Chapter 4 General Chemistry
Chapter 4 General Chemistry Chem 109
Popular in General Chemistry
Popular in Chemistry
This 11 page Class Notes was uploaded by mkennedy24 on Thursday February 11, 2016. The Class Notes belongs to Chem 109 at University of Nebraska Lincoln taught by Eric Malina in Spring 2016. Since its upload, it has received 98 views. For similar materials see General Chemistry in Chemistry at University of Nebraska Lincoln.
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Date Created: 02/11/16
I. Chapter 4: Chemical Quantities and Aqueous Reactions a. Section 4.1: Climate Changing (Not required for exam, just informational) b. Section 4.2: Reactions Stoichiometry i. Stoichiometry: The numerical relationship between chemical amounts in a balanced equation ii. Objective: Calculate mass relationships using balanced chemical equations 1. Example: TiCl (s4 + O (g2 TiO + 22l (g) 2ow many grams of TiO can2be produced from 100.0g of TiCl with excess O ? 4 2 1molTiCl4 a. 100.0g TiCl x4 189.69gTiCl4 x 1 molTiO2 x 79.88gTiO2 = 42.11g TiO 1 molTiCl4 1 molTiO2 2 c. Section 4.3: Limiting Reactant, Theoretical and Percent Yield i. Limiting Reagent/ Reactant: What is going to be used up first; What limits the reaction from continuing ii. Example: TiCl (4) + O (2) TiO +22Cl (g)2 50g of TiCl4and 10g of O ,2which one is the limiting reactant 1molTiCl4 1molO2 1. 50.00g TiCl 4 189.69gTiCl4 x 1molTiCl4 x 32.00 gO2 =8.435g O 1 molO2 2 2. Conclusion: TiCl 4s the limiting reagent because we have more O than 2e needed therefore making TiCl th4 limiting reagent iii. Objective: Determine how much of a reactant is left over 1. Using previous example: Take what you have which is 10.00g O , 2hat you get 8.435g O 2 and find the difference: 10.00g O – 2.435g O 2 = 1.56g O l2ft over iv. Objective: Determine a theoretical and percent yield 1. Theoretical Yield: Max amount of product that can be made in a chemical reaction based on the amount of limiting reactant ActualYield 2. Percent yield: ( )x 100= % TheoreticalYield yield 3. Example: If 50.00g TiCl a4d 10g O actu2lly produced 34.26 Cl , 2hen what is the percent yield? a. Find two pieces: actual and theoretical yield b. ALWAYS START WITH LIMITING REACTANT 1molTiCl4 c. 50.00g TiCl 4 x 189.69gTiCl4 2 molCl 2 70.90gCl 2 1 molTiCl4 x 1molCl 2 =37.38g Cl 2 (Theoretical Yield) ActualYield 34.26gCl2 d. ( ) x 100 TheoreticalYield 37.38gCl2 = 91.65% 4. Example: 1.00g H is 2llowed to react with 9.79g N ,2producing 1.60g NH 3 a. 3H 2g) + N (2) 2NH (g3 b. What is the theoretical yield for this reaction under the given conditions? 1 mol H 2 i. Moles H =21.00g H x 2 2.016g H 2 =0.496 mol H x 23)= 1.488 mol H 2 (**We multiplied by 3 because in the balanced reaction above it takes 3 moles of H to2react with 1 mole N *2. 1molN 2 ii. Moles N 2 9.79g N x 2 28.02gN 2 = .3493 moles N (**2e do not need to multiply the amount of N 2 moles by anything since there is only 1 mol of N r2acting so basically multiplying the amount of moles by 1 is redundant**) iii. The above conversion factors conclude that the limiting reactant is H2since we need 1.488 mol to react fully and are only given .496, therefore H w2uld be used up first iv. Moles NH = 3496 mol H x 2 2mol NH 3 3mol H 2 = .331 mol NH x 3 17.0307gNH3 1mol NH3 = 5.63g NH 3his is the theoretical yield c. What is the percent yield for this reaction under the conditions given? i. Actual yield: 1.60g NH 3 ii. Theoretical Yeild: 5.63g NH 3 Actual yield 1.60 iii. x100= Theoretical yield 5.63 28.4% d. Section 4.4: Solution Concentration and Solution Stoichiometry i. Solution: Homogeneous mixture ii. Solvent: Major component of solution iii. Solute: Minor component of solution iv. Aqueous solution: Water is solvent in the process v. Concentration: Ratio of solute to solvent or solute to total amount (mass, volume, moles, etc.) 1. Concentrated: High ratio (large amount of salt added to water) + a. Example: What is the concentration of K in .15M of K2S? i. Because there is twice as much K + ions in K2S than K you just multiply the .15M x 2 = 0.3M b. If CaCl 2s dissolved in water, what can be 2+ said about the concentration of the Ca ion? i. It has the same concentration as the Cl- ii. I-s concentration is half that of the Cl iii. Its concentration is twice that of the Cl- iv. Its concentration is one-third that of the Cl- 1. It is (ii) because there are 2 2+ Cl- ions for every Ca ion c. How many grams of Na PO wi3l b4 needed to produce 675mL of a solution that has a concentration of Na ions of 0.700M? .700mol Na+ ¿ 3mol Na+¿ i. 1L x 1 mol Na3PO4 x. ¿ ¿ 675L x 163.94g = 25.8g 1molNa3PO4 Na 3O 4 2. Dilute: Low ratio (small amount of salt added to water) a. Dilution: Lowering the concentration by adding solvent b. Stock solution: supply solution of highly concentrated solution c. Objective: Calculate concentrations after dilution or calculate the volume of stock to make specific concentration d. Example: What is the concentration of HCl after 10.00mL of an 11.65M HCl solution is diluted to 100.0 mL total volume? mol i. MxV L xL M1V 1M V2 2 ii. **Match units correctly and everything should cancel correctly! ** iii. 11.65M x 10.00mL = M x 200.0mL (Do algebra and divide the 100.0mL over to other side of the equal sign) iv. M 2 1.165 particles vi. Molarity: (M) Molessolute Lsolution(¿total volume) vii. Objective: Calculate molarity from amount of solute and volume of solution (or vice versa) viii. Example: What is the molarity of solution if 15.00g NaCl dissolved in enough water to make 500mL of solution? 1molNaCl 1. 15.00g NaCl x 58.44gNaCl = .2567 mol NaCl 1 L 2. 500mL x = .5000L 1000 mL .2567mol NaCl 3. Solution: = .5133M .5000L ix. Example: How many grams of AgNO in 100300mL of .125M AgNO so3ution? 1 L .125mol AgNO 3 1. 100.0mL x x x 1000 mL 1 L solution 169.88gAgNO3 1mol AgNO3 = 2.12g AgNO 3 x. Example: How many grams of MgCl would2be produced if .080g Mg reacted with 10.00mL of 1M HCl solution? 1. Using this reaction: Mg(s) + 2HCl(aq) MgCl 2aq) + H (2) 2. First thing is to determine which one is the limiting reagent 3. Limiting reagent: 1.00molHCl a. x .0100L solution x 1Lsolution 1mol Mg 24.31g Mg 2mol HCl x 1mol Mg = .122g Mg b. Mg is the limiting reagent 1mol Mg 1molMgCl2 c. .080g Mg x x x 24.31gMg 1mol Mg 95.21g MgCl 2 1mol MgCl 2 = .31g MgCl 2 xi. Example: A 25.0mL sample of a 1.04M potassium sulfate solution is mixed with 15.0mL of a 0.880M barium nitrate solution and this precipitation reaction occurs: K 2O (4q)+Ba(NO ) (a3 2 BaSO (s)+ 4 KNO (3q); The solid BaSO is 4ollected and found to have a mass of 2.53g. 1. Determine the limiting reactant 1L a. Moles K 2O =425.0mL x 1000mL x 1.04mol = .026 mol K SO x 1Lsol n 2 4 174.262 gK 2SO 4 = 4.53g K 2O 4 1molK 2SO 4 1 L b. Moles Ba(NO ) 3 25.0mL x x 1000 mL 0.880mol 261.35g ' x =3.4497g 1Lsol n 1molBa N(3 2 ) Ba(NO )3 2 1 mol K 2SO 4 c. 4.53g K 2O x4 x 174.262 g 1molBa (NO3 )2 261.35g x = 1molK 2SO4 1molBa N(3 2 ) 6.79g Ba(NO ) 3 2 d. So Ba(NO )3 2 the limiting reactant because the grams we get in the reaction are larger than what we actually are given so it will be used up first in the reaction 2. Determine theoretical yield a. 3.4497g Ba(NO ) x 1mol x 3 2 261.35g 1mol BaSO 4 x 233.394 g = 1 mol B( NO3)2 1 mol BaSO4 3.0807g BaSO This is the theoretical 4 yield 3. Determine the percent yield Actual yield a. Theoretical yield 2.53BaSO4 produced x100 3.0807gBaSO4theoreticallyproduced =82.1% e. Sect Table 4.2 Some Common Acids and Bases Name of Acid Formula Name of Base ion Formula 4.5: Hydrochloric HCl Sodium NaOH Acid Hydroxide Hydrobromic HBr Lithium LiOH Acid Hydroxide Hydroiodic Acid HI Potassium KOH Hydroxide Nitric Acid HNO 3 Calcium Ca(OH Hydroxide 2 Sulfuric Acid H2SO Barium Ba(OH 4 Hydroxide 2 Perchloric Acid HClO Ammonia NH 3 4 Acetic Acid HC2H 3 2 Hydrofluoric HF Acid Types of Aqueous Solutions and Solubility i. Electrolyte: Conducts electricity ions are in the solution 1. Strong Electrolytes: NaCl ionic compound: Na & Cl floating around in the solution (strong electrolytes fully ionize in solution a.k.a make a lot of ions which makes a lot of conduction 2. Non-electrolytes: Water (H 2), Sugar (C 6 O12 6 molecules so no ionization when they dissolve into solution 3. Weak Electrolytes: Give partial ionization when dissolved into solution (HC H O ↔ H ++ 2 3 2 C 2 3 )2i.e. acetic compound ii. Objective: Describe or identify the three types of electrolytes iii. How do you know the difference between strong, weak, and non-? 1. Soluble: if it will dissolve(salts, strong acids, strong bases) 2. Molecules are non-electrolytes 3. Weak acid and base weak electrolytes 4. **Red Acids and Bases are weak** 5. Solubility a. Soluble: it WILL dissolve b. Insoluble: it will NOT dissolve c. Objective: Describe the behavior of these two possibilities Table 4.1 Solubility Rules for Ionic Compounds in Water Compounds Containing the Exceptions Following Ions are Generally Soluble + + + + Li , Na , K , and NH 4 None NO 3 and C H2O 3 2 None - - - + 2+ Cl , Br , and I When these ions pair with Ag , Hg 2 , or Pb , the resulting compounds are insoluble SO 42- When SO 42-pairs with Sr , Ba , Pb , 2+ + 2+ Ag , or Ca , the resulting compound is insoluble Compounds Containing the Exceptions Following Ions are Generally Insoluble OH and S 2- When these ions pair with Li , Na , K , + + or NH 4 the resulting compounds are soluble When S pairs with Ca , Sr , or Ba , 2+ the resulting compound is soluble When OH pairs with Ca , Sr , Ba2+ 2+ the resulting compound is slightly soluble CO 3- and PO 43- When these ions pair with Li , Na , K , + + or NH 4 the resulting compounds are soluble i. Example: AgCl insoluble salt ii. Example: AgNO s3luble salt d. Objective: Can you predict solubility Compounds? i. Solubility rules need to be memorized! ii. Table 4.1 below f. Section 4.6: Precipitation Reactions i. Precipitation: Formation of a solid out of a solution ii. Objective: Write balances precipitation reactions and predict products iii. Example: KI(aq) + LiC H2O 3aq2 NO REACTION because no precipitate was formed. g. Section 4.7: Representing Reactions i. Molecular Equations: reactants and products represents as molecular/formula units; An equation showing the complete neutral formulas for each compound in the reaction if they existed as molecules 1. Example: MgCl (aq) + K CO (aq) MgCO (s) + 2 2 3 3 2KCl(aq) 2. Complete (Total) Ionic Equations: represent all strong electrolytes as ions in solution a. Example: Change this molecular equation to total ionic: AgNO (aq) + KCl(aq) 3 AgCl(s) + KNO 3 i. Total ionic: Ag (aq) + NO (a3) + K + - + (aq) + Cl (aq) AgCl(s) + K (aq) + NO 3aq) 3. Net Ionic Equations: No spectator ions i.e. ions that did not change in the reaction a. Example: Change above total ionic resulting in AgCl(s) equation into the net ionic equation i. Net Ionic: Ag (aq) + Cl (aq) AgCl(s) ii. Because the K and NO ions3had no change in the reaction they are known as spectator ions and are actually unnecessary in the net ionic equation b. Example: Mg (aq) + 2Cl (aq) + 2K (aq) + 2- + - + CO 3 (aq) MgCO (3) + 2K (aq) + 2Cl (aq) c. Since the Cl and K ions did not change they are spectator ions d. Net ionic equation would look like this: 2+ 2- 2- Mg (aq) + CO 3 (aq) MgCO 3 (s) h. Section 4.8: Acid-Base Reactions and Gas Evolution Arrhenius Reactions + Definitio i. Acid: Substance that produces H ions when ns dissolved in water ii. Base: Substance that produces OH ions when dissolved in water + - 1. Example: HCl H + Cl a. Because the hydrogen is a bare proton and bare protons associate with water molecules, the hydrogen becomes a hydronium ion + + b. H (aq) + H O2l) H O3(aq) 2. Polyprotic Acid: more than 1 ionizable proton and release them sequentially 3. H 2O i4 a Diprotic acid a. Strong in its first ionizable proton but weak in its second b. H 2O (4q) H (aq) + HSO (aq4- - + 2- c. HSO (4q) ↔ H (aq) + SO 4 (aq) iii. Gas Evolution Reaction: a gas forms resulting in bubbling (Many gas evolution are acid-base) iv. Acid-Base Reaction: Neutralization reaction 1. Example: HCl(aq) + NaOH(aq) NaCl(aq) + H O(l) 2 a. NaOH Base b. NaOH(aq) Na (aq) + OH (aq)- 2. Titration: Analytical technique to use a substance of a known concentration to determine the amount of concentration of another substance 3. Equivalence Point: When we have the stoichiometric ratio present 4. Indicator: Signals equivalence point (color change) a. Example: H S2 (a4) + Ba(OH) (aq) 2 BaSO (4) + H O2l) b. What is the concentration of HNO if 3 25.00mL of HNO sol3tion requires 32.65mL of .2421M NaOH solution to reach the equivalence point? i. M mol mol M mol ii. MxV =M Volume .2421M NaOH iii. x .03265 L x 1L 1 mol HNO3 1 x = .3162M 1mol NaOH .02500 L iv. Use this equation for above: HNO 3 + NaOH NaNO + H3O 2 i. Section 4.9: Oxidation-Reduction Reactions (Redox reactions) i. What differs redox reactions from Acid-Base reaction is that there are transfers of electrons (e )- ii. Objectives: Identify: 1. Redox reactions 2. Oxidizing agent 3. Reducing agent 4. ** All above comes through assigning oxidation states** iii. Oxidation: Loss of electrons (e ) - iv. Reduction: Gain of electrons (e ) 1. O.I.L: Oxidation Is Losing 2. R.I.G: Reduction Is Gaining v. Oxidizing Agent: Reactant causing something to undergo oxidation (same as finding what is reduced) vi. Reducing Agent: Reactant causing something to undergo reduction (same as finding what is oxidized) 1. A e- B therefore A B - 2. The oxidizing agent is B because it reduced from no charge to -1 3. The reducing agent is A because it oxidized meaning it loss electrons forming a +1-charge vii. Oxidation State: Accounting system of e viii. Rules for oxidation states Hierarchical meaning rule 1 has precedence over latter rules 1. Rule 1: Elements has oxidation state equal to zero 2. Rule 2: Monatomic ion oxidation state will equal the charge (Na (aq) oxidation state= +1) 3. Rule 3: Sum of oxidation states has to equal overall charge (CO 32- (1 Carbon)(oxidation state of carbon) + (3 Oxygen)(oxidation state of oxygen)= -2) 4. Rule 4: Metals only metals we can really predict oxidation states for is Group 1A and 2A 5. Rule 5: Nonmetals Oxidation states in Group 6,7,8A are -1,-2,-3 respectively **except when it cant be because hierarchical rules** (H+ is an exception for when it is paired with a metal NaH = Na+ and H- so oxidation state can equal overall charge of zero) ix. Examples: 1. 4Fe(s) + 3O (g2 2Fe O 2s)3 a. Rule 1: elements Fe=0 and O =0 2 b. Rule 3: Compound Fe O =2(23Iron)(x) + (3 Oxygen)(-2 oxidation state for Oxygen) (3)(-2)=-6 so x must equal 3 so that both the +6 and the -6 can cancel out to come to the overall charge of the compound of 0 c. Fe undergoes oxidation (Goes from 0 → +6 **Loses electrons**) d. O 2 undergoes reduction (Goes from 0 → -6 **Gains electrons**) e. Oxidizing agent: O (if it underwent 2 reduction it caused oxidation) f. Reducing agent: Fe (if it underwent oxidation it caused reduction) 2. 2C 2 6g) + 7O (2) 4CO (g2 + 6H O(g2 a. Rule 1: elements O =02 b. Rule 3: H 2 (2 Hydrogen)(+1 oxidation state for H) + (-2 oxidation state for O)(1 Oxygen) =0; C H2 62 Carbon)(-3 oxidation state for C) +(6 H)(+1 oxidation state for H)=0; CO 2(1 C)(+4 oxidation state for C) +(2 O)(-2 oxidation state for O)= 0 c. **When determining oxidation states, ignore coefficients: It just creates more work** d. Oxidation: C (Goes from -6 → +4 **Lost electrons**) e. Reduction: O (Goes from 0 → -4 **Gained electrons**) f. Oxidizing Agent: O 2 g. Reducing Agent: C H 2ag6nts are almost always substances ) h. This example is a combustion reaction i. Substance with Carbon and Hydrogen reacting with O an2 producing CO H O 2 2 ii. Combustion reactions are redox reactions
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