Chapter 5. Stoichiometry - PDF Free Download

Chapter 5 Stoichiometry

Chapter 5 Table of Contents (5-1) Counting by weighing (5-2) Atomic masses (5-3) Learning to solve problems (5-4) The mole (5-5) Molar mass (5-6) Percent composition of compounds (5-7) Determining the formula of a compound (5-8) Chemical equations

Chapter 5 Table of Contents (5-9) Balancing chemical equations (5-10) Stoichiometric calculations: Amounts of reactants and products (5-11) Concept of limiting reactant

Chapter 5 Questions to Consider Can atoms be counted? Does the number of atoms in a reaction affect the result? Can the mass of a sample be used to determine the number of moles it comprises?

Section 5.1 Counting by Weighing Average Mass It is used to determine the number of atoms in a set quantity of a substance It is determined using a sample of the substance Average mass = total mass of atoms in sample number of atoms in sample Atoms do not need to be identical in order to be counted by weighing Copyright Cengage Learning. All rights reserved 5

Section 5.2 Atomic Masses Atomic Masses The modern system of determining atomic masses was first used in 1961 Based on 12 C 12 atomic mass units (u) The most accurate method currently used to compare the masses of atoms involves the use of the mass spectrometer

Section 5.2 Atomic Masses Mass Spectrometer Atoms are passed into a stream of high speed electrons that convert them into positive ions Ions are passed through a magnetic field Accelerating ions create their own magnetic field, resulting in a change in the path travelled

Section 5.2 Atomic Masses Mass Spectrometer The degree of deviation depends on the mass of the ion Ions with higher masses deviate the least Deviated ions hit the detector plate Comparing the deflected position of each ion gives their mass

Section 5.2 Atomic Masses Fig 5.1 - The Mass Spectrometer

Section 5.2 Atomic Masses Determining Atomic Masses Consider the analysis of 12 C and 13 C in a mass spectrometer 13 Mass C 12 Mass C 1.0836129 Based on the definition of the atomic mass unit 13 Mass of C = (1.0836129)(12 u) = 13.003355 u Exact number by definition The average mass for an element is also referred to as the average atomic mass or atomic mass

Section 5.2 Atomic Masses Example 5.1 - The Average Mass of an Element When a sample of natural copper is vaporized and injected into a mass spectrometer, the results shown in Fig. 5-3 are obtained. Use these data to compute the average mass of natural copper. The mass values for 63 Cu and 65 Cu are 62.93 u and 64.93 u, respectively

Section 5.2 Atomic Masses Solution Information needed To calculate the average mass of natural copper Information available 63 Cu mass = 62.93 u 65 Cu mass = 64.93 u Step 1 - Determine the mass of 100 atoms of natural copper As shown by the graph, of every 100 atoms of natural copper, 69.09 are 63 Cu and 30.91 are 65 Cu

Section 5.2 Atomic Masses Solution Determining the mass of 100 atoms of natural copper (69.09 atoms u u ) (62.93 ) + (30.91 atoms)(64.93 ) = 6355 u atom atom The average mass of a copper atom is 6355 u 100 atoms = 63.55 u/atom

Section 5.2 Atomic Masses Conceptual Problem Solving It is a flexible and creative method of solving problems based on the fundamentals of chemistry An understanding of the process used to determine the answer to problems is essential Methods to approaching a problem Pigeonholing The big picture

Section 5.3 Learning to Solve Problems Steps to Solving a Problem Where are we going? Decide on the final goal How do we get there? Find the starting point by working backwards from the final goal Reality Check Cross-check the answer that has been found Copyright Cengage Learning. All rights reserved 15

Section 5.4 The Mole The Mole and Avogadro s Number The mole is defined as the number equal to the number of carbon atoms in exactly 12 grams of pure 12 C 6.022 10 23 Also called Avogadro s number One mole of a substance contains 6.022 10 23 units of that substance Copyright Cengage Learning. All rights reserved 16

Section 5.4 The Mole The Mole and Avogadro s Number A sample of a natural element with a mass equal to the element s atomic mass expressed in grams contains 1 mole of atoms

Section 5.4 The Mole Interactive Example 5.2 - Determining the Mass of a Sample of Atoms Americium is an element that does not occur naturally. It can be made in very small amounts in a device known as a particle accelerator. Compute the mass in grams of a sample of americium containing six atoms.

Section 5.4 The Mole Solution Information needed The mass of six americium atoms Information available Mass of 1 atom of Am = 243 u Step 1 - Determine the mass of six americium ions 6 atoms u 243 atom 3 = 1.46 10 u

Section 5.4 The Mole Solution Step 2 - Use Avogadro s number to convert atomic mass units to grams 23 6.022 10 u = 1 g 1 g 23 6.022 10 u Step 3 - Determine the mass of six americium ions in grams 1 g 21 = 2.42 10 g 6.022 10 u 3 1.46 10 u 23

Section 5.5 Molar Mass Molar Mass of a Compound It is the mass of one mole of the compound measured in grams Traditionally called molecular weight Formula unit Used for compounds that do not contain molecules NaCl - Sodium chloride CaCO 3 - Calcium carbonate Copyright Cengage Learning. All rights reserved 21

Section 5.5 Molar Mass Interactive Example 5.6 - Calculating Molar Mass I Juglone, a dye known for centuries, is produced from the husks of black walnuts. It is also a natural herbicide (weed killer) that kills off competitive plants around the black walnut tree but does not affect grass and other noncompetitive plants. The formula for juglone is C 10 H 6 O 3. a.) Calculate the molar mass of juglone. b.) A sample of 1.56 10 2 g of pure juglone was extracted from black walnut husks. How many moles of juglone does this sample represent? Copyright Cengage Learning. All rights reserved 22

Section 5.5 Molar Mass Solution a.) The molar mass is obtained by summing the masses of the component atoms. In 1 mole of juglone there are 10 moles of carbon atoms, 6 moles of hydrogen atoms, and 3 moles of oxygen atoms: 10 C: 10 12.01 g = 120.1 g 6 H: 6 1.008 g = 6.048 g 3 O: 3 16.00 g = 48.00 g Mass of 1 mol C H O = 174.1 g 10 6 3 The molar mass of juglone is 174.1 g

Section 5.5 Molar Mass Solution b.) The mass of 1 mole of this compound is 174.1 g; thus 1.56 10 2 g is much less than a mole. The exact fraction of a mole can be determined as follows: 2 1.56 10 g juglone 1 mol juglone 174.1 g juglone 5 = 8.96 10 mol juglone

Section 5.6 Percent Composition of Compounds Describing the Composition of a Compound In terms of the numbers of its constituent atoms In terms of the percentages (mass) of its elements Consider ethanol (C 2 H 5 OH) Mass of C = 2 mol Mass of H = 6 mol g 12.01 mol = 24.02 g g 1.008 mol = 6.048 g Mass of O = 1 mol g 16.00 mol = 16.00 g Mass of 1 mol C H OH = 46.07 g 2 5

Section 5.6 Percent Composition of Compounds Mass Percent It is calculated by comparing the mass percent of the element in one mole of the compound with the total mass of one mole of the compound and multiplying the result by 100%. Calculating the mass of carbon in ethanol mass of C in 1 mol of C H OH 2 5 Mass percent of C = 100% mass of 1 mol of C H OH 2 5 24.02g = 100 = 52.14% 46.07g

Section 5.6 Percent Composition of Compounds Interactive Example 5.9 - Calculating Mass Percent Carvone is a substance that occurs in two forms having different arrangements of atoms but the same molecular formula (C 10 H 14 O) and mass. One type of carvone gives caraway seeds their characteristic smell, and the other type is responsible for the smell of spearmint oil. Compute the mass percent of each element in carvone.

Section 5.6 Percent Composition of Compounds Solution Objective To find the mass percent of each element in carvone Information available Molecular formula, C 10 H 14 O Information needed to find the mass percent Mass of each element (1 mole of carvone) Molar mass of carvone

Section 5.6 Percent Composition of Compounds Solution Step 1 - Determine the mass of each element on one mole of C 10 H 14 O Mass of C in 1 mol = 10 mol Mass of H in 1 mol = 14 mol g 12.01 mol = 120.1 g g 1.008 mol = 14.11 g Mass of O in 1 mol = 1 mol g 16.00 mol = 16.00 g

Section 5.6 Percent Composition of Compounds Solution Step 2 - Determine the molar mass of C 10 H 14 O 120.1 g + 14.11 g 16.00 g 150.2 g C + H + O C H O 10 14 10 14 Step 3 - Determine the mass of each element 120.1 g C Mass percent of C = 100% = 79.96% 150.2 g C H O 10 14 14.11 g H Mass percent of H = 100% = 9.394% 150.2 g C H O 10 14 16.00 g O Mass percent of O = 100% = 10.65% 150.2 g C H O 10 14

Section 5.7 Determining the Formula of a Compound Formulas Determined by using a weighed sample and one of the following techniques Decomposing it into its component elements Introducing oxygen to produce substances such as CO 2, H 2 O, etc., which are collected and weighed

Section 5.7 Determining the Formula of a Compound Fig 5.5 - Analyzing for Carbon and Hydrogen

Section 5.7 Determining the Formula of a Compound Empirical and Molecular Formula Empirical formula: The simplest whole-number ratio of the various types of atoms in a compound Can be obtained from the mass percent of elements in a compound

Section 5.7 Determining the Formula of a Compound Empirical and Molecular Formula The molecular formula varies for molecules and ions For molecular substances, it is the formula of the constituent molecules Always an integer multiple of the empirical formula For ionic substances, it is the same as the empirical formula

Section 5.7 Determining the Formula of a Compound Fig 5.6 - Substances whose Empirical and Molecular Formulas are Different

Section 5.7 Determining the Formula of a Compound Empirical Formula Determination Since mass percentage gives the number of grams of a particular element per 100 grams of compound, base the calculation on 100 grams of compound Each percent will then represent the mass in grams of that element Determine the number of moles of each element present in 100 grams of compound using the atomic masses of the elements present

Section 5.7 Determining the Formula of a Compound Empirical Formula Determination Divide each value of the number of moles by the smallest of the values If each resulting number is a whole number (after appropriate rounding), these numbers represent the subscripts of the elements in the empirical formula If the numbers obtained in the previous step are not whole numbers, multiply each number by an integer so that the results are all whole numbers

Section 5.7 Determining the Formula of a Compound Interactive Example 5.10 - Determining Empirical and Molecular Formulas I Determine the empirical and molecular formulas for a compound that gives the following percentages on analysis (in mass percents): 71.65% Cl 24.27% C 4.07% H The molar mass is known to be 98.96 g/mol

Section 5.7 Determining the Formula of a Compound Solution Objective To find the empirical and molecular formulas for the given compound Information available Percent of each element Molar mass of the compound (98.96 g/mol) Information needed to find the empirical formula Mass of each element in 100.00 g of compound Moles of each element

Section 5.7 Determining the Formula of a Compound Solution Step 1 - Determine the mass of each element in 100.00 g of compound Cl 71.65 g C 24.27 g H 4.07 g Step 2 - Determine the moles of each element in 100.00 g of compound 71.65 g Cl 1 mol Cl = 2.021 mol Cl 35.45 g Cl

Section 5.7 Determining the Formula of a Compound Solution 24.27 g C 4.07 g H 1 mol C = 2.021 mol C 12.01 g C 1 mol H = 4.04 mol H 1.008 g H Step 3 - Determine the empirical formula for the compound Dividing each mole value by 2.021 (the smallest number of moles present), we find the empirical formula ClCH 2.

Section 5.7 Determining the Formula of a Compound Solution Step 4 - Determine the molecular formula for the compound Compare the empirical formula mass to the molar mass Empirical formula mass = 49.48 g/mol Molar mass is given = 98.96 g/mol Molar mass 98.96 g/mol = = 2 Empirical formula mass 49.48 g/mol Molecular Formula = (ClCH ) = Cl C H 2 2 2 2 4 The substance comprises molecules of Cl 2 C 2 H 4

Section 5.8 Chemical Equations Chemical Reactions Involve the reorganization of atoms in one or more substances In a chemical equation, the reactants are situated on the left side of an arrow, and the products are located on the right CH + O 4 2 2 2 Reactants CO + H O Products Bonds are broken and new bonds are formed Both sides possess the same number of atoms Copyright Cengage Learning. All rights reserved 43

Section 5.8 Chemical Equations Balancing a Chemical Equation Atoms are neither created nor destroyed in a chemical reaction The number of atoms on each side of the equation must be the same Consider the reaction between methane and oxygen CH + 2O CO + 2H O 4 2 2 2

Section 5.8 Chemical Equations The Meaning of a Chemical Equation It contains information on: The nature of the reactants and products The relative numbers of each Equations also provide the physical state of the reactants and products State Solid Liquid Gas Aqueous solution Copyright Cengage Learning. All rights reserved 45 Symbol (s) (l) (g) (aq)

Section 5.8 Chemical Equations Table 5.2 - Balanced Equation for the Combustion of Methane CH ( g) + 2O ( g) CO ( g) + 2H O( g) 4 2 2 2

Section 5.9 Balancing Chemical Equations The Importance of a Balanced Reaction All atoms present in the reactants must be accounted for in the products An unbalanced equation is useless While balancing equations, care must be taken to avoid altering the formulas of compounds Subscripts cannot be altered Atoms cannot be added or subtracted Copyright Cengage Learning. All rights reserved 47

Section 5.9 Balancing Chemical Equations Balancing an Equation Start with the most complicated molecules Consider the following unbalanced equation C H OH( l) + O ( g) CO ( g) + H O( g) 2 5 2 2 2 The most complicated molecule is C 2 H 5 OH Balancing carbon C H OH( l) + O ( g) 2CO ( g) + H O( g) 2 5 2 2 2 2 C atoms 2 C atoms

Section 5.9 Balancing Chemical Equations Balancing an Equation Balancing hydrogen atoms C H OH( l) + O ( g) 2CO ( g) + 3H O( g) 2 5 2 2 2 (5 +1) H (3 2) H Balancing oxygen atoms C H OH( l) + 3O ( g) 2CO ( g) + 3H O( g) 2 5 2 2 2 1O 6O (2 2)O 3O 7O 7O

Section 5.9 Balancing Chemical Equations Balancing an Equation Verifying the results C H OH( l) + 3O ( g) 2CO ( g) + 3H O( g) 2 5 2 2 2 2 C atoms 2 C atoms 6 H atoms 6 H atoms 7 O atoms 7 O atoms

Section 5.9 Balancing Chemical Equations Concept Check Which of the following is true about balanced chemical equations? I. The number of molecules is conserved II. III. The coefficients tell you how much of each substance you have Atoms are neither created nor destroyed IV. The coefficients indicate the mass ratios of the substances used V. The sum of the coefficients on the reactant side equals the sum of the coefficients on the product side

Section 5.9 Balancing Chemical Equations Interactive Example 5.13 - Balancing a Chemical Equation I Chromium compounds exhibit a variety of bright colors. When solid ammonium dichromate, (NH 4 ) 2 Cr 2 O 7, a vivid orange compound, is ignited, a spectacular reaction occurs. Although the reaction is actually somewhat more complex, let s assume here that the products are solid chromium(iii) oxide, nitrogen gas (consisting of N 2 molecules), and water vapor. Balance the equation for this reaction.

Section 5.9 Balancing Chemical Equations Solution From the description given, the reactant is solid ammonium dichromate, (NH 4 ) 2 Cr 2 O 7 (s), and the products are nitrogen gas, N 2 (g), water vapor, H 2 O(g), and solid chromium(iii) oxide, Cr 2 O 3 (s). The formula for chromium(iii) oxide can be determined by recognizing that the Roman numeral III means that Cr 3+ ions are present. For a neutral compound, the formula must then be Cr 2 O 3, since each oxide ion is O 2.

Section 5.9 Balancing Chemical Equations Solution The unbalanced equation is: (NH ) Cr O ( s) Cr O ( s) + N ( g) + H O( g) 4 2 2 7 2 3 2 2 Note that nitrogen and chromium are balanced (two nitrogen atoms and two chromium atoms on each side), but hydrogen and oxygen are not. A coefficient of 4 for H 2 O balances the hydrogen atoms (NH ) Cr O ( s) Cr O ( s) + N ( g) + 4H O( g) 4 2 2 7 2 3 2 2 (4 2) H (4 2) H

Section 5.10 Stoichiometric Calculations Stoichiometry of Reactions - Principles Consider the following reaction between propane and oxygen C H ( g) + 5O 3CO ( g) + 4H O( g) 3 8 2 2 2 Calculating the number of moles of propane present in 96.1 grams (the molar mass of propane is 44.1) 1 mol C3H8 96.1 g C3H = 2.18 mol C3H8 44.1 g C H Constructing a mole ratio 8 3 8 5 mol O2 1 mol C H 3 8 Copyright Cengage Learning. All rights reserved 55

Section 5.10 Stoichiometric Calculations Stoichiometry of Reactions - Principles Calculating the moles of O 2 needed 2.18 mol C3H8 5 mol O Converting moles into grams 2 = 10.9 mol O2 1 mol C H Therefore, 349 grams of oxygen can burn 96.1 grams of propane 3 8 32.0 g O2 10.9 mol O2 1 mol O = 349 g O2 2

Section 5.10 Stoichiometric Calculations Stoichiometry of Reactions - Principles In order to determine the mass of carbon dioxide produced, conversion between moles of propane and moles of carbon dioxide is required Forming the mole ratio The conversion is 3 mol CO 1 mol C H 2 3 8 3 mol CO2 2.18 mol C3H8 1 mol C H = 6.54 mol CO2 3 8

Section 5.10 Stoichiometric Calculations Stoichiometry of Reactions - Principles Calculating the mass of CO produced using the molar mass of CO 2 (44.0 g/mol) 44.0 g CO 6.54 mol CO = 288 g CO 2 2 2 1 mol CO2

Section 5.10 Stoichiometric Calculations Problem-Solving Strategy - Calculating Masses of Reactants and Products Balance the equation for the reaction Convert the known mass of the reactant or product to moles of that substance Use the balanced equation to set up the appropriate mole ratios Use the appropriate mole ratios to calculate the number of moles of desired reactant or product Convert from moles back to grams if required by the problem

Section 5.10 Stoichiometric Calculations Problem-Solving Strategy - Calculating Masses of Reactants and Products

Section 5.10 Stoichiometric Calculations Interactive Example 5.16 - Chemical Stoichiometry I Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide from the living environment by forming solid lithium carbonate and liquid water. What mass of gaseous carbon dioxide can be absorbed by 1.00 kg of lithium hydroxide?

Section 5.10 Stoichiometric Calculations Solution Objective To find the mass of CO 2 absorbed by 1.00 kg LiOH Information available Chemical reaction 1.00 kg LiOH LiOH( s) + CO ( g) Li CO ( g) + H O( l) 2 2 3 2 Information needed to find the mass of CO 2 Balanced equation for the reaction

Section 5.10 Stoichiometric Calculations Solution Step 1 - State the balanced equation for the reaction 2LiOH( s) + CO ( g) Li CO ( g) + H O( l) 2 2 3 2 Step 2 - Determine the moles of LiOH Determine the molar mass to find the moles of LiOH 6.941 + 16.00 + 1.008 = 23.95 g/mol Using the molar mass 1.00 kg LiOH 1000 g LiOH 1 mol LiOH = 41.8 mol LiOH 1 kg LiOH 23.95 g LiOH

Section 5.10 Stoichiometric Calculations Solution Step 3 - Determine the mole ratio between CO 2 and LiOH in the balanced equation 1 mol CO 2 2 mol LiOH Step 4 - Calculate the moles of CO 2 1 mol CO2 41.8 mol LiOH 2 mol LiOH = 20.9 mol CO2

Section 5.10 Stoichiometric Calculations Solution Step 5 - Determine the mass of CO 2 formed from 1.00 kg LiOH 44.0 g CO2 20.9 mol CO2 1mol CO Thus, 920. g of CO 2 (g) will be absorbed by 1.00 kg of LiOH(s) 2 2 = 9.20 10 g CO 2

Section 5.11 The Concept of Limiting Reactant Stoichiometric Mixture Consider the reaction between nitrogen and hydrogen, forming ammonia Each molecule of ammonia possesses: 1 N 2 molecule 3 H 2 molecules N ( g) 3H ( g) 2NH ( g) 2 2 3

Section 5.11 The Concept of Limiting Reactant Stoichiometric Mixture There are just enough molecules to ensure that all are paired

Section 5.11 The Concept of Limiting Reactant Stoichiometric Mixture A stoichiometric mixture is one that possesses equivalent amounts of reactants that match the numbers in the balanced equation The presence and quantity of the limiting reactant determines the amount of the product formed

Section 5.11 The Concept of Limiting Reactant Determination of Limiting Reactant Using Reactant Quantities Consider a reaction in which 25.0 kg of nitrogen is mixed with 5.0 kg of hydrogen to form ammonia The amount of ammonia formed needs to be calculated The balanced equation is used to identify the limiting reactant N ( g) + 3H ( g) 2NH ( g) 2 2 3

Section 5.11 The Concept of Limiting Reactant Determination of Limiting Reactant Using Reactant Quantities Determination of the moles of the reactants 25.0 kg N 2 5.00 kg H 2 1000 g N 2 1 kg N 2 1000 g H 2 1 kg H 2 1 mol N 28.0 g N Calculating the total amount of moles of H 2 that react with 8.93 10 2 moles of N 2 2 8.93 10 mol N 2 2 1 mol H 2 2 2.016 g H 3 mol H 2 1 mol N 2 2 2 8.93 10 mol N2 3 2.48 10 mol H 2 3 = 2.68 10 mol H 2

Section 5.11 The Concept of Limiting Reactant Determination of Limiting Reactant Using Reactant Quantities 2.48 10 3 moles of H 2 requires 8.27 10 2 moles of N 2 Nitrogen is in excess (8.93 10 2 moles) Hydrogen is the limiting reactant

Section 5.11 The Concept of Limiting Reactant Determination of Limiting Reactant Using Reactant Quantities In an alternate method to determine the limiting reactant, a comparison is made between: Mole ratio of substances required by the balanced equation Mole ratio of the reactants actually present 3 mol H 1 mol N This means that mol H2 3 (required) = 3 mol N 1 2 2 2

Section 5.11 The Concept of Limiting Reactant Determination of Limiting Reactant Using Reactant Quantities Adding values of the experiment 3 mol H2 2.48 10 (actual) = 2.78 2 mol N 8.93 10 2 The actual mole ratio of H 2 to N 2 is too small H 2 is the limiting factor

Section 5.11 The Concept of Limiting Reactant Determination of Limiting Reactant Using Quantities of Products Formed In this method of determining the chemical reactant, the amounts of products formed by complete consumption of the reactant are considered The limiting reactant is the reactant that produces the smallest amount of the product

Section 5.11 The Concept of Limiting Reactant Determination of Limiting Reactant Using Quantities of Products Formed Consider the reaction between 25.0 kg (8.93 10 2 moles) of nitrogen and 5.00 kg (2.48 10 3 moles) of hydrogen Determining the amount of NH 3 that can form 2 mol NH 8.93 10 mol N = 1.79 10 mol NH 2 3 3 2 3 1 mol N2

Section 5.11 The Concept of Limiting Reactant Determination of Limiting Reactant Using Quantities of Products Formed Calculating the amount of NH 3 formed upon complete consumption of H 2 2 mol NH 2.48 10 mol H = 1.65 10 mol NH 3 3 3 2 3 3 mol H2 The amount of NH 3 produced from H 2 is lesser than that from N 2 Converting moles to kilograms 17 g NH 1.65 10 mol NH = 2.80 10 g NH = 28.0 kg NH 3 3 4 3 3 3 1 mol NH3

Section 5.11 The Concept of Limiting Reactant Interactive Example 5.17 - Stoichiometry: Limiting Reactant Nitrogen gas can be prepared by passing gaseous ammonia over solid copper(ii) oxide at high temperatures. The other products of the reaction are solid copper and water vapor. If a sample containing 18.1 g of NH 3 is reacted with 90.4 g of CuO, which is the limiting reactant? How many grams of N 2 will be formed?

Section 5.11 The Concept of Limiting Reactant Solution Objective To find the limiting reactant To find the mass of N 2 produced Information available The chemical reaction 18.1 g NH 3 90.4 g CuO NH (g) + CuO(s) N (g) + CU(s) + H O(g) 3 2 2

Section 5.11 The Concept of Limiting Reactant Solution Information required Balanced equation for the reaction Moles of NH 3 Moles of CuO Step 1 - Balance the equation 2NH ( g) + 3CuO( s) N ( g) + 3CU( s) + 3H O( g) 3 2 2

Section 5.11 The Concept of Limiting Reactant Solution Step 2 - Find the moles of the reactants Determine the molar masses NH 3-17.03 g/mol CuO - 79.55 g/mol 1 mol NH3 18.1 g NH3 17.03 g NH 90.4 g CuO = 1.06 mol NH3 3 1 mol CuO = 1.14 mol CuO 79.55 g CuO

Section 5.11 The Concept of Limiting Reactant Solution A. Determining the limiting reactant by comparing the moles of reactants Step 1 - Identify the ratio between NH 3 and CuO 3 mol CuO 2 mol NH3 Step 2 - Determine the moles of CuO that react with 1.06 moles of NH 3 1.06 mol NH 3 3 mol CuO = 1.59 mol CuO 2 mol NH 3

Section 5.11 The Concept of Limiting Reactant Solution The actual amount of CuO is 1.14 The required amount is 1.59 moles of CuO Step 4 - Compare the mole ratio of CuO and NH 3 with the required ratio mol CuO 3 (required) = = 1.5 mol NH 2 Thus, CuO is the limiting reactant 3 mol CuO 1.14 (actual) = = 1.08 mol NH 1.06 3

Section 5.11 The Concept of Limiting Reactant Solution B. The limiting reactant can also be determined by calculating the moles of N 2 that would be formed by complete combustion of NH 3 and CuO 1 mol N2 1.06 mol NH3 2 mol NH 1 mol N2 1.14 mol CuO 3 mol CuO = 0.530 mol N2 3 = 0.380 mol N2 CuO is limiting as it produces a smaller amount of N 2

Section 5.11 The Concept of Limiting Reactant Solution Step 1 - Determine the mass of N 2 produced using the mole ratio between N 2 and CuO 1 mol N 2 3 mol CuO 1.14 mol CuO 1 mol N 2 = 0.380 mol N2 3 mol CuO Step 2 - Determine the mass of N 2 using its molar mass 28.02 g N 0.380 mol N = 10.6 g N 2 2 2 1 mol N2

Section 5.11 The Concept of Limiting Reactant Theoretical Yield and Percent Yield Theoretical yield is the highest quantity of product that can be obtained from a given amount of limiting reactant Percent Yield is the actual amount obtained Less than theoretical yield Actual yield 100% percent yield Theoretical yield

Section 5.11 The Concept of Limiting Reactant Problem-Solving Strategy - Problems Involving Masses of Reactants and Products Write and balance the equation for the reaction Convert the known masses of substances to moles Determine which reactant is limiting Using the amount of the limiting reactant and the appropriate mole ratios, compute the number of moles of the desired product Convert from moles to grams, using the molar mass

Section 5.11 The Concept of Limiting Reactant Concept Check Which of the following reaction mixtures produces the greatest amount of product? Each involves the reaction symbolized by the following equation 2H 2 + O 2 2H2O I. 2 moles of H 2 and 2 moles of O 2 II. 2 moles of H 2 and 3 moles of O 2 III. 2 moles of H 2 and 1 mole of O 2 IV. 3 moles of H 2 and 1 mole of O 2 V. Each produce the same amount of product

Section 5.11 The Concept of Limiting Reactant Concept Check You know that chemical A reacts with chemical B. You react 10.0 g of A with 10.0 g of B. What information do you need to know in order to determine the mass of product that will be produced? Solution We need to know: The mole ratio between A, B, and the product they form. In other words, we need to know the balanced reaction equation The molar masses of A, B, and the product they form

Section 5.11 The Concept of Limiting Reactant Problem-Solving Strategy - Problems Involving Masses of Reactants and Products

Section 5.11 The Concept of Limiting Reactant Interactive Example 5.18 - Calculating Percent Yield Methanol (CH 3 OH), also called methyl alcohol, is the simplest alcohol. It is used as a fuel in race cars and is a potential replacement for gasoline. Methanol can be manufactured by combining gaseous carbon monoxide and hydrogen. Suppose 68.5 kg CO(g) is reacted with 8.60 kg H 2 (g). Calculate the theoretical yield of methanol. If 3.57 10 4 g CH 3 OH is actually produced, what is the percent yield of methanol?

Section 5.11 The Concept of Limiting Reactant Solution Primary objective To calculate the theoretical yield of methanol To calculate the percent yield of methanol Information available The chemical reaction 68.5 kg CO(g) 8.60 kg H 2 (g) H ( g) + CO( g) CH OH( l) 2 3 3.57 10 4 g CH 3 OH is produced

Section 5.11 The Concept of Limiting Reactant Solution Information needed Balanced equation for the reaction Moles of H 2 Moles of CO Which is the limiting reactant Amount of CH 3 OH produced Finding the limiting reactant Step 1 - Balance the equation 2H ( g) + CO( g) CH OH( l) 2 3

Section 5.11 The Concept of Limiting Reactant Solution Step 2 - Identify the molar masses, which will help determine moles H 2 2.016 g/mol CO 28.02 g/mol 68.5 kg CO 1000 g CO 1 mol CO 1 kg CO 28.02 g CO 3 = 2.44 10 mol CO 8.60 kg H 2 1000 g H 2 1 kg H 2 1 mol H 2 2.016 g H 2 3 = 4.27 10 mol H 2

Section 5.11 The Concept of Limiting Reactant Solution Determination of limiting reactant using reactant quantities Step 1 - Determine the mole ratio between H 2 and CO in the balanced equation 2 mol H 2 1 mol CO

Section 5.11 The Concept of Limiting Reactant Solution Step 2 - Compare the mole ratio of H 2 and CO required by the equation with the actual mole ratio The actual mole ratio is smaller than the required ratio H 2 is limiting mol H 2 2 (required) = = 2 mol CO 1 3 mol H2 4.27 10 (actual) = = 1.75 3 mol CO 2.44 10

Section 5.11 The Concept of Limiting Reactant Solution Determination of limiting reactant using quantities of products formed Step 1 - Calculate the amounts of CH 3 OH formed by complete consumption of CO(g) and H2(g) 3 2.44 10 mol CO 1 mol CH OH 3 1 mol CO 1 mol CH OH 3 3 4.27 10 mol H2 2 mol H2 Complete consumption of H 2 produces a smaller amount of CH 3 OH 3 = 2.44 10 mol CH3OH 3 = 2.14 10 mol CH3OH

Section 5.11 The Concept of Limiting Reactant Solution Calculating the theoretical yield of methanol Step 1 - Determine the moles of CH 3 OH formed 3 4.27 10 mol H 2 1 mol CH3OH 2 mol H Step 2 - Determine the theoretical yield of CH 3 OH in grams 32.04 g CH OH 3 3 2.14 10 mol CH3OH 3 The theoretical yield of CH 3 OH is 6.86 10 4 g 2 1 mol CH OH 3 = 2.14 10 mol CH3OH 4 = 6.86 10 g CH3OH

Section 5.11 The Concept of Limiting Reactant Solution Step 3 - Determine the percent yield of CH 3 OH Actual yield (grams) Theoritical yield (grams) 4 3.57 10 g CH3OH 100 = 4 6.86 10 g CH3OH 100% = 52.0%