Download Balancing Chemical Equations

Chemistry 201
Lecture 1
Conversions
Stoichiometry
Balancing Equations
NC State University
The mass of a proton
The mass of a proton is:
1.6726231 × 10−27 kg
or
1.6726231 × 10−24 grams
We know this value accurately because of mass
spectrometry. The number cited here has eight
significant figures. We do not usually need this
precision, so we often write the value as
1.67 × 10−24 grams
to three significant figures.
Significant figures
The number of significant figures is equal to the
number of digits in a measured or calculated value that
contribute to its precision. Precision refers to the
ability to reproducibly measure or calculate a value.
If we use three significant figures, it suggests that we
can reproducibly measure the value to within about 1
part in 100 or with an accuracy of 1%. This is the most
common number of significant figures and this will be
the default in this course (unless otherwise specified).
Example
The mass of the electron is reported to be:
9.1093819 × 10-31 kg
Write this number to three significant figures.
Example
The mass of the electron is reported to be:
9.1093819 × 10-31 kg
Write this number to three significant figures.
Solution: The specified value should be as close as
possible to the true value so we should round-off the
last digit. In this case we round up to obtain
9.11 × 10-31 kg
Example
The mass of a neutron is approximately equal to sum of
the masses of an electron and a proton. Give the
neutron mass to three significant figures.
Example
The mass of a neutron is approximately equal to sum of
the masses of an electron and a proton. Give the
neutron mass to three significant figures.
Solution: Start with the accurate values:
1.6726231 × 10−27 kg
9.1093819 × 10-31 kg
Example
The mass of a neutron is approximately equal to sum of
the masses of an electron and a proton. Give the
neutron mass to three significant figures.
Solution: Start with the accurate values:
1.6726231 × 10−27 kg
9.1093819 × 10-31 kg
Add them together
1.6735340 × 10−27 kg
Example
The mass of a neutron is approximately equal to sum of
the masses of an electron and a proton. Give the
neutron mass to three significant figures.
Solution: Start with the accurate values:
1.6726231 × 10−27 kg
9.1093819 × 10-31 kg
Add them together
1.6735340 × 10−27 kg
Round of to give
1.67 × 10−27 kg
Example
The actual value of the neutron mass is:
1.6749286 × 10−27 kg
Calculate the difference between the value you
obtained by summing the proton and electron masses.
How many significant figures are possible in your
answer?
Example
The actual value of the neutron mass is:
1.6749286 × 10−27 kg
Calculate the difference between the value you
obtained by summing the proton and electron masses.
How many significant figures are possible in your
answer?
Solution:
Neutron
1.6749286 × 10−27 kg
Proton + electron
1.6735340 × 10−27 kg
Difference
1.3946 × 10−30 kg
There are 5 significant figures in the answer.
Conversion factors for atomic mass
The sum of the mass of a proton and an electron is the
mass of a hydrogen atom.
Question: how many hydrogen atoms are there in a
gram of H atoms (to 3 significant figures)?
Conversion factors for atomic mass
The sum of the mass of a proton and an electron is the
mass of a hydrogen atom.
Question: how many hydrogen atoms are there in a
gram of H atoms (to 3 significant figures)?
Answer: The calculated value actually has units
1.6735340 × 10−27 kg/atom
Or
1.6735340 × 10−24 grams/atom
Therefore, we can invert it to find,
5.97(5) × 1023 atoms/gram
Atomic mass unit
When we consider all of the atoms in the periodic
table, the average mass of a nucleon is considered to
be:
1.6605388 × 10−24 grams/nucleon
We call this the atomic mass unit. We can use this
value to convert atomic masses to grams or vice versa.
We write the conversion as,
1.6605388 × 10−24 grams/amu
To employ this value we calculate the atomic mass of
an atom or molecule and then we can calculate the
weight in grams using this formula.
Avagadro’s number
If we invert average atomic mass
___________1_____________
1.6605388 × 10−24 grams/amu
We obtained the number of particles with a given amu
per gram. This number is called Avagradro’s number
and is given the symbol NA.
NA = 6.022141 × 1023 amu/gram
This number gives the number of particles for which an
atomic mass has the the same value in grams.
1 H atom = 1 amu
NA H atoms = 1 gram
1 C atom = 12 amu
NA H atoms = 12 grams
Example
How much does a molecule of pyridine weigh?
(Ummm… all right, what is its mass?)
Example
How much does a molecule of pyridine weigh?
(Ummm… all right, what is its mass?)
Solution: First, we find the chemical formula for
pyridine. C5H5N
H
H
H
H
H
N
Example
How much does a molecule of pyridine weigh?
(Ummm… all right, what is its mass?)
Solution: Second, we look up the atomic masses in the
periodic table.
atomic mass = 5(12) + 5 + 14 = 79 amu
Third, we use the conversion factor to calculate the
mass in grams
(79 amu)x (1.66 × 10−24 grams/amu) = 1.31 × 10−23 grams
Avogadro’s number
It is a bit difficult to weigh out 10−23 grams.
H
H
H
H
H
N
Avogadro’s number
Instead, let’s ask how many molecules it takes to
convert the atomic mass to its value in grams.
For example, using the grams/amu conversion, let’s
calculate how many hydrogen atoms have the mass of
1 gram.
Avogadro’s number
Instead, let’s ask how many molecules it takes to
convert the atomic mass to its value in grams.
For example, using the grams/amu conversion, let’s
calculate how many hydrogen atoms have the mass of
1 gram.
Answer: since hydrogen weighs 1 amu, its mass is
1.6605388 × 10−24 grams/atom
We can invert this value to find the number of atoms
per gram.
6.0221417 × 1023 atoms/gram
The mole
Since this number converts from atoms to gram for
hydrogen, we can see that it can be used to give the
number of atoms for any formula weight (i.e. molecular
weight of a compound given in grams). For example,
pyridine has a formula weight of 91 grams. Therefore,
There are 6.0221417 × 1023 molecules/91 grams of
pyridine. Because of the importance of this number of
atoms or molecules we give the name, mole.
1 mole = 6.0221417 × 1023 molecules
or to 3 significant figures.
1 mole = 6.02 × 1023 molecules
Avogadro’s number
as a conversion factor
Given the definition,
1 mole = 6.02 × 1023 molecules
We can see that Avogadro’s number converts from
molecules to moles.
6.02 × 1023 molecules/mole
Atomic and molecular weight
The atomic weight is the numerical value tabulated for
the mass of each atom in the periodic table in atomic
units. The use of the word “weight” is not precise here
since weight in physics represents a force (w = mg).
However, the name atomic weight is so ingrained that
we will not attempt to change it. We use the periodic
table to calculate the molar mass as follows: for H2SO4
we find the atomic weights, H = 1, S = 32 and O = 16.
Molar mass = 2(1 g/mol) + 32 g/mol + 4(16 g/mol)
= 98 g/mol
Molecular and empirical formulae
We can use the mass (weight) of a molecule as one
constraint on the formula of a compound (molecule).
Some molecules have a simpler empirical formula,
which does not correspond to the molecular formula.
For example, all carbohydrates have the empirical
formula (CH2O). However, the various sugars and other
Carbohydrates can differing total numbers of atoms so
That the molecular formula is (CH2O)n, where n = 6 for
glucose and many simple sugars. In general,
Absolute Temperature
We will use the absolute temperature scale (Kelvin) for
all chemical calculations. Why? One important reason
is that the absolute temperature is proportional to the
kinetic energy of a substance.
K stands for kinetic energy, R is the gas constant and n
is the number of moles.
N is the number of molecules.
Temperature scales
The absolute scale in Kelvins is offset from the Celsius
scale by 273.16 degrees, meaning that 0 oC ~ 273 K to
three significant figures. This value is accurate enough
for our purposes. Therefore, we can use the formula
The Celsius scale is used by every country in the world
as the temperature scale (except the United States).
We use the Fahrenheit scale.
Conversion from Fahrenheit to Celsius
The zero of the Celsius scale occurs at 32 oF and the
The boiling point of water (100 oC) occurs as 212 oF.
This means that one degree Celsius is exacly 9/5 times
one degree Fahrenheit.
For common values it is useful to have recall that
50 oF = 10 oC, 68 oF = 20 oC and 86 oF = 30 oC.
Body temperature is exactly 98.6 oF = 37.00 oC.
Ideal gas law
The number of moles, n, appears in the ideal gas law:
This is an equation of state, which means it relates the
variables of state, pressure, volume, and temperature.
R is a constant known as the universal gas constant.
R = 8.31 J/mol-K or R = 0.08206 L-atm/mol-K
Note that the units of the ideal gas law are units of
energy.
The ideal gas law as
an energy equation
At first PV may not look like an energy. However, when
a fuel is combusted (for example in the cylinder of a car
engine), it builds up a pressure, which causes an
expansion, an increase in volume. Pressure-volume
work is the how engines propel cars. Temperature also
represents an energy. The temperature of a gas is
proportional to the kinetic energy of the gas
molecules. nRT is actually an energy term as can be
seen from the units of R.
Solving for the number of moles
We can use the ideal gas law to obtain the number of
moles of a gas, provided we are given P, V and T. The
formula is:
For example, how moles are there in 22.4 liters of gas
at 273 K and at sea level.
Solution: At sea level, P = 1 atm so
Molar gas volume
The solution to the problem is n = 1.0.
We conclude that one mole has a volume of 22.4 liters
at 273 K. We call this the molar gas volume.
Note that the molar gas volume changes with
temperature.
Problem: Calculate the molar gas volume at 373 K.
Density
The density of a material is defined as the mass per
unit volume. Density is often given in grams per cubic
centimeter (cc), which is the same thing as grams per
milliliter (mL).
Density is a material property, r.
r(H2O) = 1 gm/mL.
r(C2H5OH) = 0.789 gm/mL.
r(Pb) = 11.34 gm/mL.
r(Au) = 19.30 gm/mL.
Archimedes’ dilemma
?
Archimedes’ dilemma
It is easy to find the mass of the crown,
but how does one measure its density?
Archimedes’ solution
If he seems unnecessarily excited, remember that the
penalty for failing to solve the problem was death.
Archimedes’ principle
The crown displaces a volume of water equal to its own volume.
It is easy to figure out the volume of the displaced water. Next
Archimedes realized that this explains how ships can float.
The buoyancy of an object is equal the weight of the water it
displaces.
Concentration: molarity
The molarity is defined as the number of moles per liter.
We most often consider the molarity of the solute. For
example, if we dissolve 40 grams of NaCl in one liter of
H2O we have:
In thinking about conversions it can also be useful to
understand the molarity of the solvent. What is the
molarity of H2O?
Concentration: molality
The molality is defined as the number of moles per kg of
solvent.
You might ask why there are two different units of
concentration (molarity and molality). The answer is
that these have different applications. The molality is
used for the colligative properities while the molarity is
used for many solutions in the laboratory where the
volume is the common means of measuring and
dispensing solutions.
Concentration: mole fraction
The mole fraction is the number of moles of a given
substance divided by the total number of moles present
in the system.
The convert from one of these to the other remember
that the ratio of each of these quantities are equal:
Chemical equations
A chemical equation tells us how one set of species
converts into another. As bonds are broken and
reformed the numbers of atoms in a compound will
change. The number of molecules that combine must
account for this so that there is conservation of mass.
H2 + O2
H2O
This equation is not balanced since there is an extra O
atom on the left hand side.
Balancing chemical equations
Think of the equation as an algebraic equation
a H2 + b O2 = x H2O
In order for this equation to be satisfied, the following
must be true:
H: 2a = 2x
O: 2b = x
Actually, there are an infinite number of solutions. We
just want the most practical one.
Let x = 2, then b = 1 and a = 2.
Balancing chemical equations
The balanced equation is:
2 H2 + O2 = 2 H2O
Of course, we could have chosen x = 1. Then, the
equation would be:
H2 + ½ O2 = H2O
Often we choose the value that gives all integers for the
coefficients, but this is not required.
General Method: Example
Let’s consider the combustion of ethane:
C2H6 + O2 = H2O + CO2
Step 1. apply coefficients
a C2H6 + b O2 = x H2O + y CO2
Step 2. Write down an equation for each element
C: 2a = y
H: 6a = 2x
O: 2b = x + 2y
Step 3. Choose a value that will give a an integer
answer (if possible on the first try).
General Method: Example
We are balancing
a C2H6 + b O2 = x H2O + y CO2
We are on step 3. Make a first trial guess…
C: 2a = y
H: 6a = 2x
x = 3 gives a = 1, and y = 2
O: 2b = x + 2y
General Method: Example
We are balancing
a C2H6 + b O2 = x H2O + y CO2
We are on step 3. Make a first trial guess…
C: 2a = y
H: 6a = 2x
x = 3 gives a = 1, and y = 2
O: 2b = x + 2y
therefore 2b = 3 + 2(2) = 7
C2H6 + 7/2 O2 = 3 H2O + 2 CO2
Or
2 C2H6 + 7 O2 = 6 H2O + 4 CO2
General Method: Example
Let’s consider the combustion of pyridine:
C5H5N + O2 = H2O + CO2 + NO
Step 1. apply coefficients
a C5H5N + b O2 = x H2O + y CO2 + z NO
Step 2. Write down an equation for each element
C: 5a = y
H: 5a = 2x
O: 2b = x + 2y + z
N: a = z
Step 3. Choose a value that will give a an integer
answer (if possible on the first try).
General Method: Example
We are balancing
a C6H5N + b O2 = x H2O + y CO2 + z NO
We are on step 3. Make a first trial guess for the
coefficient.
C: 5a = y
H: 5a = 2x
O: 2b = x + 2y + z
N: a = z
General Method: Example
We are balancing
a C6H5N + b O2 = x H2O + y CO2 + z NO
We are on step 3. Make a first trial guess…
C: 5a = y
H: 5a = 2x
x = 5 gives a = 2, and z = 2, y = 10
O: 2b = x + 2y + z
N: a = z
General Method: Example
We are balancing
a C6H5N + b O2 = x H2O + y CO2 + z NO
We are on step 3. Make a first try.
C: 5a = y
H: 5a = 2x
x = 5 gives a = 2, and z = 2, y = 10
O: 2b = x + 2y + z
2b = 5 + 20 + 2 = 29, b = 27/2
N: a = z
We have acceptable values if we can live with noninteger coefficients.
2 C6H5N + 27/2 O2 = 5 H2O + 10 CO2 + 2 NO
General Method: Example
We can multiply through by 2 to eliminate the noninteger coefficient.
4 C6H5N + 27 O2 = 10 H2O + 20 CO2 + 4 NO
While there is some arbitrariness in the choice, this is
inherent since the solutions are not unique. Only the
relative values of the coefficients are determined by
the stoichiometry, but not their absolute values.
DMSO Example
The combustion of dimethyl sulfoxide is given by
C2H6SO + O2 = H2O + CO2 + SO3
DMSO Example
The combustion of dimethyl sulfoxide is given by
C2H6SO + O2 = H2O + CO2 + SO3
Step 1. Apply coefficients
a C2H6SO + b O2 = x H2O + y CO2 + z SO3
Step 2. Write an equation for each element
C: 2a = y
H: 6a = 2x
S: a = z
O: a + 2b = x + 2y + 3z
DMSO Example
We are balancing
a C2H6SO + b O2 = x H2O + y CO2 + z SO3
Step 3. Make a guess to try and get integer values
C: 2a = y
H: 6a = 2x
S: a = z
O: a + 2b = x + 2y + 3z
DMSO Example
We are balancing
a C2H6SO + b O2 = x H2O + y CO2 + z SO3
Step 3. Make a guess to try and get integer values
C: 2a = y
H: 6a = 2x
x = 3, a = 1
S: a = z
O: a + 2b = x + 2y + 3z
DMSO Example
We are balancing
a C2H6SO + b O2 = x H2O + y CO2 + z SO3
Step 3. Make a guess to try and get integer values
C: 2a = y
H: 6a = 2x
x = 3, a = 1, then y = 2, z = 1
S: a = z
O: a + 2b = x + 2y + 3z
DMSO Example
We are balancing
a C2H6SO + b O2 = x H2O + y CO2 + z SO3
Step 3. Make a guess to try and get integer values
C: 2a = y
H: 6a = 2x
x = 3, a = 1, then y = 2, z = 1
S: a = z
O: a + 2b = x + 2y + 3z
2b = x + 2y + 3z – a = 3 + 2(2) + 3(1) – 1 = 9, so b = 9/2
DMSO Example
We are balancing
a C2H6SO + b O2 = x H2O + y CO2 + z SO3
Step 3. Make a guess to try and get integer values
C: 2a = y
H: 6a = 2x
x = 6, a = 2, then y = 4, z = 2
S: a = z
O: a + 2b = x + 2y + 3z
2b = x + 2y + 3z – a = 6 + 2(4) + 3(2) – 2 = 18, so b = 9
and
2 C2H6SO + 9 O2 = 6 H2O + 4 CO2 + 2 SO3
Stoichiometric ratios
Once we have balanced a chemical equation, we have
determined the fixed ratios in which the various
compounds must react in order for the reaction to
proceed. These known as stoichiometric ratios. For
example, if we return to the reaction,
2 C2H6SO + 9 O2 = 6 H2O + 4 CO2 + 2 SO3
We see that 9 O2 must react for every 2 C2H6SO. The
ratio is:
Stoichiometric ratios
Similarly, we can state that 6 moles of H2O are
produced for every two moles of C2H6SO consumed.
Thus, the stoichiometric ratio is 3 for H2O relative to
C2H6SO.
2 C2H6SO + 9 O2 = 6 H2O + 4 CO2 + 2 SO3