And to convert a quantity expressed in terms of moles (an SI 'amount of substance'), into a quantity expressed in terms of the number of individual items that make it up (a pure number, a dimensionless quantity), the number of moles of the substance is multiplied by the conversion factor, Avogadro's constant: 1 mol / 1 mol = 1 = approx.

1. The Mole And Avogadro's Number
2. Mole And Avogadro Number In Urdu
3. The Mole And Avogadro Number Khan Academy

Skills to Develop

1. The mass of one mole of any pure substance (compound or element) Mole is The amount of a pure substance containing the same number of chemical units as there are atoms in exactly 12 grams of carbon-12.
2. Avogadro's number is 6.02 x 10 23. It is the number of particles in a mole. You can use Avogadro's number to convert between mass and the number of molecules of any pure substance. If you are given the mass of a sample (such as a snowflake), convert the mass to moles, and then use Avogadro's number to convert from moles to molecules.

Make sure you thoroughly understand the following essential ideas:

• Define Avogadro's number and explain why it is important to know.
• Define the mole. Be able to calculate the number of moles in a given mass of a substance, or the mass corresponding to a given number of moles.
• Define molecular weight, formula weight, and molar mass; explain how the latter differs from the first two.
• Be able to find the number of atoms or molecules in a given weight of a substance.
• Find the molar volume of a solid or liquid, given its density and molar mass.
• Explain how the molar volume of a metallic solid can lead to an estimate of atomic diameter.

The chemical changes we observe always involve discrete numbers of atoms that rearrange themselves into new configurations. These numbers are HUGE— far too large in magnitude for us to count or even visualize, but they are still numbers, and we need to have a way to deal with them. We also need a bridge between these numbers, which we are unable to measure directly, and the weights of substances, which we do measure and observe. The mole concept provides this bridge, and is central to all of quantitative chemistry.

Counting Atoms: Avogadro's Number

Owing to their tiny size, atoms and molecules cannot be counted by direct observation. But much as we do when 'counting' beans in a jar, we can estimate the number of particles in a sample of an element or compound if we have some idea of the volume occupied by each particle and the volume of the container. Once this has been done, we know the number of formula units (to use the most general term for any combination of atoms we wish to define) in any arbitrary weight of the substance. The number will of course depend both on the formula of the substance and on the weight of the sample. However, if we consider a weight of substance that is the same as its formula (molecular) weight expressed in grams, we have only one number to know: Avogadro's number.

Avogadro's number

Avogadro's number is known to ten significant digits:

[N_A = 6.022141527 times 10^{23}.]

However, you only need to know it to three significant figures:

[N_A approx 6.02 times 10^{23}. label{3.2.1}]

So (6.02 times 10^{23}) of what? Well, of anything you like: apples, stars in the sky, burritos. However, the only practical use for (N_A) is to have a more convenient way of expressing the huge numbers of the tiny particles such as atoms or molecules that we deal with in chemistry. Avogadro's number is a collective number, just like a dozen. Students can think of (6.02 times 10^{23}) as the 'chemist's dozen'.

Before getting into the use of Avogadro's number in problems, take a moment to convince yourself of the reasoning embodied in the following examples.

Example (PageIndex{1}): Mass ratio from atomic weights

The atomic weights of oxygen and carbon are 16.0 and 12.0 atomic mass units ((u)), respectively. How much heavier is the oxygen atom in relation to carbon?

Solution

Atomic weights represent the relative masses of different kinds of atoms. This means that the atom of oxygen has a mass that is

[dfrac{16, cancel{u}}{12, cancel{u}} = dfrac{4}{3} ≈ 1.33 nonumber]

as great as the mass of a carbon atom.

Example (PageIndex{2}): Mass of a single atom

The absolute mass of a carbon atom is 12.0 unified atomic mass units ((u)). How many grams will a single oxygen atom weigh?

Solution

The absolute mass of a carbon atom is 12.0 (u) or

[12,cancel{u} times dfrac{1.6605 times 10^{–24}, g}{1 ,cancel{u}} = 1.99 times 10^{–23} , g text{ (per carbon atom)} nonumber]

The mass of the oxygen atom will be 4/3 greater (from Example (PageIndex{1})):

[ left( dfrac{4}{3} right) 1.99 times 10^{–23} , g = 2.66 times 10^{–23} , g text{ (per oxygen atom)} nonumber]

Alternatively we can do the calculation directly like with carbon:

[16,cancel{u} times dfrac{1.6605 times 10^{–24}, g}{1 ,cancel{u}} = 2.66 times 10^{–23} , g text{ (per oxygen atom)} nonumber]

Example (PageIndex{3}): Relative masses from atomic weights

Suppose that we have (N) carbon atoms, where (N) is a number large enough to give us a pile of carbon atoms whose mass is 12.0 grams. How much would the same number, (N), of oxygen atoms weigh?

Solution

We use the results from Example (PageIndex{1}) again. The collection of (N) oxygen atoms would have a mass of

[dfrac{4}{3} times 12, g = 16.0, g. nonumber]

Exercise (PageIndex{1})

What is the numerical value of (N) in Example (PageIndex{3})?

Answer

Using the results of Examples (PageIndex{2}) and (PageIndex{3}).

[N times 1.99 times 10^{–23} , g text{ (per carbon atom)} = 12, g nonumber]

or

[N = dfrac{12, cancel{g}}{1.99 times 10^{–23} , cancel{g} text{ (per carbon atom)}} = 6.03 times 10^{23} text{atoms} nonumber ]

There are a lot of atoms in 12 g of carbon.

Things to understand about Avogadro's number

• It is a number, just as is 'dozen', and thus is dimensionless.
• It is a huge number, far greater in magnitude than we can visualize
• Its practical use is limited to counting tiny things like atoms, molecules, 'formula units', electrons, or photons.
• The value of N_A can be known only to the precision that the number of atoms in a measurable weight of a substance can be estimated. Because large numbers of atoms cannot be counted directly, a variety of ingenious indirect measurements have been made involving such things as Brownian motion and X-ray scattering.
• The current value was determined by measuring the distances between the atoms of silicon in an ultrapure crystal of this element that was shaped into a perfect sphere. (The measurement was made by X-ray scattering.) When combined with the measured mass of this sphere, it yields Avogadro's number. However, there are two problems with this:
  • The silicon sphere is an artifact, rather than being something that occurs in nature, and thus may not be perfectly reproducible.
  • The standard of mass, the kilogram, is not precisely known, and its value appears to be changing. For these reasons, there are proposals to revise the definitions of both N_A and the kilogram.

Moles and their Uses

The mole (abbreviated mol) is the the SI measure of quantity of a 'chemical entity', which can be an atom, molecule, formula unit, electron or photon. One mole of anything is just Avogadro's number of that something. Or, if you think like a lawyer, you might prefer the official SI definition:

Definition: The Mole

The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon 12

Avogadro's number (Equation ref{3.2.1}) like any pure number, is dimensionless. However, it also defines the mole, so we can also express N_A as 6.02 × 10²³ mol^–1; in this form, it is properly known as Avogadro's constant. This construction emphasizes the role of Avogadro's number as a conversion factor between number of moles and number of 'entities'.

Example (PageIndex{4}): number of moles in N particles

How many moles of nickel atoms are there in 80 nickel atoms?

Solution

[dfrac{80 ;atoms}{6.02 times 10^{23} ; atoms; mol^{-1}} = 1.33 times 10^{-22} mol nonumber]

Is this answer reasonable? Yes, because 80 is an extremely small fraction of (N_A).

Molar Mass

The atomic weight, molecular weight, or formula weight of one mole of the fundamental units (atoms, molecules, or groups of atoms that correspond to the formula of a pure substance) is the ratio of its mass to 1/12 the mass of one mole of C¹² atoms, and being a ratio, is dimensionless. But at the same time, this molar mass (as many now prefer to call it) is also the observable mass of one mole (N_A) of the substance, so we frequently emphasize this by stating it explicitly as so many grams (or kilograms) per mole: g mol^–1.

It is important always to bear in mind that the mole is a number and not a mass. But each individual particle has a mass of its own, so a mole of any specific substance will always correspond to a certain mass of that substance.

Example (PageIndex{5}): Boron content of borax

Borax is the common name of sodium tetraborate, (ce{Na2B4O7}).

1. how many moles of boron are present in 20.0 g of borax?
2. how many grams of boron are present in 20.0 g of borax?

Solution

The formula weight of (ce{Na2B4O7}) so the molecular weight is:

[(2 times 23.0) + (4 times 10.8) + (7 times 16.0) = 201.2 nonumber]

1. 20 g of borax contains (20.0 g) ÷ (201 g mol^–1) = 0.10 mol of borax, and thus 0.40 mol of B.
2. 0.40 mol of boron has a mass of (0.40 mol) × (10.8 g mol^–1) = 4.3 g.

Example (PageIndex{6}): Magnesium in chlorophyll

The plant photosynthetic pigment chlorophyll contains 2.68 percent magnesium by weight. How many atoms of Mg will there be in 1.00 g of chlorophyll?

Solution

Each gram of chlorophyll contains 0.0268 g of Mg, atomic weight 24.3.

• Number of moles in this weight of Mg: (.0268 g) / (24.2 g mol^–1) = 0.00110 mol
• Number of atoms: (0.00110 mol) × (6.02E23 mol^–1) = (6.64 times 10^{20})

Is this answer reasonable? (Always be suspicious of huge-number answers!) Yes, because we would expect to have huge numbers of atoms in any observable quantity of a substance.

Molar Volume

This is the volume occupied by one mole of a pure substance. Molar volume depends on the density of a substance and, like density, varies with temperature owing to thermal expansion, and also with the pressure. For solids and liquids, these variables ordinarily have little practical effect, so the values quoted for 1 atm pressure and 25°C are generally useful over a fairly wide range of conditions. This is definitely not the case with gases, whose molar volumes must be calculated for a specific temperature and pressure.

Example (PageIndex{7}): Molar Volume of a Liquid

Methanol, CH₃OH, is a liquid having a density of 0.79 g per milliliter. Calculate the molar volume of methanol.

Solution

The molar volume will be the volume occupied by one molar mass (32 g) of the liquid. Expressing the density in liters instead of mL, we have

[V_M = dfrac{32; g; mol^{–1}}{790; g; L^{–1}}= 0.0405 ;L ;mol^{–1} nonumber]

The molar volume of a metallic element allows one to estimate the size of the atom. The idea is to mentally divide a piece of the metal into as many little cubic boxes as there are atoms, and then calculate the length of each box. Assuming that an atom sits in the center of each box and that each atom is in direct contact with its six neighbors (two along each dimension), this gives the diameter of the atom. The manner in which atoms pack together in actual metallic crystals is usually more complicated than this and it varies from metal to metal, so this calculation only provides an approximate value.

Example (PageIndex{8}): Radius of a Strontium Atom

The density of metallic strontium is 2.60 g cm^–3. Use this value to estimate the radius of the atom of Sr, whose atomic weight is 87.6.

Solution

The molar volume of Sr is:

[dfrac{87.6 ; g ; mol^{-1}}{2.60; g; cm^{-3}} = 33.7; cm^3; mol^{–1}]

The volume of each 'box' is'

[dfrac{33.7; cm^3 mol^{–1}} {6.02 times 10^{23}; mol^{–1}} = 5.48 times 10^{-23}; cm^3]

The side length of each box will be the cube root of this value, (3.79 times 10^{–8}; cm). The atomic radius will be half this value, or

[1.9 times 10^{–8}; cm = 1.9 times 10^{–10}; m = 190 pm]

Note: Your calculator probably has no cube root button, but you are expected to be able to find cube roots; you can usually use the x^y button with y=0.333. You should also be able estimate the magnitude of this value for checking. The easiest way is to express the number so that the exponent is a multiple of 3. Take (54 times 10^{-24}), for example. Since 3³=27 and 4³ = 64, you know that the cube root of 55 will be between 3 and 4, so the cube root should be a bit less than 4 × 10^–8.

So how good is our atomic radius? Standard tables give the atomic radius of strontium is in the range 192-220 pm.

Contributors

• Stephen Lower, Professor Emeritus (Simon Fraser U.) Chem1 Virtual Textbook

Avogadro Number Calculations II
How Many Atoms or Molecules?

The value I will use for Avogadro's Number is 6.022 x 10²³ mol¯¹.

Types of problems you might be asked look something like these:

0.450 mole of Fe contains how many atoms? (Example #1)
0.200 mole of H₂O contains how many molecules? (Example #2)

0.450 gram of Fe contains how many atoms? (Example #3)
0.200 gram of H₂O contains how many molecules? (Example #4)

When the word gram replaces mole, you have a related set of problems which requires one more step.

And, two more:

0.200 mole of H₂O contains how many atoms?
0.200 gram of H₂O contains how many atoms?

When the word gram replaces mole, you have a related set of problems which requires one more step. In addition, the two just above will have even another step, one to determine the number of atoms once you know the number of molecules.

Here is a graphic of the procedure steps:

Pick the box of the data you are given in the problem and follow the steps toward the box containing what you are asked for in the problem.

Example #1: 0.450 mole of Fe contains how many atoms?

Solution:

Start from the box labeled 'Moles of Substance' and move (to the right) to the box labeled 'Number of Atoms or Molecules.' What do you have to do to get there? That's right - multiply by Avogadro's Number.

0.450 mol x 6.022 x 10²³ mol¯¹ = see below for answer

Example #2: 0.200 mole of H₂O contains how many molecules?

Solution:

Start at the same box as Example #1.

0.200 mol x 6.022 x 10²³ mol¯¹ = see below for answer

The answers (including units) to Examples #1 and #2

The unit on Avogadro's Number might look a bit weird. It is mol¯¹ and you would say 'per mole' out loud. The question then is WHAT per mole?

The answer is that it depends on the problem. In the first example, I used iron, an element. Almost all elements come in the form of individual atoms, so the correct numerator with most elements is 'atoms.' (The exceptions would be the diatomic elements plus P₄ and S₈.)

So, doing the calculation and rounding off to three sig figs, we get 2.71 x 10²³ atoms. Notice 'atoms' never gets written until the end. It is assumed to be there in the case of elements. If you wrote Avogadro's Number with the unit atoms/mol in the problem, you would be correct.

The Mole And Avogadro's Number

The same type of discussion applies to substances which are molecular in nature, such as water. So the numerator I would use in example #2 is 'molecule' and the answer is 1.20 x 10²³ molecules.

Once again, the numerator part of Avogadro's Number depends on what is in the problem. Other possible numerators include 'formula units,' ions, or electrons. These, of course, are all specific to a given problem. When a general word is used, the most common one is 'entities,' as in 6.022 x 10²³ entities/mol.

Keep this in mind: the 'atoms' or 'molecules' part of the unit is often omitted and simply understood to be present. However, it will often show up in the answer. Like this:

0.450 mol x 6.022 x 10²³ mol¯¹ = 2.71 x 10²³ atoms

It's not that a mistake was made, it's that the 'atoms' part of atoms per mole was simply assumed to be there.

Example #3: 0.450 gram of Fe contains how many atoms?

Example #4: 0.200 gram of H₂O contains how many molecules?

Look at the solution steps in the image above and you'll see we have to go from grams (on the left of the image above) across to the right through moles and then to how many atoms or molecules.

Solution to Example #3:

Step One (grams ---> moles): 0.450 g / 55.85 g/mol = 0.0080573 mol

Step Two (moles ---> how many): (0.0080573 mol) (6.022 x 10²³ atoms/mol) = 4.85 x 10²¹ atoms

Solution to Example #4:

Step One: 0.200 g / 18.015 g/mol = 0.01110186 mol

Step Two: (0.01110186 mol) (6.022 x 10²³ molecules/mol) = 6.68 x 10²¹ molecules

Example #5: Calculate the number of molecules in 1.058 mole of H₂O

Solution:

(1.058 mol) (6.022 x 10²³ mol¯¹) = 6.371 x 10²³ molecules

Example #6: Calculate the number of atoms in 0.750 mole of Fe

Solution:

(0.750 mol) (6.022 x 10²³ mol¯¹) = 4.52 x 10²³ atoms (to three sig figs)

Example #7: Calculate the number of molecules in 1.058 gram of H₂O

Solution:

(1.058 g / 18.015 g/mol) (6.022 x 10²³ molecules/mole)

Here is the solution set up in dimensional analysis style:

1 mol	6.022 x 1023
1.058 g x	–––––––––	x	––––––––––	= 3.537 x 1022 molecules (to four sig figs)
18.015 g	1 mol
↑ grams to moles ↑		↑ moles to ↑ molecules

Example #8: Calculate the number of atoms in 0.750 gram of Fe

(0.750 gram divided by 55.85 g/mole) x 6.022 x 10²³atoms/mole

1 mol	6.022 x 1023
0.750 g x	–––––––––	x	––––––––––	= 8.09 x 1021 atoms (to three sig figs)
55.85 g	1 mol

Example #9: Which contains more molecules: 10.0 grams of O₂ or 50.0 grams of iodine, I₂?

Solution:

Basically, this is just two two-step problems in one sentence. Convert each gram value to its mole equivalent. Then, multiply the mole value by Avogadro's Number. Finally, compare these last two values and pick the larger value. That is the one with more molecules.

1 mol	6.022 x 1023
10.0 g x	–––––––––	x	––––––––––	= number of O2 molecules
31.998 g	1 mol

1 mol	6.022 x 1023
50.0 g x	–––––––––	x	––––––––––	= number of I2 molecules
253.8 g	1 mol

Example #10: 18.0 g of H₂O is present. (a) How many oxygen atoms are present? (b) How many hydrogen atoms are present?

Solution:

1) Convert grams to moles:

18.0 g / 18.0 g/mol = 1.00 mol

2) Convert moles to molecules:

(1.00 mol) (6.02 x 10²³ mol¯¹) = 6.02 x 10²³ molecules

3) Determine number of atoms of oxygen present:

(6.02 x 10²³ molecules) (1 O atom / 1 H₂O molecule) = 6.02 x 10²³ O atoms

4) Determine number of atoms of hydrogen present:

(6.02 x 10²³ molecules) (2 H atoms / 1 H₂O molecule) = 1.20 x 10²⁴ H atoms (to three sig figs)

Notice that there is an additional step (as seen in step 3 for O and step 4 for H). You multiply the number of molecules times how many of that atom are present in the molecule. In one molecule of H₂O, there are 2 atoms of H and 1 atom of O.

Sometimes, you will be asked for the total atoms present in the sample. Do it this way:

(6.02 x 10²³ molecules) (3 atoms/molecule) = 1.81 x 10²⁴ atoms (to three sig figs)

The 3 represents the total atoms in one molecule of water: one O atom and two H atoms.

Example #11: Which of the following contains the greatest number of hydrogen atoms?

(a) 1 mol of C₆H₁₂O₆
(b) 2 mol of (NH₄)₂CO₃
(c) 4 mol of H₂O
(d) 5 mol of CH₃COOH

Solution:

1) Each mole of molecules contains N number of molecules, where N equals Avogadro's Number. How many molecules are in each answer:

(a) 1 x N = N
(b) 2 x N = 2N
(c) 4 x N = 4N
(d) N x 5 = 5N

2) Each N times the number of hydrogen atoms in a formula equals the total number of hydrogen atoms in the sample:

(a) N x 12 = 12N
(b) 2N x 8 = 16N
(c) 4N x 2 = 8N
(d) 5N x 4 = 20N

(d) is the answer.

Example #12: How many oxygen atoms are in 27.2 L of N₂O₅ at STP?

Solution:

1) Given STP, we can use molar volume:

27.2 L / 22.414 L/mol = 1.21353 mol

2) There are five moles of O atoms in one mole of N₂O₅:

(1.21353 mol N₂O₅) (5 mol O / 1 mol N₂O₅) = 6.06765 mol O

3) Use Avogadro's Number:

(6.06765 mol O) (6.022 x 10²³ atoms O / mole O) = 3.65 x 10²⁴ atoms O (to three sig figs)

Example #13: How many carbon atoms are in 0.850 mol of acetaminophen, C₈H₉NO₂?

Mole And Avogadro Number In Urdu

Solution:

The Mole And Avogadro Number Khan Academy

1) There are 8 moles of C in every mole of acetaminophen:

(0.850 mol C₈H₉NO₂) (8 mol C / mol C₈H₉NO₂) = 6.80 mol C

2) Use Avogadro's Number:

(6.80 mol C) (6.022 x 10²³ atoms C / mole C) = 4.09 x 10²⁴ atoms C (to three sig figs)

Example #14: How many atoms are in a 0.460 g sample of elemental phosphorus?

Solution:

Phosphorus has the formula P₄. (Not P!!)

0.460 g / 123.896 g/mol = 0.00371279 mol

(6.022 x 10²³ molecules/mol) (0.00371279 mol) = 2.23584 x 10²¹ molecules of P₄

(2.23584 x 10²¹ molecules) (4 atoms/molecule) = 8.94 x 10²¹ atoms (to three sig figs)

Set up using dimensional analysis style:

1 mol	6.022 x 1023 molecules	4 atoms
0.460 g x	––––––––	x	––––––––––––––––––	x	–––––––––	= 8.94 x 1021 atoms
123.896 g	1 mol	1 molecule

Example #15: Which contains the most atoms?

(a) 3.5 molecules of H₂O
(b) 3.5 x 10²² molecules of N₂
(c) 3.5 moles of CO
(d) 3.5 g of water

Solution:

The correct answer is (c). Now, some discussion about each answer choice.

Choice (a): You can't have half of a molecule, so this answer should not be considered. Also, compare it to (b). Since (a) is much less than (b), (a) cannot ever be the answer to the most number of atoms.

Choice (b): this is a viable contender for the correct answer. Since there are two atoms per molecule, we have 7.0 x 10²² atoms. We continue to analyze the answer choices.

Choice (c): Use Avogadro's number (3.5 x 10²³ mol¯¹) and compare it to choice (b). You should be able to see, even without the 3.5 moles, choice (c) is already larger than choice (b). Especially when you consider that N₂ and CO both have 2 atoms per molecule.

Choice (d): 3.5 g of water is significantly less that the 3.5 moles of choice (c). 3.5 / 18.0 equals a bit less that 0.2 moles of water.

Bonus Example: A sample of C₃H₈ has 2.96 x 10²⁴ H atoms.

(a) How many carbon atoms does the sample contain?
(b) What is the total mass of the sample?

Solution to (a):

1) The ratio between C and H is 3 to 8, so this:

3	y
–––––––	=	––––––––––––––––
8	2.96 x 1024 H atoms

2) will tell us the number of carbon atoms present:

y = 1.11 x 10²⁴ carbon atoms

3) By the way, the above ratio and proportion can also be written like this:

3 is to 8 as y is to 2.96 x 10²⁴

Be sure you understand that the two different ways to present the ratio and proportion mean the same thing.

Solution to (b) using hydrogen:

1) Determine the moles of C₃H₈ present.

2.96 x 10²⁴ / 8 = 3.70 x 10²³ molecules of C₃H₈

2) Divide by Avogadro's Number:

3.70 x 10²³ / 6.022 x 10²³ mol¯¹ = 0.614414 mol <--- I'll keep some guard digits

3) Use the molar mass of C₃H₈:

0.614414 mol times 44.0962 g/mol = 27.1 g (to three sig figs)