# Atomic Weights in Chemistry

These are my notes on atomic weights in chemistry.

Intro
Each atom has a definite and characteristic weight. This weight provides a way
to state the amount of a substance required for a chemical reaction.

Example 1
The notation $${^{35}_{17}Cl}$$ indicates a neutral atom of chlorine.
1. What is its atomic number? 17
2. What is its mass number? 35
3. How many protons does it have? 17
4. How many electrons? 17
5. How many neutrons? 18

Example 2
Different atoms of the same element can have different numbers of neutrons and,
therefore, different mass numbers. Here is another neutral chlorine atom.
$${^{37}_{17}Cl}$$
1. What is its mass number? 37
2. How many protons does it have? 17
3. How many neutrons? 20

Example 3
Since neutrons and protons combine to make up the mass number, two atoms of the
same element can have different mass numbers. Chlorine can exist  as
$${^{37}_{17}Cl}$$ and as $${^{35}_{17}Cl}$$. The only difference between
these atoms of chlorine is that $${^{37}_{17}C}$$ has two more neutrons than
$${^{35}_{15}Cl}$$. Antimony can exist as $${^{121}_{51}Sb}$$ and
$${^{123}_{51}Sb}$$. The only difference between these atoms of antimony is
that $${^{123}_{51}Sb}$$ contains two more neutrons than
$${^{121}_{51}Sb}$$.

Example 4
$${^{37}_{17}Cl}$$ has a greater mass than $${^{35}_{17}Cl}$$ because of
the two extra neutrons. Which of the following atoms of antimony has the greater
mass, $${^{121}_{51}Sb}$$ or $${^{123}_{51}Sb}$$? $${^{123}_{51}Sb}$$
does because it has two more neutrons.

Example 5
Atoms of the same element having different masses are called isotopes. Elements
as found in nature are usually mixtures of two or more isotopes. The atom
$${^{123}_{51}Sb}$$ is one isotope of the element antimony.
$${^{121}_{51}Sb}$$ is another isotope of antimony. The main difference
between two isotopes of the same element is the number of neutrons.

Example 6
Isotopes exist for every known element. The isotopes of the element neon were
first discovered by two English scientists. They continued in their work to
discover other isotopes through inventing the mass spectrograph or spectrometer.
In the mass spectrograph, atoms of different masses of the same element are charged and accelerated by an electron beam toward a target, such as a
photographic plate. A strong magnetic field bends the paths of the charged
atoms. Atoms of greater mass have their paths bent to a lesser degree than atoms
of lighter mass.
In the diagram of the mass spectrograph pictured, where do the
lighter atoms strike? The higher point A because the path of the lighter atoms
is bent to a greater degree.

Example 7
An analogy to the mass spectrograph would be to roll a bowling ball and a
basketball at the same speed at a target while a stiff crosswind is blowing. The
bowling ball is considerably heavier than a basketball. Look at the diagram
below. Which ball would strike the target at point B?
The basketball because it is lighter and its path is more readily changed by the
crosswind.

Example 8
In the rolling balls analogy to the spectrograph, the bowling ball and
basketball are analogous to isotopes of different mass. The strong crosswind is
analogous to the? magnetic field

Example 9
Thomson and Aston invented an instrument that detects the presence and
characteristics of isotopes. What is this instrument called?
mass spectrograph or spectrometer

Example 10
The atomic weights of the elements are listed in the periodic table. The atomic
weight of sodium, for example, is listed as 22.990. The atomic weight listed for
sodium is actually the atomic weight of a mixture of isotopes,
$${^{22}_{11}Na}$$ and $${^{23}_{11}Na}$$. The proportion of these
isotopes is generally constant wherever sodium is found.
The atomic weight of an element is the average weight of a mixture of two or
more?
isotopes

Example 11
The periodic table lists the atomic weight of Al as?
26.982

Example 12
Atomic weights are based on the carbon 12 scale. That is, carbon 12 or
$${^{12}_{6}C}$$, the most abundant isotope of carbon, is used as the
standard unit  in measuring atomic weights. By international standard, one atom
of the $${^{12}_{6}C}$$ isotope has an atomic weight of exactly 12 atomic
mass units, abbreviated amu. The atomic weight of the $${^{12}_{6}C}$$
isotope is exactly?
12 amu

Example 13
All atomic weights can be expressed in atomic mass units. By international
agreement, 12 amu would equal the mass of a single $${^{12}_{6}C}$$ atom. One
amu is equal to what fraction of a single $${^{12}_{6}C}$$ atom?
1/12 the mass of a single $${^{12}_{6}C}$$ atom

Example 14
Although the $${^{12}_{6}C}$$ isotope weights exactly 12.000 amu by
definition, the atomic weight of C as listed on the periodic table is 12.011
amu. The atomic weight of carbon as listed on the periodic table is greater than
that of the $${^{12}_{6}C}$$ isotope. Why?
The atomic weight of an element is the average weight of a mixture of two or
more isotopes.

Example 15
While the element carbon as found in nature is made up largely of the
$${^{12}_{6}C}$$ isotope (98.9%), a small quantity of $${^{13}_{6}C}$$
isotope (1.1%) is mixed uniformly as part of the element. The $${^{12}_{6}C}$$
isotope has an atomic weight of 12.000 amu. The $${^{13}_{6}C}$$ isotope has
an atomic weight of 13.003 amu. The resultant average atomic weight would be
slightly (heavier, lighter) than 12.000 amu?
Heavier.

Example 16
The atomic weights on the periodic table are the average atomic weights of the
isotope mixtures in the element. We can determine the average atomic weight of
an element if we know the approximate mass of each isotope and the proportion of
each isotope within the element.

Here are the steps for calculating the average atomic weight of the element
carbon.
1. Multiply the mass of the $${^{12}_{6}C}$$ isotope by its decimal
proportion (12.000*0.989).
2. Multiple the mass of the $${^{13}_{6}C}$$ isotope by its decimal
proportion (13.003*0.011).
Add the results to find the average atomic weight of the element C.
You should get 12.011 or 12.01 rounded.

Example 17
Calculate the atomic weight of fluorine.
For the $${^{19}_{9}F}$$ fluorine isotope, with a mass of 19.000 and
proportion of 0.997, we have: 18.943
For the $${^{18}_{9}F}$$ isotope, with a mass of 18.000 and proportion of
0.003, we get: 0.054
added together, we get: 18.997 or 19.00

Example 18
Sodium has two isotopes, $${^{23}_{11}Na}$$ and $${^{22}_{11}Na}$$. The
isotope $${^{23}_{11}Na}$$ has an atomic mass of approximately 23.000 amu,
and its proportion in the element is 99.2%. The isotope $${^{22}_{11}Na}$$
has a mass of approximately 22.000 amu and a proportion within the element of
0.8%. Determine the atomic weight of sodium.
$$23.000 * .992 =22.816$$
$$22.000 * .008 =.176$$
Add the totals together to get:
$$22.99\text{ amu }$$

Example 19
The element cobalt has an isotope $${^{60}_{27}Co}$$ that has an approximate
mass of 60.000 and constitutes 48.0% of the element. Another isotope
$${^{58}_{27}Co}$$, has an approximate mass of 58.000 and constitutes 52.0%
of the element. Calculate the atomic weight of Co using the given data.
$$60.000 * .48 =28.8$$
$$58.000 * .52 =30.16$$
Add the totals together to get:
$$58.96 \text{ amu }$$

Example 20
The percentage proportions of all the isotopes within an element must add up to
a total of 100%. The decimal proportions of all the isotopes within an element

Example 21
We have been calculating atomic weight given the mass and proportion of
isotopes. We can also determine the proportion of each individual isotope within
an element if we know the atomic weight of the element. The element Cr, which
has an overall atomic weight of 51.996 amu, has two isotopes:
$${^{52}_{24}Cr}$$ with atomic mass of 52.000 amu and $${^{51}_{24}Cr}$$
with an atomic mass of 51.000 amu.
We are given the sum as $$51.996$$
We do not know the proportions of either yet.

Example 22
To solve this equation with two unknowns, you must form a second equation
showing the relationship between A and B. You have already learned the answer to
the following question in example 20. Add the decimal proportions.
A+B=1

Example 23
Modify the equation in example 22 so that just B remains on the left side of the
equation.
B=1-A

Example 24
Here is the equation you need for calculating the sum of the isotopes
mass * proportions: (52.000A)+(51.000B)=51.996. Substitute the expression (1-A)
for B in the equation:
$$52.000A+51.000(1-A)=51.996$$

Example 25
Solve the equation derived in example 24 to determine the value of A (the
proportion of $${^{52}_{24}Cr}$$ in an average mixture of chromium) to the
nearest thousandth.
$$52.000A+51.000(1-A)=51.996$$
$$52.000A+51.000-51.000A=51.996$$
$$1.000A+51.000=51.996$$
$$1.000=.996$$
$$A=.996$$

Example 26
Since B=1-A, what is the value of B?
$$B=1-A$$
$$B=1-.996$$
$$b=.004$$

Example 27
Here is a similar problem. In the next few examples, you will determine the
proportion of the $${^{35}_{17}Cl}$$ isotope in an average mixture of
chlorine, which is made up of both $${^{35}_{17}Cl}$$ and
$${^{37}_{17}Cl}$$.
$$34.97 * A =34.97*A$$
$$36.97 * B =36.97*B$$
The values of A and B added together equal?
1

Example 28
Since A and B=1, then B=?
$$1-A$$

Example 29
The proportion $$1-A$$ has been substituted for B.
$$34.97 * A =34.97A$$
$$36.97 * B =36.97B$$
Added together is $$35.45$$
Using the table above, determine the proportion of $${^{35}_{17}Cl}$$ within
the element chlorine (find the value of A).
$$34.97A+36.97(1-A)=35.45$$
$$34.97A+36.97-36.97A=35.45$$
$$36.97-2.00A=35.45$$
$$36.97=35.45+2.00A$$
$$1.52=2.00A$$
$$A=.76$$

Example 30
Determine the proportion of $${^{37}_{17}Cl}$$ in Cl.
The proportion has been given the value of B.
$$B=1-A$$
$$B=1-.76$$
$$B=.24$$

Example 31
Neon has two isotopes: $${^{22}_{10}Ne}$$ and $${^{20}_{10}Ne}$$. The
approximate mass of $${^{22}_{10}Ne}$$ is 22.000 and the mass of
$${^{20}_{10}Ne}$$ is approximately 20.000 amu. The atomic weight of neon is
20.179 amu. Determine the proportion of the $${^{22}_{10}Ne}$$ isotope within
the element.
$$22.000A+20.000B=20.179$$
$$B=1-A$$
$$22.000A+20.000(1-A)=20.179$$
$$22.000A+20.000-20.000A=20.179$$
$$2.000A+20.000=20.179$$
$$2.000A=.179$$
$$A=.090$$

Example 32
What is the proportion of the other isotope, $${^{20}_{10}Ne}$$, within the
element?
$$B=1-A$$
$$B=1-.090$$
$$B=.910$$

Example 33
So far, we have considered all atomic weights in terms of atomic mass units.
1. An atomic weight expressed in amu represents the average weight of how many
atoms of an element?
1. An atomic weight expressed in amu represents the average weight of one single
atom of an element.
2. Carbon has an atomic weight of 12.011 amu, which represents the average
weight of how many atoms of carbon?
1

Example 34
Since it is impossible to measure the weight of one atom with a laboratory
balance, another unit for expressing atomic weight must be used. Atomic weight
can be expressed in grams as well as amu. An atomic weight expressed in grams
contains $$6.022*10^{23}$$ atoms. This number, called Avogadro's number, will be
encountered often.
1. If the atomic weight of carbon is expressed as 12.011 amu, it represents the
average weight of how many atoms?
one
2. If the atomic weight of carbon is expressed as 12.011 grams, it represents
the average weight of how many atoms?
$$6.022*10^{23}$$

Example 35
Using information in example 34. answer the following question. One gram is how
many times heavier than 1 amu?
$$6.022*10^{23}$$

Example 36
One ton is equivalent to 2000 pounds. One pound represents 1/2000 of a ton.
$$6.022*10^{23}$$ amu is equivalent to 1 gram. One amu represents what fraction of
a gram?
$$\frac{1}{6.022*10^{23}}$$

Example 37
Avogadro's number $$6.022*10^{23}$$ is written in exponential notation. It
actually represents a very large number. Exponential notation will be used
often. The exponent indicates the number of places that the decimal must be
moved. The number 645,000 has the decimal moved five places to the left. The
result is $$6.45*10^{-5}$$.

To divide two numbers with exponential notation, divide the decimal portions of
the numbers and subtract the exponent in the denominator from the exponent in
the numerator.

Example 38
Unit Factor Analysis
If the problems in earlier example had been more difficult, we would have used
the unit factor method (called factor label analysis and dimensional analysis)
for solving problems.

The unit factor method involves multiplying the given value by one or more
conversion factors.