me cchapter 4

Chapter 4

Calculations and the Chemical Equation

Denniston Topping Caret

7th Edition

Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

4.1 The Mole Concept and Atoms

• Atoms are exceedingly small– Unit of measurement for mass of an atom is

atomic mass unit (amu) – unit of measure for the mass of atoms

• carbon-12 assigned the mass of exactly 12 amu

• 1 amu = 1.66 x 10-24 g

• Periodic table gives atomic weights in amu

• What is the atomic weight of one atom of fluorine? Answer: 19.00 amu

• What would be the mass of this one atom in grams?

• Chemists usually work with much larger quantities

– It is more convenient to work with grams than amu when using larger quantities

Mass of Atoms4.

1 T

he M

ole

Con

cept

and

A

tom

s

atom F

F g10156.3

Famu 1

g101.661

atom F

Famu 19.00 23-24

• A practical unit for defining a collection of atoms is the mole

1 mole of atoms = 6.022 x 1023 atoms

• This is called Avogadro’s number– This has provided the basis for the concept

of the mole

The Mole and Avogadro’s Number

4.1

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4.1

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The Mole

• To make this connection we must define the mole as a counting unit

– The mole is abbreviated mol

• A mole is simply a unit that defines an amount of something– Dozen defines 12– Gross defines 144

F mol 1

F atom10022.6

Famu 1

F g1066.1

F atom 1

Famu 00.19 2324

=19.00 g F/mol F or 19.00 g/mol F4.1

The

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Atomic Mass• The atomic mass of one atom of an element

corresponds to:

– The average mass of a single atom in amu

– The mass of a mole of atoms in grams

– 1 atom of F is 19.00 amu 19.00 amu/atom F

– 1 mole of F is 19.00 g 19.00 g/mole F

4.1

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Molar Mass• Molar mass - The mass in grams of 1 mole of

atoms

• What is the molar mass of carbon?

12.01 g/mol C

• This means counting out a mole of Carbon atoms (i.e., 6.022 x 1023) they would have a mass of 12.01 g

• One mole of any element contains the same number of atoms, 6.022 x 1023, Avogadro’s number

4.1

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Calculating Atoms, Moles, and Mass

• We use the following conversion factors:

• Density converts grams – milliliters

• Atomic mass unit converts amu – grams

• Avogadro’s number converts moles – number of atoms

• Molar mass converts grams – moles

4.1

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Strategy for Calculations• Map out a pattern for the required

conversion

• Given a number of grams and asked for number of atoms

• Two conversions are required • Convert grams to moles

1 mol S/32.06 g S OR 32.06 g S/1 mol S

• Convert moles to atomsmol S x (6.022 x 1023 atoms S) / 1 mol S

4.1

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Practice Calculations1. Calculate the number of atoms in 1.7

moles of boron.

2. Find the mass in grams of 2.5 mol Na (sodium).

3. Calculate the number of atoms in 5.0 g aluminum.

4. Calculate the mass of 5,000,000 atoms of Au (gold)

Interconversion Between Moles, Particles, and Grams

4.1

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4.2 The Chemical Formula, Formula Weight, and Molar

Mass• Chemical formula - a combination of

symbols of the various elements that make up the compound

• Formula unit - the smallest collection of atoms that provide two important pieces of information– The identity of the atoms– The relative number of each type of atom

4.2

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sChemical Formula

Consider the following formulas:

• H2 – 2 atoms of hydrogen are chemically bonded forming diatomic hydrogen, subscript 2

• H2O – 2 atoms of hydrogen and 1 atom of oxygen, lack of subscript means one atom

• NaCl – 1 atom each of sodium and chlorine

• Ca(OH)2 – 1 atom of calcium and 2 atoms each of oxygen and hydrogen, subscript outside parentheses applies to all atoms inside

4.2

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F

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sChemical Formula

Consider the following formulas:

• (NH4)3SO4 – 2 ammonium ions and 1 sulfate ion

– Ammonium ion contains 1 nitrogen and 4 hydrogen

– Sulfate ion contains 1 sulfur and 4 oxygen

– Compound contains 2 N, 8 H, 1 S, and 4 O

• CuSO4.5H2O

– This is an example of a hydrate - compounds containing one or more water molecules as an integral part of their structure

– 5 units of water with 1 CuSO4

Comparison of Hydrated and Anhydrous Copper Sulfate

Hydrated copper sulfate Anhydrous copper sulfate

4.2

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Marked color difference illustrates the factthat these are different compounds

Formula Weight and Molar Mass• Formula weight - the sum of the atomic weights

of all atoms in the compound as represented by its correct formula

– expressed in amu

• What is the formula weight of H2O?– 16.00 amu + 2(1.008 amu) = 18.02 amu

• Molar mass – mass of a mole of compound in grams / mole– Numerically equal to the formula weight in amu

• What is the molar mass of H2O?

– 18.02 g/mol H2O

4.2

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Formula Unit

• Formula unit – smallest collection of atoms from which the formula of a compound can be established

• When calculating the formula weight (or molar mass) of an ionic compound, the smallest unit of the crystal is used

4.2

The

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For

mul

a W

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Mol

ar M

ass

What is the molar mass of (NH4)3PO4?

3(N amu) + 12(H amu) + P amu + 4(O amu)=

3(14.01) + 12(1.008) + 30.97 + 4(16.00)= 149.10 g/mol (NH4)3PO4

4.3 The Chemical Equation and the Information It Conveys

A Recipe For Chemical Change• Chemical equation - shorthand notation of a

chemical reaction– Describes all of the substances that react and all

the products that form, physical states, and experimental conditions

– Reactants – (starting materials) – the substances that undergo change in the reaction

– Products – substances produced by the reaction

4.3

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ysFeatures of a Chemical Equation1. Identity of products and reactants must

be specified using chemical symbols2. Reactants are written to the left of the

reaction arrow and products are written to the right

3. Physical states of reactants and products may be shown in parentheses

4. Symbol over the reaction arrow means that energy is necessary for the reaction to occur

5. Equation must be balanced

)(O )2Hg( )2HgO( 2 gls

ProductsProducts – written on the right

Reactants – written on the left of arrow

Products and reactants must be specified using chemical symbols

Physical states are shown in parentheses

– energy is needed

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ys Features of a Chemical Equation

4.3

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ys The Experimental Basis of a Chemical Equation

We know that a chemical equation represents a chemical change

• One or more substances changed into new substances

• Different chemical and physical properties

4.3

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ysEvidence of a Reaction Occurring

The following can be visual evidence of a reaction:

•Release of a gas– CO2 is released when acid is placed in a solution

containing CO32- ions

•Formation of a solid (precipitate)– A solution containing Ag+ ions mixed with a solution

containing Cl- ions

•Heat is produced or absorbed – Acid and base are mixed together

•Color changes

4.3

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ysSubtle Indications of a Reaction

• Heat or light is absorbed or emitted

• Changes in the way the substances behave in an electrical or magnetic field before and after a reaction

• Changes in electrical properties

Writing Chemical Reactions

• We will learn to identify the following patterns of chemical reactions:– combination– decomposition– single-replacement– double-replacement

• Recognizing the pattern will help you write and understand reactions

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A + B AB

• Examples:

2Na(s) + Cl2(g) 2NaCl(s)

MgO(s) + CO2(g) MgCO3(s)

4.3

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ysCombination Reactions

• The joining of two or more elements or compounds, producing a product of different composition

4.3

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ys Types of Combination Reactions

1. Combination of a metal and a nonmetal to form a salt

2. Combination of hydrogen and chlorine molecules to produce hydrogen chloride

3. Formation of water from hydrogen and oxygen molecules

4. Reaction of magnesium oxide and carbon dioxide to produce magnesium carbonate

AB A + B

• Examples:

2HgO(s) 2Hg(l) + O2(g)

CaCO3(s) CaO(s) + CO2(g)

4.3

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ys Decomposition Reactions

• Produce two or more products from a single reactant

• Reverse of a combination reaction

4.3

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ys Types of Decomposition Reactions

1. Heating calcium carbonate to produce calcium oxide and carbon dioxide

2. Removal of water from a hydrated material

1. Single-replacement

• One atom replaces another in the compound producing a new compound

• Examples:

Cu(s)+2AgNO3(aq) 2Ag(s)+Cu(NO3)2(aq)

2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g)

A + BC B + AC

4.3

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ysReplacement Reactions

1. Replacement of copper by zinc in copper sulfate

2. Replacement of aluminum by sodium in aluminum nitrate

4.3

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ys Types of Replacement Reactions

2. Double-replacement• Two compounds undergo a “change

of partners”

• Two compounds react by exchanging atoms to produce two new compounds

4.3

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ysReplacement Reactions

AB + CD AD + CB

AB + CD AD + CB4.3

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• Reaction of an acid with a base to produce water and salt

HCl(aq)+NaOH(aq) NaCl(aq)+H2O(l)

• Formation of solid lead chloride from lead nitrate and sodium chloride

Pb(NO3)2(aq) + 2NaCl(aq)

PbCl2(s) + 2NaNO3(aq)

Types of Chemical Reactions

Precipitation Reactions

• Chemical change in a solution that results in one or more insoluble products

• To predict if a precipitation reaction can occur it is helpful to know the solubilities of ionic compounds

4.3

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Predicting Whether Precipitation Will Occur

• Recombine the ionic compounds to have them exchange partners

• Examine the new compounds formed and determine if any are insoluble according to the rules in Table 4.1

• Any insoluble salt will be the precipitate

Pb(NO3)2(aq) + NaCl(aq)

PbCl2 (?) + NaNO3 ( ?)(s) (aq)

4.3

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4.3

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ys Predict Whether These Reactions Form Precipitates

• Potassium chloride and silver nitrate

• Potassium acetate and silver nitrate

4.3

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ysReactions with Oxygen

• Reactions with oxygen generally release energy

• Combustion of natural gas

– Organic compounds CO2 and H2O are usually the products

CH4+2O2CO2+2H2O

• Rusting or corrosion of iron

4Fe + 3O2 2Fe2O3

The H+ on HCl was transferred to the oxygen in OH-, giving H2O

4.3

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ys Acid-Base Reactions

• These reactions involve the transfer of a hydrogen ion (H+) from one reactant (acid) to another (base)

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

Two electrons are transferred from Zn to Cu2+

4.3

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ys Oxidation-Reduction Reactions

• Reaction involves the transfer of one or more electrons from one reactant to another

Zn(s) + Cu2+(aq) Cu(s) + Zn2+(aq)

Writing Chemical Reactions

Consider the following reaction:

hydrogen reacts with oxygen to produce water

• Write the above reaction as a chemical equation

H2 + O2 H2O

• Don’t forget the diatomic elements

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4.3

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ysLaw of Conservation of Mass

• Law of conservation of mass - matter cannot be either gained or lost in the process of a chemical reaction

– The total mass of the products must equal the total mass of the reactants

A Visual Example of the Law of Conservation of Mass

4.4

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4.4 Balancing Chemical Equations

• A chemical equation shows the molar quantity of reactants needed to produce a particular molar quantity of products

• The relative number of moles of each product and reactant is indicated by placing a whole-number coefficient before the formula of each substance in the chemical equation

)(O )2Hg( )2HgO( 2 gls

Coefficient - how many of that substance are in the reaction

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Balancing

• The equation must be balanced – All the atoms of every reactant must also

appear in the products• Number of Hg on left? 2

– on right 2• Number of O on left? 2

– on right 2

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Examine the Equation

H2 + O2 H2O

• Is the law of conservation of mass obeyed as written? NO

• Balancing chemical equations uses coefficients to ensure that the law of conservation of mass is obeyed

• You may never change subscripts!

• WRONG: H2 + O2 H2O2

Step 1. Count the number of moles of atoms of each element on both product and reactant sides

Reactants Products 2 mol H 2 mol H 2 mol O 1 mol O

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Steps in Equation Balancing

The steps to balancing:

H2 + O2 H2O

H2 + O2 H2O

H2 + O2 2H2O

This balances oxygen, but is hydrogen still balanced?

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Step 2. Determine which elements are not balanced – do not have same number on both sides of the equation

– Oxygen is not balanced

Step 3. Balance one element at a time by changing the coefficients


Reactants Products 4 mol H 4 mol H

2 mol O 2 mol O4.4

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H2 + O2 2H2O

How will we balance hydrogen?

2H2 + O2 2H2O

Step 4. Check! Make sure the law of conservation of mass is obeyed

Balancing an Equation4.

4 B

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Practice Equation Balancing

Balance the following equations:

1. C2H2 + O2 CO2 + H2O

2. AgNO3 + FeCl3 Fe(NO3)3 + AgCl

3. C2H6 + O2 CO2 + H2O

4. N2 + H2 NH3

4.5 Calculations Using the Chemical Equation

• Calculation quantities of reactants and products in a chemical reaction has many applications

• Need a balanced chemical equation for the reaction of interest

• The coefficients represent the number of moles of each substance in the equation

4.5

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General Principles

1. Chemical formulas of all reactants and products must be known

2. Equation must be balanced to obey the law of conservation of mass

• Calculations of an unbalanced equation are meaningless

3. Calculations are performed in terms of moles

• Coefficients in the balanced equation represent the relative number of moles of products and reactants

Using the Chemical Equation

• Examine the reaction:

2H2 + O2 2H2O

• Coefficients tell us?

– 2 mol H2 reacts with 1 mol O2 to produce 2 mol H2O

• What if 4 moles of H2 reacts with 2 moles of O2?

– It yields 4 moles of H2O4.5

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2H2 + O2 2H2O

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Using the Chemical Equation

• The coefficients of the balanced equation are used to convert between moles of substances

• How many moles of O2 are needed to react with 4.26 moles of H2?

• Use the factor-label method to perform this calculation

2H2 + O2 2H2O

2

22 H mol__

O __molH mol 26.4 1

22.13 mol O2

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Use of Conversion Factors

• Digits in the conversion factor come from the balanced equation

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Conversion Between Moles and Grams

• Requires only the formula weight

• Convert 1.00 mol O2 to grams

– Plan the path

– Find the molar mass of oxygen

• 32.0 g O2 = 1 mol O2

– Set up the equation

– Cancel units 1.00 mol O2 x 32.0 g O2

1 mol O2 – Solve equation 1.00 x 32.0 g O2 = 32.0 g O2

moles ofOxygen

grams ofOxygen

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Conversion of Mole Reactants to Mole Products

• Use a balanced equation

• C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

• 1 mol C3H8 results in: – 5 mol O2 consumed 1 mol C3H8 /5 mol O2

– 3 mol CO2 formed 1 mol C3H8 /3 mol CO2

– 4 mol H2O formed 1 mol C3H8 /4 mol H2O

• This can be rewritten as conversion factors

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Calculating Reacting Quantities

• Calculate grams O2 reacting with 1.00 mol C3H8

• Use 2 conversion factors – Moles C3H8 to moles O2

– Moles of O2 to grams O2

– Set up the equation and cancel units

– 1.00 mol C3H8 x 5 mol O2 x 32.0 g O2 = 1 mol C3H8 1 mol O2

– 1.00 x 5 x 32.0 g O2 = 1.60 x 102 g O2

moles Oxygen

grams Oxygen

moles C3H8

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Calculating Grams of Product from Moles of Reactant

• Calculate grams CO2 from combustion of 1.00 mol C3H8

• Use 2 conversion factors – Moles C3H8 to moles CO2– Moles of CO2 to grams CO2


– 1.00 mol C3H8 x 3 mol CO2 x 44.0 g CO2 = 1 mol C3H8 1 mol CO2

– 1.00 x 3 x 44.0 g CO2 = 1.32 x 102 g CO2

moles CO2

grams CO2

moles C3H8

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Relating Masses of Reactantsand Products

• Calculate grams C3H8 required to produce 36.0 grams of H2O

• Use 3 conversion factors – Grams H2O to moles H2O– Moles H2O to moles C3H8– Moles of C3H8 to grams C3H8


36.0 g H2O x 1 mol H2O x 1 mol C3H8 x 44.0 g C3H8 18.0 g H2O 4 mol H2O 1 mol C3H8

– 36.0 x [1/18.0] x [1/4] x 44.0 g C3H8 = 22.0 g C3H8

moles H2O

grams C3H8

moles C3H8

grams H2O

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Calculating a Quantity of Reactant

• Ca(OH)2 neutralizes HCl• Calculate grams HCl neutralized by 0.500 mol

Ca(OH)2 – Write chemical equation and balance

• Ca(OH)2(s) + 2HCl(aq) CaCl2(s) + 2H2O(l)

– Plan the path


0.500 mol Ca(OH)2 x 2 mol HCl x 36.5 g HCl 1 mol Ca(OH)2 1 mol HCl

Solve equation 0.500 x [2/1] x 36.5 g HCl = 36.5 g HCl

molesCa(OH)2

grams HCl

molesHCl

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General Problem-solving Strategy

Na + Cl2 NaCl

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Sample Calculation

1. Balance the equation

2. Calculate the moles Cl2 reacting with 5.00 mol Na

3. Calculate the grams NaCl produced when 5.00 mol Na reacts with an excess of Cl2

4. Calculate the grams Na reacting with 5.00 g Cl2

2Na + Cl2 2NaCl

%100yield ltheoretica

yield actual yield %

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Theoretical and Percent Yield

• Theoretical yield - the maximum amount of product that can be produced – Pencil and paper yield

• Actual yield - the amount produced when the reaction is performed– Laboratory yield

• Percent yield:

= 125 g CO2 actual x 100% = 97.4% 132 g CO2 theoretical

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Sample Calculation

If the theoretical yield of iron was 30.0 g and actual yield was 25.0 g, calculate the percent yield:

2 Al(s) + Fe2O3(s) Al2O3(aq) + 2Fe(aq)

• [25.0 g / 30.0 g] x 100% = 83.3%

• Calculate the % yield if 26.8 grams iron was collected in the same reaction

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