chapter 2

5/28/2018 Chapter 2

1/57

CHAPTER 2

Atoms, Molecules & Stoichiometry

5/28/2018 Chapter 2

2/57

Learning Outcomes

define the terms relative atomic, isotopic, molecular and formula

masses, based on the 12C scale

define the term mole in terms of the Avogadro constant

analyse mass spectra in terms of isotopic abundances

calculate the relative atomic mass of an element given the relativeabundances of its isotopes, or its mass spectrum

define the terms empirical and molecular formulae

calculate empirical and molecular formulae, using combustion data

or composition by mass

write and/or construct balanced equations

5/28/2018 Chapter 2

3/57

Learning Outcomes

perform calculations, including use of the mole concept, involving:

(i) reacting masses (from formulae and equations)

(ii) volumes of gases (e.g. in the burning of hydrocarbons)

(iii) volumes and concentrations of solutions

deduce stoichiometric relationships from calculations

describe and explain redox processes in terms of electron transfer

and/or of changes in oxidation number (oxidation state)

(calculation of oxidation number is required)

5/28/2018 Chapter 2

4/57

Relative Mass

Relative Mass

Relative Atomic

Mass

-Ratio of the average massof one atom of the element

to 1/12 the mass of an atom

of 12C isotope,expressed on

the12C scale

Relative IsotopicMass

- Ratio of the mass of one

atom of the isotope to 1/12

the mass of an atom of 12C

isotope,expressed on the12C scale

Relative Molecular

Mass- Ratio of the average mass of

one molecule of the substance

to 1/12 the mass of an atom of12C isotope,expressed on the12C scale

Relative FormulaMass

- Ratio of the average

mass of one formula unit of

the compound to 1/12 the

mass of an atom of 12C

isotope,expressed on the

12C scale

5/28/2018 Chapter 2

5/57

A Mole of Atoms

A moleis a collection that contains

6.02 x 1023atoms of an element (Avogadros

number).

1 mole element Number of Atoms

1 mole C = 6.02 x 1023C atoms

1 mole Na = 6.02 x 1023Na atoms

1 mole Au = 6.02 x 1023Au atoms

5/28/2018 Chapter 2

6/57

A Mole of a Compound

A mole

Of a covalent compound has Avogadros number of

molecules.

1 mole CO2= 6.02 x 1023CO2molecules

1 mole H2O = 6.02 x 1023H2O molecules

Of an ionic compound contains Avogadros number

of formula units.

1 mole NaCl = 6.02 x 1023NaCl formula units

1 mole K2SO4 = 6.02 x 1023K2SO4formula units

5/28/2018 Chapter 2

7/57

5/28/2018 Chapter 2

8/57

One Mole of Four

Elements

One mole each of helium,

sulfur, copper, and mercury.

How many atoms of helium

are present? Of sulfur? Of

copper? Of mercury?

5/28/2018 Chapter 2

9/57

Using Avogadros Number

Avogadros number is used toconvert molesof a

substance toparticles.

How many Cu atoms are in 0.50 mole Cu?

0.50 mole Cu x 6.02 x 1023Cu atoms

1 mole Cu= 3.0 x 1023Cu atoms

5/28/2018 Chapter 2

10/57

Using Avogadros Number

Avogadros number is used to convertparticlesof a

substance tomoles.

How many moles of CO2are in 2.50 x 1024molecules CO2?

2.50 x 1024

molecules CO2x 1 mole CO2

6.02 x 1023molecules CO2

= 4.15 moles CO2

5/28/2018 Chapter 2

11/57

Molar Mass

The molar massis

The mass of one mole of a substance.

The atomic mass of an element expressed in grams.

5/28/2018 Chapter 2

12/57

The mass spectrometer is an instrument used to measure

the masses and relative (natural) abundances of the

isotopes present in a sample of an element

AVERAGE ATOMIC MASS

5/28/2018 Chapter 2

13/57

Mass Spectrometer

In a mass spectrometer, charged particles are passed through a magnetic

field.

The path of the stream of particles will be bent by the magnetic field

depending on the mass and charge of the particles.

When naturally occurring samples of most elementsare charged andpassed through a mass spectrometer, the spectrum indicates more than

one form of the element.

These are isotopes of the element:the number of protons is the same but

the number of neutrons/atom differs for each isotope giving each isotope aslightly different mass.

5/28/2018 Chapter 2

14/57

boron-10 23

boron-11 100

There are 2 isotopes for boron:

Suppose you had 123 typical atoms of boron.

23 of these would be 10B and 100 would be 11B.

The total mass of these would be (23 x 10) + (100 x 11) = 1330

The average mass of these 123 atoms would be 1330 / 123 = 10.8 (to 3 significant

figures).

10.8 is the relative atomic mass of boron.

Mass Spectrometer

5/28/2018 Chapter 2

15/57

The data can be summarized as follows:

Isotope Mass` Abundance

90Zr 90.00 amu 51.5 %91Zr 91.00 amu 11.2 %92Zr 92.00 amu 17.1 %94Zr 94.00 amu 17.4 %96Zr 96.00 amu 2.80 %

atomic mass of isotope 100 %

5/28/2018 Chapter 2

16/57

Calculate the weighted average mass of zirconium using the data below.

Change each percent to a decimal by dividing by 100.

Multiply by the mass.

Add it all together.

Isotope Mass` Abundance90Zr 90.00 amu 51.5 %91Zr 91.00 amu 11.2 %92Zr 92.00 amu 17.1 %94Zr 94.00 amu 17.4 %96Zr 96.00 amu 2.80 %

0.515(90.00) + 0.112(91.00) + 0.171(92.00) + 0.174(94.00) +0.0280(96.00)

= 91.3 amu

5/28/2018 Chapter 2

17/57

Calculate the average atomic mass for Germanium

20.52 %

69.92428 amu

27.43 %

71.92174 amu

7.76 %

72.9234 amu

36.54 %

73.92115 amu

7.76 %

75.9214 amu

5/28/2018 Chapter 2

18/57

Chlorine consists of molecules, not individual atoms. Whenchlorine is passed into the ionisation chamber, an electron is

knocked off the molecule to give a molecular ion , Cl2+.

These ions won't be particularly stable, and some will fall apart to

give a chlorine atom and a Cl+ion.

The mass spectrum of chlorine

5/28/2018 Chapter 2

19/57

Empirical and Molecular Formula

The empir ical formu la of a compound is the simplest formula which shows

the ratio of the atoms of the different elements in the compound.

Question:

69.58% Ba, 6.090% C, 24.32% O.

What is the empirical formula?

Assume you have 100 g of sample,

1: 69.58 g Ba, 6.090 g C, 24.32 g O

2: Ba: 69.58 g 137.33 g/mol = 0.50666 mol Ba

C: 6.090 g 12.01 g/mol = 0.50708 mol C

O: 24.32 g 16.00 g/mol = 1.520 mol O

5/28/2018 Chapter 2

20/57

3.

4: the simplest formula is BaCO3

mol (reduced)

mol

1.520/

0.50666

= 3.000

0.50708/

0.50666

= 1.001

0.50666/

0.50666

= 1

1.5200.507080.50666

OCBa

Empirical Formula

5/28/2018 Chapter 2

21/57

Empirical Formula

CxHy+ (x+ )O2(g) x CO2(g) + H2O(g)y

2

y

2

5/28/2018 Chapter 2

22/57

PROCEDURE FOR...

Obtaining an Empirical

Formula from Combustion

Analysis

EXAMPLE 3.19

Obtaining an Empirical

Formula from Combustion

Analysis

1. Write down asgiven the masses of

each combustion product and the

mass of the sample (if given).

2. Convert the masses of CO2and

H2O from step 1 to moles by usingthe appropriate molar mass for each

compound as a conversion factor.

GIVEN: 1.83 g CO2, 0.901 g H2O

FIND: empirical formula

3. Convert the moles of CO2

and

moles of H2Ofrom step 2 to moles

of C and moles of H using the

conversion factors inherent in the

chemical formulas of CO2and

H2O.

Upon combustion, a compound containing only carbon andhydrogen produces 1.83 g CO2 and 0.901 g H2O. Find the

empirical formula of the compound.

2011 Pearson Education, Inc.

continued

5/28/2018 Chapter 2

23/57

4. If the compound contains an element other

than C and H, find the mass of the other

element by subtracting the sum of the

masses of C and H (obtained in step 3) from

the mass of the sample. Finally, convert themass of the other element to moles.

5. Write down a pseudoformula for the

compound using the number of moles of each

element (from steps 3 and 4) as subscripts.

6. Divide all the subscripts in the formula

by the smallest subscript. (Round allsubscripts that are within 0.1 of a whole

number.)

No other elements besides C and H, so

proceed to next step.

7. If the subscripts are not whole

numbers, multiply all the subscripts

by a small whole number to get

whole-number subscripts.

C0.0416H0.100

2011 Pearson Education, Inc.

The correct empirical formula is C5H12.

5/28/2018 Chapter 2

24/57

Molecular Formula

The molecular formula of a compound is one which shows the actual

number of atoms of each element present in one molecule of a

compound.

The molecular formula can be obtained if the empirical formula and Mrare known.

Example: A compound has the empirical formula CH2Br. Its relative

molecular mass is 187.8. Deduce the molecular formula of this

compound.n(CH2Br) = 187.8

n(93.9) = 187.8

n = 2

Molecular formula = C2H

4Br

2

5/28/2018 Chapter 2

25/57

The subscriptsin a formula give

The relationship of atoms in the formula.

The moles of each element in 1 mole of

compound.

Glucose

C6H12O6

In 1 molecule: 6 atoms C 12 atoms H 6 atoms O

In 1 mole: 6 moles C 12 moles H 6 moles O

Chemical Formula and Mole Concept

5/28/2018 Chapter 2

26/57

Chemical equation

Uses chemical symbols and chemical formulas to

describe the changes occur in a chemical reaction

Reactant is a starting material

Written on left side

Product is a substance produced

Written on right sideCH4+ 2O2 CO2+ 2H2O

Writing and Balancing Chemical Equation

Reactant Product

5/28/2018 Chapter 2

27/57

Law of Conservation of Mass

The Law of Conservation of Massindicates that in

an ordinary chemical reaction,

Matter cannot be created or destroyed.

No change in total mass occurs in a reaction.

Mass of products is equal to mass of reactants.

5/28/2018 Chapter 2

28/57

Example:2Ag + S Ag2S

in this equation the number of reactants are equal to

that of the products.(2 Ag and 1 S)

Balance the following chemical equations.

Fe(s) + S(s) Fe2S3(s)

NH3+ O2 N2+ H2O

HCl + O2 Cl2+ H2O

Writing and Balancing Chemical Equation

5/28/2018 Chapter 2

29/57

a chemical equation indicates the relative amounts

of each reactant and product in a given chemical

reaction.

Example:

CH4 + 2O2 CO2 + 2H2O

The equation shows that 1 mol of methane reactswith 2 mol of oxygen to produce 2 mol of water.

Chemical Equations and The Mole Concept

5/28/2018 Chapter 2

30/57

A balanced chemical equation can be used to calculate the

relative amounts of substances involved in chemical reactions.

Example:

When 18.6 g ethane gas C2H6burns in oxygen, how

many grams of CO2are produced?

2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(g)

18.6 g ? g

Chemical Calculations Using Chemical Equations

5/28/2018 Chapter 2

31/57

Chemical Calculations Using Chemical Equations

Solution:g C2H6 mole C2H6 mole CO2 g CO2

molar mole-mole molar

mass C2H6 factor mass CO2

= 18.6 g C2H6x 1 mole C2H6x 4 moles CO2x 44.0 g CO2

30.1 g C2H6 2 moles C2H6 1 mole CO2

molar mole-mole molarmass C2H6 factor mass CO2

= 54.4 g CO2

5/28/2018 Chapter 2

32/57

Balancing Ionic Equations

When ionic compounds dissolve in water, the ions separate from each

other.

Example : NaSO4 (s) + aq Na+(aq) + SO4

-2 (aq)

The ions that play no part in the reaction are called spectator ions.

Molecular Equation:

K2CrO4+ Pb(NO3)2 PbCrO4 + 2 KNO3

Soluble Soluble Insoluble Soluble

Total Ionic Equation:(cancell ing spectator ions)

2 K++ CrO4-2+ Pb+2+ 2 NO3

-

PbCrO4(s) + 2 K++ 2 NO3

-

Net Ionic Equation:

CrO4-2+ Pb+2 PbCrO4(s)

5/28/2018 Chapter 2

33/57

Limiting Reactants

You can make cookies

until you run out of one

of the ingredients Sugar is the limiting

ingredient in making

the cookies (present in

small amount). Sugar will limit the

amount of cookies you

can make

Cookies Analogy:

5/28/2018 Chapter 2

34/57

Limiting Reactants

The limiting reactant is the reactant present in

the smallest stoichiometric amount

In other words, its the reactant youll run out of first (in

this case, the H2) O2would be the excess reagent

5/28/2018 Chapter 2

35/57

Theoretical yield

Maximum amount of product that can be

obtained in a reaction from the given amounts

of reactant.

Percentage yield

A comparison of the amount actually obtainedto the amount it was possible to make.

Actual Yield

Theoretical YieldPercent Yield= x 100

5/28/2018 Chapter 2

36/57

Silicon dioxide (quartz) is usually quite unreactive but reacts

readily with hydrogen fluoride according to the following

equation.

SiO2(s) + 4HF(g) SiF4(g) + 2H2O(l)

2.0 mol HF x 1 mol SiO24 mol HF

= 0.5 mol SiO2

2.0 mol ? mol

If 2.0 mol of HF are exposed to 4.5 mol of SiO2, which is the limiting reactant?

4.5 mol

4.5 mol SiO2 x 4 mol HF1 mol SiO2 = 18 mol HF

LR

LR

(smaller number)

5/28/2018 Chapter 2

37/57

Calculations Involving Volumes of Gases

One mole of any gas occupies a volume of 22.4 dm3at

s.t.p. (or 24 dm3at r.t.p)

This is the molar volume of gases.

Example:

1. What volume is occupied by 4.21 moles of ammonia gas, NH3at STP?

4.21 mol 22.4 L = 94.3 L

1 mole

5/28/2018 Chapter 2

38/57

Calculations Involving Volumes of Gases

2. What volume of hydrogen, measured at STP, can be released by 42.7

g of zinc as it reacts with hydrochloric acid?

Zn (s )+ 2 HCl (aq) H2(g )+ ZnCl2(aq)

42.7 g Zn 1 mol Zn 1 mol H2 22.4 L H2 = 14.6 L H265.4 g Zn 1 mol Zn 1 mol H2

39

5/28/2018 Chapter 2

39/57

PrenticeHall 2005General Chemistry4thedition, Hill, Petrucci, McCreary, Perry

Chapter Three

Solute: the substance being dissolved.

Solvent: the substance doing the dissolving.

Concentrationof a solution: the quantity of a

solute in a given quantity of solution (or solvent).

A concentratedsolution contains a relatively large

amount of solute vs. the solvent (or solution).

A dilutesolution contains a relatively smallconcentration of solute vs. the solvent (or solution).

Concentrated and dilute arent very quantitative

Solutions and

Solution Stoichiometry

5/28/2018 Chapter 2

40/57

CONCENTRATION OF AQUEOUS

SOLUTION

The concentration of aqueous solution may be expressed either as:

a) mass of solute per dm3of solution (units: g dm-3) or

b) mol of solute per dm3of solution (units: moldm-3)

41

5/28/2018 Chapter 2

41/57


Chapter Three

Molarity(M), or molar concentration, is the amount

of solute, in moles, per liter of solution:

A solution that is 0.35 M sucrose contains 0.35

moles of sucrose in each liter of solution.

Keep in mind that molarity signifies moles of

solute per liter of solution, not liters of solvent.

Molar Concentration

moles of soluteMolarity=

liters of solution

42

5/28/2018 Chapter 2

42/57


Chapter Three

Preparing 0.01000 M KMnO4

Weigh 0.01000

mol (1.580 g)

KMnO4.

Dissolve in water. How much

water? Doesnt matter, as long as

we dont go over a liter.

Add more water

to reach the

1.000 liter mark.

43

5/28/2018 Chapter 2

43/57


Chapter Three

Dilutionis the process of preparing a more dilute

solution by adding solvent to a more concentrated

one.

Addition of solvent does not change the amount ofsolutein a solution but does change the solution

concentration.

It is very common to prepare a concentrated stocksolution of a solute, then dilute it to other

concentrations as needed.

Dilution of Solutions

44

Vi li i th Dil ti f S l ti

5/28/2018 Chapter 2

44/57


Chapter Three

Visualizing the Dilution of a Solution

We start and

end with the

same amount of

solute.

Addition of

solvent has

decreased the

concentration.

45

5/28/2018 Chapter 2

45/57


Chapter Three

Dilution Calculations

couldnt be easier.

Moles of solute does not change on dilution.

Moles of solute =MV Therefore

MconcVconc= MdilVdil

46Example 3 26

5/28/2018 Chapter 2

46/57


Chapter Three

Example 3.26

How many milliliters of a 2.00 M CuSO4stocksolution are needed to prepare 0.250 L of 0.400 M

CuSO4?

5/28/2018 Chapter 2

47/57

Acid-base titration

A neutralization reaction in which a measured volume of an

acid or a base of known concentration is completely reacted

with a measured volume of a base or an acid of unknownconcentration

Acid-base indicator

A compound that exhibits different colours depending on the

pH of its solution

Acid-Base Titrations

S t f tit ti id ith b

5/28/2018 Chapter 2

48/57

Setup for titrating an acid with a base

5/28/2018 Chapter 2

49/57

The unknown concentration of the base or acid can be

calculated by using this formula:

M1V1 = M2V2

Acid-Base Titrations

5/28/2018 Chapter 2

50/57

SAMPLE PROBLEM

In an acid-base titration, 17.45 mL of 0.180 M

nitric acid, HNO3, were completely neutralized

by 14.76 mL of aluminium hydroxide, Al(OH)3.

Calculate the concentration of the aluminiumhydroxide.

5/28/2018 Chapter 2

51/57

SAMPLE ANSWER

The balanced equation for the reaction is:

3HNO3(aq) + Al(OH)3(aq) Al(NO3)3(aq) + 3H2O(l)

The number of moles of nitric acid used is:

y mol = 0.180 mol/L x 0.01745 L = 3.14 x 10-3mol HNO3

From the stoichiometry of the reaction, the number of moles ofaluminium hydroxide reacted is:

3.14 x 10-3mol HNO3 x 1 mol Al(OH)3= 1.05 x 10-3mol

3 mol HNO3

Therefore, the concentration of the aluminium hydroxide is:

1.05 x 10-3mol Al(OH)3= 0.0711 M

0.01476 L

EXERCISE

5/28/2018 Chapter 2

52/57

EXERCISE

If 30 mL of 0.5 M HCl neutralizes 20.40 mL of Ba(OH)2

solution, what is the molarity of the Ba(OH)2solution? You

have to write a balanced equation for the neutralization.

5/28/2018 Chapter 2

53/57

Redox Reactions

Oxidation-reduction (redox) reaction:

A chemical reaction in which there is a transfer of

electrons from one reactant to another reactant.

Oxidation number

of an atom is the charge that atom would have if

the compound was composed of ions.

5/28/2018 Chapter 2

54/57

Rules for determining oxidation numbers:

Zero for element in its elemental state

Equal to the charge for the monatomic ion

For Groups IA and IIA in compounds always +1, +2

+1 for hydrogen in most compounds

-2 for oxygen in most compounds

In binary compounds, the more electronegative element is

assigned a negative oxidation number (as in its binary ioniccompounds)

The sum of the individual oxidation numbers is equal to zero for

a neutral compound, equal to the charge on the polyatomic ion

Redox Reactions

Oxidation number

5/28/2018 Chapter 2

55/57

Oxidation number

The charge the atom would have in a molecule (or an

ionic compound) if electrons were completely transferred.

1. Free elements (uncombined state) have an oxidation

number of zero.

Na, Be, K, Pb, H2, O2, P4 = 0

2. In monatomic ions, the oxidation number is equal to

the charge on the ion.

Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2

3. The oxidation number of oxygen isusually2. In H2O2and O2

2-it is 1.

4 The oxidation number of hydrogen is +1 except when

5/28/2018 Chapter 2

56/57

4. The oxidation number of hydrogen is 1exceptwhen

it is bonded to metals in binary compounds. In these

cases, its oxidation number is1.

6. The sum of the oxidation numbers of all the atoms in a

molecule or ion is equal to the charge on the

molecule or ion.

5. Group IA metals are +1, IIA metals are +2and fluorine is

always1.

HCO3-

O = -2 H = +1

3x(-2)+ 1+ ?= -1

C = +4

Oxidation numbers of allthe elements in HCO3

-?

5/28/2018 Chapter 2

57/57

NaIO3

Na = +1 O = -2

3x(-2) + 1+ ?= 0

I = +5

K2Cr2O7

O = -2 K = +1

7x(-2) + 2x(+1) + 2x(?)= 0

Cr = +6

Oxidation numbers of all

the elements in the

following ?

chapter 2

Documents