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Summary (CAIE) Cambridge A Level Chemistry (9701)
Summary & Examples with Solutions

Stoichiometry

1
...
By international agreement, atomic mass is the mass of the atom in atomic
mass unit (amu)
...
Carbon-12 is the carbon isotope that has six
protons and six neutrons
...
For example, on average,
a hydrogen atom is only 8
...
Thus, if the
mass of onecarbon-12 atom is exactly 12 amu, the atomic mass of hydrogen must be
0
...
00 amu or 1
...
Similar calculations show that the atomic mass of iron
is 55
...
Thus, although we do not know just how much an average iron atom’s mass
is, we know that it is approximately 56 times as massive as a hydrogen atom
...
This
means that when we measure the atomic mass of an element, we must generally settle
for the average mass of the naturally occurring mixture of isotopes
...
90 percent and 1
...
The atomic mass of carbon-13 has been determined to be 13
...

Thus, the average atomic mass of carbon can be calculated as follows:
(0
...
00000 amu) + (0
...
00335) = 12
...
01 amu, we are referring to the
average value
...
In every sample of this element,
75
...
23% are atoms of 37Cl
...
9689 amu and that of 37Cl is 36
...
From these data,
calculate the average atomic mass of chlorine
...
77% of the mass is contributed by atoms of 35Cl and 24
...
Thus, when we calculate the mass of the “average atom” we have to
take into account both of the masses of the isotopes and their relative abundances
...
9689 amu x 75
...
496 amu

Mass contribution of 37Cl = 36
...
23% 37Cl = 8
...
”
26
...
957 amu = 35
...
45 amu

2
...
in the sample
...
Based on this definition and the fact that the average atomic masses in the
periodic table are relative values, we can deduce that we will have a mole of atoms of
any element if weigh an amount equal to the atomic mass in gram units (this often called
the gram atomic mass)
...
06 amu and one mole of sulfur will weigh
32
...


The Mole Concept Applied to Compounds
The gram molecular mass of a molecular substance (the mass in grams equal to the
molecular mass) is also equal to one mole of those molecules
...
02 amu, the sum of the atomic masses
of two H atoms and one O atom
...

1 mole of ionic compound X = gram formula mass of X
The ionic compound Al2O3 has two aluminum atoms with an atomic mass of 26
...
00 amu each
...
96
amu, and one mole of Al2O3 has a gram formula mass of 101
...

To simplify, we will often use the following relationship between moles and mass unless
one of the other, equivalent, definitions provides more clarity
...
In an experiment to prepare TiO2, we start with a 23
...
How
many moles of Ti do we have?
Solution
We have a tool for converting the mass into moles that states 1 mol X is equal to the
atomic mass of X with gram units
...
867 g Ti
Start by setting up the problem in the form of an equation showing the number and
its units that we start with and the units we want when we are finished
...
5 g Ti = ? mol Ti
Now use the quality between mass and moles to set up the ratio that will cancel the
grams of titanium as shown below
1 𝑚𝑜𝑙 𝑇𝑖
23
...
491 mol Ti
47
...
We need 0
...
How many grams do we
need to weigh?
Solution
We need the tool for calculating the molar mass of FeCl3
...

Molar mass FeCl3 = 55
...
453 g/mol) = 162
...
204 g FeCl3
Then we construct the conversion factor from the quality between mass and moles
to perform the conversion:
𝟏𝟔𝟐
...
254 mol FeCl3 x (
) = 41
...
Avogadro’s Number and the Molar Mass of an Element
In the SI system the mole (mol) is the amount of a substance that contains as many
elementary entities (atoms, molecules, or other particles) as there are atoms in exactly
12 g (or 0
...
The actual number of atoms in 12 g of
carbon-12 is called Avogadro’s Number (NA), in an honor of the Italian scientist Amadeo
Avogadro, currently accepted value is

NA = 6
...
022 x 1023 units of X
Thus, just as one mole of hydrogen atoms contains 6
...

We have seen that 1 mole of carbon-12 atoms has a mass of exactly 12 g and contains
6
...
This mass of carbon-12 is its molar mass (ℳ), defined as the mass
(in grams or kilograms) of 1 mole of units (such as atoms or molecules) of a substance
...
99 amu and its molar mass is 22
...
97 amu and its molar mass is 30
...

Knowing the molar mass and Avogadro’s number, we can calculate the mass of a single
atom in grams
...
00 g and there
are 6
...
00 g carbon−12 atoms
6
...
993 x 10-23 g

The relationship between mass (m in grams) of an element and number of moles of an elements (n) and
between number of moles of an element and number of atoms (N) of an element
...


We can use the preceding result to determine the relationship between atomic mass
units and grams
...
993 𝑥 10−23 𝑔

= 6
...
022 x 1023 amu
And
1 amu = 1
...

The notions of Avogadro’s number and molar mass enable us to carry out conversions
between mass and moles of atoms and between moles and number of atoms
...
022 𝑥 1023 𝑋 𝑎𝑡𝑜𝑚𝑠

Where X represents the symbol of an element
...


Example
1
...
Its percent isotopic abundance among K isotopes is 0
...

How many 40K atoms are present in 225 mL of whole milk containing 1
...
There is no
single conversion factor that allow us to complete this conversion in one step, so we
anticipate having to complete several steps or conversions
...
65 mg
K/mL = 1
...
We can carry out the conversions g K
mol K
atoms K by
using conversion factors based on the molar mass of K and the Avogadro constant
...

The required conversions can be carried out in a stepwise fashion, or they can be
combined into a single line calculation
...
First, we convert
from mL milk to g K
...
65 𝑚𝑔 𝐾

x

1𝑔𝐾

1 𝑚𝐿 𝑚𝑖𝑙𝑘 1000 𝑚𝑔 𝐾

= 0
...
371 g K x

1 𝑚𝑜𝑙 𝐾
39
...
49 x 103 mol K

And then we convert from mol K to atoms K,
9
...
022 𝑥 1023 𝑎𝑡𝑜𝑚𝑠 𝐾
1 𝑚𝑜𝑙 𝐾

= 5
...

5
...
012 𝑎𝑡𝑜𝑚𝑠 40𝐾
100 𝑎𝑡𝑜𝑚𝑠 𝐾

= 6
...
An aspirin tablet contains 325 mg of acetylsalicyclic acid (C9H8O4)
...
You need both the molar mass of acetylsalicyclic
acid and Avogadro’s number as conversion factors
...


RELATIONSHIP USED
C9H8O4 molar mass = 9(12
...
008) + 4(16
...
15 g/mol
325 mg C9H8O4 x

10−3 𝑔
1 𝑚𝑔

x

1 𝑚𝑜𝑙 𝐶9 𝐻8 𝑂4
180
...
022 𝑥 1023 𝐶9 𝐻8 𝑂4 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠
1 𝑚𝑜𝑙 𝐶9 𝐻8 𝑂4

= 1
...
Formula Mass
Formula Mass for Covalent Substances
For covalent substances, the formula represents the numbers and types of atoms
composing a single molecule of the substance; therefore, the formula mass may be
correctly referred to as a molecular mass
...
The average molecular mass of a chloroform molecule
is therefor equal to the sum of the average atomic masses
...
37 amu, which is the sum of
the average atomic masses of each of its constituent atoms
...


Formula Mass for Ionic Compounds
Ionic compounds are composed of discrete cations and anions combined in ratios to
yield electrically neutral bulk matter
...
Keep
in mind, however, that the formula for an ionic compound does not represent the

composition of a discrete molecule, so it may not correctly be referred to as the
‘molecular mass’
...
The formula mass
for this compound is calculated as 58
...


Example
1
...
The
formula for juglone is C10H6O3
...
Calculate the molar mass of juglone
...
A sample of 1
...

How many moles of juglone does this sample represent?
Solution
a
...
In 1
mole of juglone, there are 10 moles of carbon atoms, 6 moles of hydrogen atoms, and
3 moles of oxygen atoms:
10 C: 10 x 12
...
1 g

6 H: 6 x 1
...
048 g

3 O: 3 x 16
...
00 g

Mass of 1 mol C10H6O3 = 174
...
1 g, which is the molar mass
...
Thus, 1
...
The exact fraction of amole can be
determined as follows:
1 𝑚𝑜𝑙 𝑗𝑢𝑔𝑙𝑜𝑛𝑒
1
...
96 x 105 mol juglone
174
...
The volatile liquid ethyl mercaptan, C2H6S, is one of the most odoriferous substance
known
...
0 L sample? The density
of liquid ethyl mercaptan is 0
...


Solution
Because the density is given in g/mL, it will be helpful to convert the measured volume
to milliliters
...
Finally,
the Avogadro constant can be used to convert the amount in moles to the number of
molecules
...

Convert from volume to mass
1
...
84 𝑔 𝐶2 𝐻6 𝑆
x
x
= 8
...
4 x 10 4 g C2H6S x

1 𝑚𝑜𝑙 𝐶2 𝐻6 𝑆
62
...
4 x 105 mol C2H6S

Convert from moles to molecules
1
...
02 𝑥 1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝐶2 𝐻6 𝑆
1 𝑚𝑜𝑙 𝐶2 𝐻6 𝑆

= 8
...

1
...
84 𝑔 𝐶2 𝐻6 𝑆
1 𝑚𝐿

x

1 𝑚𝑜𝑙 𝐶2 𝐻6 𝑆
62
...
1 x 1018 molecules C2H6S

6
...
Determining Empirical and Molecular Formulas
Percent Composition
When a compound’s formula is unknown, measuring the mass of each of its constituent
elements is often the first in the process of the determining the formula experimentally
...
The
percent composition of this compound could be represented as follows:
%H=

%C=

𝑚𝑎𝑠𝑠 𝐻
𝑚𝑎𝑠𝑠 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑
𝑚𝑎𝑠𝑠 𝐶
𝑚𝑎𝑠𝑠 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑

x 100%

x 100%

If analysis of a 10
...
5 g H and 7
...
5 𝑔 𝐻
10
...
5 𝑔 𝐻
10
...
A molecule of NH 3 contains one N atom
weighing 14
...
008 amu) = 3
...

The formula mass of ammonia is therefore (14
...
024 amu) = 17
...
01 𝑎𝑚𝑢 𝑁
17
...
024 𝑎𝑚𝑢 𝐻
17
...
27%

x 100% = 17
...
Percent composition is obtained by dividing the mass of each element in 1
mole of the compound by the molar mass of the compound and multiplying by 100
percent
...
For example,
in 1 mole of hydrogen peroxide (H2O2) there are 2 moles of H atoms and 2 moles of O
atoms
...
02 g, 1
...
00 g,
respectively
...
008 𝑔 𝐻
34
...
00 𝑔 𝑂
34
...
926%

x 100% = 94
...
926% + 94
...
99%
...


Example
Analysis of a 12
...
34 g C, 1
...
85 g N
...
34 𝑔 𝐶
12
...
85 𝑔 𝐻
12
...
85 𝑔 𝑁
12
...
0%

x 100% = 15
...
7%

The analysis results indicate that the compound is 61
...
4% H, and 23
...


Determination of Empirical Formulas
Chemical formulas represent the relative numbers, not masses, of atoms in substance
...
To accomplish this, we can
use molar masses to convert the mass of each element to a number of moles
...
Consider a sample of compound determined to contain 1
...
287 g
H
...
17 g C x

1 𝑚𝑜𝑙 𝐶
12
...
287 g C x

= 0
...
008 𝑔 𝐻

= 0
...
142H0
...
Of
course, per accepted convention, formulas contain whole-number subscripts, which can
be achieved by dividing each subscript by the smaller subscript:

𝐶0
...
248 or CH2
0
...
142

The empirical formula for this compound is thus CH2
...


Example
The bacterial fermentation of grain to produce ethanol forms a gas with a percent
composition of a 27
...
71% O
...
The calculation is “most convenient”
because, per the definition for percent composition, the
mass of a given element in grams is numerically equivalent to
the element’s mass percentage
...
The mass
percentages provided may be more conveniently expressed
as fractions:
27
...
71% O =

27
...
71 𝑔 𝐶
100 𝑔 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑

The molar amounts of carbon and hydrogen in a 100 g sample
are calculated by dividing each element’s mass by its molar
mass:
27
...
01 𝑔

) = 2
...
71 g O(
) = 4
...
00 𝑔

Procedure for calculating the empirical
formula of a compound from its percent
compositions

Coefficient for the tentative empirical formula are derived by dividing each molar
amount by the lesser of the two:
2
...
272
4
...
272

=1
=2

Since the resulting ratio is one carbon to two oxygen atoms, the empirical formula is CO2

Experimental Determination of Empirical Formulas
When ethanol is burned, carbon dioxide (CO2) and water(H2O) are given off
...

(Molecular oxygen was added in the combustion process, but some of the oxygen may
also have come from the original ethanol sample)
...
Suppose that in one experiment the
combustion of 11
...
0 g of CO2 and 13
...
We can
calculate the mass of carbon and hydrogen in the original 11
...
5 g ethanol contains 6
...
51 g of hydrogen
...
5 g – (6
...
51 g) = 4
...
5 g ethanol is:

The formula of ethanol is therefore C0
...
5O0
...
Because the number of atoms must be an integer, we divide the
subscripts by 0
...


In summary, empirical formulas are derived from experimentally measured element
masses by:
1
...
Dividing each element’s molar amount by the smallest molar amount to yield
subscripts for a tentative empirical formula
3
...
Empirical formula mass is the sum of the average atomic
masses of all the atoms represented in an empirical formula
...
The empirical formula mass for this compound is approximately 30 amu(the
sum of 12 amu for one C atom, 2 amu for two H atoms, and 16 amu for one O atom)
...
In this case, we are merely considering one mole of
empirical formula units and molecules, as opposed to single units and molecules
...
02% C, 8
...
27% N
...
57 g
of nicotine contains 0
...
As the first step, use the
percent composition to derive the compound’s empirical formula
...
02 g C)(

1 𝑚𝑜𝑙 𝐶
12
...
710 g H)(
(17
...
01 𝑔 𝐻

) = 6
...
624 mol H

1 𝑚𝑜𝑙 𝑁
14
...
233 mol N

Next, we calculate the molar ratios of these elements relative to the least abundant
element, N
6
...
233 mol N = 5
8
...
233 mol N = 7
1
...
233 mol N = 1

1
...
233
6
...
233
8
...
233

= 1
...
998 mol C
= 6
...
The empirical formula mass for this compound is therefore
81
...
13 g/mol formula unit
...
57 𝑔 𝑛𝑖𝑐𝑜𝑡𝑖𝑛𝑒
0
...
3 g
𝑚𝑜𝑙

Comparing the molar mass and empirical formula mass indicates that each nicotine
molecule contains two formula units:
162
...
13 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑢𝑛𝑖𝑡

= 2 formula units/molecule

Thus, we can derive the molecular formula for nicotine from the empirical formula by
multiplying each subscript by two:

(C5H7N)2 = C10H14N2

6
...

Often, though not always, a solution contains one component with a concentration that
is significantly greater than that of all components
...
Solution in which water is the solvent is called an aqueous solution
...
Solute concentration are often describe with
qualitative terms such dilute (of relatively low concentration) and concentrated (of
relatively high concentration)
...
Molarity
is defined as the number of moles of solute in exactly 1 liter (1 L) of the solution:
M=

𝒎𝒐𝒍 𝒔𝒐𝒍𝒖𝒕𝒆
𝑳 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏

Example
A 355 mL soft drink sample contains 0
...
What is the
molar concentration of sucrose in the beverage?
Solution
Since the molar amount of solute and the volume of solution are both given, the molarity
can be calculated using the definition of molarity
...
133 𝑚𝑜𝑙
355 𝑚𝐿 𝑥

1𝐿
1000 𝑚𝐿

= 0
...

A simple mathematical relationship can be used to relate the volumes and
concentration of a solutions before and after the dilution process
...
Since the dilution process does not change the amount of solute in the
solution, n1 = n2
...
Reflecting this
versatility, the dilution equation is often written:
C1V1 = C2V2
Where C and V are concentration and volume, respectively
...
850 L of a 5
...
80
L by the addition of water, what is the molarity of the diluted solution?
Solution
We are given the volume and concentration of a stock solution, V 1 and C1, and the
volume of the resultant diluted solution, V2
...
We thus rearrange the dilution equation in order to isolate C2:
C1V1 = C2V2
C2 =

𝐶1 𝑉1
𝑉2

Since the stock solution is being diluted by more than two-fold (volume is increased
from 0
...
80 L), we would expect the diluted solution’s concentration to be less
than one-half 5 M
...
Substituting the given values for the terms
on the right side of this equation yields:
𝑚𝑜𝑙

0
...
00 𝐿
C2 =
= 2
...
80 𝐿

Other Units for Solution Concentrations
Mass Percentage
The mass percentage of a solution component is defined as the ratio of the component’s
mass to the solution’s mass, expressed as a percentage:
mass percentage =

mass of component
mass of solution

x 100%

Mass percentage is also referred to by similar names such as percent mass, percent
weight, and other variations on this theme
...


Example
A 5
...
75 mg (0
...
What is the
percent by mass of glucose in spinal fluid?
Solution
The spinal fluid sample contains roughly 4 mg of glucose in 5000 mg of fluid, so the mass
fraction of glucose should be a bit less than one part in 1000, or about 0
...
Substituting
the given mass into the equation defining mass percentage yields:
1𝑔

% glucose =

3
...
0 𝑔 𝑠𝑝𝑖𝑛𝑎𝑙 𝑓𝑙𝑢𝑖𝑑

= 0
...
1%)

Volume Percentage
The concentration of a solution formed by dissolving a liquid solute in a liquid solvent is
therefore often expressed as a volume percentage, %vol or (v/v)%:
volume percentage =

volume solute
volume solution

x 100%

Example
Rubbing alcohol (isopropanol) is usually sold as a 70%vol aqueous solution
...
785 g/mL, how many grams of isopropyl alcohol are
present in a 355 mL bottle of rubbing alcohol?
Solution
Per definition of volume percentage, the isopropanol volume is 70% of the total solution
volume
...
785 𝑔 𝑖𝑠𝑜𝑝𝑟𝑜𝑝𝑦𝑙 𝑎𝑙𝑐𝑜ℎ𝑜𝑙

)(

100 𝑚𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
1 𝑚𝐿 𝑖𝑠𝑜𝑝𝑟𝑜𝑝𝑦𝑙 𝑎𝑙𝑐𝑜ℎ𝑜𝑙
isopropyl alcohol

) = 195 g

Mass-Volume Percentage
A mass-volume percent is a ratio of a solute’s mass to the solution’s volume expressed
as percentage
...
9% mass/volume (m/v), indicating that the composition
is 0
...


Parts per Million and Parts per Billion
Very low solute concentrations are often expressed using approximately small units
such as parts per million (ppm) or part per billion (ppb)
...
The mass-based definitions of ppm and ppb are given here:
ppm =

ppb =

𝑚𝑎𝑠𝑠 𝑠𝑜𝑙𝑢𝑡𝑒
𝑚𝑎𝑠𝑠 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑚𝑎𝑠𝑠 𝑠𝑜𝑙𝑢𝑡𝑒
𝑚𝑎𝑠𝑠 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

x 106 ppm

x 109 ppb

Example
According to the EPA, when the concentration of lead in tap water reaches 15 ppb,
certain remedial actions must be taken
...
Comparing this two unit definitions shows that ppm is 1000 times
greater than ppb (1 ppm = 103 ppb):
15 ppb x

1 𝑝𝑝𝑚
103 𝑝𝑝𝑏

= 0
...
However, only the volume of solution (300 mL) is given, so
we must use the density to derive the corresponding mass
...
Rearranging the
equation defining the ppb unit and substituting the given quantities yields:
ppb =

𝑚𝑎𝑠𝑠 𝑠𝑜𝑙𝑢𝑡𝑒
𝑚𝑎𝑠𝑠 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

mass solute =

mass solute =

x 109 ppb

𝑝𝑝𝑏 𝑥 𝑚𝑎𝑠𝑠 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
109 𝑝𝑝𝑏

15 𝑝𝑝𝑏 𝑥 300 𝑚𝐿 𝑥

1
...
5 x 10-6 g

Finally, convert this mass to the requested unit of micrograms:
1 𝑔

4
...
5 g

7
...


This example illustrates the fundamental aspects of any chemical equation:
1
...

2
...

3
...

4
...


Balancing Equations
An equation is balanced meaning that equal numbers of atoms for each element
involved in the reaction are represented on the reactant and product sides
...


1
...
Organize the formulas in the pattern of an equation
with plus sign and an arrow (think of the arrow as an equal sign because we need to
end up with the same number of each atom on both sides)
...

2
...
When doing step 2, make no changes in the formulas, either in the atomic
symbols or their subscripts
...
It often helps to start the process with most complex formula,
leaving elements and simple compounds to the end
...
The
reactants are zinc, Zn(s), and hydrochloric acid, an aqueous solution of the gas hydrogen
chloride, HCl, symbolize as HCl(aq)
...

1
...

Zn(s) + HCl(aq)

ZnCl2(aq) + H2(g)

(unbalanced)

2
...

Some Guidelines for Balancing Equations
1
...
Elements, particularly H 2
and O2, should be left until the end
...
Balance atoms that appear in only two formulas: one as a reactant and the other as
a product
...

3
...

We will leave the Zn(s) and H2(g) until later
...
Because there are two Cl to the right of the arrow but only one to the left,
we put a 2 in front of the HCl on the left side
...


Example
When aluminum metal is exposed to air, a protective layer of aluminum oxide (Al 2O3)
forms on its surface
...
Write a
balance equation for the formation of Al2O3
...
We can balance the Al atoms by placing a coefficient of 2 in front of
Al on the reactants side
...
We can balance the O atoms by placing a coefficient of 2 in front of O2 on
the reactant side
...
However, equations are normally balanced with the smallest
set of whole number coefficients
...

3

2 (2 Al + 2 O2

Al2O3)

or
4 Al + 3 O2

2 Al2O3

Equations for Ionic Reactions
When aqueous solutions of CaCl2 and AgNO3 are mixed, a reaction takes place
producing aqueous Ca(NO3)2 and solid AgCl2
...

Explicitly representing all dissolved ions result in a complete ionic equation
...
These spectator ions-ions whose
presence is required to maintain charge neutrally-are neither chemically nor physically
changed by the process, and so they may be eliminated from the equation to yield a
more succinct representation called a net ionic equation:
Ca2+(aq) + 2 Cl(aq) + 2 Ag+(aq) + 2 NO3(aq)
2 Cl(aq) + 2 Ag+(aq)

Ca2+(aq) + 2 NO3(aq) + 2 AgCl(s)

2 AgCl(s)

Following the convention of using the smallest possible integers as coefficients, this
equation is then written:
2 Cl(aq) + 2 Ag+(aq)

2 AgCl(s)

Example
When carbon dioxide is dissolved in an aqueous solution of sodium hydroxide, the
mixture reacts to yields aqueous sodium carbonate and liquid water
...


Solution
Begin by identifying formulas for the reactants and products and arranging them
properly:
CO2(aq) + NaOH(aq)

Na2CO3(aq) + H2O(l)

(unbalanced)

Balance is achieved easily in this case by changing the coefficient for NaOH to 2,
resulting in the molecular equation for this reaction:
CO2(aq) + 2 NaOH(aq)

Na2CO3(aq) + H2O(l)

The two dissolved ionic compounds, NaOH and Na2CO3, can be represented as
dissociated ions to yield the complete ionic equation:
CO2(aq) + 2 Na+ (aq) + 2 OH (aq)

2 Na+(aq) + CO32 (aq) + H2O(l)

Finally, identify the spectator ion(s), in this case Na + (aq), and remove it from each side
of the equation to generate the next ionic equation:
CO2(aq) + 2 Na+ (aq) + 2 OH (aq)
CO2(aq) + 2 OH (aq)

2 Na+(aq) + CO32 (aq) + H2O(l)
CO32 (aq) + H2O(l)

8
...
Many reactions of this type involve the exchange of ions between
ionic compounds in aqueous solution and are sometimes referred to as double
displacement, double replacement, or metathesis reactions
...
Substances with relatively
large soluble are said to be soluble
...
Substances with
relatively low solubility are said to be insoluble, and these are the substances that
readily precipitate from solution
...
If precipitation is expected, write a balanced net ionic equation for the
reaction
...
Potassium sulfate and barium nitrate
b
...
Lead nitrate and ammonium carbonate
Solution
a
...
The solubility
guidelines indicate BaSO4 is insoluble, and so a precipitation is expected
...
The two possible products for this combination are LiC 2H3O2 and AgCl
...

The net ionic equation for this reaction:
Ag+(aq) + Cl(aq)

AgCl(s)

c
...
The
solubility guidelines indicate PbCO3 is insoluble, and so a precipitation is expected
...

oxidation = loss of electron
reduction = gain of electron
In this reaction, then, sodium is oxidized and chlorine undergoes reduction
...
Likewise, chlorine functions as an oxidizing agent (oxidant), as it effectively
removes electron from (oxidizes) sodium
...
For example, a
reaction similar to the one yielding NaCl:
H2(g) + Cl2(g)

2 HCl(g)

The product of this reaction is a covalent compound, so transfer of electrons in the
explicit sense is not involved
...
The oxidation number (or oxidation state) of an element in
a compound is the charge its atoms would possess if the compound was ionic
...

1
...

2
...

3
...
The sum of oxidation numbers for all atoms in a molecule or polyatomic ion equals
the charge on the molecule or ion
...
H2S
b
...
Na2SO4
Solution
a
...

Using this oxidation number and the compound’s formula, guideline 4 may then be
used to calculate the oxidation number for sulfur:
charge on H2S = 0 = (2 x (+1)) + (1 x S)
S = 0 – (2 x (+1)) = - 2
b
...

Using this oxidation number and the ion’s formula, guideline 4 may then be used to
calculate the oxidation number for sulfur:

charge on SO32 = 2 = (3 x (2)) + (1 x S)
S = 2 – (3 x (2)) = + 4
c
...

According to guideline 2, the oxidation number for sodium is +1
...
Definition for the complementary
process of this reaction class are correspondingly revised as:
oxidation = increase in oxidation number
reduction = decrease in oxidation number
In the reaction between sodium and chlorine to yield sodium chloride, sodium is
oxidized (its oxidation number increases from 0 in Na+ to 1 in NaCl) and chlorine is
reduced (its oxidation number decreases from 0 in Cl2 to 1 in NaCl)
...


Example
Identify which equations represent redox reactions, providing a name for the reaction
if appropriate
...

a
...
2 Ga(l) + 3 Br2(l)
c
...
BaCl2(aq) + K2SO4(aq)
e
...
This is not redox reaction, since oxidation numbers remain unchanged for all
elements
...
This is a redox reaction
...
The reducing agent is Ga(l)
...
The oxidizing agent
is Br2(l)
...
This is a redox reaction
...
Oxygen is
oxidized, its oxidation number increasing from 1 in H2O2(aq) to 0 in O2(g)
...

For disproportionation reactions, the same substance functions as an oxidant and a
reductant
...
This is not redox reaction, since oxidation numbers remain unchanged for all
elements
...
This is a redox reaction (combustion)
...
The reducing agent (fuel) is C2H4(g)
...

The oxidizing agent is O2(g)
...
Although this species are not oxidizes or
reduced, they do participate in chemical change in other ways ( e
...
, by providing the
elements required to form oxyanions)
...
One very useful approach is to use the method of
half-reactions, which involves the following steps:
1
...

3
...

5
...


Write the two half-reactions representing the redox process
...

Balance oxygen atoms by adding H2O molecules
...

Balance charge by adding electrons
...


7
...

8
...
Add OH ions to both sides of the equation in numbers equal to the number of H+
ions
...
On the side of the equation containing both H + and OH ions, combine this ions
to yield water molecules
...
Simplify the equation by removing any redundant water molecules
...
Finally, check to see that both the number of atoms and the total charges are
balanced
...

Solution
Cr2O72 + Fe2+

Cr3+ + Fe3+

Step 1
...
Balance all elements except oxygen and hydrogen
...
Changing the coefficient on the right side of the equation to 2
achieves balance with regard to Cr atoms
...
Balance oxygen atoms by adding H2O molecules
...
The chromium half-reaction shows seven O atoms on the left and none
on the right, so 7 water molecules are added to the right side
...
Balance the hydrogen atoms by adding H+ ions
...
The chromium half-reaction shows 14 H atoms on the right and none
on the left, so 14 hydrogen ions are added to the left side
...
Balance charge by adding electrons
...
Adding one electron
to the right side bring that side’s total charge (3+) + (1-) = 2+, and charge balance is
achieved
...
The total charge on the right side is (2 x 3+) = 6+ (2
Cr3+ ions)
...

Fe3+ + e

Fe2+
Cr2O72 + 6 e + 14 H+

2 Cr3+ + 7 H2O

Step 6
...
To be consistent with mass conservation,
and the idea that redox reactions involve the transfer of electrons, the iron halfreaction’s coefficient must be multiplied by 6
...
Add the balanced half-reactions and cancel species that appear on both sides of
the equation
...
Removing them from each side of the
equation yields the simplified, balance equation here:
6 Fe2+ + Cr2O72 + 14 H+

6 Fe3+ + 2 Cr3+ + 7 H2O

9
...
Chemical formulas provide the identities of the reactants and products involved
in the chemical change, allowing classification of the reaction
...
These quantitative relationships are known as the reaction’s stoichiometry
...
The coefficients in the balance equation are used
to derived stoichiometric factors that permit computation of the desired quantity
...

Numerous variations on the beginning and ending computational steps are possible
depending upon what particular quantities are provided and sought (volumes, solution
concentration, and so forth)
...


The flowchart depicts the various computational steps involved in most reaction
stoichiometry calculations
...
Reaction Yields
Limiting Reactant
Consider this concept now with regard to a chemical process, the reaction of hydrogen
with chlorine to yield hydrogen chloride:
H2(g) + Cl2(g)

2 HCl(g)

The balanced equation shows the hydrogen and chlorine react in a 1:1 stoichiometric
ratio
...
This substance is the limiting reactant, and the other substance is the excess
reactant
...
This represent a 3:2 (or
1
...
Hydrogen, therefore, is present in excess, and chlorine is the

limiting reactant
...

An alternative approach to identifying the limiting reactant involves comparing the
amount of product expected for the complete reaction of each reactant
...
The reactant yielding the lesser amount of product is the
limiting reactant
...

Since enough hydrogen was provided to yield 6 moles of HCl, there will be unreacted
hydrogen remaining once this reaction is complete
...


When H2 and Cl2 are combined in nonstoichiometric amounts, one of these reactants
will limit the amount of HCl that can be produced
...
72 mol of triethylene
glycol, C6H14O4, in an excess of O2?
Solution
An excess of O2 means that there is more than enough O2 available to permit the
complete conversion of the triethylene glycol to CO2 and H2O
...

2 C6H14O4 + 15 O2

12 CO2 + 14 H2O

Thus, 12 mol CO2 are produced for every 2 mol C6H14O4 burned
...

? mol CO2 = 2
...
3 mol CO2

What mass of CO2 is formed in the reaction of 4
...
Using a stepwise approach:
Convert from grams of C6H14O4 to moles of C6H14O4 by using the molar mass of
C6H14O4 as a conversion factor
...
16 g C6H14O4 x

1 𝑚𝑜𝑙 𝐶6 𝐻14 𝑂4
150
...
0277 mol C6H14O4

Convert from moles of C6H14O4 to moles of CO2 by using the stoichiometric factor
...
0277 mol C6H14O4 x

12 𝑚𝑜𝑙 𝐶𝑂2
2 𝑚𝑜𝑙 𝐶6 𝐻14 𝑂4

= 0
...

0
...
01 𝑔 𝐶𝑂2
1 𝑚𝑜𝑙 𝐶𝑂2

= 7
...

P4(s) + 6 Cl2(g)

4 PCl3(l)

What is the maximum mass of PCl3 that can be obtained from 125 g P4 and 323 g Cl2?
The following conversion are required: mol limiting reactant
mol PCl 3
g PCl3
The key to solving this problem is to correctly identify the limiting reactant
...
If more than 6 mol Cl2 is available per mole of P4, chlorine is excess
and P4 is the limiting reactant
...


323 g Cl2 x
125 g Cl2 x

1 𝑚𝑜𝑙 𝐶𝑙2
70
...
9 𝑔 𝑃4

= 4
...
01 mol P4

Since there is less than 6 mol Cl2 per mole of P4, chlorine is the limiting reactant
...

Convert from grams of Cl2 to moles of Cl2 by using the molar mass of Cl2
...
91 𝑔 𝐶𝑙2

= 4
...

4
...
04 mol PCl3

Convert from moles of PCl3 to grams of PCl3 by using the molar mass of PCl3
...
04 mol PCl3 x

137
...
In practice, the amount of product obtained is
called the actual yield, and it is often less than the theoretical yield for a number of
reasons
...

percent yield =

actual yield
𝑡ℎ𝑒𝑜𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑

x 100%

Actual and theoretical yields may be expressed as masses or molar amounts (or any
other appropriate property; e
...
, volume, if the product is a gas)
...
0 g of phosphorus
with 35
...
4 g of solid phosphorus trichloride
...

2 P(s) + 3 Cl2(g)

2 PCl3(s)

Solution
The first step is to figure out which reactant, P or Cl 2, is the limiting reactant, because
we must bass all calculations on the limiting reactant
...
Finally we calculate
the percentage yield
...
97 g P
1 mol Cl2 = 70
...
We’ll choose phosphorous and see
whether there is enough to react with 35
...

12
...
97 g P

x

3 mol 𝐶𝑙2
2 mol P

x

70
...
2 g Cl2

Thus, with 35
...
2 g of Cl2 needed, there is not enough Cl2 to
react with all 12
...
The Cl2 will be all used up before the P is used up, so Cl 2 is the
limiting reactant
...

To find the theoretical yield of PCl3, we calculate how many grams of PCl3 could be made
from 35
...
0 g Cl2 x

1 mol 𝐶𝑙2
70
...
32 g P𝐶𝑙3
1 mol P𝐶𝑙3

= 45
...
4 g of PCl3, not 45
...
𝟒 𝒈 𝒐𝒇 𝐏𝑪𝒍𝟑
𝟒𝟓
...
8%



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