles of solute = (molarity)(liters of solution) Moles of solute = (0.500 mol/L)(2.50 L) = 1.25 mol 1.25 mol Na2SO4x 142.037 g Na2SO4/1 mol Na2SO4= 178 g We can use the equation for molarity to solve for any of the three variables. Molarity = moles of solute/liters of solution Moles of solute = (molarity)(liters of solution) Liters of solution = moles of solute/molarity Dilution A concentrated solution can be diluted by adding more solvent. This changes the concentration and the volume of the solution, but the number of moles of solute in the container stays the same. Moles ofsolute = (molarity)(liters of solution) Moles ofsolute = (Cinitial)(Vinitial) Moles ofsolute = (Cfinal)(Vfinal) (Cinitial)(Vinitial) = (Cfinal)(Vfinal) C1V1 = C2V2 C1V1 = C2V2 How many liters of a 5.00 M HCl solution are neededto make 1.25 liters of 0.200 M HCl? C1V1 = C2V2 V1 = C2V2 = (0.200 mol/L HCl) /C1(5.00 mol/L HCl) x 1.25 L HCl V1 = 0.050 L of the 5.00 M HCl solution Solution Stoichiometry We use masses (grams) and numbers of particles (moles) when calculating amounts of reactants and products for a reaction. In many reactions the reactants and products are in solution. For these species, we can use volumes (liters) and concentrations (molarity) to calculate the amounts (moles) of each material. We can use the equation for molarity to solve for any of the three variables. Molarity = moles of solute/liters of solution Moles of solute = (molarity)(liters of solution) Liters of solution = moles of solute/molarity Solubilities of Ionic Compounds in Water An electrolyte is a substance that dissolves in water to give separate ions. The solubility of an ionic compound in water is very difficult to predict with a theoretical model. The best way to predict solubilities is to remember empirical (experimentally-observed) trends. An ionic compound is made up of positively-charged cations and negatively charged anions. These are held together in the solid state by the force of attraction between the opposite charges. The larger the charges, the larger the force of attraction. In order for an ionic compound to dissolve in water, the water molecules must be able to separate the cations from the anions. If the charges on the ions are large, the force of attraction between them is strong, and the water molecules are unable to separate them (the compound does not dissolve). Charge of Cation Charge of Anion Solubility 1+ 1- Very Soluble 1+ 2- Fairly Soluble 2+ 1- Fairly Soluble 1+ 3- Slightly Soluble 3+ 1- Slight Soluble 2+ 2- Insoluble 2+ 3- Insoluble 3+ 2- Insoluble 3+ 3- Insoluble Exception 1: Halides of Group 11 metals (Cu, Ag, and Au) and mercury (Hg) and lead (Pb) tend to have very low solubilities, or are insoluble. These compounds have significant covalent character, and do not separate readily into ions in solution. Exception 2: Fluorides (F-) and hydroxides (OH-) of the smaller Group 2 and Group 3 elements are usually insoluble. The concentration of charge on these small ions increases the force of attraction between them, and makes them less likely to separate in water. Is AlPO4 soluble, slightly soluble, or insoluble in water? Al3+ has a charge of 3+ and PO43- has a charge of 3-. There will be a strong force of attraction between the ions, and the water molecules will not be able to separate them. The compound should be insoluble. Precipitation Reactions When two or more ionic compounds dissolve in water, the cations and anions separate and become surrounded with water molecules. If there is some combination of one of the cations with one of the anions that forms a compound that is not soluble, that compound will form a solid precipitate and come out of solution. CaCl2 + Na2SO4 CaSO4(s) + 2Na+(aq) + 2Cl-(aq) Both CaCl2 and Na2SO4are soluble in water. The CaCl2 dissolves to form Ca2+ ions and Cl- ions, each of which is surrounded by water molecules. Likewise, the Na2SO4dissolves to form Na+ions and SO42- ions, also surrounded by water molecules. However, when the two solutions are mixed a solid precipitate of CaSO4 forms and the Na+and Cl- ions stay in solution. CaSO4precipitates with some water in the ionic lattice it forms gypsum (two waters per CaSO4), which is the rock-like material in “drywall” boards used in buildings. If the gypsum is heated, some of the water is lost to form Plaster of Paris (0.5 water per CaSO4). Soluble Examples: NH4Cl + KNO3 NH4+(aq) + Cl-(aq) + K+(aq) + NO3-(aq) All combinations of the cations with the anions in this solution are soluble. Therefore, everything stays in solution and there is no solid precipitate. (NH4Cl, KNO3,NH4NO3,and KClare all soluble) NetIonic Equations Equations that describe precipitation reactions can be written several ways. CaCl2(aq) + Na2SO4(aq) CaSO4(s) + 2NaCl(aq) can also be written as Ca2+(aq) + 2Cl-(aq) + 2Na+(aq) + SO42-(aq) CaSO4(s) + 2Na+(aq) + 2Cl-(aq) Ca2+(aq) + 2Cl-(aq) + 2Na+(aq) + SO42-(aq) CaSO4(s) + 2Na+(aq) + 2Cl-(aq) However, some of the ions do not change during the reaction. The Na+and Cl- ions are in solution before the reaction and they are still in solution after the reaction. These are called spectator ions (they do not participate in the reaction). Ca2+(aq) + 2Cl-(aq) + 2Na+(aq) + SO42-(aq) CaSO4(s) + 2Na+(aq) + 2Cl-(aq) We could leave out the spectator ions and just write a net ionic equation that contains only the species that change during the course of the reaction. Ca2+(aq) + SO42-(aq) CaSO4(s) Quantitative Chemical Analysis The amount of a particular substance in a mixture can be determined by adding a reagent that reacts with that substance but does not react with the other components of the mixture. By measuring the amount of the reagent added to just achieve complete reaction, or by measuring the amount of product formed, the amount of the original substance in the mixture can be determined. The reaction must be one that goes to completion (uses up all of the substance). Example The amount of a particular metal ion in solution can be determined by adding a substance that forms an insoluble compound with the metal ion. When the insoluble metal compound precipitates from solution, it is collected by filtration and weighed. By knowing the formula of the insoluble compound, we can calculate how much of the metal ion was in solution. A 0.123 g sample of the mineral thenardite contains sodium sulfate, along with other insoluble compounds. The finely powdered sample is added to water, which dissolves the sodium sulfate, and the solution is filtered away from the insoluble solids. Aqueous barium chloride solution is then added to the filtered solution until no more barium sulfate precipitates. The barium sulfate is removed by filtration, dried, and weighed. Its mass is found to be 0.177 g. What is the mass percent of sodium sulfate in the sample? Na2SO4(aq) + BaCl2 (aq) BaSO4 (s) + 2NaCl (aq) Molar mass of BaSO4 = (1)(137.327) + (1)(32.066) + (4)(15.9994) = 233.39 g mol-1 0.177 g BaSO4 x 1 mol BaSO4/ 233.39 g BaSO4 = 7.5839 x 10-4 mol BaSO4 7.5839 x 10-4mol BaSO4 x 1 mol Na2SO4/1BaSO4 = 7.5839 x 10-4mol Na2SO4 Molar mass of Na2SO4 = (2)(22.9898) + (1)(32.066) + (4)(15.9994) = 142.04 g mol-1 7.5839 x 10-4mol Na2SO4x 142.04 g Na2SO4/1 mol Na2SO4 = 0.1077 g Na2SO4 Mass percent of Na2SO4in the sample = 0.1077 g Na2SO4/0.123g sample total x 100= 87.6% Acid – Base Definitions "... an acid is a substance that produces H+ ions ... when it is dissolved in water.” An Arrhenius acid is a combination of an H+ ion with an anion (An-) that dissolves in water to give aqueous H+ ions and An- ions. H2SO4 is an example of an Arrhenius acid. "... a base is a substance that produces hydroxide (OH-) ions in water.” An Arrhenius base is a combination of a cation (Cn+) with hydroxide ions (OH-) that dissolves in water to give aqueous Cn+ ions and OH- ions. Ca(OH)2 is an example of an Arrhenius base. pH A Special Concentration Unit pH = - log10 [H3O+] When an acid is dissolved in water it releases H+ ions that attach to water molecules to form H3O+ ions (which are called hydronium ions). The higher the concentration of H3O+ ions, the more acidic the solution. In pure water, a very small number of water molecules break apart to form equal numbers of H3O+ ions and OH- ions (hydroxide ions). These continuously recombine to form water molecules and then break apart again to reform the ions. The rate of the forward reaction to form H3O+ ions and OH- ions is equal to the rate of the reverse reaction in which the ions recombine to form water molecules (the system is in equilibrium). 2 H2O ⇌ H3O+ + OH- 2 H2O ⇌ H3O+ + OH- For pure water at 25 ºC: [H3O+] = [OH-] = 1.0 x 10-7M pH = - log10 [H3O+] = 7 2 H2O ⇌ H3O+ + OH- When acid is added to the water, the concentration of H3O+ ions goes up and the OH- concentration goes down (the larger number of H3O+ ions react with some of the OH- ions to form more water molecules). 2 H2O ⇌ H3O+ + OH- Acidic solutions have a pH value less than 7. As [H3O+] increases, the exponent for the 10 becomes a smaller negative number, and therefore - log10 [H3O+] becomes a smaller positive number. 2 H2O ⇌ H3O+ + OH- Basic solutions have a pH value greater than 7. When [OH-] increases, [H3O+] decreases. As [H3O+] decreases, the exponent for the 10 becomes a larger negative number, and therefore - log10 [H3O+] becomes a larger positive number. Trends in Acid Strengths Acid strength increases going left to right across a row and increases going down a column for binary hydrides, HxEy, of the p-block elements. HI is the strongest of the binary hydride acids. Acid strength PH3 < H2S < HCl For the hydrogen halides (HX, where X = F, Cl, Br, I) HF is a weak acid, HCl is a strong acid, and HI is a very strong acid. Acid strength HF < HCl < HBr < HI For Oxyacids (HxEOy): Acid strength increases as the ratio of oxygen atoms to hydrogen atoms increases. For HxEOy, the hydrogens are usually attached to oxygen atoms. To determine the strength of the acid, rewrite the formula as E(OH)x(O)y-x. The value of y – x (the number of oxygens that do not have hydrogens attached) indicates the strength of the acid. y – x Formula Acid Strength 0 E(OH)x Very weak 1 E(OH)x(O) Weak 2 E(OH)x(O)2 Strong 3 E(OH)x(O)3 Very strong Practice Problem Formula y – x Acid Strength H3PO4 1 Weak HClO4 3 Very Strong H2SO4 2 Strong H3BO3 0 Very Weak There are many organic acids in which a hydrocarbon is attached to a -CO2H group. These are almost always weak acids. Base + Acid a Salt + Water AOH + HB AB + HOH (H2O) A salt is an ionic compound made up of an anion and a cation other than H+. The term "salt" is used to refer to ionic compounds in general. NaCl is just one member of a large family of salts. According to the Arrhenius definition, an acid can be thought to contain a piece of a water molecule (an H+ ion) and a base contains the other piece of the water molecule (an OH- ion). When a solution of an acid and a solution of a base are brought together, the H+ and the OH- ions reunite to form water molecules. The ions that accompanied the H+ and OH- ions form a salt. HA (an acid) + B(OH) (a base) BA (a salt) + H2O An acid produces H+ ions in water. A base produces OH- ions in water. A salt is an ionic compound that does not contain H+ or OH- ions. An acid and a base will react to give a salt and water. All of the species must be present in order for the reaction to be classified as an acid-base reaction. The reactants must be an acid and a base, and the products must be a salt and water. For instance, the reaction Zn + 2 HCl ZnCl2 + H2 is not an acid-base reaction even though an acid is one of the reactants and a salt is one of the products. Titrations One general method of analysis involves the controlled addition of a known reagent to an unknown compound to carry out a reaction that can be monitored in a way that clearly shows when a stoichiometric amount of the known reagent has been added. This process is called titration, and it is a form of quantitative analysis. It can be used to determine the concentration of a species in solution or to determine a molecular weight. Many reactions that take place in solution can be performed as titrations. One of the most common is the reaction between an acid and a base in aqueous solution. In a typical acid/base titration, a solution with a known concentration of a known strong base is slowly added to a measured sample of the acid until the equivalence point is reached. At the equivalence point, the number of moles of base added is equal to the number of moles of H+ ions produced by the acid in the sample. By knowing the number of moles of base added to reach the equivalence point, we can calculate how many moles of acid were in the sample. At the equivalence point: (conc. base)(volume base) = moles of base moles of base = moles of H+ from the acid If one mole of acid produces one mole of H+ ions, then the number of moles of base added is the same as the number of moles of acid in the sample. If one mole of acid produces two moles of H+ ions (as in H2SO4), then the number of moles of acid in the sample is half the number of moles of base added. Once we know how many moles of acid were present in the sample, we can calculate the concentration of the sample if it is a solution. moles of acid = conc. (moles/liter)volume of solution If the initial sample was pure acid rather than a solution, we can use the mass of the acid sample and the number of moles of acid that were determined by the titration to calculate the molecular weight of the acid. mass of acid/moles of acid =molecular weight (g/mole) A reverse acid/base titration works equally well, in which a solution with a known concentration of a known strong acid is slowly added to a measured sample of an unknown base until the equivalence point is reached. Combustion Reactions A + O2 A combustion reaction is one in which a substance reacts with O2. This is often accompanied by flames. How to recognize a combustion reaction: Look for O2 as a reactant. C3H8 + 5 O2 3 CO2 + 4 H2O Charts: Chapter 5 Things to remember and practice Pressure: The United States National Institute of Standards and Technology uses one atmosphere as the standard pressure. Pressure is force per unit area. The S. I. unit of force is the newton. Force = mass x acceleration 1 newton = 1 kg m s-2 The S. I. unit of pressure is the pascal. 1 pascal (Pa) = 1 newton / meter2 1 pascal = 1 kg m s-2m-2 = 1 kg s-2m-1 1 bar = 1 x 105 pascals (Pa) 1 bar = 100 kilopascals (kPa) 1 bar = 1 x 105 kg s-2m-1 1 bar = 0.9872 atmospheres 1 atmosphere = 760 mm Hg = 760 torr 1 atmosphere = 14.6960 pounds per square inch 1 atmosphere = 1.01325 bar 1 atmosphere = 1.01325 x 10^5pascals Memorize the following conversion factor: 1 atmosphere = 760 mm Hg = 760 torr General Gas Law: P1V1/(T1) = P2V2/(T2) This equation can be used for any problem in which the amount (number of moles) of gas stays the same. If any variable (P, V, or T) stays the same, it can be cancelled from both sides, and the equation is then reduced to Boyle’s, Charles’s, or Gay-Lussac’s Law. P V = n R T R is the ideal gas constant. R = 0.082057 L atm K-1mol-1 R = 8.314472 J K-1mol-1 Note that Liters x atmospheres (V x P) are units of energy in the form of work. PV Work: Work is done when a gas undergoes a volume change at constant pressure. Work = P*Delta V Pressure is force per unit area. Force = mass x acceleration = (kg) (m s-2) Pressure = force / area = (kg m s-2) / m2 = kg m-1 s-2 Pressure x Volume = (kg m-1 s-2) (m3) = kg m2 s-2 Work = Force x displacement = (kg m s-2) (m) = kg m2 s-2 Remember that kg m2 s-2 are units of energy (Joules). Compare to the formula for kinetic energy: Kinetic Energy = ½ m v2 Energy units are kg (m s-1)2 = kg m2 s-2= 1 Joule Standard Condition: Many physical properties of gases are listedin tables of data. In order to be able to compare values for different gases,the values are given for gases that are at the same temperature and pressure. Standard temperature for gases is 0oC (273.15 K) Standard pressure for gases is 1atm. Gas Densityand Molar Mass: n (# of moles) = (mass) / (molar mass) ‘P V = n R T P V = (m / M) R T This equation can be used to determine the molar mass (M) of a gas if the pressure, volume, P V = (m / M) R Tture are known. The equation can be rearranged to show the relationship between the density of a gas and the other variables. density= m = ( )( ) V R T ( )( ) To calculate Molar Mass: P M density= m = ( )( ) V ( )( ) (m)(R)(T) density R (T) M (V)(P) P Gas Mixtures and Partial Pressures: gas in the mixture behaves independently of all the others and obeys the ideal gas law.cules. Each For a mixture of ideal gases A, B, C,… in a container of volume V: PA = nA PB = nB R T / V; etc. components (A, B, C, …) in the mixture.s equal to the sum of all of the “partial pressures” of the Ptotal = PA + PB + PC + … Since PA = nA R T / V; PB = nB R T / V; etc. Ptotal = PA + PB + PC + … Ptotal = (nA + nB + nC + …) (R T / V) Ptotal = (ntotal) (R T / V) The mole fraction of each gas in the mixture is the number of moles of that gas divided by the total number of moles of all of the gases in the mixture. XA = nA / (nA + nB + nC + …) = nA / ntotal nA = (XA) (ntotal) The partial pressure of gas A is therefore: PA = XA Ptotal The Kinetic-Molecular Theory of Gases: kinetic energy = ½ m u2 where u is the speed of the molecule. average kinetic energy = ½ m ù2 where ù2 is the average of the squares of the speeds of the molecules. From experiment, it is known that: average kinetic energy = (3/2) R T When using this formula, make sure that R has units of energy (joules). R = 8.314472 J K-1mol-1
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