Titrimetric Analysis PDF

Cebu Doctors’ University College of Arts and Sciences Physical Sciences Department MODULE 3 – TITRIMETRIC ANALYSIS Compiled by: Joselito R. Tumulak Jr., RChT, MS (cand.) Analytical Chemistry Professor INTENDED LEARNING OUTCOMES At the end of this module, you will be able to: 1. Define titration and...

Cebu Doctors’ University College of Arts and Sciences Physical Sciences Department MODULE 3 – TITRIMETRIC ANALYSIS Compiled by: Joselito R. Tumulak Jr., RChT, MS (cand.) Analytical Chemistry Professor INTENDED LEARNING OUTCOMES At the end of this module, you will be able to: 1. Define titration and explains its importance to analytical chemistry; 2. Understand the concept of equivalence point and the different types of titrations; 3. Explain the relationship between equivalent weight and normality; 4. Differentiate normality from molarity; 5. Use titration data for calculation of titrant concentrations and quantity of analytes; 6. Describe the preparation and standardization of the standard solutions used in neutralization titrations; 7. Explain the mechanisms of some common acid/base indicators; and 8. Link the principles of neutralization titrations to some of its applications in the determination of various analytes. UNIT OUTLINE Topic I. Basic Concepts in Titration A. Requirements for Volumetric Titrations B. Types of Volumetric Titrations II. Normality A. Equivalent Weight B. Normality III. Working with Titration Data A. Calculating Molar Concentrations from Standardization Data B. Calculating the Quantity of Analyte from Titration Data IV. Standard Solutions in Neutralization Titration A. Preparation and Standardization of Acid Solutions B. Preparation and Standardization of Base Solutions V. Acid/Base Indicators VI. Some Applications of Neutralization Titrations A. Elemental Analysis B. Determination of Organic Functional Groups Page 1 3 6 8 11 12 I. BASIC CONCEPTS IN TITRATION Titrations are analytical methods that are widely used in analytical chemistry to determine acids, bases, oxidants, reductants, metal ions, proteins, and many other species. They are based on a reaction between the analyte and a standard reagent known as the titrant. The reaction is of known and reproducible stoichiometry. The volume, or the mass, of the titrant needed to react completely with the analyte is determined and used to calculate the quantity of analyte. Titration in which the titrant's volume is measured is referred to as volumetric titration, whereas titration in which the titrant's mass is measured is referred to as gravimetric titration. Page 1 of 15 A volume-based titration is shown in this figure in which the standard solution is added from a burette, and the reaction occurs in the Erlenmeyer flask. In any titration, the point of chemical equivalence, called the end point when determined experimentally, is signaled by an indicator color change or a change in an instrumental response. This is the most common type of titrations used in many laboratories. A. Requirements for Volumetric Titrations Volumetric titration involves dissolving a weighed amount of sample and titrating the resulting solution with a standard solution (titrant). § Titration – progressive addition of a standard solution to a solution with which it reacts. § Standard solution – a solution with accurately known concentration. Conditions to be met for volumetric titrations: 1. The sample must be soluble in a solvent. 2. The reaction between the titrant and the sample must be rapid. 3. No side reaction should occur. 4. The reaction must be complete. 5. There should be an abrupt change in some properties at or near the equivalence point. 6. An indicator should be available. Requirements for volumetric titrations: 1. Accurate balance with which to weigh materials for analysis 2. Graduated instruments to measure volumes of solutions used, such as burettes and pipettes. 3. Titrants (standard solutions) 4. Indicators § Types of indicators i. Internal indicator – added to a system. ii. External indicator – indicator is allowed to react in a separate container with a few drops of the solution being titrated. Standard solutions must undergo standardization before it can be used in titrimetric analysis. Standardization is the process of determining the concentration of a standard solution. Three ways of standardization: 1. Direct Standardization – dissolving a weighed amount of pure dry chemical and diluting the solution to an exactly measured volume. Example: In preparing 1M AgNO3, weigh 169.89 g of pure AgNO3 and add water to give exactly 1 L solution. 2. Titration of a weighed primary standard by the solution to be standardized. § Primary standard – a pure dry solid substance of known chemical composition and purity. § Requirements for a primary standard: i. It must be a substance of known purity and definite composition. ii. It must be stable at the temperature employed for drying. Page 2 of 15 iii. iv. Its composition must not be affected by changes in humidity. It must react quantitatively and in a known way with the solution to be standardized. v. There must be a suitable indicator for the titration. 3. Titration of a measured volume of a solution that has itself been standardized previously. The previously standard solution is known as the secondary standard. B. Types of Volumetric Titrations Types of volumetric titrations according to the way the titrant is used: 1. Direct titration – the standard solution is titrated directly with the sample in question. 2. Indirect titration – an excess reagent is added to the sample in question and the excess reagent is titrated with another standard solution. Types of volumetric titrations according to the type of reaction used: 1. Neutralization methods – uses the reaction between an acid and a base. Acid + Base à Salt + Water a. Alkalimetry – an acid is titrated with a standard solution of a base. b. Acidimetry – a base is titrated with a standard solution of an acid. 2. Redox methods – utilizes redox reaction between an oxidizing agent and a reducing agent. The titrant is often oxidizing agents. Oxidizing agent + Reducing agent à Reduced substance + Oxidized substance a. Permangonometry – uses standard solution of potassium permanganate (KMnO4) as titrant. The titrant is its own indicator. b. Dichrometry - uses standard solution of potassium permanganate (KMnO4) as titrant. c. Cerimetry - uses standard solution of ceric sulfate (Ce(SO4)2) as titrant. d. Iodimetry – uses standard iodine solution (I2) to titrate reducing agents. e. Iodometry – a solution of iodide ions are added to a solution containing an oxidizing agent and the iodine produced is titrated with a standard sodium thiosulfate solution (Na2S2O3). 3. Precipitimetry – utilizes precipitation reactions and differences in Ksp. 4. Complexometric method – utilizes EDTA as titrant; used for metals analysis. II. NORMALITY A. Equivalent Weight The term equivalent weight is often seen in in volumetric analysis literature. The definition of this term depends on its behavior in a specific chemical reaction. Thus, we define equivalent weight in terms of the following reaction types: Neutralization Reactions § Acid – the weight of acid which furnishes 1 mole of H+ in a neutralization reaction, i.e. 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑒𝑞. 𝑤𝑡. (𝑎𝑐𝑖𝑑) = 𝑛𝑜. 𝑜𝑓 𝑟𝑒𝑝𝑙𝑎𝑐𝑒𝑎𝑏𝑙𝑒 𝐻! § Page 3 of 15 Base – the weight of base which reacts with 1 mole of H+ in a neutralization reaction, i.e. 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑒𝑞. 𝑤𝑡. (𝑎𝑐𝑖𝑑) = 𝑛𝑜. 𝑜𝑓 𝑠𝑝𝑒𝑐𝑖𝑒𝑠 𝑟𝑒𝑎𝑐𝑡𝑖𝑛𝑔 𝑤𝑖𝑡ℎ 𝐻! Example: o o HCl + NaOH à NaCl + H2 O 1 mol HCl furnishes 1 mol H+; thus, eq. wt. (HCl) = molar mass HCl/1 1 mol NaOH reacts with 1 mol H+; thus, eq. wt. (NaOH) = molar mass HCl/1 o o H2SO4 + 2 NaOH à Na2SO4 + 2 H2O 1 mol H2SO4 furnishes 2 mol H+; thus, eq. wt. (H2SO4) = molar mass H2SO4/2 1 mol of NaOH reacts with 1 mol H+; thus, eq. wt. (NaOH) = molar mass HCl/1 Oxidation-Reduction (Redox) Reactions § Oxidizing agent – weight of oxidizing agent which consumes one mole of electrons (e-) in a redox reaction, i. e. 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑒𝑞. 𝑤𝑡. (𝑜𝑥𝑖𝑑𝑖𝑧𝑖𝑛𝑔 𝑎𝑔𝑒𝑛𝑡) = 𝑛𝑜. 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑒 " 𝑔𝑎𝑖𝑛𝑒𝑑 § Reducing agent – weight of oxidizing agent which furnishes one mole of electrons (e-) in a redox reaction, i. e. 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑒𝑞. 𝑤𝑡. (𝑟𝑒𝑑𝑢𝑐𝑖𝑛𝑔 𝑎𝑔𝑒𝑛𝑡) = 𝑛𝑜. 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑒 " 𝑙𝑜𝑠𝑡 Example: Cu2+ + Zn à Zn2+ + Cu o o o o B. Normality Cu2+ à Cu (Cu2+ changes in oxidation number from +2 to 0; therefore, it gained 2 moles e-) 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑒𝑞. 𝑤𝑡. (𝐶𝑢#! ) = 2 Zn à Zn2+ (Zn changes in oxidation number from 0 to +2; therefore, it lost 2 moles e-) 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑒𝑞. 𝑤𝑡. (𝑍𝑛) = 2 MnO4- + 5 Fe2+ + 8H+ → Mn2+ + 5 Fe3+ + 4 H2O MnO4- à Mn2+ (The Mn changes in oxidation number from +7 to +2 since Mn in MnO4- has an oxidation number of +7. Therefore, it gained 5 moles e-.) 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑒𝑞. 𝑤𝑡. (𝑀𝑛𝑂$ " ) = 5 Fe2+ à Fe3+ (Fe2+ changes in oxidation number from +2 to +3; therefore, it lost 1 mole e-) 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑒𝑞. 𝑤𝑡. (𝐹𝑒 #! ) = 1 Normality is a concentration unit that utilizes equivalent weight. Its unit is normal, symbolized by N. Normality is expressed as: 𝑛𝑜. 𝑜𝑓 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡𝑠 𝑠𝑜𝑙𝑢𝑡𝑒 𝑁𝑜𝑟𝑚𝑎𝑙𝑖𝑡𝑦 = (1) 𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 Where no. of equivalent solute expressed as: 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑛𝑜. 𝑜𝑓 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡𝑠 𝑠𝑜𝑙𝑢𝑡𝑒 = (2) 𝑒𝑞. 𝑤𝑡. 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 Incorporating the second equation to the first equation, we can express normality as: %&'()* ,-.,/0*& 3 𝑁𝑜𝑟𝑚𝑎𝑙𝑖𝑡𝑦 = ( &1.%*.,-.,/0*& )(4.,/0*',5) Page 4 of 15 Examples: 1. What is the normality of a solution prepared by dissolving 15.8 g KMnO4 in enough water to prepare 500 mL solution? The molar mass of KMnO4 is 158 g/mol, and the pertinent reaction it undergoes is MnO4- à Mn2+. Given: wt. solute = 15.8 g volume of solution = 500 mL molar mass of KMnO4 = 158 g/mol Solution: Step 1: Solve for the equivalent weight of solute. As previously established, when MnO4- is converted to Mn2+, it resulted to a net gain of 5 e-. Therefore, 𝑒𝑞. 𝑤𝑡. (𝑀𝑛𝑂$ " ) = 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 158 = = 31.6 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡𝑠 5 5 Step 2: Solve for the normality. 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 1 15.8 𝑔 1 𝑁𝑜𝑟𝑚𝑎𝑙𝑖𝑡𝑦 = L ML M=L ML M=𝟏𝑵 𝑒𝑞. 𝑤𝑡. 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 31.6 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡𝑠 0.5 𝐿 2. How many grams of Ba(OH)2 should be dissolved in 500 mL to prepare 0.25 N Ba(OH)2 solution? The molar mass of Ba(OH)2 is 171 g/mol. Given: Normality of Ba(OH)2 solution = 0.25 N Volume of solution = 500 mL = 0.5 L Solution: Step 1. Solve for the equivalent weight of solute. Since there are 2 OH- in Ba(OH)2, one mole of it can react with 2 moles H+. Therefore, its equivalent weight is: 𝑒𝑞. 𝑤𝑡. (𝐵𝑎(𝑂𝐻)# ) = 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 171 = = 85.5 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡𝑠 2 2 Step 2. Solve for the weight solute from normality formula. 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 1 𝑁𝑜𝑟𝑚𝑎𝑙𝑖𝑡𝑦 = L ML M 𝑒𝑞. 𝑤𝑡. 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 = (𝑁)(𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛) (𝑒𝑞. 𝑤𝑡. 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒) = (0.25 𝑁)(0.5 𝐿)(85.5 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡𝑠) = 𝟏𝟎. 𝟕 𝒈 𝑩𝒂(𝑶𝑯)𝟐 Page 5 of 15 III. WORKING WITH TITRATION DATA Two types of volumetric calculations are discussed here. In the first, we compute concentrations of solutions that have been standardized against either a primary standard or another standard solution. In the second, we calculate the amount of analyte in a sample from titration data. Both types of calculation are based on three algebraic relationships. A. Calculating Molar Concentrations from Standardization Data 1. A 50.00-mL portion of an HCl solution required 29.71 mL of 0.01963 M Ba(OH)2 (previously standardized) to reach an end point with bromocresol green indicator. Calculate the molar concentration of the HCl. Ba(OH)2 + 2 HCl à BaCl2 + 2 H2O Solution: Step 1. Find the stoichiometric ratio. 1 mole of Ba(OH)2 reacts with 2 moles HCl, so the stoichiometric ratio is: 2 𝑚𝑜𝑙 𝐻𝐶𝑙 𝑠𝑡𝑜𝑖𝑐ℎ𝑖𝑜𝑚𝑒𝑡𝑟𝑖𝑐 𝑟𝑎𝑡𝑖𝑜 = 1 𝑚𝑜𝑙 𝐵𝑎(𝑂𝐻)# Step 2. Find the no. of moles of Ba(OH)2 used: 𝑛𝑜. 𝑜𝑓 𝑚𝑜𝑙 𝐵𝑎(𝑂𝐻)# = 29.71 𝒎𝑳 𝑩𝒂(𝑶𝑯)𝟐 × 1 𝑳 𝑩𝒂(𝑶𝑯)𝟐 𝑚𝑜𝑙 𝐵𝑎(𝑂𝐻)# × 0.01963 1000 𝒎𝑳 𝑩𝒂(𝑶𝑯)𝟐 𝐿 𝑩𝒂(𝑶𝑯)𝟐 Step 3. Find the number of moles HCl: 𝑛𝑜. 𝑜𝑓 𝑚𝑜𝑙 𝐻𝐶𝑙 = 𝑛𝑜. 𝑜𝑓 𝑚𝑜𝑙 𝐵𝑎(𝑂𝐻)# × 𝑠𝑡𝑜𝑖𝑐ℎ𝑖𝑜𝑚𝑒𝑡𝑟𝑖𝑐 𝑟𝑎𝑡𝑖𝑜 1 𝑳 𝑩𝒂(𝑶𝑯)𝟐 𝒎𝒐𝒍 𝑩𝒂(𝑶𝑯)𝟐 2 𝑚𝑜𝑙 𝐻𝐶𝑙 = 29.71 𝒎𝑳 𝑩𝒂(𝑶𝑯)𝟐 × × 0.01963 × 1000 𝒎𝑳 𝑩𝒂(𝑶𝑯)𝟐 𝑳 𝑩𝒂(𝑶𝑯)𝟐 1 𝒎𝒐𝒍 𝑩𝒂(𝑶𝑯)𝟐 Step 4. Find the molarity of HCl: 𝑛𝑜. 𝑜𝑓 𝑚𝑜𝑙 𝐻𝐶𝑙 𝑀 𝑜𝑓 𝐻𝐶𝑙 = 𝑉𝑜𝑙 𝑜𝑓 𝐻𝐶𝑙 (𝑖𝑛 𝐿) 1 𝑳 𝑩𝒂(𝑶𝑯)𝟐 𝒎𝒐𝒍 𝑩𝒂(𝑶𝑯) 2 𝑚𝑜𝑙 𝐻𝐶𝑙 29.71 𝒎𝑳 𝑩𝒂(𝑶𝑯)𝟐 × 1000 𝒎𝑳 𝑩𝒂(𝑶𝑯) × 0.01963 𝑳 𝑩𝒂(𝑶𝑯) 𝟐 × 1 𝒎𝒐𝒍 𝑩𝒂(𝑶𝑯) 𝟐 𝟐 𝟐 = 1𝐿 50 𝑚𝐿 × 1000 𝑚𝐿 = 𝟎. 𝟎𝟐𝟑𝟑𝟑 𝑴 𝑯𝑪𝒍 2. Titration of 0.2121 g of pure Na2C2O4 (134.00 g/mol) required 43.31 mL of KMnO4. What is the molar concentration of the KMnO4 solution? The chemical reaction is: 2 MnO4- + 5 C2O4- + 16H+ à 2 Mn2+ + 10 CO2 + 8 H2O Solution: Step 1. Find the stoichiometric ratio. 2 moles of KMnO4 reacts with 5 moles Na2C2O4, so the stoichiometric ratio is: 2 𝑚𝑜𝑙 𝐾𝑀𝑛𝑂$ 𝑠𝑡𝑜𝑖𝑐ℎ𝑖𝑜𝑚𝑒𝑡𝑟𝑖𝑐 𝑟𝑎𝑡𝑖𝑜 = 5 𝑚𝑜𝑙 𝑁𝑎# 𝐶# 𝑂$ Step 2. Find the number of moles of the primary standard used which is Na2C2O4: 𝑛𝑜. 𝑜𝑓 𝑚𝑜𝑙 𝑁𝑎# 𝐶# 𝑂$ = 0.2121 𝒈 𝑵𝒂𝟐 𝑪𝟐 𝑶𝟒 × Page 6 of 15 1 𝑚𝑜𝑙 𝑁𝑎# 𝐶# 𝑂$ 134 𝒈 𝑵𝒂𝟐 𝑪𝟐 𝑶𝟒 Step 3. Find the number of moles KMnO4: 𝑛𝑜. 𝑜𝑓 𝑚𝑜𝑙 𝐾𝑀𝑛𝑂$ = 0.2121 𝒈 𝑵𝒂𝟐 𝑪𝟐 𝑶𝟒 × 1 𝒎𝒐𝒍 𝑵𝒂𝟐 𝑪𝟐 𝑶𝟒 2 𝑚𝑜𝑙 𝐾𝑀𝑛𝑂$ × 134 𝒈 𝑵𝒂𝟐 𝑪𝟐 𝑶𝟒 5 𝒎𝒐𝒍 𝑵𝒂𝟐 𝑪𝟐 𝑶𝟒 Step 4. Find the concentration of KMnO4: 1 𝒎𝒐𝒍 𝑵𝒂 𝑪 𝑶 2 𝑚𝑜𝑙 𝐾𝑀𝑛𝑂$ 0.2121 𝒈 𝑵𝒂𝟐 𝑪𝟐 𝑶𝟒 × 134 𝒈 𝑵𝒂 𝟐𝑪 𝟐𝑶 𝟒 × 5 𝒎𝒐𝒍 𝑵𝒂 𝑪 𝑶 𝟐 𝟐 𝟒 𝟐 𝟐 𝟒 𝑀 𝑜𝑓 𝐾𝑀𝑛𝑂$ = = 𝟎. 𝟎𝟏𝟒𝟔𝟐 𝑴 𝑲𝑴𝒏𝑶𝟒 1𝐿 43.1 𝑚𝐿 × 1000 𝑚𝐿 B. Calculating the Quantity of Analyte from Titration Data 1. A 0.8040-g sample of an iron ore is dissolved in acid. The iron is then reduced to Fe2+ and titrated with 47.22 mL of 0.02242 M KMnO4 solution. Calculate the results of this analysis in terms of % Fe (55.847 g/mol). The chemical reaction is: MnO4- + 5 Fe2+ + 8 H+ à Mn2+ + 5 Fe3+ + 4 H2O Step 1. Find the stoichiometric ratio. 5 moles of Fe2+ reacts with 1 mole KMnO4, so the stoichiometric ratio is: 5 𝑚𝑜𝑙 𝐹𝑒 #! 𝑠𝑡𝑜𝑖𝑐ℎ𝑖𝑜𝑚𝑒𝑡𝑟𝑖𝑐 𝑟𝑎𝑡𝑖𝑜 = 1 𝑚𝑜𝑙 KMn𝑂$ Step 2. Find the number of moles KMnO4 used: 𝑛𝑜. 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝐾𝑀𝑛𝑂$ = 47.22 𝒎𝑳 𝑲𝑴𝒏𝑶𝟒 × 1 𝑳 𝑲𝑴𝒏𝑶𝟒 𝑚𝑜𝑙 𝐾𝑀𝑛𝑂$ × 0.02242 1000 𝒎𝑳 𝑲𝑴𝒏𝑶𝟒 𝑳 𝑲𝑴𝒏𝑶𝟒 Step 3. Find the number of moles Fe2+ in the sample: 𝑛𝑜. 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝐹𝑒 #! 1 𝑳 𝑲𝑴𝒏𝑶𝟒 𝒎𝒐𝒍 𝑲𝑴𝒏𝑶𝟒 5 𝑚𝑜𝑙 𝐹𝑒 #! = 47.22 𝒎𝑳 𝑲𝑴𝒏𝑶𝟒 × × 0.02242 × 1000 𝒎𝑳 𝑲𝑴𝒏𝑶𝟒 𝑳 𝑲𝑴𝒏𝑶𝟒 1 𝒎𝒐𝒍 𝑲𝑴𝒏𝑶𝟒 Step 4. Find the mass of Fe2+: 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐹𝑒 #! = 47.22 𝒎𝑳 𝑲𝑴𝒏𝑶𝟒 × × 55.847 𝑔 𝐹𝑒 #! 1 𝒎𝒐𝒍 𝑭𝒆𝟐! 1 𝑳 𝑲𝑴𝒏𝑶𝟒 𝒎𝒐𝒍 𝑲𝑴𝒏𝑶𝟒 5 𝒎𝒐𝒍 𝑭𝒆𝟐! × 0.02242 × 1000 𝒎𝑳 𝑲𝑴𝒏𝑶𝟒 𝑳 𝑲𝑴𝒏𝑶𝟒 1 𝒎𝒐𝒍 𝑲𝑴𝒏𝑶𝟒 Step 5. Find the percent Fe2+: % 𝐹𝑒 55.847 𝑔 𝐹𝑒 #! 1 𝑳 𝑲𝑴𝒏𝑶𝟒 𝒎𝒐𝒍 𝑲𝑴𝒏𝑶 5 𝒎𝒐𝒍 𝑭𝒆𝟐! 47.22 𝒎𝑳 𝑲𝑴𝒏𝑶𝟒 × 1000 𝒎𝑳 𝑲𝑴𝒏𝑶 × 0.02242 𝑳 𝑲𝑴𝒏𝑶 𝟒 × 1 𝒎𝒐𝒍 𝑲𝑴𝒏𝑶 × 1 𝒎𝒐𝒍 𝑭𝒆𝟐! 𝟒 𝟒 𝟒 = 0.8040 𝑔 𝑆𝑎𝑚𝑝𝑙𝑒 × 100 = 𝟑𝟔. 𝟕𝟕% Page 7 of 15