# Avogadro's Number Particles

Avogadro's number, or Avogadro's constant, is the number of particles found in one mole of a substance. It is the number of atoms in exactly 12 grams of carbon-12.This experimentally determined value is approximately 6.0221 x 10 23 particles per mole. Jan 23, 2020 Avogadro's number, or Avogadro's constant, is the number of particles found in one mole of a substance. It is the number of atoms in exactly 12 grams of carbon -12. This experimentally determined value is approximately 6.0221 x 10 23 particles per mole.

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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.

## Avogadro's Number Molar Mass

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. But 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, 6.022141527 × 10^{23}, usually designated by *N*_{A}.

**Amadeo Avogadro (1766-1856) never knew his own number!**

Avogadro only originated the *concept* of this number, whose actual value was first estimated by Josef Loschmidt, an Austrian chemistry teacher, in 1895.

You should know it to three significant figures:*N*_{A} = 6.02 × 10^{23}

6.02 × 10^{23} of *what*? Well, of anything you like: apples, stars in the sky, burritos. But 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.

Think of 6.02 × 10^{23} as the 'chemist's dozen'.

The basic idea (CurtisWang, 4 min) ****

The Mole Explained (utaustinchem, 9 min) ****

The Mole, Avogadro's no., counting by weight (dcaulf, 13½ min) *****

Simple mole calculations (bozemanscience, 5m) ****

Mole and Avogadro's number (Khan, 9½ m) **

The mole concept (IsaacsTeach, 5 min) ****

Mole problems review (LindaHanson, 23 min) ****

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

Problem Example 1: mass ratio from atomic weightsThe atomic weights of oxygen and of carbon are 16.0 and 12.0, 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 16/12 =

**4/3**≈ 1.33 as great as the mass of a carbon atom.

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

* Solution:* The absolute mass of the carbon atom is 12.0

**u**,

or 12 × 1.6605 × 10

^{–27}g = 19.9 × 10

^{–27}kg. The mass of the oxygen atom will be 4/3 greater, or

**2.66 × 10**.

^{–26 }kgAlternatively: (12 g/mol) ÷ (6.022 × 10^{23} mol^{–1}) × (4/3) = **2.66 × 10 ^{–23} g**.

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:* The mass of an oxygen atom (16 u) is 16/12 = 4/3 that of a carbon atom (12 u), so the collection of

*N*oxygen atoms would have a mass of

4/3 × 12 g =

**16.0 g**.

(If the foregoing problems don" t make sense, you should review the previous lesson.)< p>

**Things to understand about Avogadro's number N_{A}**

• 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; see here for some interesting comparisons with other huge numbers.

• 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. But there are two problems with this: 1) The silicon sphere is an artifact, rather than being something that occurs in nature, and thus may not be perfectly reproducible. 2) 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. See here for more, and stay tuned!

Wikipedia has a good discussion of Avogadro's number

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 mol of anything is just Avogadro's number of that something. Or, if you think like a lawyer, you might prefer the official SI definition:

**Some useful mole links**

Wikipedia article on the mole

Mystified by the Mole? Stop it!

A short history of Avogadro's Number

National Mole Day site

Avogadro's number *N*_{A} = 6.02 × 10^{23}, like any pure number, is dimensionless. However, it also defines the mole, so we can also express *N*_{A} as

6.02 × 10^{23} 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'.

*N*particles

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

* Solution:* (80 atoms) / (6.02E23 atoms mol

^{–1}) =

**1.33E–22**

**mol**

Is this answer reasonable? Yes, because 80 is an extremely small fraction of *N*_{A}.

Molar mass and its uses (IsaacsTeach, 6½ m) ****

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 C^{12} 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}.

Don't let this confuse you; it is very 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 *have a* mass unique to that substance.

Borax is the common name of sodium tetraborate, Na_{2}B_{4}O_{7}. In 20.0 g of borax,*(a)* how many moles of boron are present?*(b)* how many grams of boron are present?

* Solution:* The formula weight of Na

_{2}B

_{4}O

_{7}is (2 × 23.0) + (4 × 10.8) + (7 × 16.0) = 201.2.

*a)* 20 g of borax contains (20.0 g) ÷ (201 g mol^{–1}) = 0.10 mol of borax, and thus **0.40 mol** of B.

*b)* 0.40 mol of boron has a mass of (0.40 mol) × (10.8 g mol^{–1}) = **4.3 g**.

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: (.00110 mol) × (6.02E23 mol

^{–1}) =

**6.64E20**

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 of a pure substance

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.

Problem Example 6 : Molar volume of a liquidMethanol, CH_{3}OH, 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} = (32 g mol^{–1}) / (790 g L^{–1}) = **0.0405 L mol ^{–1}**

Problem Example 7: 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 (87.6 g mol

^{–1}) / (2.60 g cm

^{–3}) = 33.7 cm

^{3}mol

^{–1}

The volume of each 'box' is (33.7 cm^{3} mol^{–1}) / (6.02E23 mol^{–1}) = 5.48E–23 cm^{3}

The side length of each box will be the cube root of this value, 3.79E–8 cm. The atomic radius will be half this value, or 1.9E–8 cm = 1.9E–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 54.8E–24, for example. Since 3

^{3}=27 and 4

^{3}= 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, depending on how it is defined.

Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially important that you know the precise meanings of all the highlighted terms in the context of this topic.

- 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.

Contrary to the beliefs of generations of chemistry students, Avogadro’s number—the number of particles in a unit known as a mole—was not discovered by Amadeo Avogadro (1776-1856). Avogadro was a lawyer who became interested in mathematics and physics, and in 1820 he became the first professor of physics in Italy. Avogadro is most famous for his hypothesis that equal volumes of different gases at the same temperature and pressure contain the same number of particles.

## Avogadro's Number Particles Are Made

The first person to estimate the actual number of particles in a given amount of a substance was Josef Loschmidt, an Austrian high school teacher who later became a professor at the University of Vienna. In 1865 Loschmidt used kinetic molecular theory to estimate the number of particles in one cubic centimeter of gas at standard conditions. This quantity is now known as the Loschmidt constant, and the accepted value of this constant is 2.6867773 x 10^{25} m^{-3}.

The term “Avogadro’s number” was first used by French physicist Jean Baptiste Perrin. In 1909 Perrin reported an estimate of Avogadro’s number based on his work on Brownian motion—the random movement of microscopic particles suspended in a liquid or gas. In the years since then, a variety of techniques have been used to estimate the magnitude of this fundamental constant.

Accurate determinations of Avogadro’s number require the measurement of a single quantity on both the atomic and macroscopic scales using the same unit of measurement. This became possible for the first time when American physicist Robert Millikan measured the charge on an electron. The charge on a mole of electrons had been known for some time and is the constant called the Faraday. The best estimate of the value of a Faraday, according to the National Institute of Standards and Technology (NIST), is 96,485.3383 coulombs per mole of electrons. The best estimate of the charge on an electron based on modern experiments is 1.60217653 x 10^{-19} coulombs per electron. If you divide the charge on a mole of electrons by the charge on a single electron you obtain a value of Avogadro’s number of 6.02214154 x 10^{23} particles per mole.

## Avogadro's Number Of Particles

Another approach to determining Avogadro’s number starts with careful measurements of the density of an ultrapure sample of a material on the macroscopic scale. The density of this material on the atomic scale is then measured by using x-ray diffraction techniques to determine the number of atoms per unit cell in the crystal and the distance between the equivalent points that define the unit cell (see *Physical Review Letters,* 1974, 33, 464).