3.3 chem myp mole formula

M5 Stoichiometry II:M5 Stoichiometry II:
Mole ConceptMole Concept

M5 chem Stoichiometry: Mole Concept
Slide 2 of 49
Learning ObjectivesLearning Objectives
• Concepts:
– mole, Avogadro’s number, molar mass
– relative atomic mass, relative molecular mass, relative formula mass
• Skills:
– Given a chemical formula be able to state the elements present and their
proportion
– Be able to calculate the molar mass of a substance given its formula and
table of relative atomic masses
– Given moles of a certain substance (element or compound) determine the
mass of the substance and vice versa
– Given moles of a certain gaseous substance (element or compound)
determine the volume of the substance and vice versa
– Given moles of a certain substance (element or compound) determine the
number atoms/molecules of the substance and vice versa

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Relative Molecular Mass/Formula MassRelative Molecular Mass/Formula Mass
• Relative atomic mass (AR or RAM): Mass of an atom of the element
relative to that of hydrogen.
– It essentially tells us much heavier the atom of the element is than a hydrogen
atom.
• Relative molecular mass (MR or RMM): This tells us how much heavier
the molecule of the compound is than a hydrogen atom.
– It is the sum of RAM of all the atoms that make up the molecule.
• Relative formula mass (or RFM): sum of the relative atomic masses of
all the ions in a single formula unit of an ionic substance.
– This tells us how much heavier a formula unit of a compound is than a hydrogen
atom.
• Notice then that relative mass, whether atomic or molecular or formula, has no
unit.
Obviously because the measurement is relative to the mass of a hydrogen atom.

Slide 4 of 49
Mole and the Avogadro numberMole and the Avogadro number
• Mole (abbreviated to mol) is the unit of measurement of amount
(number of elementary particles) of chemical substance.
In chemistry amount is measured in numbers.
That is amount refers to the number of elementary particles—the
smallest particles.
While a couple of things consists of 2 things; a dozen consists of 12
things; a gross 144; a ream of paper 500 sheets of paper, a mole of
anything consists of a very very VERY large number.
A mole of a chemical substance is 6 × 1023
of the elementary particles—
atoms, molecules, ions—of that substance.
This number is referred to as the Avogadro Constant (or number).

Slide 5 of 49
Magnitude of a moleMagnitude of a mole
• That number is such as large number that it is said that if the whole
population of the world wished to count up to this number between
them, and they all worked at counting without any breaks at all, it
would take six million years for them to finish!
Further, a line 6 × 1023
mm long would stretch from the earth to the Sun
and back two million times!!!
If under room temperature and pressure, however, a box the size of 24
dm3
(box of dimension 29 cm × 29 cm × 29 cm) contains that many
number of molecules of gas that make up air, then atoms and
molecules must be very very small!
– That is, at room temperature and pressure, the volume of 1 mole of any
gas (6 × 1023
atoms of He, or 6 × 1023
molecules of H2 or 6 × 1023
molecules of CO2) is always 24 dm3
, which is referred to as the molar
volume of gas.

Slide 6 of 49
Mole of SubstanceMole of Substance
• While 1 dozen apples would represent 12 apples, 1 mole apples
would represent 6 × 1023
apples.
Similarly, 1 mole of helium would represent 6 × 1023
atoms of helium.
1 mole of sodium would represent 6 × 1023
atoms of sodium.
1 mole of hydrogen gas (H2) would represent 6 × 1023
molecules of H2.
1 mole of bromine (Br2) would represent 6 × 1023
molecules of bromine.
1 mole of iodine (I2) would represent 6 × 1023
molecules of iodine.
1 mole of carbon monoxide would represent 6 × 1023
molecules of CO.
1 mole of sulfur dioxide would represent 6 × 1023
molecules of SO2.
1 mole of sodium chloride would represent 6 × 1023
pairs of NaCl.
1 mole of magnesium chloride would represent 6 × 1023
units of MgCl2.

Slide 7 of 49
Mole and the Avogadro numberMole and the Avogadro number
• Atoms, molecules and ions are very very small, therefore, instead of counting
them individually, they are counted by measuring their mass and volume!
– The quantity of a substance which contains the Avogadro constant equivalent of
elementary particles (atoms, molecules or ions) is the relative mass of the
substance expressed in grams.
– If we were to measure out 1 g of hydrogen atoms it turns out it would contain 6 ×
1023
atoms of hydrogen atoms.
• It follows than that,
– since relative molecular mass of hydrogen gas (H2) is 2, 2 g of hydrogen gas will
contain 6 × 1023
molecules of H2.
– In other words, since a hydrogen molecules is twice as heavy as a hydrogen atom,
twice as many grams of hydrogen molecules would have to me weighed out in order
to have the same number of hydrogen molecules as in 1 g of hydrogen atom.

Slide 8 of 49
Molar MassMolar Mass
– A sodium atom is 23 times as heavier than an atom of hydrogen (ie. Relative tomic
mass of sodium is 23).
– Therefore 23 g of Sodium will contain 6 × 1023
atoms of sodium.
– 28 g of carbon monoxide will contain 6 × 1023
molecules of CO.
– 18 g of water will contain will contain 6 × 1023
molecules of H2O.
– 62 g of Na2O will contain 6 × 1023
units of Na2O
– In other words 2 × 6 × 1023
Na+
ions and 6 × 1023
O2-
ions.
• Mass of a mole of a substance is referred to as the molar mass of that
substance.
– 1 g is the molar mass of (elemental) hydrogen atoms
– 2 g is the molar mass of hydrogen gas
– 23 g is the molar mass of sodium
– 28 g molar mass of CO etc.

Slide 9 of 49
A Mole of WaterA Mole of Water

Slide 10 of 49
Changing between the different quantitiesChanging between the different quantities
• Chemical substances can be characterized and quantified in a few different
ways.
– Mass, moles, molar mass, number of atoms, (if gaseous) volume, molar volume
• You need to be able to convert between a few of these quantities for elements
and compounds.
Though there are whole variety of methods of converting between those
quantities, you will be introduced to conversions
– using Factor-label method,
– using ratio,
– using arithmetic and
– using formula.
• We’ll first discuss factor-label method and then look at the rest.

Slide 11 of 49
Changing between the different quantities:Changing between the different quantities:
ElementsElements
4 g He 1 mole He
or
1 mole He 4 g He
– Expressing the relationship as a fraction allows us to use them in
calculations requiring the interconversion between the two quantities.
• Let’s take helium:
– The relationship between mole, mass, number of atoms and volume for
this gas is as follows: 1 mole of helium weighs 4 g, contains 6 × 1023
atoms, and has a volume of 24 dm3
.
• Let’s take the first relationship and express it as an equality:
1 mole of helium gas = 4 g of helium.
– The relationship between mole and mass for Helium can also be
expressed as a fraction:

Slide 12 of 49
Element: Relationship between mole & massElement: Relationship between mole & mass
4 g He
1 mole He
• The first fraction allows us to go from moles of
He to mass of He.
Here’s an example:
• How many grams of helium would have to weight out so that we
have 15 moles of the gas?
4 g He
15 mole He 60 g He
1 mole He
 
⇒ ÷
 
• This method of solving quantitative problems in chemistry is
variously referred to as Factor-labeled method or dimensional
analysis (dimension meaning units).

Slide 13 of 49
• The above expression can be derived using ratios or arithmetic.
Using ratio:
4 g He x g He
1 mole He 15 mole He
4 g He
x 15 mole He
1 mole He
=
 
⇒ =  ÷
 
• Using arithmetic:
If 1 mole of He weighs 4 g,
Then 15 moles must weigh 4 × 15 g (which is essentially the same as
above).

Slide 14 of 49
• How many moles of helium does a cylinder with 155 gram of the gas
contain?
1 mole He
155 g He 38.8 mole He
4 g He
 
= ÷
 
1 mole He
4 g He
• The second fraction of course allows us to go
from mass of He to moles of helium.

Slide 15 of 49
molar mass of A
1 mole A
1 mole A
molar mass of A
mole of element A mass of element A
 
 ÷
 
 
 ÷
 
→¬ 
• The same relationship for molecular elements
2
2
2
2
molar mass of A
1 mole A
2 2
1 mole A
molar mass of A
mole of element A mass of element A
 
 ÷
 
 
 ÷
 
→¬ 
• 1. mole ⇔ mass
– In general the relationship between the given quantity (mass or mole) of
an element and the unknown quantity (the other one of the two) and the
conversion factor can be summarized as shown below:

Slide 16 of 49
Practice Questions: MolePractice Questions: Mole ⇒⇒ Mass of ElementMass of Element
• 1. Convert the following into mass:
a) 5.0 mole of helium (He)
4 g He
5.0 mole He 20 g He
1 mole He
 
= ÷
 
23 g He
0.50 mole Na 11.5 g He
1 mole Na
 
= ÷
 
c) 0.050 mole of potassium (K)
b) 0.50 mole of sodium (Na)
39 g K
0.050 mole K 1.95 g He
1 mole K
 
= ÷
 

Slide 17 of 49
d) 0.00050 mole of lead (Pb)
207 g Pb
0.00050 mole Pb 0.104 g Pb
1 mole Pb
 
= ÷
 
e) 5.0 mole of hydrogen gas (H2)
f) 0.25 mole of chlorine gas (Cl2)
2
2 2
2
2 g H
5.0 mole H 10 g H
1 mole H
 
= ÷
 
2
2 2
2
71 g Cl
0.25 mole Cl 17.7 g Cl
1 mole Cl
 
= ÷
 

Slide 18 of 49
g) 0.25 mole of bromine (Br2)
h) 0.50 mole of iodine (I2)
i) 0.25 mole of iron (Fe)
2
2 2
2
160 g Br
0.25 mole Br 40 g Br
1 mole Br
 
= ÷
 
2
2 2
2
254 g I
0.50 mole I 127 g I
1 mole I
 
= ÷
 
56 g Fe
0.25 mole Fe 14 g Fe
1 mole Fe
 
= ÷
 

Slide 19 of 49
Practice Questions: MassPractice Questions: Mass ⇒⇒ Mole of ElementMole of Element
• 2. Convert the following into mole:
a) 25.0 g of helium (He)
b) 25.0 g of sodium (Na)
c) 25.0 g of potassium (K)
1 mole He
25.0 g He 6.25 mole He
4 g He
 
= ÷
 
1 mole Na
25.0 g Na 1.09 mole Na
23 g Na
 
= ÷
 
1 mole K
25.0 g K 0.64 mole K
39 g K
 
= ÷
 

Slide 20 of 49
d) 25.0 g of lead (Pb)
1 mole Pb
25.0 g Pb 0.12 mole Pb
207 g Pb
 
= ÷
 
e) 25.0 g of hydrogen gas (H2)
f) 25.0 g of chlorine gas (Cl2)
2
2 2
2
1 mole H
25.0 g H 12.5 mole H
2 g H
 
= ÷
 
2
2 2
2
1 mole Cl
25.0 g Cl 0.35 mole Cl
71 g Cl
 
= ÷
 

Slide 21 of 49
g) 25.0 g of bromine (Br2)
2
2 2
2
1 mole Br
25.0 g Br 0.156 mole Br
160 g Br
 
= ÷
 
h) 25.0 g of iodine (I2)
i) 25.0 g of iron (Fe)
2
2 2
2
1 mole I
25.0 g I 0.0984 mole I
254 g I
 
= ÷
 
1 mole Fe
25.0 g Fe 0.446 mole Fe
56 g Fe
 
= ÷
 

Slide 22 of 49
Element: Relationship between Mole & # ofElement: Relationship between Mole & # of
atomsatoms
23
23
1 mole He 6 10 atoms He
or
1 mole He6 10 atoms He
×
×
– Using one of these two fractions, we can easily convert between mole
and number of atoms.
• Going back to our relationships between mole, mass, number of
atoms and volume for helium:
– 1 mole of helium weighs 4 g, contains 6 × 1023
atoms (Avogadro number
equivalent), and has a volume of 24 dm3
.
Let’s take the relationship between mole and number of atoms and
expressing it as an equality:
1 mole of helium gas = 6 × 1023
atoms of helium.
– Expressed as a fraction:

Slide 23 of 49
Two questionsTwo questions
• Which of the two fractions would be appropriate for conversion from
moles of a substance to number of particles of the substance? Give
the fraction.
Which of the two fractions would be appropriate for conversion from
number of particles of a substance to moles of the substance? Give
the fraction.

Slide 24 of 49
Element: Relationship between Mole & # ofElement: Relationship between Mole & # of
atomsatoms
2. mole ⇔ number of atoms
• Atomic elements
Avogadro# of atoms of A
1 mole A
1 mole A
Avogadro# of atoms of A
mole of element A atoms of element A
 
 ÷
 
 
 ÷
 
→¬ 
• Molecular elements
2
2
2
2
Avogadro# of molecules of A
1 mole A
2 2
1 mole A
Avogadro# of molecules of A
mole of element A molecules of element A
 
 ÷
 
 
 ÷
 
→¬ 

Slide 25 of 49
Practice Questions: MolePractice Questions: Mole ⇒⇒ # of# of
Atoms/Molecules of ElementAtoms/Molecules of Element
• 1. Convert the following into atoms/molecules:
a) 2.0 moles of helium (He)
23
236 10 atoms He
2.0 mole He 12 10 atoms He
1 mole He
 ×
= × ÷ ÷
 
b) 2.0 moles of neon (Ne)
c) 0.30 moles of sodium (Na)
23
236 10 atoms Ne
2.0 mole Ne 12 10 atoms Ne
1 mole Ne
 ×
= × ÷ ÷
 
23
226 10 atoms Na
0.03 mole Na 1.8 10 atoms Ne
1 mole Na
 ×
= × ÷ ÷
 

Slide 26 of 49
Practice Questions: MolePractice Questions: Mole ⇒⇒ # of# of
Atoms/Molecules of ElementAtoms/Molecules of Element
d) 0.25 mole of bromine (Br2)
23
232
2 2
2
6 10 molecules Br
0.25 mole Br 1.5 10 molecules Br
1 mole Br
 ×
= × ÷
 ÷
 
e) 0.50 mole of iodine (I2)
f) 0.020 moles of hydrogen gas (H2)
23
232
2 2
2
6 10 molecules I
0.50 mole I 3 10 molecules I
1 mole I
 ×
= × ÷
 ÷
 
23
222
2 2
2
6 10 molecules H
0.020 mole H 1.2 10 molecules H
1 mole H
 ×
= × ÷
 ÷
 

Slide 27 of 49
Practice Questions: # of Atoms/Molecules ofPractice Questions: # of Atoms/Molecules of
ElementElement ⇒⇒ MoleMole
• 2. Convert the following into moles:
a) 3 × 1024
atoms of He
24
23
1 mole He
3 10 atoms He 5 moles He
6 10 atoms He
 
× = ÷
× 
b) 3 × 1024
atoms of Ne
c) 9 × 1022
atoms of Na
24
23
1 mole Ne
3 10 atoms Ne 5 moles Ne
6 10 atoms Ne
 
× = ÷
× 
22
23
1 mole Na
9 10 atoms Na 0.15 mole Na
6 10 atoms Na
 
× = ÷
× 

Slide 28 of 49
Practice Questions: # of Atoms/Molecules ofPractice Questions: # of Atoms/Molecules of
ElementElement ⇒⇒ MoleMole
d) 9 × 1022
molecules of Bromine (Br2)
22 2
2 223
2
1 mole Br
9 10 molecules Br 0.15 mole Br
6 10 molecules Br
 
× = ÷
 ÷× 
e) 9 × 1024
molecules of iodine (I2)
f) 9 × 1024
molecules of hydrogen gas (H2)
24 2
2 223
2
1 mole I
9 10 molecules I 15 mole I
6 10 molecules I
 
× = ÷
 ÷× 
24 2
2 223
2
1 mole H
9 10 molecules H 15 mole H
6 10 molecules H
 
× = ÷
 ÷× 

Slide 29 of 49
Element: Relationship between Mole & VolumeElement: Relationship between Mole & Volume
of gaseous elementof gaseous element
3
3
1 mole He 24 dm He
or
1 mole He24 dm He
– These two fractions allow the interconversions between mole and
volume of gas.
• Going back to our given relationships between mole, mass, number of
atoms and volume for helium:
– 1 mole of helium weighs 4 g, contains 6 × 1023
atoms, and has a volume of
24 dm3
.
• Let’s take the relationship between mole and volume expressing it as
an equality:
1 mole of helium gas =24 dm3
of helium.
– Expressed as a fraction:

Slide 30 of 49
Element: Relationship between Mole & VolumeElement: Relationship between Mole & Volume
of gaseous elementof gaseous element
3. mole ⇔ volume
• Atomic elements
3
3
24 dm A
1 mole A
1 mole A
24 dm A
mole of element A volume of element A
 
 ÷
 
 
 ÷
 
→¬ 
• Molecular elements
3
2
2
2
3
2
24 dm A
1 mole A
2 2
1 mole A
24 dm A
mole of element A volume of element A
 
 ÷
 
 
 ÷ ÷
 
→¬ 

Slide 31 of 49
Practice Questions: MolePractice Questions: Mole ⇒⇒ Volume of gaseousVolume of gaseous
elementelement
• 1. Convert the following into volume:
a) 5.0 mole of helium (He)
3
324 dm He
5.0 mole He 120 dm He
1 mole He
 
= ÷ ÷
 
b) 5.0 mole of Neon (Ne)
c) 5.0 mole of hydrogen gas (H2)
3
324 dm Ne
5.0 mole Ne 120 dm Ne
1 mole Ne
 
= ÷ ÷
 
3
32
2 2
2
24 dm H
5.0 mole H 120 dm H
1 mole H
 
= ÷
 ÷
 

Slide 32 of 49
elementelement
d) 2.5 mole of Krypton (Kr)
3
324 dm Kr
2.5 mole Kr 60 dm Kr
1 mole Kr
 
= ÷ ÷
 
e) 2.5 mole of Argon (Ar)
f) 2.5 mole of chlorine gas (Cl2)
3
324 dm Ar
2.5 mole Ar 60 dm Ar
1 mole Ar
 
= ÷ ÷
 
3
32
2 2
2
24 dm Cl
2.5 mole Cl 60 dm Cl
1 mole Cl
 
= ÷
 ÷
 

Slide 33 of 49
Practice Questions: Volume of gaseous elementPractice Questions: Volume of gaseous element
⇒⇒ MoleMole
a) 24.0 dm3
of helium (He)
3
3
1 mole He
24 dm He 1 mole He
24 dm He
 
= ÷
 
b) 34.0 dm3
of Neon (Ne)
c) 34.0 dm3
of hydrogen gas (H2)
3
3
1 mole Ne
34 dm Ne 1.41 mole Ne
24 dm Ne
 
= ÷
 
3 2
2 23
2
1 mole H
34 dm H 1.41 mole H
24 dm H
 
= ÷
 ÷
 

Slide 34 of 49
Practice Questions: Volume of gaseous elementPractice Questions: Volume of gaseous element
⇒⇒ MoleMole
d) 2.0 dm3
of Krypton (Kr)
3
3
1 mole Kr
2.0 dm Kr 0.083 mole Kr
24 dm Kr
 
= ÷
 
e) 2.0 dm3
of Argon (Ar)
f) 2.0 dm3
of chlorine gas (Cl2)
3
3
1 mole Ar
2.0 dm Ar 0.083 mole Ar
24 dm Ar
 
= ÷
 
3 2
2 23
2
1 mole Cl
2.0 dm Cl 0.083 mole Cl
24 dm Cl
 
= ÷
 ÷
 

Slide 35 of 49
Quantitative Relationships: CompoundsQuantitative Relationships: Compounds
• In the case of a gaseous compound such as carbon dioxide, the
relationship between mole, mass, number of atoms and volume for
this gas is as follows: 1 mole of CO2 weighs 44 g (ie. the molar mass is
44 g), contains 6 × 1023
molecules of CO2, and has a volume of 24
dm3
.
Be able to perform the following conversion for compounds:
1. mole ⇔ mass
x
x
x
x
molar mass of AB
1 mol AB
x x
1 mol AB
molar mass of AB
mol of AB mass of AB
 
 ÷
 
 
 ÷
 
→¬ 

Slide 36 of 49
Quantitative Relationships: CompoundsQuantitative Relationships: Compounds
3
x
x
x
3
x
24 dm AB
1 mol AB
x x
1 mol AB
24 dm AB
mol AB Volume of AB
 
 ÷
 
 
 ÷ ÷
 
→¬ 
2. mole ⇔ volume

Slide 37 of 49
Practice Questions: MolePractice Questions: Mole ⇒⇒ Mass of CompoundMass of Compound
• 1. Convert the following into mass:
a) 5.0 mole of carbon monoxide (CO)
28 g CO
5.0 mole CO 140 g CO
1 mole CO
 
= ÷
 
b) 0.50 mole of sulfur dioxide (SO2)
c) 0.050 mole of water (H2O)
2
2 2
2
64 g SO
0.50 mole SO 32 g SO
1 mole SO
 
= ÷
 
2
2 2
2
18 g H O
0.050 mole H O 0.90 g H O
1 mole H O
 
= ÷
 

Slide 38 of 49
d) 0.050 mole of N2O5
2 5
2 5 2 5
2 5
108 g N O
0.050 mole N O 5.4 g N O
1 mole N O
 
= ÷
 
e) 5.0 mole of hydrogen chloride gas (HCl)
f) 0.25 mole of magnesium oxide (MgO)
36.5 g HCl
5.0 mole HCl 182.5 g HCl
1 mole HCl
 
= ÷
 
40 g MgO
0.25 mole MgO 10 g MgO
1 mole MgO
 
= ÷
 

Slide 39 of 49
g) 0.50 mole of sodium carbonate (Na2SO4)
2 4
2 4 2 4
2 4
142 g Na SO
0.50 mole Na SO 71 g Na SO
1 mole Na SO
 
= ÷
 
h) 0.25 mole of ammonium phosphate ( (NH4)3PO4 )
i) 0.25 mole of copper sulfate pentahydrate (CuSO4•5H2O)
4 3 4
4 3 4 4 3 4
4 3 4
149 g (NH ) PO
0.25 mole (NH ) PO 37.3 g (NH ) PO
1 mole (NH ) PO
 
= ÷
 
4 2
4 2 4 2
4 2
250 g CuSO 5H O
0.25 mole CuSO 5H O 62.5 g CuSO 5H O
1 mole CuSO 5H O
 •
• = • ÷
• 

Slide 40 of 49
Practice Questions: MassPractice Questions: Mass ⇒⇒ Mole of CompoundMole of Compound
a) 25.0 g of methane (CH4)
b) 25.0 g of ammonia (NH3)
c) 25.0 g of sulfur dioxide (SO2)
4
4 4
4
1 mole CH
25.0 g CH 1.56 mole CH
16 g CH
 
= ÷
 
3
3 3
3
1 mole NH
25.0 g NH 1.47 mole NH
17 g NH
 
= ÷
 
2
2 2
2
1 mole SO
25.0 g SO 0.39 mole SO
64 g SO
 
= ÷
 

Slide 41 of 49
Practice Questions: MassPractice Questions: Mass ⇒⇒ Mole of CompoundMole of Compound
d) 25.0 g of copper sulfate pentahydrate (CuSO4•5H2O)
e) 25.0 g of magnesium chloride (MgCl2)
f) 25.0 g of iron oxide (Fe2O3)
4 2
4 2 4 2
4 2
1 mole CuSO 5H O
25.0 g CuSO 5H O 0.1 g CuSO 5H O
250 g CuSO 5H O
 •
• = • ÷
• 
2
2 2
2
1 mole MgCl
25.0 g MgCl 0.263 mole MgCl
95 g MgCl
 
= ÷
 
2 3
2 3 2 3
2 3
1 mole Fe O
25.0 g Fe O 0.156 mole Fe O
160 g Fe O
 
= ÷
 

Slide 42 of 49
CompoundCompound
• 3. Convert the following into volume:
a) 5.0 mole of carbon monoxide (CO)
b) 1.2 mole of sulfur dioxide (SO2)
c) 0.25 mole of hydrogen chloride gas (HCl)
3
324 dm CO
5.0 mole CO 120 dm CO
1 mole CO
 
= ÷ ÷
 
3
32
2 2
2
24 dm SO
1.2 mole SO 28.8 dm SO
1 mole SO
 
= ÷
 ÷
 
3
324 dm HCl
0.25 mole HCl 6 dm HCl
1 mole HCl
 
= ÷ ÷
 

Slide 43 of 49
Practice Questions: Volume of gaseousPractice Questions: Volume of gaseous
compoundcompound ⇒⇒ MoleMole
a) 24.0 dm3
of ammonia (NH3)
b) 4.0 dm3
of carbon monoxide (CO)
c) 0.20 dm3
of sulfur dioxide (SO2)
3 3
3 33
3
1 mole NH
24.0 dm NH 1 mole NH
24 dm NH
 
= ÷
 ÷
 
3
3
1 mole CO
4.0 dm CO 0.167 mole CO
24 dm CO
 
= ÷
 
3 2
2 23
2
1 mole SO
0.20 dm SO 0.00833 mole SO
24 dm SO
 
= ÷
 ÷
 

Slide 44 of 49
Practice Questions: Volume of gaseousPractice Questions: Volume of gaseous
compoundcompound ⇒⇒ MoleMole
d) 24.0 dm3
of hydrogen sulfide gas (H2S)
e) 2.0 dm3
of methane (CH4)
f) 0.25 dm3
of carbon dioxide (CO2)
3 2
2 23
2
1 mole H S
24.0 dm H S 1 mole H S
24 dm H S
 
= ÷
 ÷
 
3 4
4 43
4
1 mole CH
2.0 dm CH 0.0833 mole CH
24 dm CH
 
= ÷
 ÷
 
3 2
2 23
2
1 mole CO
0.25 dm CO 0.0104 mole CO
24 dm CO
 
= ÷
 ÷
 

Slide 45 of 49
Quantitative Relationship: Formula for AtomicQuantitative Relationship: Formula for Atomic
ElementsElements
Relationship between mass, moles and molar mass:
Quantitative properties that the
element can be characterized by
For exampleElement
can be…
Atomic Na (s), Hg (l), He (g) Mass, moles, molar mass, number of atoms,
(and if gaseous) volume, molar volume
Mass
Moles
Molar
mass
Mass = Moles Molar mass×
Mass
Molar mass =
Moles
Mass
Moles =
Molar mass

Slide 46 of 49
Quantitative Relationship: Formula for AtomicQuantitative Relationship: Formula for Atomic
ElementsElements
• Relationship between number of atoms, moles and Avogadro’s
constant.
# of
atoms
Moles
Avogadro’s
constant
# of atoms = Moles Avogadro's constant×
# of atoms
Moles =
Avogadro's constant
# of atoms
Avogadro's Constant =
Moles

Slide 47 of 49
Quantitative Relationship: Formula for GaseousQuantitative Relationship: Formula for Gaseous
SubstancesSubstances
• If the element is gaseous, relationship between moles, volume and
molar volume can be summarized as follows:
Volume
Moles
Molar
volume
volume = Moles Molar volume×
Volume of gas
Moles =
Molar volume
Volume (of gas)
Molar volume =
Moles

Slide 48 of 49
Quantitative Relationship: Formula for MolecularQuantitative Relationship: Formula for Molecular
Elements & CompoundsElements & Compounds
For exampleElement can
also be…
H2 (g), Br2 (l), I2 (s)
(S8 (s), P4(s))
Molecular
Quantitative properties that the
element can be characterized by
Mass, moles, molar mass, number of
molecules, number of atoms,
(and if gaseous) volume, molar volume
• In this case, the relationship between mass, moles and molar mass is
exactly the same as that for atomic element.
In the case of a gaseous molecular element, relationship between moles,
volume and molar volume is also the same as for atomic elements.
As for compounds, the relationship between mole, mass and molar mass
as well as that between mole, volume and molar volume is the same
as that for elements.

Slide 49 of 49
Quantitative Relationship: Formula for MolecularQuantitative Relationship: Formula for Molecular
Elements & CompoundsElements & Compounds
• As for the relationship between number of molecules and moles, you
just need to change number of atoms to number of molecules.
# of
molecules
Moles
Avogadro’s
constant
# of molecules = Moles Avogadro's constant×
# of molecules
Moles =
Avogadro's constant
# of atoms
Avogadro's Constant =
Moles

3.3 chem myp mole formula

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