Lecture 2: Stoichiometry

Using the information in chemical equations

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Expressing mass in chemistry◦ Formula weight◦ Molecular weight◦ Molar mass

Concept of the mole Atomic mass % Empirical formula (simplest formula) Molecular formula (actual formula) Combustion analysis

Part I: Basic Skills

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Learning Catalytics

Formula weight (FW) ◦ sum of masses of all atoms in the chemical formula

e.g. NaCl FW = 23.0 amu + 35.5 amu = 58.5 amu

Molecular weight (MW)◦ sum of masses of atoms in the molecular formula

e.g. H2O MW = 2×1.0 amu + 16.0 amu = 18.0 amu

Molecular compounds have FW and MW Non-molecular compounds have only FW

Formula Weight & Molecular Weight

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Number of atoms in 12 g of 12C◦ 1 mole = 6.022×1023

Avogadro’s number (NA) ◦ 6.022x1023 mol-1 (note the units)

Molar mass (M) ◦ mass of 1 mol of molecular formula ◦ e.g. MHCl = 36.5 g mol-1

NOTE: Molar mass (in g/mol) is numerically equal to the FW (in amu)

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The Mole (abbrev.= mol)

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One Mole Each:

Mass Moles: use molar mass (n = m/M)e.g. How many moles in 168 g of water?

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Mass Moles Number of Particles

1-mol g 02.18

g 168Mm

n g 02.18

mol 1g 168 mol 9.32

Moles Particles: use Avogadro’s numbere.g. How many molecules in 168 g of water?

g 18.02mol 1

g 168 1-23 mol 10022.6 24105.61

How many moles of nitrate present when 5.00 g of aluminum nitrate dissolved in water?

Aluminum Nitrate = Al(NO3)3 (or AlN3O9)

Molar Mass = MAl + 3MN + 9MO

= (26.982) + 3×(14.0067) + 9×(15.9994) = 212.997 g/molMoles of Al(NO3)3 = (5.00 g)/(212.997 g/mol)

= 0.02347 molMoles NO3

- = 3 × Moles Al(NO3)3 = 0.0704 mol

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Another example

Empirical formula: Relative numbers of atoms in a substance◦ can be obtained from elemental analysis data

Elemental analysis data◦ relative masses of elements in compound◦ can convert to relative numbers of atoms using

the atomic masses

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Application of the Mole Concept

A compound contains 75.9% carbon, 17.7% nitrogen, and 6.4% hydrogen by mass. What is the empirical formula?

Step 1: Assume 100 g of sample. 100 g contains

◦ 75.9 g of C◦ 17.7 g of N◦ 6.4 g of H.

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Example

Step 2: Convert the masses of atoms in the sample to moles, using the molar masses of the atoms

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Example

C mol 6.32C g 12.011

C mol 1 C g 9.75

N mol 26.1N g 14.0067

N mol 1 N g 7.71

H mol 3.6H g 1.0079

H mol 1 H g .46

Step 3: Calculate the mole ratio Divide moles from step 2 by the smallest

value

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Example

26.126.1

:26.13.6

:26.132.6

:: NHC :15.05.01:

The empirical formula is C5H5N Note:

◦ 5.01 is not exactly 5◦ Remember:

It is a measurement Measurements always have errors

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Example

A compound contains C (63.8%), H (6.4%) and N (29.8%). What is the empirical formula?

A. C2.5H3N1

B. C64H6N30

C. C5.3H6.3N2.1

D. C5H6N2

E. C10H12N4

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LC: Empirical Formula

A compound contains C (63.8%), H (6.4%) and N (29.8%). What is the empirical formula?

In 100 g of sample there are: 63.8 g C(1 mol/12.011 g) = 5.31 moles C 6.4 g H(1 mol/1.0079 g) = 6.3 moles H 29.8 g N(1 mol/14.007 g) = 2.13 moles N

The mole ratio is therefore:C : H : N = 2.49 : 3.0 : 1

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Solution: Empirical Formula

2.49, is not a whole number ratio of N◦ In between 2 & 3◦ Not likely experimental error◦ Close to 2.5, or 5/2

Multiply the ratio by 2 to get whole numbers2 x ( C : H : N ) = 5 : 6 : 2

The empirical formula is C5H6N2

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Solution: Empirical Formula

Strategy:

Mass % Analysis

Data

Work out molesof atoms in

100g of sample

Divide by lowestnumber of molesto get mole ratio

Ismole ratio

simple, wholenumbers

?Multiply by smallestfactor that will result

in whole-number ratioNo

Report result(empirical formula)

Yes

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Need one more thing: Molar mass of cmpd

e.g. MW of compound is 136.15 g/mol. Empirical formula is C4H4O. What is molecular formula?

FW = (4x12.01)+(4x1.01)+16.00= 68.08 g mol-1

About half the molar massTherefore, the molecular formula is C8H8O2

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Molecular Formula from Empirical Formula

Used to find empirical formulae of molecules containing only C, H & O◦ C converted to CO2

◦ H is converted to H2O

CO2 and H2O trapped and weighed◦ mass % of C and H are found from stoichiometry◦ 1 C atom per CO2 trapped; 2 H atoms per H2O

Remaining mass attributed to O

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e.g. Combustion Analysis

Stoichiometric equation for combustion of a compound of C, H and O is:

CXHYOZ + {X+Y/4-Z/2} O2 X CO2 +Y/2 H2O

moles CO2 (RHS) = moles C atom (LHS) moles H2O (RHS) = 1/2 moles H atom (LHS)

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Combustion of CHO Compounds

1.663 g of a C-H-O compound is combusted, producing 3.48 g of CO2 and 0.712 g of H2O. What is the empirical formula of the compound?

1 mole of C atom produces 1 mole of CO2, therefore,

Amt. C =

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Example

2CO g 48.3

= 0.0791 mol C2

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CO g 01.44CO mol 1

2CO mol 1C mol 1

1 mole of H2O produced implies 2 moles of H in the compound

Amt H =

H and C are in a 1:1 ratio in the original cmpd Combined mass of H and C is:

◦ (0.0791 mol)(12.01 g/mol) C +(0.0790 mol)(1.008 g/mol)

H = 1.03 g

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Example

OH g .7120 2

= 0.0790 mol HOH g 8.021

OH mol 1

2

2OH mol 1

H mol 2

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The remainder of the compound is O

Mass of O = Mass of Cmpd - Mass of C+H = 1.663 g - 1.030 g

= 0.633 g

Moles of O = (0.633 g)/(16.00 g mol-1)= 0.0396 mol

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Example

The mole ratio, C:H:O is 0.0791 : 0.0790 : 0.0396

Or, dividing by 0.0396, 2.00 : 1.99 : 1

Therefore, the empirical formula is C2H2O

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Example

Suggested Chapter 3 Exercises Review Questions: 16-22. Problems by Topic, Cumulative Problems,

Challenge Problems: 87, 89, 91, 95, 99, 107, 109, 113, 115, 117, 121, 125, 131, 139, 149.

Note: answers to all odd-numbered problems are found in Appendix III.

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“Molar” interpretation of stoichiometric coefficients

Stoichiometric ratios Limiting reagents Theoretical Yield Percentage yield

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Part II: Stoichiometry

Please review § 4.2 (self-study) to remind yourself how to write balanced chemical equations.

The balanced chemical equation is key to stoichiometry.

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Balancing Chemical Equations

Quantitative aspects of reactions Relative moles of species in the reaction Key: balanced chemical equation From the equation can derive mole ratios

◦ Like “conversion factors” in dimensional analysis◦ Convert reagent consumed into product formed

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Stoichiometry

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Found from stoichiometric equation Convert between equivalent molar amounts

of reactants and products in a particular reaction

Treated like unit conversion factors◦ e.g., in the previous example, we used

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Stoichiometric Ratios

OH mol 1

H mol 2

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Top and bottom are equivalentfactor equivalent to 1

Before constructing stoichiometric ratios:◦ Ensure equation is balanced

Interpret values as moles ofe.g. for H2(g) + 1/2O2(g) H2O(l)

1 mol H2 0.5 mol O2 1 mol H2O

Ratios:

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Warning! Warning! Danger!

2

2

H mol 1O mol .50

2

2

O mol 5.0OH mol 1

OH mol 1H mol 1

2

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Students sometimes use stoichiometric ratios with mass data, without converting to moles

e.g. Find mass O2 req’d to form 16.7 g Fe2O3 from Fe3O4?

Fe3O4(s) + O2(g) Fe2O3(s)

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Typical Pitfall

2 31/2

Balance First

Amateur error: Do not do this in a test!

mass O2 = 16.7 g Fe2O3 32

2

OFe 3O 5.0

= 2.78 g O2 WR

ONG

Stoichiometric ratios are MOLAR ratios Convert moles of one substance into moles of

another substance

moles O2 = 16.7g Fe2O3

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The Right Way

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OFe mol 3O mol 5.0

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OFe g 59.71OFe mol 1

Convert tomoles Fe2O3

Convert tomoles O2

= 1.7410-2 mol O2

mass O2 = 1.7410-2 mol O2

2

2

O mol 1O g 32.00

= 0.557 g O2

In chemistry, the fundamental unit of “amount of a substance” is the mole

Stoichiometry shows quantitative relationship between reactants and products in a reaction

Stoichiometric info is summarised in the balanced chemical equation

Data in mass units usually has to be converted to moles before it is of ANY use

The conversion factor between mass and moles is the molar mass (units: g mol-1)

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Take-Home Message

Concept: Limiting Reagents

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For Those Who Don’t Bake:

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Limiting reagent (reactant) is completely consumed in a reaction◦ Other reagents are said to be in excess

Limiting reagent limits amount of product that can form◦ Basis for “theoretical yield”

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Limiting Reagent & Theoretical Yield

Consider the following reaction: 2 SO2 (g) + O2 (g) + 2 H2O (l) 2 H2SO4 (aq)

Molar masses are: MSO2 = 64.06; MH2SO4 = 98.08; MO2

= 32.00.In this reaction, 21.0 g of SO2 reacts with 5.00 g of

O2 and an excess amount of H2O.

Answer two LC questions:1. What is the limiting reagent?2. What mass of H2SO4 is produced?

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LC: Limiting Reagent & Yield

Consider the following reaction:2 SO2 (g) + O2 (g) + 2 H2O (l) 2 H2SO4 (aq)

Molar masses are: MSO2 = 64.06; MH2SO4 = 98.08; MO2 = 32.00.

In this reaction, 21.0 g of SO2 reacts with 5.00 g of O2 and an excess amount of H2O.

What is the limiting reagent?A. SO2

B. O2

C. H2O

D. H2SO4

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LC: Limiting Reagents

Convert masses to moles:◦ amt. SO2 = (21.0 g)/(64.06 g mol-1) = 0.328 mol◦ amt. O2 = (5.00 g)/(32.00 g mol-1) = 0.156 mol◦ amt. H2O = excess (given)

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Solution

Determine whether O2 or SO2 is the limiting reagent from the stoichiometry.

(0.328 mol SO2)(1mol O2)/(2mol SO2) = 0.164 mol O2

Need 0.164 mol O2 to react with available SO2

◦ Have only 0.156 mol O2

◦ Therefore, O2 is limiting reagent

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Solution

Consider the following reaction:2 SO2 (g) + O2 (g) + 2 H2O (l) 2 H2SO4 (aq)

Molar masses are: MSO2 = 64.06; MH2SO4 = 98.08; MO2 = 32.00.

In this reaction, 21.0 g of SO2 reacts with 5.00 g of O2 and an excess amount of H2O.

What mass of H2SO4 is produced?

A. 0.312 g H2SO4

B. 0.328 g H2SO4

C. 26.0 g H2SO4

D. 30.6 g H2SO4

E. 32.2 g H2SO4

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LC: Yield

How much H2SO4 if all the O2 is consumed?(0.156 mol O2)×(2 mol H2SO4)/(1 mol O2)

= 0.312 mol H2SO4

Convert to mass:(0.312 mol H2SO4)×(98.07 g/mol) = 30.6 g H2SO4

This is the Theoretical Yield of H2SO4

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Solution

Convert masses to moles Divide moles of each reagent by its

stoichiometric constant Lowest ratio is limiting reagent. e.g. In the preceding example,

◦ SO2: (0.328 mol)/(2 mol) = 0.164

◦ O2: (0.156 mol)/(1 mol) = 0.156

(limiting reagent)

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Determining Limiting Reagents the Easy Way

Theoretical yield: Mass of product formed if a reaction goes to completion (limiting reagent completely consumed)

Percent yield: Ratio of actual mass of product formed, divided by the theoretical yield, expressed as a percentage.

N.B. To find the theoretical yield, you must first determine the limiting reagent.

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Theoretical Yield

4.21g of cyclohexene (C6H10) reacts in presence of 10.0 g Br2 to form 12.1 g of 1,2-dibromocyclohexane (C6H10Br2) as:

C6H10 (l) + Br2 (l) C6H10Br2 (l)

Find the theoretical and percent yields:McHx= 82.14 g/mol

MBr2= 159.8 g/mol

McHxBr2 = 241.9 g/mol

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Example

Step 1: Convert masses to molesC6H10: (4.21 g)(1 mol/82.14 g) = 0.0512 molBr2: (10.0 g)(1 mol/159.8 g) = 0.0625 mol

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Solution

Step 2: Find the limiting reagentC6H10 (both stoichiometric constants are 1)

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Solution

Step 3: Find theoretical yield of C6H10Br2 amount: (0.0512 mol C6H10)(1mol C6H10Br2/1mol

C6H10)= 0.0512 mol C6H10Br2

mass: (0.0512 mol)(241.9 g/mol) = 12.4 g

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Solution

Step 4: Find % Yield = [(12.1 g)/(12.4 g)]x100% = 97.6%

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Solution

Stoichiometry is the quantitative relationship between the reactants and products

Think in MOLES! ◦ molar mass converts mass to moles and vice

versa Limiting reagent determines theoretical

yield◦ Know how to determine the limiting reagent

quickly

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Summary

Suggested Chapter 4 Exercises Review Questions: 1-4, 16. Problems by Topic, Cumulative Problems,

Challenge Problems: 25, 29, 33, 35, 37, 61, 65, 67, 71, 73, 75, 77, 79, 83, 85, 107, 109, 125.

Note: answers to all odd-numbered problems are found in Appendix III.

A solutions manual (with full solutions to odd-numbered problems) can be found through Mastering Chemistry: log in, click on Study Area, then click on Selected Solutions Manual, then choose the chapter.

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