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CHAPTER 4: CHEMICAL QUANTITIES and AQUEOUS REACTIONS REACTION STOICHIOMETRY: Amounts of Reactants & Products Chemical equation represents the number of molecules involved in a reaction, not the mass of the molecule. However, the mass reactants should be equal to the mass of products. Thus, if you know the mass or reactants, you can calculate the mass of products. Mass to Mass Conversion: Steps involved in stoichiometric calculation 4. Write the equation 5. Balance the chemical equation Mole to Mole Conversion: 6. Find out the moles of desired reactants and products (using the ratios derived from the balanced equation) Steps involved in stoichiometric calculation 7. Convert moles to grams 1. Write the equation 2. Balance the chemical equation 3. Find out the moles of desired reactants or products (using the ratios derived from the balanced equation) Example: How many moles of ammonia are produced when 0.6 moles of N2 react with H2? 1. Write the equation N2 + H2 → NH3 2. Balance the chemical equation N2 + 3H2 → 2NH3 3. From the balanced equation, find out the moles of desired reactants and products. 1 moles of N2 gives 2 moles of NH3 4. Find the number of moles of Product formed. 0.6 moles N2 x 2 mole NH3 1 mole N2 8. Calculate the amount of desired reactant or product Example 209 g of methanol (CH3OH) burns in air to give CO2 and H2O. What is the mass of water? 5. Write the equation CH3OH + O2 → CO2 + H2O 6. Balance the chemical equation 2CH3OH + 3O2 → 2CO2 + 4H2O 7. From the balanced equation, find out the moles of desired reactants and products. 2 moles of CH3OH gives 4 moles of H2O. 8. Convert moles to grams. Molar mass of CH3OH = (1 x Mass of C) + (4 x Mass of H) + . (1xMass of O) in g = (1 x 12.01) + (4 x 1.008) + (1 x 16) g = 32 g Molar mass of H2O = (2 x Mass of H) + (1xMass of O) in g = 1.2 moles of NH3 produce = (2 x 1.008) + (1 x 16) g = 18 g General Chemistry – Handout (Page: 1) 2 moles of CH3OH produce 4 moles of H2O. Therefore, Solving limiting reactant problems 1. Write the equation. 2 x 32 g of CH3OH produce 4 x 18 g of H2O 2. Balance the chemical equation. 64 g of CH3OH produce 72 g of H2O 3. Convert grams of reactants to moles 9. Calculate the amount of desired reactant or product. 64 g of CH3OH give 72 g of H2O Amount of H2O produced from 209 g of CH3OH = 209 g CH3OH x 72 g of H2O 64 g CH3OH = 235.12 g of H2O 235.12 g of H2O is produced from 209 g of methanol. LIMITING REACTANT 4. Find out the moles of product by each reactant. The reactant giving the least no. of product will be limiting reactant. 5. Calculate the mass of the product from the Limiting reactant. That will be the theoretical yield. Example: Methanol (CH3OH) is produced form 1.00 g H2 and 6.44 g CO. Which is the limiting reactant? What is the Limiting reactant and theoretical yield? Reaction: H2 + CO → CH3OH Balanced equation: 2H2 + CO → CH3OH Convert grams to moles: The limiting reactant is the reactant that is consumed first, limiting the amounts of products formed. Molar mass of H2 = 2 x 1.008 g = 2 g/mole Moles of H2 = 1.00 g x (1 mol /2 g) = 0.5 mol Molar mass of CO = 12.01 g + 16 g = 28 g Moles of CO = 6.44 g x (1 mol /28 g) = 0.23 mol Limiting reactant: Moles of CH3OH from H2: 0.5 mol. H2 x 1 mol CH3OH 2 mol H2 = 0.25 mol CH3OH Moles of CH3OH from CO: 6 moles of H + 12 moles of Cl → 6 moles of HCl Cl – Limiting reactant 0.23 mol. CO x 1 mol CH3OH = 0.23 mol CH3OH 1 mol CO As CO is giving less amount (i.e. 0.23mol) of CH3OH therefore, it is the limiting reactant and H2 is in Excess. General Chemistry – Handout (Page: 2) Theoretical Yield (mass of product): Molar mass of CH3OH: 12 1 16 = (1 x Mass of C) + (4 x Mass of H) + (1 x Mass of O) in grams = (1 x 12) + (4 x 1) + (1 x 16) = 32 grams/mol. Molar mass of aspirin (C9H8O4) = 12 1 16 (9 x Mass of C) + (8 x Mass of H) + (4 x Mass of O) in grams = (9 x 12) + (8 x 1) + (4 x 16) = 180 grams 3. Calculate the amount of desired reactant or product Mass of CH3OH: 0.23 mol. CH3OH x 32 g CH3OH 1 mol CH3OH = 7.36 g CH3OH 180 g of aspirin produced from 138 g of salicylic acid (limiting reactant) Amount of aspirin formed from 14.4 grams of salicylic acid (S.A) = 14.4 g S.A x (180 g aspirin / 138 g S.A) = 18.78 g aspirin PERCENT YIELD Theoretical Yield and actual yield is required to calculate percent yield. Theoretical Yield is the amount of product that would result if the entire limiting reagent reacted. Its amount is calculated using the balanced equation. 4. Calculate the percent yield Theoretical yield = 18.78 g Actual yield = 6.26 g Percent yield = Actual Yield is the amount of product actually obtained from a reaction. It is usually given. Percent yield = Actual Yield Theoretical Yield = 33.33% SOLUTION 14.4 g of salicylic acid (limiting reactant) gives actual yield of 6.26 g of aspirin. Calculate the percent yield of aspirin using the following equation. Salicylic acid + C4H6O3 (l) Acetic anhydride → C9H8O4 (s) Aspirin x 100% = (6.26/18.78) x 100% x 100% Example C7H6O3 (s) Actual Yield Theoretical Yield + CH3CO2H (l) A homogeneous mixture of two or more substances, which may be solids, liquids, gases. Solvent: Used to dissolve (mix homogenously) solid, liquid or gas. So it is called dissolving medium. Mostly liquids are used as solvent. - In general, solvent is present in greater amount. Acetic acid Solute: Solids, liquids or gases that are dissolved in the solvent. 1. 1 mole of salicylic acid (limiting reactant) produce 1 mole of aspirin 2. Convert moles to grams. Molar mass of salicylic acid (C7H6O3) = (7 x Mass of 12C) + (6 x Mass of 1H) + (3 x Mass of 16O) in grams = (7 x 12) + (6 x 1) + (3 x 16) = 138 grams - In general, solute is present in lesser amount. Solid (or) Liquid (or) Gas + Liquid Solute → Solvent Liquid Solution General Chemistry – Handout (Page: 3) Moles of solute = Molarity x Liter of solution Example Sugar (solid solute) + Water (liquid solvent) → Sugar solution (liquid) = 4 moles/L Amount of solute in grams x 2 L = 8 moles = Mole x Molar mass = 8 x 40 g (molar mass of NaOH) = 320 g Exercise 1. Calculate the molarity of a solution prepared by dissolving 11.5 g of solid NaOH in enough water to make 1.5 L of a solution. SOLUTION CONCENTRATION AND SOLUTION STOICHIOMETRY Gram → Moles → Molarity In solutions, the solute is dissolved in a solvent. The amount of solute in a solution is measured using a ‘concentration unit’ called molarity. Gram to Mole conversion: Molarity is defined as moles of solute per liter of solution. Molarity (M) = Moles Liters of solution 1 M = 1 mole of solute dissolved in 1 L of solution Example 1 moles of NaOH (40 gram) is present in 2 liters of solution. What is the molarity? Molarity (M) = Moles = 1 mole = 0.5 M L. of soln. 2L Using molarity (M), you can calculate the moles of solute (amount of solute) present in the solution. 1 mole of NaOH (Molar mass) = 23 + 16 + 1 = 40 g 11.5 g NaOH x 1 mole = 0.288 mol 40 g NaOH Molarity = Moles of solute = 0.288 mol = 0.192 M Liters of solution 1.5 L Exercise 2. How many mL of a 0.155 M KCl solution contains 2.55 g KCl? Molarity = number of Moles of solute Liters of solution ⇒ Liters of Solution = no. of moles of solute Molarity 19 17 Moles of solute = Molarity x Liter of solution (Volume) Molar mass of KCl = (1 x Mass of K) + (1 x Mass of Cl) in grams = (1 x 39.10) + (1 x 35.45) = 74.55 g/mol Mole to gram conversion: no. of moles of solute Amount of solute (in grams) = Moles of solute x Molar mass Example What is amount of NaOH present in 2 liters of 4M solution? Molarity → Moles → Amount of solute = mass in g / molar mass = 2.55 g / 74.55 g/mol. = 0.0342 mol ⇒ Liters of Solution = no. of moles of solute Molarity 3 = 0.0342 mol. = 2.206 L = 2.206 x 10 mL 0.155 M General Chemistry – Handout (Page: 4) DILUTION Exercise Dilution means reducing educing the concentration of a solution (reducing the molarity of a solution). - The concentration of a solution can be reduced by adding solvent into the solution. - Dilution changes only the concentration of the solution, but does not change the number of moles present. What is the volume of 16 M H2SO4 must be used to prepare 1.5 L of a 0.10 M H2SO4 solution? Concentrated solution: M1 = 16 M, V1 = ? Diluted solution: M2 = 0.10 M, V2 = 1.5 L M1 V1 = M2V2 Moles of solute before dilution = Moles of solute after dilution 16 x V1 = 0.10 x 1.5 In the lab, concentrated solution is called as stock solution and diluted solution is called as working solution. Formula for dilution related problems: M1 V1 = M2V2 V1 = (0.10 x 1.5) / 16 = 0.0094 L = 9.4 ml Example: What volume (in L) of 2.50 M NaCl solution is required to completely react with 0.150 L of a 0.150 M AgNO3 solution according to the reaction: NaCl(aq) + AgNO3(aq) AgCl(ss) + NaNO3(aq) M1 , V1 = Molarity and volume lume of concentrated solution (before dilution) Strategy: M2 , V2 = Molarity and volume of diluted solution (after dilution) L of AgNO3 soln mol AgNO3 mol NaCl L of NaCl soln Example 2 M solution (2 moles of solute/1 L solution) Dilution 0.150 mol AgNO3 1 L solution M [AgNO3 ] = 1 mol NaCl 1 mol AgNO3 1 L solution 0.250 mol NaCl 0.150 mol AgNO3 1 L solution Add 1 L solvent 1 M solution (2 moles of solute / 2 L solution) 1 mol AgNO3 = 1 mol NaCl M [NaCl] = Concentration: Changed from 2 M to 1 M Number of moles: Does not change (2 moles) M1V1 = 2 (M) x 1 (L) = 2 M2V2 = 1 (M) x 2 (L) = 2 = 0.150 L × 0.250 mol NaCl 1 L solution 0.150 mol AgNO3 1 L × 1 mol NaCl 1L × 1 mol AgNO3 0.250 mol NaCl = 0.0900 L of NaCl Therefore, M1 V1 = M2V2 General Chemistry – Handout (Page: 5) + Aqueous solution The meaning of ‘Aqua’ is water. If water is the solvent, the solution is called aqueous solution (solute can be solid, liquid or gas). - Finally positive (Na ) and negative (Cl ) ions were separated from the molecule (NaCl), and surrounded by water. This process is called hydration. The ions surrounded by water are called hydrated ions. Example: 1) Sugar + Water → Sugar solution (or) aqueous solution of sugar 2) NaCl (salt) + Water → Salt solution (or) aqueous solution of salt H2O WATER: The common solvent - Structure of Water (H2O) Cl surrounded by H2O NaCl Two hydrogen atoms share electrons unequally with one oxygen atom to form two polar covalent bonds. + Na surrounded by H2O Because of the unequal sharing of electron (‘O’ has more electrons than ‘H’), Example 2: NH4NO3 + Hydrogen has partial positive charge (Delta positive; δ ). Oxygen has partial negative charge (δ ). NH4NO3 dissolves in water to produces hydrated ions (NH4+ and NO3-). - (Please note that some ionic compounds are not soluble in water). Thus, water is a polar molecule. Water dissolves Polar Molecules Water dissolves Ionic molecules H2O also dissolve polar (non-ionic) molecules which has partial positive (δ+) and partial negative (δ+) charge (like water). Ionic molecules (salt) have positive ions and negative ions. Example 1: NaCl Example: Ethanol (C2H5OH) is a polar molecule. + - When NaCl dissolved in water, it produces Na and Cl ions Negative end (δ ) of polar solute molecule (ethanol) is attracted to + water's hydrogen (δ ). Positive end (δ ) of polar solute molecules (ethanol) is attracted to water's oxygen (δ ). H2O (l) NaCl (s) →→→→ Na+ (aq) + Cl- (aq) + - Positive ions (Na ) attracted to the oxygen end (δ ) of water. Negative ions (Cl ) attracted to the hydrogen end (δ ) of water. - - + + General Chemistry – Handout (Page: 6) 1. Strong electrolytes All the solutes (100%) are dissolved in water to produce separate ions. The solution strongly conducts electric current. Example: H2O i. Ionic compounds: NaCl H2O (l) + NaCl (s) →→→→ Na (aq) + Cl (aq) SOLUBILITY ii. Strong acids: HCl, H2SO4, HNO3 Solubility is the property of a solute (solid, liquid, or gas) to dissolve in a solvent (liquid). H2O (l) + HCl →→→→ H (aq)+ Cl (aq) "Like Dissolves Like" Rule H2O (l) + H2SO4 →→→→ H (aq)+ HSO4 (aq) 1. Polar solvents dissolve polar and ionic compounds. Example: Ethanol (polar) and NaCl (ionic) dissolve in water (polar). H2O (l) H+ (aq)+ NO - (aq) HNO3 →→→→ 3 iii. Strong base: NaOH, KOH H2O (l) Na+ (aq) + OH- (aq) NaOH →→→→ 2. Non-polar solvents dissolves non-polar compounds. Example: Fat (non-polar) dissolve in benzene (non-polar). H2O (l) K+ (aq) + OH- (aq) KOH →→→→ 2. Weak electrolytes ELECTROLYTES Solute dissolved in water solvent to produce aqueous solution. Solutes are dissolved in water, but only a small percentage of dissolved molecules produce ions. The solution weakly conducts electric current. Solute (solid, liquid or gas) + Water (solvent) → Aqueous solution. If the aqueous solution conducts electric current, the solute is called as electrolytes. Electrolytes are classified into 3 types. Example: i. 1. Strong electrolyte Weak acids: Acetic acid (HC2H3O2) H2O (l) H O+ (aq) + C H O - (aq) + HC H O (aq) HC2H3O2 →→→→ 3 2 3 2 2 3 2 2. Weak electrolyte 3. Non-electrolyte Ions present in the solution (+ve and –ve ions produced form the solute) are responsible to conduct electric current. ii. Weak base: Ammonia (NH3) H2O (l) + NH3 →→→→ NH4 (aq) + OH (aq) + NH3 (aq) 3. Non-electrolytes General Chemistry – Handout (Page: 7) + Solutes are dissolved in water, but do not produce any ions. The solution does not conduct an electric current. Example: i. - - + AgCl (s) + K (aq) + NO3 (aq) 3) Net ionic equation (show only components that actually react) Ethanol: C2H5OH + - Ag (aq) + Cl (aq) → AgCl (s) H2O (l) C2H5OH →→→→ C2H5OH (aq) ii. + - Ag (aq) + NO3 (aq) + K (aq) + Cl (aq) → - + K and NO3 are present in both sides of the reaction (see reaction 2). They are called spectator ions. Sugar 1) PRECIPITATION REACTION TYPES OF CHEMICAL REACTIONS Solution + Solution → Insoluble solid (precipitate) Chemical reaction is a change in which substances (reactants) are changed into one or more new substances (products). Solution reaction: If the reactants are solutions, the reaction is called solution reaction. Important solution reactions 1) Precipitation reactions Solution + Solution → Insoluble solid (precipitate) Description of the Acid-Base reaction: 1) Formula equation AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaNO3 (aq) 2) Complete ionic equation + - + - Ag (aq) + NO3 (aq) + Na (aq) + Cl (aq) → - + 2) Acid-base reactions or Neutralization reactions Transfer of proton from one to another. Acid solution + Base solution → Salt solution + Water AgCl (s) + Na (aq) + NO3 (aq) 3) Net ionic equation + 3) Oxidation-reduction reactions Solution + Solution → Transfer of electron from one to another - Ag (aq) + Cl (aq) → AgCl (s) Simple rules for the solubility of salts in water - Describing solution reactions We can describe the solution reaction in 3 ways. 1) Formula equation (reactants and products are represented in molecular formula) 1. Most nitrate (NO3 ) salts are soluble. + + + + 2. Most salts containing the alkali metals (Group 1A: Li , Na , K , Cs , + + Rb ) and ammonium (NH4 ) ions are soluble. - Ionic compounds separated into +ve and –ve ions in the solution. - + 2+ 2+ 4. Most sulfate salts are soluble (NOT BaSO4, PbSO4, HgSO4, CaSO4) AgNO3 (aq) + KCl (aq) → AgCl (s) + KNO3 (aq) 2) Complete ionic equation (all strong electrolytes shown as ions) - 3. Most Cl , Br , and I salts are soluble (NOT Ag , Pb , Hg2 ) - 5. Most OH salts are not soluble (NaOH, KOH are soluble, Ba(OH)2, Ca(OH)2 are marginally soluble) 2- 2- 2- 3- 6. Most S , CO3 , CrO4 , PO4 salts are not soluble. General Chemistry – Handout (Page: 8) Description of the Acid-Base reaction: Exercise 1: Calculate the mass of PbSO4 formed when 1.25 L of 0.0500 M Pb(NO3)2 and 2.00 L of 0.0250 M Na2SO4 are mixed. The reaction is 1) Formula equation HCl (aq) + NaOH (aq) → NaCl (s) + H2O (aq) Na2SO4 (aq) + Pb(NO3)2 (aq) → PbSO4 (s) + 2NaNO3 (aq) 2) Complete ionic equation + - + - + - H (aq) + Cl (aq) + Na (aq) + OH (aq) → Na (aq) + Cl (aq) + H2O (l) Moles of Pb(NO3)2 = 1.25 L X 0.0500 moles = 0.0625 moles 1L 3) Net ionic equation + Moles of Na2SO4 = 2.00 L X 0.0250 moles = 0.0500 moles 1L - H (aq) + OH (aq) → H2O (l) Na2SO4 is the limiting reactant. Acid-Base Titrations Therefore, 0.0500 mol PbSO4 is formed. + - In the neutralization reaction (H + OH → H2O): + Exercise 2: On the basis of solubility rules, predict whether aluminum nitrate and iron (III) phosphate are soluble in water or not. Aluminum nitrate – Soluble (rule 1) Iron (III) phosphate – Not soluble (rule 6) - Moles of Acid (H ) = Moles of Base (OH ) Therefore, if the concentration of one reactant is known, we can find out the concentration another reactant required for complete neutralization. This can be measured by ‘titration’ (with the use of a chemical indicator). The point at which the indicator changes color is called ‘end point’. 3) OXIDATION-REDUCTION REACTIONS Exercise 3: When Al(NO3)3 and Ba(OH)3 solutions are mixed, what precipitate (if any) will form? Oxidation = Lose of electrons Reduction = Gain of electrons Possible reaction: Al(NO3)3 + Ba(OH)3 → Ba(NO3)3 + Al(OH)3 As per solubility rule 5, Al(OH)3 is not soluble. So it form precipitate. 2Mg (s) + O2 (g) → 2MgO (s) Oxidation: 2Mg → 2Mg - 2) ACID-BASE REACTION (OR) NEUTRALIZATION REACTION + 2+ Reduction: O2 + 4e → 2O - An acid produces H when dissolved in water, and a base produced OH . Brønsted -Lowry definition: ACID is a proton donor, and BASE is a proton acceptor - + 4e 2- 2Mg2+ + 2O2- → 2MgO In this reaction, Mg is oxidized and O2 is reduced. Mg is reducing agent and O2 is oxidizing agent. Acid solution + Base solution → Salt solution + Water General Chemistry – Handout (Page: 9) Oxidation state Oxidation number of the compound = 0 Oxidation state (or oxidation number) is an indicator of the degree of oxidation of an atom in a chemical compound. Therefore, (2 x +1) + (2 x oxidation number of Cr) + (7 x -2) = 0 + 2 + Oxidation number of 2Cr -14= 0 S.No Oxidation state (or oxidation number) 1 The oxidation number of the atom of a free element is zero Element (H, Na, Cl) = 0 2 In ionic compounds: The oxidation number of a monatomic ion equals its charge Na in Nacl = +1 In covalent compounds: The oxidation number of the atoms depends on sharing of electrons. O in H2O = -2 3 4 The sum of the oxidation states for an electrically neutral compound must be zero Example Oxidation number of 2Cr = -2+14 = +12 Oxidation number of Cr = +6 + - Cl in NaCl = -1 H in H2O = +1 Examples: Identify the oxidizing agent and reducing agent in the following reactions: 1. 2Li (s) + Cl2 (g) → 2LiCl (s) F in HF = -1 NaCl = +1-1 = 0 H2O = (2 x +1) -2 = 0 Example 1 2 Li0 (s) + Cl20 (g) → 2 Li+1Cl-1 Li0 (s) → Li+1 (Loss of electron – Oxidation – Li is Reducing Agent) 0 -1 Cl2 (g) → Cl (Gain of electron – Reduction – Cl2 is Oxidizing agent) In IF7 (compound), the oxidation number of F is -1. What is the oxidation number of I? 2. 2Al (s) + 3 Sn 2+ 3+ (aq) → 2 Al (aq) + 3 Sn (s) Oxidation number of F = -1 2Al0 (s) + 3 Sn2+ (aq) → 2 Al3+ (aq) + 3 Sn0 (s) The oxidation number of the compound = 0 Al → Al (Loss of electrons – Oxidation – Al is Reducing Agent) Therefore, Sn 0 2+ +3 0 2+ → Sn (Gain of electrons – Reduction – Sn is Oxidizing agent) Oxidation number of I + Oxidation number of F7 = 0 Oxidation number of I + (-7) = 0 Combustion Reaction: Oxidation number of I = +7 These are the redox reaction in which substance reacts with oxygen (O2) to form oxygen-containing compound. Example 2 In K2Cr2O7 (compound), the oxidation number of O is -2 & K = +1. What is the oxidation number of Cr? O = -2, K = +1 General Chemistry – Handout (Page: 10)