chapter 5 stoichiometry · questions to consider ... injected into a mass spectrometer, the results...
TRANSCRIPT
Chapter 5
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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
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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
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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 5.1Counting by Weighing
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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
Atoms do not need to be identical in order to be counted by weighing
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Average Mass
total mass of atoms in sampleAverage mass =
number of atoms in sample
Section 5.2Atomic Masses
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The modern system of determining atomic masses was first used in 1961
Based on 12C
12 atomic mass units (u)
The most accurate method currently used to compare the masses of atoms involves the use of the mass spectrometer
Atomic Masses
Section 5.2Atomic Masses
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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 5.2Atomic Masses
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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 5.2Atomic Masses
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Fig 5.1 - The Mass Spectrometer
Section 5.2Atomic Masses
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Determining Atomic Masses
Consider the analysis of 12C and 13C in a mass spectrometer
Based on the definition of the atomic mass unit
The average mass for an element is also referred to as the average atomic mass or atomic mass
13
12
Mass C 1.0836129
Mass C
13Mass of C = (1.0836129)(12 u) = 13.003355 u
Exact number by definition
Section 5.2Atomic Masses
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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 63Cu and 65Cu are 62.93 u and 64.93 u, respectively
Section 5.2Atomic Masses
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Solution
Information needed
To calculate the average mass of natural copper
Information available
63Cu mass = 62.93 u
65Cu 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 63Cu and 30.91 are 65Cu
Section 5.2Atomic Masses
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Solution
Determining the mass of 100 atoms of natural copper
The average mass of a copper atom is
(69.09 atomsu
) (62.93 atom
u) + (30.91 atoms)(64.93
atom) = 6355 u
6355 u = 63.55 u/atom
100 atoms
Section 5.2Atomic Masses
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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 5.3Learning to Solve Problems
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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
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Section 5.4The Mole
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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 12C
6.022 × 1023
Also called Avogadro’s number
One mole of a substance contains 6.022 × 1023 units of that substance
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Section 5.4The Mole
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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 5.4The Mole
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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 5.4The Mole
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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 atomsu
× 243atom
3 = 1.46 × 10 u
Section 5.4The Mole
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Solution
Step 2 - Use Avogadro’s number to convert atomic mass units to grams
Step 3 - Determine the mass of six americium ions in grams
23
23
6.022 10 u = 1 g
1 g
6.022 10 u
31.46 × 10 u23
1 g ×
6.022 × 10 u
–21 = 2.42 × 10 g
Section 5.5Molar Mass
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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
CaCO3 - Calcium carbonate
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Section 5.5Molar Mass
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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 C10H6O3.
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?
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Section 5.5Molar Mass
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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:
The molar mass of juglone is 174.1 g
10 6 3
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
Section 5.5Molar Mass
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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:
–21.56 10 g juglone1 mol juglone
174.1 g juglone
–5 = 8.96 10 mol juglone
Section 5.6Percent Composition of Compounds
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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 (C2H5OH)
Mass of C = 2 molg
× 12.01 mol
= 24.02 g
Mass of H = 6 molg
×1.008 mol
= 6.048 g
Mass of O = 1 molg
×16.00 mol
2 5
= 16.00 g
Mass of 1 mol C H OH = 46.07 g
Section 5.6Percent Composition of Compounds
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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
2 5
2 5
mass of C in 1 mol of C H OHMass percent of C = 100%
mass of 1 mol of C H OH
24.02g = × 100 = 52.14%
46.07g
Section 5.6Percent Composition of Compounds
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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 (C10H14O) 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 5.6Percent Composition of Compounds
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Solution
Objective
To find the mass percent of each element in carvone
Information available
Molecular formula, C10H14O
Information needed to find the mass percent
Mass of each element (1 mole of carvone)
Molar mass of carvone
Section 5.6Percent Composition of Compounds
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Solution
Step 1 - Determine the mass of each element on one mole of C10H14O
Mass of C in 1 mol = 10 molg
× 12.01 mol
= 120.1 g
Mass of H in 1 mol = 14 molg
× 1.008 mol
= 14.11 g
Mass of O in 1 mol = 1 molg
× 16.00 mol
= 16.00 g
Section 5.6Percent Composition of Compounds
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Solution
Step 2 - Determine the molar mass of C10H14O
Step 3 - Determine the mass of each element
10 14 10 14
120.1 g + 14.11 g 16.00 150.2 g
C + H + O C H O
g
10 14
10 14
10 14
120.1 g CMass percent of C = × 100% = 79.96%
150.2 g C H O
14.11 g HMass percent of H = × 100% = 9.394%
150.2 g C H O
16.00 g OMass percent of O = × 100% = 10.65%
150.2 g C H O
Section 5.7Determining the Formula of a Compound
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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 CO2, H2O, etc., which are collected and weighed
Section 5.7Determining the Formula of a Compound
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Fig 5.5 - Analyzing for Carbon and Hydrogen
Section 5.7Determining the Formula of a Compound
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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 5.7Determining the Formula of a Compound
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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 5.7Determining the Formula of a Compound
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Fig 5.6 - Substances whose Empirical and Molecular Formulas are Different
Section 5.7Determining the Formula of a Compound
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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 5.7Determining the Formula of a Compound
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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 5.7Determining the Formula of a Compound
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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 5.7Determining the Formula of a Compound
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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 5.7Determining the Formula of a Compound
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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 Cl1 mol Cl
35.45 g Cl
= 2.021 mol Cl
Section 5.7Determining the Formula of a Compound
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Solution
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 ClCH2.
24.27 g C1 mol C
12.01 g C
= 2.021 mol C
4.07 g H1 mol H
1.008 g H
= 4.04 mol H
Section 5.7Determining the Formula of a Compound
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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
The substance comprises molecules of Cl2C2H4
2 2 2 2 4
Molar mass 98.96 g/mol = = 2
Empirical formula mass 49.48 g/mol
Molecular Formula = (ClCH ) = Cl C H
Section 5.8Chemical Equations
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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
Bonds are broken and new bonds are formed
Both sides possess the same number of atoms
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4 2 2 2CH + O CO + H O
Reactants Products
Section 5.8Chemical Equations
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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
4 2 2 2CH + 2O CO + 2H O
Section 5.8Chemical Equations
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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
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State Symbol
Solid (s)
Liquid (l)
Gas (g)
Aqueous solution
(aq)
Section 5.8Chemical Equations
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Table 5.2 - Balanced Equation for the Combustion of Methane
4 2 2 2CH ( ) + 2O ( ) CO ( ) + 2H O( )g g g g
Section 5.9Balancing Chemical Equations
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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
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Section 5.9Balancing Chemical Equations
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Balancing an Equation
Start with the most complicated molecules
Consider the following unbalanced equation
The most complicated molecule is C2H5OH
Balancing carbon
2 5 2 2 2C H OH( ) + O ( ) CO ( ) + H O( )l g g g
2 5 2 2 2
2 C atoms 2 C atoms
C H OH( ) + O ( ) 2CO ( ) + H O( )l g g g
Section 5.9Balancing Chemical Equations
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Balancing an Equation
Balancing hydrogen atoms
Balancing oxygen atoms
2 5 2 2 2
(5 +1) H (3 2) H
C H OH( ) + O ( ) 2CO ( ) + 3H O( )l g g g
2 5 2 2 2
7O 7O
1O 6O (2 2)O 3O
C H OH( ) + 3O ( ) 2CO ( ) + 3H O( )l g g g
Section 5.9Balancing Chemical Equations
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Balancing an Equation
Verifying the results
2 5 2 2 2
2 C atoms 2 C atoms
6 H atoms 6 H atoms
C H OH( ) + 3O ( ) 2CO ( ) + 3H O( )l g g g
7 O atoms 7 O atoms
Section 5.9Balancing Chemical Equations
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Concept Check
Which of the following is true about balanced chemical equations?
I. The number of molecules is conserved
II. The coefficients tell you how much of each substance you have
III. 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 5.9Balancing Chemical Equations
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Interactive Example 5.13 - Balancing a Chemical Equation I
Chromium compounds exhibit a variety of bright colors. When solid ammonium dichromate, (NH4)2Cr2O7, 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 N2
molecules), and water vapor. Balance the equation for this reaction.
Section 5.9Balancing Chemical Equations
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Solution
From the description given, the reactant is solid ammonium dichromate, (NH4)2Cr2O7(s), and the products are nitrogen gas, N2(g), water vapor, H2O(g), and solid chromium(III) oxide, Cr2O3(s). The formula for chromium(III) oxide can be determined by recognizing that the Roman numeral III means that Cr3+ ions are present. For a neutral compound, the formula must then be Cr2O3, since each oxide ion is O2–.
Section 5.9Balancing Chemical Equations
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Solution
The unbalanced equation is:
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 H2O balances the hydrogen atoms
4 2 2 7 2 3 2 2(NH ) Cr O ( ) Cr O ( ) + N ( ) + H O( )s s g g
4 2 2 7 2 3 2 2
(4 2) H (4 2) H
(NH ) Cr O ( ) Cr O ( ) + N ( ) + 4H O( )s s g g
Section 5.10Stoichiometric Calculations
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Stoichiometry of Reactions - Principles
Consider the following reaction between propane and oxygen
Calculating the number of moles of propane present in 96.1 grams (the molar mass of propane is 44.1)
Constructing a mole ratio
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3 8 2 2 2C H ( ) + 5O 3CO ( ) + 4H O( )g g g
396.1 g C H 3 8
83 8
1 mol C H ×
44.1 g C H3 8= 2.18 mol C H
2
3 8
5 mol O
1 mol C H
Section 5.10Stoichiometric Calculations
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Stoichiometry of Reactions - Principles
Calculating the moles of O2 needed
Converting moles into grams
Therefore, 349 grams of oxygen can burn 96.1 grams of propane
3 82.18 mol C H 2
3 8
5 mol O
1 mol C H 2 = 10.9 mol O
210.9 mol O 2
2
32.0 g O
1 mol O 2 = 349 g O
Section 5.10Stoichiometric Calculations
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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
2
3 8
3 mol CO
1 mol C H
3 82.18 mol C H 2
3 8
3 mol CO
1 mol C H 2 = 6.54 mol CO
Section 5.10Stoichiometric Calculations
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Stoichiometry of Reactions - Principles
Calculating the mass of CO produced using the molar mass of CO2 (44.0 g/mol)
22 2
2
44.0 g CO6.54 mol CO = 288 g CO
1 mol CO
Section 5.10Stoichiometric Calculations
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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 5.10Stoichiometric Calculations
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Problem-Solving Strategy - Calculating Masses of Reactants and Products
Section 5.10Stoichiometric Calculations
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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 5.10Stoichiometric Calculations
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Solution
Objective
To find the mass of CO2 absorbed by 1.00 kg LiOH
Information available
Chemical reaction
1.00 kg LiOH
Information needed to find the mass of CO2
Balanced equation for the reaction
2 2 3 2LiOH( ) + CO ( ) Li CO ( ) + H O( )s g g l
Section 5.10Stoichiometric Calculations
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Solution
Step 1 - State the balanced equation for the reaction
Step 2 - Determine the moles of LiOH
Determine the molar mass to find the moles of LiOH
Using the molar mass
2 2 3 22LiOH( ) + CO ( ) Li CO ( ) + H O( )s g g l
6.941 + 16.00 + 1.008 = 23.95 g/mol
1.00 kg LiOH1000 g LiOH
1 kg LiOH
1 mol LiOH
23.95 g LiOH = 41.8 mol LiOH
Section 5.10Stoichiometric Calculations
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Solution
Step 3 - Determine the mole ratio between CO2 and LiOH in the balanced equation
Step 4 - Calculate the moles of CO2
21 mol CO
2 mol LiOH
41.8 mol LiOH 21 mol CO
2 mol LiOH 2 = 20.9 mol CO
Section 5.10Stoichiometric Calculations
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Solution
Step 5 - Determine the mass of CO2 formed from 1.00 kg LiOH
Thus, 920. g of CO2 (g) will be absorbed by 1.00 kg of LiOH(s)
220.9 mol CO 2
2
44.0 g CO
1mol CO 2
2 = 9.20 10 g CO
Section 5.11The Concept of Limiting Reactant
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Limiting Reactant
The reactant that is used the
most, limiting the quantity
of the product formed
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Section 5.11The Concept of Limiting Reactant
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Stoichiometric Mixture
Consider the reaction between nitrogen and hydrogen, forming ammonia
Each molecule of ammonia possesses:
1 N2 molecule
3 H2 molecules
2 2 3N ( ) 3H ( ) 2NH ( )g g g
Section 5.11The Concept of Limiting Reactant
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Stoichiometric Mixture
There are just enough molecules to ensure that all are paired
Section 5.11The Concept of Limiting Reactant
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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 reactantdetermines the amount of the product formed
Section 5.11The Concept of Limiting Reactant
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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
2 2 3N ( ) + 3H ( ) 2NH ( )g g g
Section 5.11The Concept of Limiting Reactant
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Determination of Limiting Reactant Using Reactant Quantities
Determination of the moles of the reactants
Calculating the total amount of moles of H2 that react with 8.93 × 102 moles of N2
225.0 kg N21000 g N
21 kg N
2
2
1 mol N
28.0 g N 2
2
2
8.93 10 mol N
5.00 kg H
21000 g H
21 kg H2
2
1 mol H
2.016 g H 3
2 2.48 10 mol H
228.93 10 mol N 2
2
3 mol H
1 mol N 3
2 = 2.68 10 mol H
Section 5.11The Concept of Limiting Reactant
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Determination of Limiting Reactant Using Reactant Quantities
2.48 × 103 moles of H2 requires 8.27 × 102 moles of N2
Nitrogen is in excess (8.93× 102 moles)
Hydrogen is the limiting reactant
Section 5.11The Concept of Limiting Reactant
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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
This means that
2
2
3 mol H
1 mol N
2
2
mol H 3(required) = 3
mol N 1
Section 5.11The Concept of Limiting Reactant
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Determination of Limiting Reactant Using Reactant Quantities
Adding values of the experiment
The actual mole ratio of H2 to N2 is too small
H2 is the limiting factor
32
22
mol H 2.48 10(actual) = 2.78
mol N 8.93 10
Section 5.11The Concept of Limiting Reactant
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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 5.11The Concept of Limiting Reactant
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Determination of Limiting Reactant Using Quantities of Products Formed
Consider the reaction between 25.0 kg (8.93 × 102
moles) of nitrogen and 5.00 kg (2.48 × 103 moles) of hydrogen
Determining the amount of NH3 that can form
2 332 3
2
2 mol NH8.93 10 mol N = 1.79 10 mol NH
1 mol N
Section 5.11The Concept of Limiting Reactant
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Determination of Limiting Reactant Using Quantities of Products Formed
Calculating the amount of NH3 formed upon complete consumption of H2
The amount of NH3 produced from H2 is lesser than that from N2
Converting moles to kilograms
3 332 3
2
2 mol NH2.48 10 mol H = 1.65 10 mol NH
3 mol H
3 433 3 3
3
17 g NH1.65 10 mol NH = 2.80 10 g NH = 28.0 kg NH
1 mol NH
Section 5.11The Concept of Limiting Reactant
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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 NH3 is reacted with 90.4 g of CuO, which is the limiting reactant? How many grams of N2 will be formed?
Section 5.11The Concept of Limiting Reactant
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Solution
Objective
To find the limiting reactant
To find the mass of N2 produced
Information available
The chemical reaction
18.1 g NH3
90.4 g CuO
3 2 2NH (g) + CuO(s) N (g) + CU(s) + H O(g)
Section 5.11The Concept of Limiting Reactant
Copyright © 2016 Cengage Learning. All Rights Reserved.
Solution
Information required
Balanced equation for the reaction
Moles of NH3
Moles of CuO
Step 1 - Balance the equation
3 2 22NH ( ) + 3CuO( ) N ( ) + 3CU( ) + 3H O( )g s g s g
Section 5.11The Concept of Limiting Reactant
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Solution
Step 2 - Find the moles of the reactants
Determine the molar masses
NH3 - 17.03 g/mol
CuO - 79.55 g/mol
318.1 g NH 3
3
1 mol NH
17.03 g NH 3 = 1.06 mol NH
90.4 g CuO1 mol CuO
79.55 g CuO
= 1.14 mol CuO
Section 5.11The Concept of Limiting Reactant
Copyright © 2016 Cengage Learning. All Rights Reserved.
Solution
A. Determining the limiting reactant by comparing the moles of reactants
Step 1 - Identify the ratio between NH3 and CuO
Step 2 - Determine the moles of CuO that react with 1.06 moles of NH3
3
3 mol CuO
2 mol NH
31.06 mol NH3
3 mol CuO
2 mol NH = 1.59 mol CuO
Section 5.11The Concept of Limiting Reactant
Copyright © 2016 Cengage Learning. All Rights Reserved.
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 NH3 with the required ratio
Thus, CuO is the limiting reactant
3
3
mol CuO 3 (required) = = 1.5
mol NH 2
mol CuO 1.14 (actual) = = 1.08
mol NH 1.06
Section 5.11The Concept of Limiting Reactant
Copyright © 2016 Cengage Learning. All Rights Reserved.
Solution
B. The limiting reactant can also be determined by calculating the moles of N2 that would be formed by complete combustion of NH3 and CuO
CuO is limiting as it produces a smaller amount of N2
31.06 mol NH 2
3
1 mol N
2 mol NH 2 = 0.530 mol N
1.14 mol CuO 21 mol N
3 mol CuO 2 = 0.380 mol N
Section 5.11The Concept of Limiting Reactant
Copyright © 2016 Cengage Learning. All Rights Reserved.
Solution
Step 1 - Determine the mass of N2 produced using the mole ratio between N2 and CuO
Step 2 - Determine the mass of N2 using its molar mass
21 mol N
3 mol CuO
1.14 mol CuO 21 mol N
3 mol CuO 2 = 0.380 mol N
22 2
2
28.02 g N 0.380 mol N = 10.6 g N
1 mol N
Section 5.11The Concept of Limiting Reactant
Copyright © 2016 Cengage Learning. All Rights Reserved.
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 yield100% percent yield
Theoretical yield
Section 5.11The Concept of Limiting Reactant
Copyright © 2016 Cengage Learning. All Rights Reserved.
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 5.11The Concept of Limiting Reactant
Copyright © 2016 Cengage Learning. All Rights Reserved.
Concept Check
Which of the following reaction mixtures produces the greatest amount of product? Each involves the reaction symbolized by the following equation
I. 2 moles of H2 and 2 moles of O2
II. 2 moles of H2 and 3 moles of O2
III. 2 moles of H2 and 1 mole of O2
IV. 3 moles of H2 and 1 mole of O2
V. Each produce the same amount of product
2 2 22H + O 2H O
Section 5.11The Concept of Limiting Reactant
Copyright © 2016 Cengage Learning. All Rights Reserved.
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 5.11The Concept of Limiting Reactant
Copyright © 2016 Cengage Learning. All Rights Reserved.
Problem-Solving Strategy - Problems Involving Masses of Reactants and Products
Section 5.11The Concept of Limiting Reactant
Copyright © 2016 Cengage Learning. All Rights Reserved.
Interactive Example 5.18 - Calculating Percent Yield
Methanol (CH3OH), 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 H2(g). Calculate the theoretical yield of methanol. If 3.57 × 104 g CH3OH is actually produced, what is the percent yield of methanol?
Section 5.11The Concept of Limiting Reactant
Copyright © 2016 Cengage Learning. All Rights Reserved.
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 H2(g)
3.57 × 104 g CH3OH is produced
2 3H ( ) + CO( ) CH OH( )g g l
Section 5.11The Concept of Limiting Reactant
Copyright © 2016 Cengage Learning. All Rights Reserved.
Solution
Information needed
Balanced equation for the reaction
Moles of H2
Moles of CO
Which is the limiting reactant
Amount of CH3OH produced
Finding the limiting reactant
Step 1 - Balance the equation
2 32H ( ) + CO( ) CH OH( )g g l
Section 5.11The Concept of Limiting Reactant
Copyright © 2016 Cengage Learning. All Rights Reserved.
Solution
Step 2 - Identify the molar masses, which will help determine moles
H2 2.016 g/mol
CO 28.02 g/mol
68.5 kg CO1000 g CO
1 kg CO
1 mol CO
28.02 g CO 3
2
= 2.44 10 mol CO
8.60 kg H
2 1000 g H
21 kg H2
2
1 mol H
2.016 g H 3
2 = 4.27 10 mol H
Section 5.11The Concept of Limiting Reactant
Copyright © 2016 Cengage Learning. All Rights Reserved.
Solution
Determination of limiting reactant using reactant quantities
Step 1 - Determine the mole ratio between H2 and CO in the balanced equation
22 mol H
1 mol CO
Section 5.11The Concept of Limiting Reactant
Copyright © 2016 Cengage Learning. All Rights Reserved.
Solution
Step 2 - Compare the mole ratio of H2 and CO required by the equation with the actual mole ratio
The actual mole ratio is smaller than the required ratio
H2 is limiting
2
32
3
mol H 2 (required) = = 2
mol CO 1
mol H 4.27 10 (actual) = = 1.75
mol CO 2.44 10
Section 5.11The Concept of Limiting Reactant
Copyright © 2016 Cengage Learning. All Rights Reserved.
Solution
Determination of limiting reactant using quantities of products formed
Step 1 - Calculate the amounts of CH3OH formed by complete consumption of CO(g) and H2(g)
Complete consumption of H2 produces a smaller amount of CH3OH
32.44 10 mol CO 31 mol CH OH
1 mol CO 3
3
32
= 2.44 10 mol CH OH
4.27 10 mol H
3
2
1 mol CH OH
2 mol H 3
3 = 2.14 10 mol CH OH
Section 5.11The Concept of Limiting Reactant
Copyright © 2016 Cengage Learning. All Rights Reserved.
Solution
Calculating the theoretical yield of methanol
Step 1 - Determine the moles of CH3OH formed
Step 2 - Determine the theoretical yield of CH3OH in grams
The theoretical yield of CH3OH is 6.86 × 104 g
324.27 10 mol H 3
2
1 mol CH OH
2 mol H 3
3 = 2.14 10 mol CH OH
332.14 10 mol CH OH 3
3
32.04 g CH OH
1 mol CH OH 4
3 = 6.86 10 g CH OH