Conversions Stoichiometry

Balancing Equations

Chemistry 201

NC State University

Lecture 1

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

N

H

H

H

H H

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.

N

H

H

H

H H

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?

If he seems unnecessarily excited, remember that the

penalty for failing to solve the problem was death.

Archimedes’ solution

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 non-

integer coefficients.

2 C6H5N + 27/2 O2 = 5 H2O + 10 CO2 + 2 NO

General Method: Example We can multiply through by 2 to eliminate the non-

integer 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