# 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

must add up to 1.**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.