Chemistry 115 Week 2 Notes
Chemistry 115 Week 2 Notes Chem 115
Popular in Fundamentals of Chemistry 1
Popular in Chemistry
This 16 page Class Notes was uploaded by Danielle Gibson on Friday August 19, 2016. The Class Notes belongs to Chem 115 at West Virginia University taught by Erin Battin in Fall 2016. Since its upload, it has received 9 views. For similar materials see Fundamentals of Chemistry 1 in Chemistry at West Virginia University.
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Date Created: 08/19/16
Lecture Notes (Set 2) Chapter 3 Chemical Equations o Chemical equations are used to represent changes that occur when chemicals react to form new compounds. o General Setup of a Chemical Reaction Reactant + Reactant + … Product + Product (phase) (phase) (phase) (phase) 2K (s) + Cl (g) 2KCl (aq) 2 Any symbol over the arrow means something o Coefficient – Number written before each reactant or product o Subscript – Indicate the number of each atom within a compound o Every chemical equation must be balanced! Mass: # of reactant atoms = # of product atoms Charge: Net charge of the reaction must be equal Balancing Chemical Equations o Guidelines for Balancing Chemical Equations: 1. Generally, assign the most complicated compound a coefficient of 1 Compound with the most atoms 2. Balance the atoms on each side of the equation by changing the coefficients until # of reactant atoms = # of product atoms 3. Check that you have the lowest set of whole number coefficients No fractions…for now o Example: ____Al + ____ S ____ Al S 2 3 o Note: There is no setinstone way to balance a chemical reaction, you simply have to practice and mess with the coefficients o Examples: _____ C 3 8 _____ O 2 _____ CO + _2___ H O 2 _____ Ca(OH) + _____ HCl _____ CaCl + H O 2 2 2 _____ K C2O2 +4_____ Ca (As3 ) 4 2____ K AsO 3 ____4 CaC O 2 _____ FeC O2 •42H O +2_____O __2__ FeO + _____ H O + _____2O 2 o Example: Upon properly balancing the following chemical equation, determine the SUM of the coefficients for the REACTANTS. MnO (2) + KClO (aq3 +KOH(aq) KMnO (aq) + KC4(aq) +H O(l) 2 What to Do After Balancing Chemical Reactions o After balancing chemical reactions, you can then complete calculations using the balanced reactions o Four Common Conversion Factors Avogadro’s Number (N ) a Chemical Formula Molar Mass Stoichiometry o Note: For Calculations always determine the units given and wanted and the chemical compound given and wanted. o Note: Let the units guide you when choosing which conversion factor(s) to use! Avogadro’s Number o We use convenient units to describe things in everyday life… Miles for large distances & Feet for shorter distances Dollars for small amounts of money & Billions for larger amounts of money o We also use convenient chemical units to describe atoms, molecules, ions, formula units, photons, etc. Avogadro’s Number – 6.022 x 10 (Know this constant/number!) o Mole Amount of a substance o Conversion Factor #1: Avogadro’s Number & Moles 23 1 mole = 6.022 x 10 atoms, molecules, photons, formula unit, etc. o Example: How many iron atoms are present in sample of 3.0 mol iron? o Example: How many moles of calcium is present in 5.00 molecules of calcium? Chemical Formulas and Moles o We can use the compound to determine our conversion factor. o Conversion Factor #2: Chemical Formula & Moles H O 2 Na 2O 3 o Example: How many moles of carbon atoms, hydrogen atoms, and oxygen atoms are there in 2.5 moles of C6H 12?6 o Example: Calculate the number of atoms of sodium in 2.53 moles of Na CO 2 3 +2 o Example: Calculate the ions of Ca if you have 3.4 moles of calcium phosphate. Molar Mass o Sometimes we want to relate moles to mass (grams) because mass is our “working” unit. o Molecular Mass – Sum of mass numbers (A) of all atoms in the molecule amu units o Molar Mass – Sum of mass numbers (A) of all atoms in the molecule g/mol unit o Example: Calculate the molar mass of ammonium permanganate and FeC O • 2H O 2 4 2 o Conversion Factor #3: Mass (grams) & Moles 1 mole (mol) of compound = Molar Mass (g) of compound o Example: What is the mass (in g) of 3.01 moles of SO 2 o Example: Calculate the number of moles in 32 g of methane Stoichiometry o Sometimes we want to relate one compound to another compound o You must have a balanced chemical equation! _____ H (2) + _____ O (g)2 _____ H O (g) 2 o Conversion Factor #4: Moles of compound #1 & Moles of compound #2 _____ mole (mol) of compound #1 = _____ mole (mol) of compound #2 o Example: If 0.347 moles of sulfuric acid are reacted with an excess of aluminum oxide, how many moles of aluminum sulfate will be produced? _____ H SO + _____ Al O _____ H O + _____ Al (SO ) 2 4 2 3 2 2 4 3 o Example: How many moles of N are requi2ed to produce 2.89 moles of N O if you 2 3 have excess O ? 2 _____ N 2+ _____ O 2 _____ N O 2 3 Stoichiometry: Mass Problems o We ALWAYS combine conversion factors. o Steps for EVERY mass problem 1) Grams of compound #1 to moles of compound #1 2) Moles of compound #1 to moles of compound #2 3) Moles of compound #2 to grams of compound #2 o Example: What mass of aluminum oxide can be produced by reacting 12 g of aluminum with excess oxygen? _____ Al + _____ O 2_____ Al O 2 3(C o Example: In the combustion analysis of 0/1127 g of glucose (C H O )6 w12t6mass, in grams, of CO 2would be produced? The other product is dihydrogen monoxide. Stoichiometry: Limiting Reagent Problems o In every chemical reaction, not all of the reactants are used up when they form the products We see some of the reactants left over on the product side of the reaction… but we don’t write it. o How do you know if you need to complete a limiting reagent calculation? You will see amounts for each reactant o Limiting Reagent: A reactant that limits and controls the amount of product that is formed. Completely used up in the reaction o Excess Reagent: A reactant that does not limit the amount of product formed. Not all of the excess reactant is consumed/used up in the chemical reaction… you can determine the amount of excess reagent left over. o Example: You are trying to make as many cars as possible. You know that each car requires 1 motor and 4 tires. You have 12 tires and 2 motors. What are the limiting and excess reactants? o There are two methods to consider when solving limiting reagent problems: Method #1: Reactant Method Use this method when determining the amount of excess reagent left over. Method #2: Product Method This method is most commonly used Both methods will get you the same answer. o Method #1: Reactant Method Example: How many moles of iron are formed if 0.200 moles of Fe S are 2 3 reacted with 0.400 moles of hydrogen gas? _____ Fe S2 3 _____ H __2 Fe + _____ H S 2 o Method #2: Product Method Example: How many moles of iron are formed if 0.200 moles of Fe Fe S are 2 3 reacted with 0.400 moles of hydrogen gas? _____ Fe S2 3 _____ H __2 Fe + _____ H S 2 Stoichiometry: Limiting Reagent Mass Problems o When doing limiting reagent mass problems, you are doing the same thing but with a few more conversion factors! o Example: How much lead chloride can be produced from the reaction of 8.00 g of lead nitrate and 2.67 g of aluminum chloride? NH + O NO + H O 3 2 2 o Example: How many grams of NO are produced from the reaction of 30.0 g of ammonia with 40.00 g of oxygen gas? NH + O NO + H O 3 2 2 o Example: From the previous example, how many grams of ammonia are needed to completely react with 40.00 g of Oxygen gas? Stoichiometry: Limiting Reagent Mass Problems: o Aluminum reacts with oxygen to produce aluminum oxide which can be used as an absorbent, desiccant, or catalyst for organic reactions. 4Al(s) + 3O (2) 2Al O 2s)3 Balanced o A mixture of 82.49 g of aluminum (= 26.98 g/mol) and 117.65 g of oxygen (= 32.00 g/mol) is allowed to react. Identify the limiting reactant and determine the mass of the excess reactant present in the vessel when the reaction is complete. Percent Yield o Percent Yield: The amount of product actually formed in a reaction as compared to the amount that is theoretically possible. o Formula: % Yield = Actual Yield * 100 KNOW THIS FORMULA!! Theoretical Yield o Actual Yield: Amount of product obtained experimentally. o Theoretical Yield: Maximum amount of product that could theoretically be obtained if the reaction goes to completion This is essentially a mass problem o Example: What is the percent yield of a reaction when 32.8 g of C H O is obta5ne12 from a reaction of 26.3 g of C H 4wi8h excess CH OH? 3 C H + CH OH C H O 4 8 3 5 12 o Example: How many grams of CH Cl is produ2ed 2rom the reaction of 1.85 of methane if the percent yield is 43.1% o Example: A 5.95 g sample of AgNO is reacted3with excess BaCl according to 2he equation below, and 4.00 g of AgCl are experimentally produced. What is the percent yield of AgCl? (Molar Mass of AgNO = 169.39 g/mol; BaCl = 208.2 g/2ol; AgCl = 143.32 g/mol; Ba(NO ) =3 21.32 g/mol) AgNO +3BaCl A2Cl + Ba(NO ) (3 2ALANCED) Mass Percent o It is important to know how much of an element is present in a given amount of compound… we can use mass percent! o Mass % = (Atomic Mass of X(amu)) * (Atoms of X in formula) *100 Molecular mass of Total Compound (amu) o Mass % = (mols of X in formula) * (molar mass of X (g/mol)) *100 Mass of 1 mol of Total Compound (g) o This is a fancy way of saying: Mass % = PART * 100 KNOW THIS FORMULA!! TOTAL o Example: Calculate the mass percent of each element in C H O 6 12 6 Mass Percent… Sort of o You can determine the mass of an element in a sample from mass percent. o Mass of element = mass of compound * mass of element in 1 mol of cmpd Mass of 1 mol of cmpd o This is a fancy way of saying: Mass = SAMPLE * PART Know This Formula! TOTAL o Example: From the previous example, determine the mass of carbon in 34.87 g of glucose Empirical Formulas o Empirical Formulas: Chemical formulas that represent the smallest whole number ratios of atoms in a compound… You have seen this with ionic compounds. o Steps to Determine Empirical Formulas: 1) Want to go from grams to moles Do these steps for every 2) Divide by the smallest # of moles empirical formula problem! 3) These become your subscripts They must be whole numbers… if not multiply to obtain the nearest whole number o Example: A sample of a compound containing aluminum and sulfur is found to contain 35.6 g of aluminum and 63.4 g of sulfur. What is its empirical formula? o Example: Ascorbic Acid contains 40.92% carbon, 4.58% hydrogen and 54.50% oxygen by mass. What is its empirical formula? o Example: A 25.0 g sample of an unknown compound containing phosphorus and oxygen was decomposed and found to contain 10.9 g of phosphorus. What is the empirical formula of the compound? o Determine the empirical formulas from the given molecular formulas. Which compounds have the same empirical formulas? Molecular Formula: I) C12 O18 II)6C 9 O6 III8 C12 6 IV)8C 12O8 Molecular Formula o You can obtain the molecular formula from the empirical formula Step 1: n = Mass Mass of Cmpd Know these steps when solving Empirical Mass of Cmpd for a molecular formula Step 2: C x *y = C & nx ny Example: A compound has a molar mass of 282.54 g/mol and an empirical formula of C H10h21. s the molecular formula of this compound? Liquid Phase Chemical Reactions in Solution o Most chemical reactions take place in solution… so now we are dealing with volumes and concentrations! o Most reactions need mobile ions to form products o Example: Pb(NO )3 2s) + Kl (s) Pb(NO )3 2(aq) + Kl (aq) o Solution: Homogenous mixture in which 2 or more compounds freely mix Solute: The substance that is dissolved in a solvent to produce a solution The substance that is in the smaller amount Solvent: The dissolving medium in which the other components are dissolved in resulting in a solution The substance in the larger amount Solutions: Concentration Calculations o Concentration: Used to express the amount of a substance present per unit volume of solvent Usually expressed as Molarity (M or mol/L) Mol/L are the only acceptable units for Molarity o Formula: [Concentration (M)] = Amount of solute (mol) KNOW THIS Volume (L) FORMULA!!!! o Example: 0.12 M KCl = 0.12 mol KCl 1 L o Example: What is the concentration of a solution made by dissolving 0.066 moles of potassium chloride in 0.022 L of water? Solutions: Molarity Calculations o Example: How many moles of hydrochloric acid are contained in 175 mL of a 55 M hydrochloric acid solution? o Example: What is the molarity of a solution prepared by dissolving 0.440 g KSCN in enough water to make 340.0 mL of solution? o Example: What volume of water is needed to produce a 0.75 M sodium chloride solution if you use 15 g of sodium chloride? o How many grams of NaOH are required to prepare 240. mL of a 3.50 M NaOH (MW = 40.0 g/mol) solution? Solutions: Dilution Calculations o When diluting something, you are NOT making new compounds but just decreasing the concentration. o Formula: M V =1M 1 2 2 KNOW THIS FORMULA!!! o Note: Only use this formula when dealing with the same compound!!! o Example: How would you prepare 250.0 mL of a 0.509 M sulfuric acid solution if the stock solution is 18.0 M? Can you use M1V1=M2V2 for this problem? o What volume of a 0.300 M Ba(OH) is requ2red to neutralize 48.0 mL of a .200 M HClO 3olution? Ba(OH) 2 (aq)2HClO 3 (aq) H 2 +(l) ClO ) 3 2 (aq) Solution Calculations o Just like with mass problems, we had conversion factors that we can use for problems dealing with concentration and volumes! o For solutions calculations, we are going to progress as follows: 1 concentration 1 concentration & 1 volume 2 concentrations & 1 volume (Solution Stoichiometry) 1 concentration & 2 volumes (Titrations) o Note: Most solution calculation problems start the same way! o 1 Concentration Example: What is the concentration of calcium ions in 0.500 M calcium chloride solution? What is the concentration of chlorine ions? +2 CaCl 2 Ca + 2Cl Example: A magnesium chloride solution was found to have a 0.275 M chlorine ion concentration. What is the concentration of the magnesium chloride solution? o 1 Concentration and 1 Volume Example: How many moles of each ion are present in 200.0 mL of 5.0 M potassium phosphate? Example: How many moles of each ion are present in 25 mL of 0.150 M magnesium chloride? Example: What mass of potassium ion is contained in 38.2 mL of 3.3 M potassium sulfate? Solution Stoichiometry Calculations o 2 Calculations and 1 volume Cannot use M V1 =1M V 2 2 Because we are doing stoichiometry, we have to use our balanced chemical reaction as one our conversion factor o Steps for Solution Stoichiometry Calculations: Solving for an Unknown Volume 1) Convert milliliters to liters (if necessary) 2) Multiply by the appropriate concentration 3) Convert moles of Compound # 1 to moles of Compound # 2 (i.e. Conversion Factors) 4) Divide by the appropriate concentration o Note: Know these steps when solving solution stoichiometry calculations! o Example: How many milliliters of 0.125 M NaHCO solution 3re needed to neutralize 18.0 mL of 0.100 M HCl? HCl + NaHCO N3Cl + H O + CO2 2 • Example: What is the volume of 0.250 M sulfuric acid needed to react with 50.0 mL of 0.100 M sodium hydroxide? H 2O 4+ NaOH Na SO 2 H 4 2 Solutions: Titration Calculations o Titration: An experimental technique used to determine the concentration of a solution by allowing a carefully measured volume with a known concentration to react with a solution whose concentration is unknown Usually have to use an indicator to show the color change that occurs upon neutralization Endpoint: Point in the titration where the indicator changes color Equivalence Point: Where stoichiometric amounts of both reagents are present o Neutralization Reaction: When an acid and base react to form a salt and water o Example: HCl (aq) + NaOH (aq) NaCl (aq) + H O (l) 2 o 1 Concentration & 2 volumes Steps for Solution Stoichiometry Calculations: Solving for an Unknown Concentration 1) Convert milliliters to liters (if necessary) 2) Multiply by the appropriate concentration 3) Convert moles of Compound # 1 to moles of Compound # 2 (i.e. Conversion Factors) 4) Divide by the appropriate volume (in liters) Example: It took 48.6 mL of 0.100 M sodium hydroxide solution to react with 20.0 mL of unknown hydrochloric acid. What is the concentration of the acid? Example: In an acidbase neutralization reaction, 32.89 mL of 0.800 M KOH reacts with 150.0 mL of H SO2. 4hat is the concentration of the acid (H 2O )4 KOH + H SO K SO + H O 2 4 2 4 2 Solution Calculation o Aqueous potassium iodate (KIO ) and 3otassium iodide (KI) react in the presence of dilute hydrochloric acid, as shown below. o KIO (aq) + 5KI(aq) + 6HCl(aq) 3I (aq) + 6KCl(aq) + 3H O(l) 3 2 2 o What mass of iodine (I ) i2 formed when 50.0 mL of 0.020 M KIO solution 3eacts with an excess of KI and HCl? Another Aspect of Liquid Phase Solubility Classifications o With every chemical equation, phases should be included….but how do we determine those? o There are rules for determining solid (s) and aqueous (aq) phases o You cannot use these rules to determine liquid (l) or gaseous (g) phases….but you are responsible for knowing the common ones! o Soluble: Dissolving significantly in a solvent Solubility > 0.1 M Ionic compounds that are soluble dissociate into their individual ions Aqueous phase (aq) Ions move around independently, but have some water molecules loosely attached The ions are evenly distributed within the solution Example: NaCl o Insoluble – Does not dissolve significantly in a solvent Solubility < 0.1 M Ionic compounds that are insoluble do not dissociate into their individual ions Solid phase (s) Example: CaCO3 o Saturated Solution – A solution where no more solute can be dissolved at the current temperature Not an insoluble compound o KNOW ALL SOLUBILITY RULES!! o Example: : Predict the solubility of the following compounds in water KCl Na 2O 3 Fe(OH) 2 NH N4 3 PbSO 4 o Choose the compound(s) that is/are SOLUBLE in water: I. BaB2 II. 2g S4 III. FeCl3 Predicting the Products of a Chemical Reaction o Sometimes, as chemist, we have to determine the products that will form in a chemical reaction AND the phases o Steps for Predicting Products: 1) Separate all compounds into ions with the charges Don’t take the coefficients Don’t take subscripts for monatomic ions/Do take for polyatomic ions 2) Crisscross partners 3) Balance the compound using subscripts Remember to use the lowest ratio of coefficients & subscripts Compounds must be balanced by charge 4) Determine the phase of the product using the solubility rules o KNOW THESE STEPS WHEN PREDICTING PRODUCTS o Example: Predict the products for the following reactions, determine the product phases and balance the reactions K 2O 4aq) + CaCl 2aq) Cu(NO ) 3 2 K C2 3 K 3O 4+ Ba(ClO ) 4 2 BaCl 2 + AgNO 3 ZnSO 4 Types of Equations o Molecular Equation – Demonstrates the full chemical formula of all species present in the reaction o Full Ionic Equation – Demonstrates the true state of matter for all reactants and products in a chemical reaction How they exist in solution Soluble ionic compounds dissociate Strong acids and bases fully dissociate Insoluble ionic compounds and covalent compounds do not dissociate Weak acids and bases do not dissociate completely o Net Ionic Equation – Demonstrates only chemical species that are involved in the actual reaction in their true state of matter Reactive Ions: Ions involved in the reaction Change in physical state Spectator Ions: Ions present in the whole chemical reaction, but do not take part in the reaction itself No change in physical state Steps for Net Ionic Equations o 1) Determine the phases for all reactants and products Use solubility rules o 2) Dissociate all aqueous compounds to ions Weak acids and bases do not dissociate Diatoms do not dissociate Be sure to include the correct charges for all ions o 3) Cancel all spectator ions Where the species did not change phase from reactants to products o 4) Rewrite the remaining equation leaving behind only the reactive ions/compounds All ions must have a charge o 5) Balance the equation o Do not try to do this at the beginning Net Ionic Equations o Example: Write the balanced net ionic equation for the following Na P3 + 4 O A3 PO + 3aNO 4 3 HCl + MnS H S (g2 + MnCl 2 Nal + Ca(NO ) 3 2al + 2aNO 3 Nal + Cl 2 NaCl + I 2 o Identify the spectator ions when aqueous solutions of sodium carbonate and magnesium chloride are combined. Acids and Bases o We have already discussed acids and bases a bit, mostly with naming and solubility o Acids: H or H O (aq)+1 3 Strong acids completely dissociate…. no molecular form + Example: HCl H + Cl Weak acids do not completely dissociate…some molecular form o Bases: OH (aq) Strong bases completely dissociate…no molecular form + Example: NaOH Na + OH Weak bases do not completely dissociate…some molecular form Example: NH 3H + 4H o Make sure you know ALL of the acids and bases from the handout!!! Net Ionic Equations with Acids and Bases o Example: Write the balanced net ionic equation for the following acid and base reactions: HCl + NaOH HF + NaOH NaF + H O 2 o Write the baanced net ionic equation for a reaction between acetic acid (C H O H) 2 3 2 and potassium hydroxide. BronsteadLowry Acids and Bases o There is more than one definition for an acid and a base! Based on whether a proton (H+) was gained or lost o Acids lose a proton o Bases gain a proton o Example: Identify which reactant is the acid and the base HCN + OH CN + H O 2 + NH 3 H O 2 OH + NH 4 H2CO 3+ CO 3+2 2HCO 3 Electrolytes o MAKE SURE YOU KNOW THE DEFINITION OF AN ELECTROLYTE & CAN IDENTIFY THE VARIOUS TYPES!! o Electrolyte: A chemical compound that ionizes when dissolved to produce an electrically conductive medium o Strong Electrolyte: Conducts strong electrical current in aqueous solution High concentration of mobile ions Complete dissociation of compounds Strong acids and bases, and soluble ionic compounds o Weak Electrolyte: Conducts weak electrical current in aqueous solution Low concentration of mobile ions Do not have complete dissociation of compounds Weak acids and bases o Nonelectrolyte – Conducts no electrical current in aqueous solution No mobile ions o Example: Determine whether the following are strong, weak, or nonelectrolytes. What substances will cause a light bulb to light? DI (l) HCl (aq) NH3 (aq) NaCl (aq) NaOH (aq) CH3COOH (aq) NaCl (s) o Which one of the following substances, when dissolved in water at equal molar concentrations, will give the solution with the LOWEST electrical conductivity? CaCl 2 Nitric acid Ammonia C6H 12 6glucose) Potassium oxide
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