Chapter 5 Stoichiometry · Questions to Consider ... injected into a mass spectrometer, the results...

Chapter 5

Stoichiometry

Chapter 5

Copyright © 2016 Cengage Learning. All Rights Reserved.

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 5.1Counting by Weighing


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

Copyright © Cengage Learning. All rights reserved 5

Average Mass

total mass of atoms in sampleAverage mass =

number of atoms in sample

Section 5.2Atomic Masses


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



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



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



Fig 5.1 - The Mass Spectrometer



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



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



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



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



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


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


Section 5.4The 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 12C

6.022 × 1023

Also called Avogadro’s number

One mole of a substance contains 6.022 × 1023 units of that substance


Section 5.4The 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 5.4The 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 5.4The Mole


Solution

Information needed

The mass of six americium atoms


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


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


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




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?




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



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


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



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



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.



Solution

Objective

To find the mass percent of each element in carvone


Molecular formula, C10H14O

Information needed to find the mass percent

Mass of each element (1 mole of carvone)

Molar mass of carvone



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



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


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



Fig 5.5 - Analyzing for Carbon and Hydrogen



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



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



Fig 5.6 - Substances whose Empirical and Molecular Formulas are Different



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



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



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



Solution

Objective

To find the empirical and molecular formulas for the given compound


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



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



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



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


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


4 2 2 2CH + O CO + H O

Reactants Products



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



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 Symbol

Solid (s)

Liquid (l)

Gas (g)

Aqueous solution

(aq)



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


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




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




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




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



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



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.



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–.



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


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


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




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




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




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



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



Problem-Solving Strategy - Calculating Masses of Reactants and Products



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?



Solution

Objective

To find the mass of CO2 absorbed by 1.00 kg LiOH


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



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



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



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


Limiting Reactant

The reactant that is used the

most, limiting the quantity

of the product formed




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




There are just enough molecules to ensure that all are paired




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



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




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




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




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




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



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




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




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



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?



Solution

Objective

To find the limiting reactant

To find the mass of N2 produced


The chemical reaction

18.1 g NH3

90.4 g CuO

3 2 2NH (g) + CuO(s) N (g) + CU(s) + H O(g)



Solution

Information required


Moles of NH3

Moles of CuO

Step 1 - Balance the equation

3 2 22NH ( ) + 3CuO( ) N ( ) + 3CU( ) + 3H O( )g s g s g



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



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



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



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



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



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



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



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



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



Problem-Solving Strategy - Problems Involving Masses of Reactants and Products



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?



Solution

Primary objective

To calculate the theoretical yield of methanol

To calculate the percent yield of methanol


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



Solution

Information needed


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



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



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



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



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



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



Solution

Step 3 - Determine the percent yield of CH3OH

433.57 10 g CH OHActual yield (grams)

100 = Theoritical yield (grams)

436.86 10 g CH OH

100% = 52.0%

Chapter 5 Stoichiometry · Questions to Consider ... injected into a mass spectrometer, the results...

Documents

Transcript of Chapter 5 Stoichiometry · Questions to Consider ... injected into a mass spectrometer, the results...