Chapter 1-mole concept.ppt

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1. The Mole Concept
(The counting unit of chemistry)
Chapter 1

2. •Atomsandmoleculesaretoosmallto
keeptrackofindividually!Itiseasierto
counttheminpackages.
Counting
•We count pieces in MOLES
•That’swhychemistscreatedtheirown
countingunitcalledthemole.

3. Avogadro’s Number and The Mole
•Atoms come in moles
6.02 x 10
23
= 1 mole = N
A
N
Ais known as Avogadro’s number

4. Mole
Definition:
•Amoleisthenumberofatomsinexactly12gofthepure
carbon-12isotope.
•ThemoleisNOTjustacountingunit,likethedozenora
gross,whichidentifiesonlythenumberofobjects.The
definitionofamolesignifiesthenumberofobjectsina
fixedmassofsubstance.
•Massspectrometrically,themassofacarbon-12atomis
1.9926×10
-23
g.
No. of carbon-12 atoms = atomic mass (g)
mass of one atom (g)
= 12 g _
1.9926×10
-23
g
= 6.022 ×10
23
atoms

5. Other definitions of the Mole
One mole contains Avogadro’s Number (6.022 x 10
23
)
A mole is the amount of a substance of a system which
contains as many elementary entities as there are atoms in
0.012kg (or12g) of Carbon-12
A mole is the quantity of a substance whose mass in grams is
the same as its formula weight
For example: Fe (55.85 g)
Iron has an atomic mass or 55.85 g mol
-1
, so one mole of iron
has a mass or 55.85 g.

6. For modern systems,
12
C is taken as a standard.
So,
12
C is assumed to be 12 a.m.u.
1
12
C atom = 12 a.m.u.
1 a.m.u.=1/12 of the mass of a single
12
Catom
=1/12 (12/6.02 X 10
23
)
=1/6.02 X 10
23
grams
=mass of 1
1
H atom
Thenbycomparingthemassesofeachatomwiththemassof
12
C,
atomicmassesofotheratomscanbeeasilycalculatedas“RELATIVE
ATOMICMASSES”.RelativeatomicmasseseasilycalculatedbyMASS
SPECTROMETER.
1
16
O = 16 a.m.u
1
40
Ca = 40 a.m.u.
1
1
H = 1 a.m.u.

7. One mole of any object alwaysmeans 6.022 ×10
23
units of those objects.
For example,1 molof H
2Ocontains 6.022 ×10
23
molecules
1 molof NaClcontains 6.022 ×10
23
formula units
Avogadro’snumberisusedtoconvertbetweenthenumberofmoles
andthenumberofatoms,ionsormolecules.
Example
0.450 molof iron contains how many atoms?
Number of atoms = number of moles ×Avogadro’s number (N
A)
Therefore,
No. of atoms = (0.450 mol) ×(6.022 ×10
23
)
= 2.7 ×10
23
atoms
Calculating the number of particles

8. Example
How many molecules are there in 4 moles of hydrogen peroxide
(H
2O
2)?
No. of molecules = no. of moles ×Avogadro’s number (N
A)
= 4 mol ×(6.022 ×10
23
mol
-1
)
= 24 ×10
23
molecules
= 2.4 ×10
24
molecules
Question
How many atoms are there in 7.2 moles of gold (Au)?
Answer: 4.3 ×10
24
atoms

9. How many calcium atoms are in 0.250 moles of calcium?
Answer: 1.51 x 10
23
Caatoms
Calculate the moles of 3.75 x 10
27
molecules of oxygen gas?
Answer : 6230 molO
2
How many Cu atoms are in 0.50 mole of Cu?
Answer: 3.0 x 10
23
Cu atoms
How many moles of CO
2are in 2.50 x 10
24
molecules of CO
2?
Answer: 4.15 moles of CO
2

10. Mass and the Mole
•Themassingramsofonemoleofanypuresubstanceiscalled
itsmolarmass.
•Themolarmassofanyelementisnumericallyequaltoits
atomicmassandhastheunitsg/mol.
•molarmass=massof1moleofsubstance
•Molarmasscanbedeterminedbyaddinguptheatomic
massesfromtheperiodictable(atomicmassgoesto1
decimalplace).

11. Q. Find the molar mass of CH
4.
= 1C + 4H
= 1 x (12.0) + 4 x (1.0)
= 16.0 g/mol
Q. Find the molar mass of Mg(OH)
2.
=1Mg + 2O + 2H
=1 x (24.3) + 2 x (16.00) + 2 x (1.0)
=58.3 g/mol
Q. Find the molar mass of MgSO
4•7H
2O.
=1Mg + 1S + 4O + 7(H
2O)
=1 x (24.3) + 1 x (32.1) + 4 x (16.00) + 7 x [2 x (1.0) + 16.00]
=246.4 g/mol

12. Converting between mass and moles
Inthelaboratory,wemeasurethemassofreactantsingramsusinga
balance.However,whenthesereacttheydosoinaratioofmoles.
Therefore,weneedtoconvertbetweenthemasswemeasureandthe
numberofmoleswerequire.
The expression relating mass and number of moles is:
Mass of sample (g) = no. of moles (mol) ×molar mass (gmol
-1
)
Example
Calculate the mass in grams in 0.75 molof sodium hydroxide, NaOH
Step 1: Find the molar mass of the compound
Na:22.99 gmol
-1
O: 16.00 gmol
-1
H: 1.008 gmol
-1
Molar mass of NaOH: 40.00 gmol
-1
Step 2: Substitute into the above expression
Mass of sample = 0.75mol ×40.00 gmol
-1
= 30g

13. Questions
Calculate the mass in grams present in:
(a) 0.57mol of potassium permanganate (KMnO
4)
Answer: Molar mass KMnO
4= 158.03 gmol
-1
Mass in grams = 0.57mol ×158.03 gmol
-1
= 90.07 g
(b) 1.16mol of oxalic acid (H
2C
2O
4)
Answer: Molar mass H
2C
2O
4= 90.04 gmol
-1
Mass in grams = 1.16mol ×90.04 gmol
-1
= 104.44 g
(c) 2.36mol of calcium hydroxide Ca(OH)
2
Answer:Molar mass Ca(OH)
2= 74.1 gmol
-1
Mass in grams = 2.36mol ×74.1 gmol
-1
= 174.87 g

14. Converting between moles and mass
Number of moles = mass of sample (g)
molar mass (gmol
-1
)
Example
Convert 25.0g of KMnO
4to moles
Step 1: Calculate the molar mass
K
Mn
O
1 ×39.10 gmol
-1
1 ×54.93 gmol
-1
4 ×16.00 gmol
-1
= 39.10 gmol
-1
= 54.93 gmol
-1
= 64.00 gmol
-1
MM = 158.03 gmol
-1
Step 2: Substitute into above expression
No. of moles =
25.0g .
158.03gmol
-1
= 0.158 mol

15. Questions
Calculate the number of moles in:
(a) 1.00g of water (H
2O)
Answer:Molar mass water = 18.02 gmol
-1
1.00g H
2O = 0.055mol
(b) 3.0g of carbon dioxide (CO
2)
Answer:Molar mass carbon dioxide = 44 gmol
-1
3.0g CO
2= 0.068mol
(c) 500g of sucrose (C
12H
22O
11)
Answer:Molar mass sucrose = 342.30 gmol
-1
500g C
12H
22O
11= 1.46mol
(d) 2.00g of silver chloride (AgCl)
Answer:Molar mass silver chloride = 143.38 gmol
-1
2.00g AgCl= 0.014mol

16. Important formulae so far….
Number of moles = mass of sample (g)
molar mass (gmol-1)
No. of carbon-12 atoms = atomic mass (g)
mass of one atom (g)
No. of atoms = No. of moles ×Avogadro’s number (N
A)
No. of molecules = No. of moles ×Avogadro’s number (N
A)
Mass of one molecule = Molar mass
Avogadro’s no.
Mass of sample (g) = no. of moles (mol) ×molar mass (gmol
-1
)
Defining the mole:
Calculating the number of atoms or molecules, given the number of moles:
Most important equation:
Calculating the mass of an individual molecule:

17. Calculating mass percentage from a chemical formula
Many elements in the periodic table occur in combination with other elements to
form compounds.
A chemical formula of a compound expresses the composition of that compound in
terms of the number of atoms of each element present in it.
The mass percentage composition allows you to determine the fraction of the total mass
each element contributes to the compound.
Example
Ammonium nitrate (NH
4NO
3) is an important compound in the fertiliser industry.
What is the mass % composition of ammonium nitrate?
Step 1: Calculate the molar mass of ammonium nitrate
Molar mass NH
4NO
3= 80.05 gmol
-1
Two N atoms: 28.016 gmol
-1
Four H atoms: 4.032 gmol
-1
Three O atoms: 48.00 gmol
-1

18. Step 2: Determine the mass % composition for each element
Nitrogen: 28.016g N in one mol of ammonium nitrate
Mass fraction of N = 28.016g
80.05g
Mass % composition of N = 28.016g×100%
80.05g
= 34.99% ≈ 35%
Hydrogen: 4.032g H in one molof ammonium nitrate
Mass fraction of N = 4.032g
80.05g
Mass % composition of H = 4.032g×100%
80.05g
= 5.04% ≈ 5%
Oxygen: 48.00g O in one molof ammonium nitrate
As above, the mass % composition of O is found to be 60%

19. Therefore,themass%compositionofammoniumnitrate(NH
4NO
3)is:
% Nitrogen:35%
% Hydrogen:5%
% Oxygen:60%
To check your answer, make sure it adds up to 100%
Question
What is the mass % composition of C
12H
22O
11?
Answer: % Carbon: 42.1%
% Hydrogen:6.5%
% Oxygen: 51.4%

20. Determining empirical formula from mass
Theempiricalformulaofacompoundreferstotherelativenumberof
atomsofeachelementpresentinthatcompound.Itgivesthesimplestratio
oftheelementsinthecompound.
For example, the empirical formula of glucose (C
6H
12O
6) is CH
2O, giving the
C:H:O ratio of 1:2:1
If you know the mass % composition and the molar mass of elements present
in a compound, you can work out the empirical formula1
Example
What is the empirical formula of a compound which has a mass %
composition of 50.05% S and 49.95% O?
Step 1: Find the atomic masses of the elements present
Sulfur(S) : 32.066 gmol
-1
Oxygen (O) : 16.000 gmol
-1

21. Step 2:Determine the number of moles of each element present
Since we are dealing with percentages, we can express the mass % as
grams if we assume we have 100g of the compound.
21
Therefore, 100g of our compound contains 50.05g of sulfurand 49.95g
of oxygen.
Convert number of grams to number of moles
Number of molSulfur= mass of sulfurin sample (g)
atomic mass of sulfur(gmol
-1
)
= 50.05g .
32.066 gmol
-1
= 1.56 mol
Similarly, the no. of molof Oxygen is found to be 3.12mol
Step 3:Determining the ratios of elements
Sulfur: 1.56mol
Oxygen: 3.12mol
Ratio 1.56 : 3.12
Ratio must be in whole numbers. Here we must divide across by 1.56
Therefore, we have a ratio of 1:2 giving an empirical formula of SO
2

22. Question
Determine the empirical formula of a compound that contains 27.3
mass% Carbon and 72.7 mass% Oxygen.
Answer: No. of molCarbon = 2.27mol
No. of molOxygen = 4.54mol
Ratio1:2Empirical formula CO
2
Monosodium glutamate (MSG) has the following mass percentage
composition: 35.51% C, 4.77 % H, 37.85% O, 8.29% N, and 13.60% Na.
What is its molecular formula if its molar mass is 169 gmol
-1
?
Answer: C
5H
8O
4NNa

23. Molarity
Some chemical reactions involve aqueous solutions of reactants
The concentration of a solution is the amount of solute present in a
given quantity of solvent or solution
This concentration may be expressed in terms of molarity (M) or
molar concentration:
M = Molarity = no. of moles
volume in Litres
Molarity is the number of moles of solute in 1 Litre (L) of solution

24. What is molarity of an 85.0mL ethanol (C
2H
5OH) solution containing 1.77g of
ethanol?
Step 1: Determine the number of moles of ethanol
Example
Molar mass of ethanol, C
2H
5OH:
2 ×carbon atoms
1 ×oxygen atom
6 ×hydrogen atoms
2 ×12.01 gmol
-1
1 ×16.00 gmol
-1
6 ×1.008 gmol
-1
24.02 gmol
-1
16.00 gmol
-1
6.048 gmol
-1
46.07 gmol
-1
No. of moles = mass in g
molar mass
No. of moles ethanol = 1.77g .
46.07 gmol
-1
= 0.038 mol

25. Step 2: Convert to molarity
Have 85.0mL ethanol
1 L = 1000mL
Have 0.085 L of ethanol
Molarity = no. of moles
volume in L
= 0.038 mol
0.085 L
= 0.45 molL
-1
≡ 0.45 M
Questions
Calculate the molarities of each of the following solutions:
(a) 2.357g of sodium chloride (NaCl) in 75mL solution
Answer: 0.5378 M
(b) 1.567mol of silver nitrate (AgNO
3) in 250mL solution
Answer: 6.268 M
(c) 10.4g of calcium chloride (CaCl
2) in 2.20 ×10
2
mL of solution
Answer: 0.426 M

26. And if it is a gas at STP, 1 mole = 22.4 L
1 mole = 6.02 x 10
23
= molar mass= 22.4 L
•Gases are mostly empty space so they
ALL have the same volume despite
different mass
•22.4 L per 1 mole of any gas at STP
•STP is Standard Temperature and Pressure
•0°C and 1 atm

27. Example 1:
Determine the volume, in liters, of 0.600 mol of SO
2
gas at STP.
Answer: 13.4 L SO
2
Example 2:
Determine the number of moles in 33.6 L of He gas
at STP.
Answer: 1.50 moles

28. Combustion Stoichiometry
•Balancing any chemical reaction requires equating the
number of atoms on both the reactant and product side of
the reaction equation.
•In combustion reactions, one of the reactants is air
•Air is approximately 20.9% O
2and 79.1% N
2by volume air
also has argon, CO
2and trace amounts of many other
species but we will ignore these for now
•This means 79.1/20.9 = 3.78 moles of N
2 per mole of O
2
•Consider the (unbalanced) combustion of methane:
CH
4+a (O
2+3.78 N
2) CO
2+ bH
2O + 3.78a N
2

29. •Inthisrepresentation,Nisinert,sowhybother
includingit?
•Therearetworeasons.
a.Ifyouwanttocalculatethemolefractionofeach
combustionproduct,youneedtoknowhowmuchN
2is
presentsinceitwillbethepredominantspecies.
b.Duringactualcombustion,atmosphericN
2canform
NO
xandhencenitrogenisNOTalwaysinert-more
aboutNO
xformationlater.
The complete combustion of methane and air is
now balanced.
CH
4+ 2 (O
2+3.78 N
2) CO
2+ 2 H
2O + 2 x 3.78 N
2

30. •In general, a simple HC fuel with the composition C
nH
mwill
undergo complete oxidation to form CO
2and H
2O.
C
nH
m+ (n +m/4)(O
2+3.78 N
2) n CO
2+m/2 H
2O + 3.78 (n+m/4)N
2
•For each mole of fuel burned, (n + m/4) x (1 + 3.78) = 4.78
x (n + m/4) moles of O
2and N
2are involved, and moles of
combustion products are generated is given as:
Moles of combustion product generated
=n + m/2 + 3.78n + 3.78m/4
=4.78n + m/4 + m/4 + 3.78m/4
=4.78(n + m/4) + m/4
The molarstoichiometric fuel-to-air ratio is
1 / [4.78 x (n + m/4) ]

31. The product mole fractions for complete combustion of this
simple HC are4/)4/(78.4
)4/(78.3
4/)4/(78.4
2/
4/)4/(78.4
2
2
2
mmn
mn
x
mmn
m
x
mmn
n
x
N
OH
CO








32. 066.0
1516
100
)2878.332(11
100











sa
f
m
m %4.73734.0
58.56
5.41
%1.14141.0
58.56
8
%4.12124.0
58.56
7
2
2
2



N
OH
CO
x
x
x Example: Calculate the stoichiometric fuel/air mass ratio and product gas
composition for the combustion ofheptane in air.
Solution
Heptane is C-C-C-C-C-C-C or C
7
H
16
.
C
7
H
16
+ 11 (O
2
+3.78 N
2
) ===> 7 CO
2
+ 8 H
2
O + 11 x 3.78 N
2
For each mole of heptane burned, 11 x (1 + 3.78) = 52.5 moles of O
2
and N
2
are involved.
The molar mass of heptane is 7 x 12 + 16 x 1 = 100. Hence, the fuel/air
mass ratio is
The total number of moles of combustion products is 7 + 8 + 11 x 3.78
= 7 + 8 + 41.5 = 56.58
The product gas composition, on a mole fraction basis, is

33. References
1.Generalchemistry-Principles,patternsandApplications.Bruce
Averill,StrategicEnergySecuritySolutions.PatriciaEldredge,R.H.
Hand,LLCISBN13:9781453322307.SaylorFoundation.Pages:
186-222