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Lecture 5. Chapter 3.
Chemical Equations:
2 H2 + O2 → 2 H2O
Stoichiometry: Calculations with
Chemical Formulas and Equations.
Chemical Equations:
The equation for hydrogen burning in
oxygen is represented as:
2 H2 + O2 → 2 H2O
The ‘+’ sign means ‘reacts with’
The arrow means ‘produces’
2 H2 + O2 → 2 H2O
The chemical substances on the left of
the equation are the ‘reactants’.
The chemical substances on the right
of the equation are the ‘products’.
The numbers in front of the formulas
are the coefficients.
Balanced equations:
Because of the Law of conservation of
matter, the numbers of each type of
atom must be the same on the left and
the right side of the equation. The
equation must be BALANCED.
(4 H’s on each side, 2 O’s on each side).
2 H2 + O2 → 2 H2O
2x2=4H 2O
Note: The coefficient
multiplies through
everything in the
substance that follows:
2 x 2 = 4H and 2 O
View of reaction at the molecular
level:
4 H’s and 2 O’s
Before reaction
4 H’s and 2 O’s
after reaction
‘produces’
O
H
H
H
H
+
O
O
O
two hydrogen
molecules
one oxygen
molecule
two water
molecules
Balancing equations:
Once we know the reactants and
products in the reaction, we can write
the unbalanced equation.
e.g. CH4 + O2
→
CO2 + H2O
To balance the equation, we must find
the coefficients that will lead to a
balanced equation.
Balancing equations: Change
only the coefficients:
In balancing an equation, only
coefficients can be changed.
Changing subscripts is not
allowed because this would
lead to a change in the nature
of the substances involved.
It is usually best to start by balancing the
element(s) that occur in the fewest chemical
formulas on either side of the equation.
There always has to be the same number of
carbon atoms on either side of the equation.
Each C atom needs one O2 molecule to form
CO2.
The 4 H’s of CH4 need two O atoms = one O2
molecule to form 2 H2O:
CH4 + 2 O2 → CO2 + 2 H2O
(Note spaces between coefficients and
formulas for substances).
Molecular view of reaction:
one C atom
four H atoms
four O atoms
+
one methane
molecule
one C atom
two O-atoms
two O-atoms
four H-atoms
+
two oxygen
molecules
‘before’
one carbon
dioxide molecule
‘after’
two water
molecules
Combination and Decomposition
reactions.
Combination reaction: Here two or
more substances react to form one
product:
2 Mg + O2(g) → 2 MgO(s)
In a decomposition reaction, one
substance breaks down into two or
more:
CaCO3(s) → CaO(s) +
CO2(g)
Combustion in air:
Combustion reactions are rapid reactions in
air that produce a flame.
Hydrocarbons consist only of C and H.
When balancing an equation with
combustion of these molecules, start with
the C. The number of CO2 molecules
produced is the same as in the
hydrocarbon:
C3H8
+
O2 →
(not yet balanced)
3 CO2
+ H2O
Next balance the H-atoms. The number
of waters produced (4) is half the
number of H’s (8) in the hydrocarbon:
C3H8
8H
+ O2 → 3 CO2
+
4 H 2O
4x2=
8H
Balance the O-atoms
Finally, add up the number of O-atoms
needed on the r.h.s. = 6 + 4 = 10.
Therefore need 5 O2 molecules.
C3H8
+
5 O2 → 3 CO2 + 4 H2O
5 x 2 = 10 O’s
10 O
3 x 2 = 6 O’s 4 O’s
10 O
Molecular view of reaction:
3 C atoms
8 H atoms
10 O atoms
3 C atoms
6 O-atoms
+
+
C3H8 + 5 O2
4 O-atoms
8 H-atoms
→
3 CO2 + 4 H2O
Combustion of O-containing derivatives:
One needs to remember here that there is an O on
the left hand side, so the numbers of O2’s needed is
going to be less by that amount:
C2H5OH + x O2 → 2 CO2 +
3 H 2O
On r.h.s. are 4 + 3 = 7 O-atoms. But ethanol already
has one O-atom, so need only 3 O2 molecules:
C2H5OH + 3 O2 → 2 CO2 + 3 H2O
1O
6 O’s
7 O’s
4 O’s
3 O’s
7 O’s
3.3. Formula weights.
Chemical formulas and chemical equations
contain quantitative information. They tell us
how much of each reactant is needed for the
reaction.
H = 1.0 amu
O = 16.0 amu
Weight of H2O =
16.0 + 1.0 + 1.0 = 18.0 amu
Formula and molecular weights:
The formula weight (F. Wt.) is the
sum of the atomic weights of each
atom in its chemical formula. If the
chemical formula is that of a
molecule, then the formula weight
is referred to as a molecular weight
(M. Wt.).
Formula weights:
e.g. M. Wt. of glucose:
C6H12O6:
C = 12.0 amu H = 1.0 amu,
O = 16.0 amu
So F. Wt. =
(6 x 12.0) + (12 x 1.0) + (6 x
16.0)
= 180 amu.
glucose
Exercise 3.5: F. Wt. of sucrose: C12H22O11: (342.0) Ca(NO3)2: 164.1 amu.
Percentage composition from formulas:
When a new compound is analyzed to obtain its
elemental composition, this is compared with
the percentage composition obtained from its
formula:
% element =
(no. atoms of element) x (At. Wt. element) x 100
F. Wt. of compound
C12H22O11 (sucrose): F. Wt. = 342.0 amu.
% C = (12 x 12.0) x 100/342 = 42.1%
% H = (22 x 1.0) x 100/342 = 6.4%
% O = (11 x 16.0) x 100/342 = 51.5%
check: 42.1 + 6.4 + 51.5 = 100.0%.
The Mole.
3.4. Avogadro’s number and
the Mole.
Amu are far too small to be
useful in everyday chemistry.
We use instead the Mole
(abbreviated mol) which is
basically the F. Wt. or M. Wt.
expressed in grams. e.g.
Water has a formula weight of
18.0 amu, so
1 mole of water weighs 18.0
grams.
Lorenzo Romano
Amedeo Carlo
Avogadro, conte
di Quaregna e di
Cerreto (1776 1856)
One mole of water
18.0 grams
of water
Water has a formula weight of 18.0
amu, so 1 mole of water weighs 18.0
grams. How do we arrive at this?
Conversion factors: 1 amu = 1.6 x 10-24 g,
1 mole contains 6.022 x 1023 molecules,
1 molecule of H2O weighs 18.0 amu:
18.0 amu x 6.022 x 1023 molecules x 1.6 x 1024 g
1 molecule
1 mol
1 amu
= 18.0 g/mol
Avogadro’s number
Note: 1 mol of 12C contains 6.022 x 1023 12C
atoms. 1 mol of C2H4 contains 6.022 x 1023
C2H4 molecules. 1 mol of glucose contains
6.022 x 1023 molecules
1 mol of water =
18.0 grams of water
1 mol of water contains 6.022 x 1023 water
molecules. How many H-atoms does it contain?
Think about dozens instead of moles:
One dozen water molecules…..
…contains one dozen oxygen atoms, but
two dozen hydrogen atoms.
One mol of water contains one mol of oxygen
atoms but two moles of Hydrogen atoms
Numbers of atoms in molecules:
One mol of ethylene (C2H4) contains 6.022 x 1023
ethylene molecules, but:
H
H
C
H
C
ethylene
C2H4
H
One mol of ethylene (C2H4)
contains 4 x 6.022 x 1023
= 2.4088 x 1024 H-atoms
One mol of ethylene (C2H4)
contains 2 x 6.022 x 1023
= 1.2044 x 1024 C-atoms
Counting BB’s
Suppose you sell BB’s for air-guns in
boxes each containing 1000 BB’s. How
would you put 1000 BB’s in each box?
Box must contain
1000 BB’s
(one BB weighs 0.34 g)
Weighing as a means of counting:
Avogadro’s number (6.022 x 1023) is a very
large number, but it is still just a number, like
a dozen is a number. Just as with BB’s, there
are too many molecules in even a tiny
amount of a substance for us to count them.
We have to weigh them. And we know that in
one mol of substance there are 6.022 x 1023
molecules. So chemists work in moles,
knowing 1 mole contains 6.022 x 1023
molecules.
Fractions of moles and numbers of atoms:
Exercise 3.8. Calculate the number of H atoms in
0.350 moles of C6H12O6.
Conversion: 1 mole = 6.022 x 1023 molecules
1 molecule = 12 H-atoms
0.35 moles x 6.022 x 1023 molecules x 12 atoms =
1 mole
1 molecule
= 25.3 x 1023 = 2.53 x 1024 H-atoms
Table 3.2: Mole Relationships:
Substance
F.Wt. Molar Mass No. of particles
(in one mole)
14.0 14.0 g/mol 6.022 x 1023
28.0 28.0 g/mol 6.022 x 1023
Atomic N
N2 gas
N atoms in
1 mole N2
Ag(s)
Ag+ ions
BaCl2
107.9 107.9 g/mol
107.9 107.9 g/mol
208.2 208.2 g/mol
AlCl3
133.3
133.3 g/mol
1.2044 x 1024 N’s
6.022 x 1023
6.022 x 1023
6.022 x 1023 Ba2+
1.2044 x 1024 Cl6.022 x 1023 Al3+
1.806 x 1024 Cl-
Some problems:
How many moles of H atoms are there
in 1.0 moles of NH4Cl ?
How many H atoms are there in 1.0
moles of NH4Cl ?
How many H atoms are there in 2 moles
of NH4Cl ?
How many H atoms are there in 0.3724
moles of NH4Cl ?
Interconverting masses and
numbers of particles:
e.g. How many Cu atoms in
a Cu penny, weight = 3.0 g
(assume the penny is pure
copper).
copper penny
Conversion: Cu = 63.5 g/mol; 1 mol = 6.022 x 1023 atoms
(note units)
3.0 g x 6.022 x 1023 atoms x 1 mol =
1 mol
63.5 g
2.8 x 1022 atoms
3.5 Empirical Formulas from Analyses.
When Chemists discover new
compounds, they may analyze them to get
their percentage elemental composition.
Thus, if we analyze a compound and find
that it contains 73.9% Hg and 26.1% Cl by
weight, we can work out the molar ratio.
Note that when we say something is
73.9% Hg, that means 100.0 g of the
substance would contain 73.9 g Hg, and
similarly 26.1 g Cl.
Assume that we have 100g sample.
Thus, the percentages mean that we
have 73.9 g Hg and 26.1 g Cl.
Conversion factors (from periodic
table):
Hg, 1 mol = 200.6 g (= atomic wt.)
Cl, 1 mol = 35.5 g (= atomic wt.)
So the molar ratio is given by:
Hg: 73.9 g x 1 mol/200.6 g = 0.368 mol
Cl: 26.2 g x 1 mol/35.5 g = 0.735 mol
To get the molar ratio, divide through by
the lowest molar ratio present:
Hg: 0.368 mol/0.368 mol = 1.00
Cl: 0.735 mol/0.368 mol = 1.997
(~ 2.00)
We can say that within experimental
error, the empirical formula of the
compound is HgCl2.
Example 3.13: Ascorbic acid is 40.92% C, 4.58% H, and 54.40% O
by mass. What is the empirical formula?
Example 3.13: Ascorbic acid
Conversion factors: C, 1 mol = 12.0 g; H, 1 mol = 1.0
g; O, 1 mol = 16.0 g.
C = 40.92 g x 1 mol/12.0 g
H = 4.58 g x 1 mol/1.0 g
O = 54.40 g x 1 mol/16.0 g
= 3.407 mol
= 4.54 mol
= 3.406 mol
Divide through by lowest number of moles (3.406)
C = 3.407 mol/3.406 mol
=
1.00
H = 4.54 mol /3.406 mol
=
1.33
O = 3.406 mol /3.406 mol =
1.00
Multiply through by 3 to get all near whole numbers.
Empirical formula is C3H4O3.
Molecular formula from empirical Formula.
If we know the molecular weight of a
compound, we can then convert the
empirical formula into a molecular formula .
We calculate the empirical formula weight,
which is the formula weight calculated from
the empirical formula. We then obtain a
whole number multiple by dividing the
molecular weight by the empirical formula
weight, and it is this multiple that we use to
multiply through the subscripts in the
empirical formula.
Mesitylene has an empirical formula of C3H4. The experimentally
determined M.Wt. is 121.0 amu. What is the molecular formula?
An example: mesitylene
Mesitylene has an empirical formula of C3H4. The
experimentally determined M.Wt. is 121.0 amu.
What is the molecular formula of mesitylene?
C3H4 empirical formula weight
= 3 x 12 + 4 x 1 = 40 amu
Multiple
=
=
=
Experimental M.Wt
empirical formula weight
121amu /40 amu
3.02 (~ 3.0 )
Multiple = 3.0, so molecular formula = C9H12