## Atomic Mass, Average Atomic Mass, & Relative Isotopic Mass

*Here are some similar Terms for Different Quantities*

Mass of one atom ⇒ Expressed in Kg, or gram, or a.m.u, or u**Atomic Mass**(m_{a}or m):Relative mass of one isotope of an element ⇒ Dimensionless**Relative Isotopic Mass**:A standard average value of relative isotopic masses of all the isotops of an Element on earth. (that you see on periodic table) ⇒ Dimensionless(A_{Standard}Relative Atomic Mass_{r},_{standard}(E)) or_{Satndard }Average Atomic Mass:Same as standard average atomic mass but we use this term for a specific given element sample from a location ⇒ Dimensionless**Relative Atomic Mass**(A_{r}) or Average Atomic Mass:

## Atomic Mass

The atomic mass is the **mass of a single atom** of an element. Mass of an atom (atomic mass) can be expressed in **Kilogram** or **Gram** or **a.m.u** or **u**.

The SI unit of mass is the kilogram (kg), but the atomic mass is most often expressed in the non-SI unit **a.m.u** (atomic mass unit), also known as **u** (unified mass).

Where **1 a.m.u** (or **u**) is defined as ** ^{1}⁄_{12} absolute mass of a single Carbon^{12} atom** at rest.

*This is a standard scale unit to calculate the mass of one atom of any element in a.m.u*.

**1 a.m.u = Absolute mass of one Carbon atom ÷ 12**

Absolute mass of one Carbon^{12} atom (in kg) **=** 1.99264687992 × 10^{−26} kg **=** 1.99264687992 × 10^{−23} g. (*We will see how it was calculated*.)

So, **1 a.m.u** **=** (1.99264687992 × 10^{−23} g) **÷** (12)**1 a.m.u** = **1.66053907 x 10 ^{-24} g** =

*(1 Dalton)*

⇒ 1 a.m.u. = 1 u = 1 Da (dalton) =

**=**

^{1}⁄_{12}x mass of a single Carbon-12 atom**1.66053907 x 10**

^{-24}g* 1.66053907 x 10^{-24} gram mass known as Atomic Mass Constant (m_{u})* and considered as

**by convention**

*1 a.m.u.*- When the Mass of an atom is expressed in
**grams**or**kilograms**, it is called**absolute mass**of the atom. - Mass of Carbon
^{12}atom is 12 a.m.u, which is also the mass number (no. of protons + no. of neutrons) of Carbon^{12}. - That means
.**1 a.m.u**is the**average mass of one proton or one neutron**of an element

### Gram to A.M.U. Conversion Factor

** 1.66053907 x 10^{–24} gram mass** is considered as

**1 a.m.u.**and it is a conventional scale to measure the mass of an atom of various elements.

We know,

1 a.m.u = 1.66053907 x 10^{-24} g (Atomic Mass Constant)

So,**1g** = (1 a.m.u) ** ÷** (1.66053907 x 10

^{-24})

=

**6.02214075 x 10**

^{23}a.m.u.Now, If we know the mass of an object in Kilogram or Gram then we can easily calculate its mass in a.m.u.

**6.02214075 x 10 ^{23}** known as Avogadro’s Number (N

_{A}) or

**1 mol**

So,

1 a.m.u. = (1g) ÷

**6.02214075 x 10**

^{23}1 a.m.u. = (1g) ÷ N

_{A}

### Calculate Atomic Masses in A.M.U.

#### Mass of Proton in a.m.u.

**The mass of a Proton** = 1.67262779 x 10^{-24} g**∵** 1g = 6.02214075 x 10^{23} a.m.u.**∴** 1.67262779 x 10^{-24} g = (1.67262779 x 10^{-24}) x (6.02214075 x 10^{23} a.m.u.) = **1.00728 a.m.u.**

#### Mass of Neutron in a.m.u.

**The mass of a Neutron** = 1.67493594 x 10^{-24} g**∵** 1g = 6.02214075 x 10^{23} a.m.u.**∴** 1.67493594 x 10^{-24} g = (1.67493594 x 10^{-24}) x (6.02214075 x 10^{23} a.m.u.) = **1.00867 a.m.u.**

#### Mass of Electron in a.m.u.

**The mass of an Electron** = 9.1 x 10^-28 g**∵** 1g = 6.02214075 x 10^{23} a.m.u.**∴** 9.1 x 10^{-28} g = (9.1 x 10^{-28}) x (6.02214075 x 10^{23} a.m.u.) = **0.00055 a.m.u.** = **5.5 x 10 ^{-4}** a.m.u

### Calculate Atomic Mass of an Element

- The
**protons**and**neutrons**of the nucleus account for nearly all of the total mass of an atom. - The
**electrons**make negligible contributions to mass of the atom. - Thus, the numeric value of the atomic mass
*when expressed in a.m.u*has nearly the same value as the mass number (Protons + Neutrons).

#### Atomic Mass of Oxygen (O^{16})

Oxygen (O^{16}) has 8 protons, 8 neutrons, and 8 electrons

and above we have calculated the mass one proton, one neutron, and one electron

Therefore,

= (8 x 1.00728 a.m.u) + (8 x 1.00867 a.m.u) + (8 x 0.00055 a.m.u)

= 16.1319886 a.m.u, *should be the atomic mass of oxygen (O ^{16}) but*

- The mass of nucleus of an atom is
than the sum of masses of all nucleons making it =*less*.*Mass Defect*- That is, some
**energy releases**when the nucleons (protons and neutrons) fuse to form a nucleus, thus the**mass decreases**(E = MC^{2}).

- That is, some
- The amount of energy released during the fusion of protons and neutrons will be the same as the amount of energy required to separate the nucleons from the nucleus =
**Binding Energy**.- Binding Energy of elements determined experimentally.

- So, Now we have to calculate mass defect of Oxygen
^{16}Nucleus

*You might thinkIt should be, More binding energy == More stable nucleus butIron (*

^{56}Fe

_{26}) has a binding energy = 492 M.e.v (Mega electron volt) andUranium (

^{238}U

_{92}) has a binding energy = 1800 M.e.v

*but Iron ( ^{56}Fe_{26}) is the most stable nucleus, that is because *

for iron, the binding energy per nucleon = 8.8 M.e.v while

for Uranium, the binding energy per nucleon = 7.6 M.e.v

*More binding energy per nucleon == More stable nucleus*

Now, If we first calculate the ** energy of 1 a.m.u mass**, and then if we are given the binding energy of a nucleus, it will be easy to calculate the mass defect for that atomic nucleus.

We know,**1 a.m.u**. = 1.66053907 x 10^{-27} Kg**1** = 1.66053907 x 10^{-27} Kg/a.m.u.

E = MC^{2}

= (1.66053907 x 10^{-27} Kg/a.m.u.) x (3 x 10^{8} m/s)^{2}

= (1.66053907 x 10^{-27}) x (3 x 10^{8})^{2} **Joule/a.m.u.***Converting Joule to electron volt* (1 J = 1.6 x 10^{19} e.v.)

= (1.66053907 x 10^{-27}) x (3 x 10^{8})^{2} x (1.6 x 10^{19}) **e.v./a.m.u.***Converting electron volt to Mega electron volt (e.v to M.e.v)*

= (1.66053907 x 10^{-27}) x (3 x 10^{8})^{2} x (1.6 x 10^{19}) x 10^{-6}**≈ 931.5 M.e.v/a.m.u.****1 a.m.u ≈ 931.5 M.e.v**

##### Actual Atomic Mass of Oxygen (O^{16})

Above we have calculated the atomic mass of oxygen = 16.1319886 a.m.u *but* during the fusion of protons and neutrons, the nucleus of the oxygen atom releases some energy hence will lose some mass.

*Mass defect of Oxygen (O ^{16})*

For Oxygen, the binding energy per nucleon = 8 M.e.v. (Determined experimentally)

Oxygen has 16 nucleons (Protons + Neutrons)

**Mass defect**= (8 M.e.v

**x**16) ÷ 931.5 M.e.v/a.m.u. =

**0.1374 a.m.u**

So the ** atomic mass of one oxygen atom** (O

^{16}) in a.m.u.

= 16.1319886 a.m.u

**–**Mass defect

= 16.1319886 a.m.u

**–**0.1374 a.m.u =

**15.9946 a.m.u**

And, the * Absolute/Actual Atomic Mass or Mass of one oxygen atom* (O

^{16})

= 15.9946 a.m.u

**x**(Atomic Mass Constant/a.m.u.)

= 15.9946 a.m.u

**x**(1.66053907 x 10

^{-24}gram/a.m.u)

=

**2.655965 x 10**

^{-23}gram= 2.655965 x 10

^{-26}Kg

#### Atomic Mass of Carbon (C^{12})

**Mass of carbon atom in a.m.u.**

Carbon has 6 protons, 6 neutrons, and 6 electrons

= **[**(6 × 1.00728 amu) + (6 × 1.00867 amu) + (6 × 0.00055 amu)**]** – Mass defect

= 12.09900 amu – Mass defect

*Mass defect of Carbon ^{12} atom*

For Carbon the binding energy per nucleon = 7.68 M.e.v. (Determined experimentally)

Carbon

^{12}has 12 nucleons (Protons + Neutrons)

Mass defect = (7.68 M.e.v

**x**12) ÷ 931.5 M.e.v/a.m.u = 0.0989371981 a.m.u ≈ 0.099 a.m.u

= 12.09900 amu – 0.099 amu

= **12 a.m.u**.

And, the *Absolute/Actual Atomic Mass or Mass of one Carbon ^{12} atom*

12 a.m.u × (Atomic Mass Constant/a.m.u.)

12 a.m.u. x (1.66053907 x 10

^{-24}gram/a.m.u.)

**=**

**1.99264688 × 10**

^{−23}g## Relative Isotopic Mass

Relative Isotopic Mass simply means ** HOW MANY TIMES** the absolute mass of one atom (in grams or Kilograms) is heavier than the

**absolute mass of one carbon**

^{1}⁄_{12}^{12}atom.

### Relative Isotopic Mass of Carbon^{12}

To determine the ** Relative Isotopic Mass** of Carbon

^{12}, we divide the mass of one C

^{12}atom by the

**mass of C**

^{1}⁄_{12}^{12}(Atomic Mass Constant).

Relative Isotopic Mass of Carbon^{12}

= (Actual Mass of C^{12} isotope) ÷ (Atomic mass constant)

= (1.99264688 × 10^{−23} g) ** ÷** (1.66053907 x 10

^{-24}g)

**=**12

So, the Atomic Mass of Carbon^{12} is 12 a.m.u but the Relative Isotopic Mass of one Carbon^{12} atom is simply 12 with no dimensions.

**Relative Isotopic Mass**representsthe absolute mass of one atom of an element is heavier then*HOW MANY TIMES*mass of one Carbon^{1}⁄_{12}^{12}atom.- Relative Isotopic Mass obtained by dividing the Atomic Mass (m
_{a})of an isotope by the Atomic Mass Constant (m_{u}) yielding a dimensionless value. - The word “relative” in the term “relative isotopic mass” refers to the scaling relative to
mass of one Carbon^{1}⁄_{12}^{12}atom.

### Relative Isotopic Mass of Oxygen^{16}

Relative Isotopic Mass of Oxygen^{16}

= (Mass of O^{16} isotope) ÷ (Atomic mass constant)

= (2.655965 x 10^{-23} gram) ** ÷** (1.66053907 x 10

^{-24}gram)

= 15.9946

*The Atomic Mass and the Relative Isotopic Mass refer to a certain specific isotope of an element. Because substances are usually not isotopically pure, it is convenient to use the Standard Elemental Atomic Mass*

_{Standard} Relative Atomic Mass (A_{r}, _{standard}(E))

The *Standard Relative Atomic Mass* is the elemental atomic mass which is the *average (mean) relative isotopic mass* of an element, weighted by the *abundance of its isotopes* on earth.

** On the Periodic Table**, the atomic mass of carbon is reported as

**12.011**. This is the

**(Standard Average Relative Atomic Mass or Standard Atomic Weight or Standard**

*Standard Relative Atomic Mass***Average Atomic Mass/Weight**) of carbon.

No single carbon atom has a relative mass of 12.011, but a handful of Carbon atoms (including its isotopes: C^{12}, C^{13}, & C^{14}) have an average mass of 12.011.

### Calculate Relative Atomic Mass

Ar,_{ standard} = [Relative isotopic mass x (% abundance / 100)] + [Relative isotopic mass x (% abundance / 100)] + ……

#### Standard Relative Atomic Mass of Chlorin

isotope ** ^{35}Cl** constitutes 75.77% of the copper on Earth and the rest 24.23% being

**, so**

^{37}ClA

_{r},

_{standard}(Cl) = [

**relative isotopic mass x (% abundance / 100)] + [**

^{35}Cl**relative isotopic mass x (% abundance / 100)]**

^{37}ClA

_{r},

_{standard}(Cl) = [34.96885269 x (75.77 / 100)] + [36.96590258 x (24.23 / 100)]

A

_{r},

_{standard}(Cl) = [34.96885269 x 0.7577] + [36.96590258 x 0.2423]

A

_{r},

_{standard}(Cl) = 35.4527379 ≈ 35.5

Because relative isotopic masses are dimensionless quantities, this weighted mean is also dimensionless. It can be converted into a measure of mass by multiplying it with the atomic mass constant.

## Relative Atomic Mass (A_{r})

Relative Atomic Mass or Atomic Weight or Average Atomic Weight or Average Atomic Mass: It is the **same as Standard Relative Atomic Mass**, this term is used for a single specific given sample from a particular location.

The relative atomic mass of a given element sample is the weighted arithmetic mean of the masses of the individual atoms (including their isotopes) that are present in the sample.

This quantity can vary substantially between samples because the sample’s origin may have produced unique combinations of isotopic abundances.

Standard Relative Atomic Mass is a standardized value obtained with the various element samples being taken from Earth and is a more common, and more specific quantity.

**References:**

https://goldbook.iupac.org/terms/view/A00496

https://goldbook.iupac.org/terms/view/A00497

https://goldbook.iupac.org/terms/view/U06554

https://goldbook.iupac.org/terms/view/R05258

https://goldbook.iupac.org/terms/view/BT07001

https://en.wikipedia.org/wiki/Atomic_mass

https://en.wikipedia.org/wiki/Relative_atomic_mass

https://courses.lumenlearning.com/introchem/chapter/nuclear-binding-energy-and-mass-defect/

https://www.westfield.ma.edu/PersonalPages/cmasi/gen_chem1/Atomic%20and%20molar%20mass/atomic_and_molar_mass.htm