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07th May 2019 @ 12 min read

1. Avogadro's Number Full
2. Avogadro's Number Definition

The Avogadro constant or (the Avogadro number earlier) is the number of elementary units in one mole of any substance. The Avogadro constant is denoted as N_A. It has the dimension of the reciprocal amount of substance (mol⁻¹). The approximate value of N_A is 6.022 × 10²³ mol⁻¹. This means one mole of any substance contains 6.022 × 10²³ elementary particles. The Avogadro constant is named after Italian scientist Amedeo Avogadro.

Experimentally, the weight of Avogadro's number (N A = 6.023 × 10 23) of carbon atoms, each containing six protons and six neutrons, equals 12.00000 g, where 12.00000 is the atomic weight. One also speaks about atomic mass units (amu): 1 amu is one-twelfth the mass of the most common isotope of carbon, 12 C. Determine the number of atoms, N, in the given amount of substance. We do this by multiplying the number of moles, n, by Avogadro's number, A,. See full answer below. Unit of the number of substances: –At present, the use of ‘moles’ as a unit in chemical calculations has been very simple and reasonable.One mole molecule, one mole atom or one mole ion means Avogadro-number N; That is, 6.023 x 1023 refers to the total amount of molecules, atoms or ions, which, when expressed in the gram, the gram molecular mass in the molecule, the gram-atomic mass in.

These elementary units in one mole can be anything like atoms, molecules, ions, electrons, protons, neutrons, particles of sand. So, when we say one mole of sodium chloride, it means 6.022 × 10²³ molecules of sodium chloride.

Values of Avogadro's Constant

The value of the Avogadro constant is revised over a period of time. As of the 2019 redefinition of the SI base units, the value of the Avogadro constant is fixed to 6.02214076×1023mol−1. This is the exact value of the constant. The table below mentions the value of the constant in different units.

History of Avogadro's Constant

The Avogadro constant has a long history. The constant is named in honour of Avogadro, but he did not discover it. In 1811, Avogadro discovered the relationship between the volume of gas and the amount of gas through his experiments. He was the first to proposed the volume of a gas is directly proportional to the amount of the gas at constant pressure and temperature. This is today we call Avogadro's law or Avogadro's hypothesis. His work does not mention Avogadro's constant.

Perrin's Work

French Nobel Laureate Jean Baptiste Perrin estimated the Avogadro number with several methods. And he credited the naming of the number to Avogadro in 1909. Perrin named the number Avogadro's number, not Avogadro's constant. This name had continued till 1971. In 1971, the International System of Unit (SI) introduced a new quantity called Avogadro's constant. The Avogadro constant has the same numerical value as the Avogadro number, but they differ in the unit which will be explained later in this article.

Perrin defined the Avogadro number as the number of atoms in one gram of hydrogen (one gram-molecule). This definition was later revised to the number of atoms in 12 grams of carbon-12 (¹²C).

Loschmidt's Estimation

Before Perrin, Loschmidt also made a significant contribution to the number. Josef Loschmidt was an Austrian scientist who is notable for his work on estimation of the diameter of the molecules in the air. Through his method, it is possible to calculate the number density (the number of molecules or atoms per unit volume). This quantity is closely relative to the Avogadro constant. The relationship between them is discussed later in this article. The number density of an ideal gas is called as the Loschmidt constant. In many of German literature, these two constants are interchangeable. They can easily be distinguished from their units. The Avogadro constant is also denotated as L in honour of Loschmidt.

Other Efforts

Robert Millikan was an American physicist and Nobel Laureate. He successfully measured the charge on an electron in 1910. The electric charge per mole (the Faraday constant) of electrons had already known at that time. With the help of these two quantities, the electric charge on an electron and the Faraday constant, it is possible to calculate the number of electrons per mole. The value of this number of electrons per mole is the same as Avogadro's constant.

One of a modern method to estimate the value of the constant is X-ray crystallography. This method estimates the constant by determining the number of silicon atoms in a crystal cell, the volume per unit cell, and the molar volume.

The measurement of the accurate value of the Avogadro's constant is always troublesome. Over the period, the new methods were developed, and the Avogadro constant has continuously been improvised. From 2019, the international committee fixed the value of the Avogadro's constant exactly to 6.022 140 76 × 10²³ mol⁻¹.

2019 Redefinition and Prior Definition of Avogadro's Constant

As discussed above, the 2019 redefinition of the Avogadro constant is 6.022 140 76 × 10²³ mol⁻¹. The consequence of this redefinition is the prior definition of the constant is no longer valid. Before the 2019 redefinition, the value of the constant was defined as the amount of atoms presents in 12g of carbon-12 (¹²C). Also, because of the definition the molar mass constant (M_u) is no longer exactly equal to 1 g mol⁻¹. Instead, it is approximately equal to 1 g mol⁻¹. This is summarised in the table below.

Note: The difference in the value of the Avogadro constant before and after the 2019 definition is very small. The redefinition would not affect most of the calculations unless the high degree of precision is needed. For practical calculations, we can take N_A = 6.022 × 10²³ mol⁻¹.

Avogadro's Constant and Mole

The Avogadro constant and the mole are related quantities. In fact, the Avogadro constant is defined in terms of the mole. The value of Avogadro's constant is the number of elementary units in one mole of any substance. The definition is universally true. The below equation establishes the relation between both.

Avogadro's Constant and Molar Mass

We can use the Avogadro constant to determine the mass of any atom if we know the molar mass of that atom. This statement is also true for molecules. The molar mass is the mass of one mole of a given sample. It is expressed in g mol⁻¹. The relation between both is as follows:

where m_i is the mass of atom i and M_i is molar mass of atom i.

Avogadro's Constant and Avogadro's Number

The Avogadro constant and the Avogadro number have the same numerical value. They only differ in the unit. The Avogadro number is a dimensionless quantity, but the Avogadro constant has the dimension of the reciprocal amount of substance (mol⁻¹). The below table describes the same.

Avogadro's Constant and Boltzmann's constant

The Boltzmann constant is an important physical constant which plays a vital role in classical statistical mechanics. It is denoted as k_B or simply k. The Avogadro constant is related to the Boltzmann constant by the gas constant R.

Avogadro's Constant and Loschmidt's Constant

The Loschmidt constant is the number density (the number of molecules per unit volume). For an ideal gas, the relationship between the Loschmidt constant and the Avogadro constant at STP (P₀ = 1 atm, T₀ = 273.15 K) is described in the equation below.

Avogadro's Constant and Faraday's Constant

The Faraday constant (F) is the Avogadro constant times the elementary charge (e).

Avogadro's Constant and Unified Mass Unit

The unified mass unit or the dalton (u) is the ratio of the molar mass constant (M_u) and the Avogadro constant.

where m_u is the atomic mass constant.

The value precise value of M_u is 0.999 999 999 65(30) g mol⁻¹. But for practical purposes, we can say M_u ≈ 1 g mol⁻¹.

Examples

Example 1: To Determine Calcium Atoms

Statement: For 100 g of calcium in a beaker, calculate the number of calcium atoms in the beaker?

Solution: The molecular weight of calcium is 40.1 g mol⁻¹. The number of moles of calcium in the beaker is

The number of calcium atoms in the beaker is calculated as:

Therefore, the number of calcium atoms is 1.50 × 10²⁴.

Example 2: To Determine Total Molecules in Sodium Chlorine Solution

Statement: Consider 50.0 g of NaCl is dissolved in 200 g of water. Estimate the total molecules in the solution?

Solution: The molecular weight of NaCl and water is 58.44 g mol⁻¹ and 18.01 g mol⁻¹.

The moles of NaCl in 50.0 g:

The moles of H₂O in 100 g:

When 1 mol of NaCl dissociates, 1 mol of Na⁺ and 1 mol of Cl⁻ are formed. So, when 0.855 5 mol of NaCl dissociates, 0.855 5 mol of Na⁺ and 0.855 5 mol of Cl⁻ are formed.

Thus, the total number of moles after the dissociation is the sum of the moles of Na⁺, Cl⁻, and H₂O.

The total number of molecules in the solution is

Therefore, the total number of moles in NaCl solution is 4.374 × 10²⁴ mol.

Example 3: To Determine Mass of Sodium Atom

Statement: The molar mass of sodium-23 is 22.989 g mol⁻¹. Calculates the mass of a sodium atom?

Solution: Let m_Na and M_Na be the atomic mass and molar mass of sodium-23. Thus, M_Na = 22.989 g mol⁻¹.

Now, m_Na can be determined using the formula below.

Therefore, the mass of a sodium-23 atom is 3.817 × 10⁻²³ g.

Example 4: To Determine Molecular Mass of Iodine gas

Statement: The atomic mass of iodine is 126.9 g mol⁻¹. Determine the molecular mass of iodine gas?

Solution: The iodine gas is a diatomic gas. The molecular formula is I₂. So, the molar mass of I₂ is twice the molar mass of I.

Now, m_I2 can be determined as:

Therefore, the mass of a iodine molecule is 4.208 × 10⁻²² g.

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Avogadro's Law:
Ten Examples

Discovered by Amedo Avogadro, of Avogadro's Hypothesis fame. The ChemTeam is not sure when, but probably sometime in the early 1800s.

Gives the relationship between volume and amount when pressure and temperature are held constant. Remember amount is measured in moles. Also, since volume is one of the variables, that means the container holding the gas is flexible in some way and can expand or contract.

If the amount of gas in a container is increased, the volume increases.

If the amount of gas in a container is decreased, the volume decreases.

Why?

Suppose the amount is increased. This means there are more gas molecules and this will increase the number of impacts on the container walls. This means the gas pressure inside the container will increase (for an instant), becoming greater than the pressure on the outside of the walls. This causes the walls to move outward. Since there is more wall space the impacts will lessen and the pressure will return to its original value.

The mathematical form of Avogadro's Law is:

This means that the volume-amount fraction will always generate a constant if the pressure and temperature remain constant.

Let V₁ and n₁ be a volume-amount pair of data at the start of an experiment. If the amount is changed to a new value called n₂, then the volume will change to V₂.

We know this:

And we know this:

Since k = k, we can conclude:

This equation will be very helpful in solving Avogadro's Law problems. You will also see it rendered thusly:

V₁ / n₁ = V₂ / n₂

Sometimes, you will see Avogadro's Law in cross-multiplied form:

V₁n₂ = V₂n₁

Avogadro's Law is a direct mathematical relationship. If one gas variable (V or n) changes in value (either up or down), the other variable will also change in the same direction. The constant K will remain the same value.

Example #1: 5.00 L of a gas is known to contain 0.965 mol. If the amount of gas is increased to 1.80 mol, what new volume will result (at an unchanged temperature and pressure)?

Solution:

(5.00 L) (1.80 mol) = (x) (0.965 mol)

x = 9.33 L (to three sig figs)

Example #2: A cylinder with a movable piston contains 2.00 g of helium, He, at room temperature. More helium was added to the cylinder and the volume was adjusted so that the gas pressure remained the same. How many grams of helium were added to the cylinder if the volume was changed from 2.00 L to 2.70 L? (The temperature was held constant.)

Solution:

1) Convert grams of He to moles:

2.00 g / 4.00 g/mol = 0.500 mol

2) Use Avogadro's Law:

2.00 L / 0.500 mol = 2.70 L / x

x = 0.675 mol

3) Compute grams of He added:

(0.175 mol) (4.00 g/mol) = 0.7 grams of He added

Example #3: A balloon contains a certain mass of neon gas. The temperature is kept constant, and the same mass of argon gas is added to the balloon. What happens?

Avogadro's Number Full

(a) The balloon doubles in volume.
(b) The volume of the balloon expands by more than two times.
(c) The volume of the balloon expands by less than two times.
(d) The balloon stays the same size but the pressure increases.
(e) None of the above.

Solution:

V₁ / n₁ = V₂ / n₂

Let's assign V₁ to be 1 L and V₂ will be our unknown.

Let us assign 1 mole for the amount of neon gas and assign it to be n₁.

The mass of argon now added is exactly equal to the neon, but argon has a higher gram-atomic weight (molar mass) than neon. Therefore less than 1 mole of Ar will be added. Let us use 1.5 mol for the total moles in the balloon (which will be n₂) after the Ar is added. (I picked 1.5 because neon weighs about 20 g/mol and argon weighs about 40 g/mol.)

1 / 1 = x / 1.5

x = 1.5

answer choice (c).

Example #4: A flexible container at an initial volume of 5.120 L contains 8.500 mol of gas. More gas is then added to the container until it reaches a final volume of 18.10 L. Assuming the pressure and temperature of the gas remain constant, calculate the number of moles of gas added to the container.

Solution:

Avogadro's Number Definition

x = 30.05 mol <--- total moles, not the moles added

30.05 − 8.500 = 21.55 mol (to four sig figs)

Notice the specification in the problem to determine moles of gas added. The Avogadro Law calculation gives you the total moles required for that volume, NOT the moles of gas added. That's why the subtraction is there.

Example #5: If 0.00810 mol neon gas at a particular temperature and pressure occupies a volume of 214 mL, what volume would 0.00684 mol neon gas occupy under the same conditions?

Solution:

1) Notice that the same conditions are the temperature and pressure. Holding those two constant means the volume and the number of moles will vary. The gas law that describes the volume-mole relationship is Avogadro's Law:

2) Substituting values gives:

3) Cross-multiply and divide for the answer:

When I did the actual calculation for this answer, I used 684 and 810 when entering values into the calculator.

4) You may find this answer interesting:

V₁/V₂ = n₁/n₂

V₂ = V₁n₂/n₁

V₂ = [(214 mL) (0.00684 mol)] / 0.00810 mol

V₂ = 181 mL

In case you don't know, PV = nRT is called the Ideal Gas Law. You'll see it a bit later in your Gas Laws unit, if you haven't already.

Example #6: A flexible container at an initial volume of 6.13 L contains 7.51 mol of gas. More gas is then added to the container until it reaches a final volume of 13.5 L. Assuming the pressure and temperature of the gas remain constant, calculate the number of moles of gas added to the container.

Solution:

1) Let's start by rearranging the Ideal Gas Law (which you'll see a bit later or you can go review it right now):

V/n = RT / P

R is, of course, a constant.

2) T and P are constant, as stipulated in the problem. Therefore, we can write this:

where k is some constant.

3) Therefore, this is true:

V/n = k

4) Given V and n at two different sets of conditions, we have:

V₁ / n₁ = k
V₂ / n₂ = k

5) Since k = k, we have this relation:

V₁ / n₁ = V₂ / n₂

6) Insert data and solve:

(6.13) (n) = (13.5) (7.51)

n = [(13.5) (7.51)] / 6.13

n = 16.54 mol (this is not the final answer)

7) Final step:

16.54 − 7.51 = 9.03 mol (this is the number of moles of gas that were added)

Example #7: A container with a volume of 25.47 L holds 1.050 mol of oxygen gas (O₂) whose molar mass is 31.9988 g/mol. What is the volume if 7.210 g of oxygen gas is removed from the container, assuming the pressure and temperature remain constant?

Solution #1:

1) Initial mass of O₂:

(1.050 mol) (31.9988 g/mol) = 33.59874 g

2) Final mass of O₂:

33.59874 − 7.210 = 26.38874 g

3) Final moles of O₂:

26.38874 g / 31.9988 g/mol = 0.824679 mol

4) Use Avogadro's Law:

25.47 L / 1.050 mol = V₂ / 0.824679 mol

V₂ = 20.00 L

Solution #2:

1) Let's convert the mass of O₂ removed to moles:

7.210 g / 31.9988 g/mol = 0.225321 mol

2) Subtract moles of O₂ that got removed:

1.050 mol − 0.225321 mol = 0.824679 mol

3) Use Avogadro's Law as above.

Solution #3:

1) This solution depends on seeing that the mass ratio is the same as the mole ratio. Allow me to explain by using Avogadro's Law:

2) Replace moles with mass divided by molar mass:

3) Since the molar mass is of the same substance (oxygen in this case), they cancel out leaving us with this:

4) Solve using the appropriate values

V₂ = 20.00 L

Example #8: What volume (in L) will 5.5 g of oxygen gas occupy if 2.2 g of the oxygen gas occupies 3.0 L? (Under constant pressure and temperature.)

Solution:

1) State the ideal gas law:

Note that it is the full version which includes the moles of gas. Usually a shortened version with the moles not present is used. Since grams are involved (which leads to moles), we choose to use the full version.

2) The problem states that P and T are constant:

3) Cross-multiply and rearrange to isolate V₂:

V₂ = (V₁) (n₂ / n₁)

4) moles = mass / molecular weight:

V₂ = (V₁) [(mass₂ / mw) / (mass₁ / mw)]

5) mw is a constant (since they are both the molecular weight of oxygen), which means it can be canceled out:

V₂ = (V₁) (mass₂ / mass₁)

6) Solve:

V₂ = 7.5 L

Example #9: At a certain temperature and pressure, one mole of a diatomic H₂ gas occupies a volume of 20 L. What would be the volume of one mole of H atoms under those same conditions?

Solution:

One mole of H atoms has 6.022 x 10²³ H atoms.

The number of independent 'particles' in each sample is the same.

Therefore, the volumes occupied by the two samples are the same. The volume of the H atoms sample is 20 L.

By the way, I agree that one mole of H₂ has twice as many atoms as one mole of H atoms. However, the atoms in H₂ are bound up into one mole of molecules, which means that one molecule of H₂ (with two atoms) counts as one independent 'particle' when considering gas behavior.

Example #10: A flexible container at an initial volume of 6.13 L contains 8.51 mol of gas. More gas is then added to the container until it reaches a final volume of 15.5 L. Assuming the pressure and temperature of the gas remain constant, calculate the number of moles of gas added to the container.

Solution:

1) State Avogadro's Law in problem-solving form:

2) Substitute values into equation and solve:

x = 21.5 mol

3) Determine moles of gas added:

21.5 mol − 8.51 mol = 13.0 mol (when properly rounded off)

Bonus Example: A cylinder with a movable piston contains 2.00 g of helium, He, at room temperature. More helium was added to the cylinder and the volume was adjusted so that the gas pressure remained the same. How many grams of helium were added to the cylinder if the volume was changed from 2.00 L to 2.50 L? (The temperature was held constant.)

Solution:

1) The two variables are the volume and the amount of gas (temp and press are constant). The gas law that relates these two variables is Avogadro's Law:

2) We convert the grams to moles:

2.00 g / 4.00 g/mol = 0.500 mol

3) Now, we use Avogadro's Law:

x = [(0.500 mol) (2.50 L)] / 2.00 L

x = 0.625 mol <--- this is the ending amount of moles, not the moles of gas added

4) This is the total moles to create the 2.50 L. We need to convert back to grams:

Notice that I subtracted 0.500 mol from 0.625 mol and used 0.125 mol in the calculation. This is because I want the amount added, not the final ending amount.