• Avogadro's Number And Mole Concept Analysis
• Avogadro's Number And Mole Concept Chart
• Mole Avogadro's Number Worksheet
• Avogadro's Number And Mole Concept Pdf
• Avogadro's Number And Mole Concept Book

Learn the basics about Avogadro's Number, as a part of chemical calculations.Avogadro’s Number or the Avogadro Constant is 6.02 X 1023 Mol-1.

• A mole is a quantity that contains 6.02.10 23 atoms, molecules and ions. Avogadro’s number is the number of particles in a mole 6.02.10 23 The mole (or mol) represents a certain number of objects. The amount of a substance that contains the same number of entities as there are atoms in 12 g of carbon-12.
• A mole is a quantity that contains 6.02.10 23 atoms, molecules and ions. Avogadro’s number is the number of particles in a mole 6.02.10 23. The mole (or mol) represents a certain number of objects. The amount of a substance that contains the same number of entities as there are atoms in 12 g of carbon-12.
• Oct 16, 2020 Hence, the 6.02 ×l0 23 number of atoms, molecules, or A formula units is called Avogadro’s number that is equivalent to one ‘mole’ of the respective compound. In basic words, 6.02 ×l0 23 particles amount to one mole as twelve eggs are equal to one dozen.
• Avogadro’s number is defined as the number of elementary particles (molecules, atoms, compounds, etc.) per mole of a substance. It is equal to 6.022×10 23 mol -1 and is expressed as the symbol N A. Avogadro’s number is a similar concept to that of a dozen or a gross. A dozen molecules is 12 molecules.

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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.

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 × 10²³, 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 × 10²³

6.02 × 10²³ 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 × 10²³ 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.

Avogadro's Number And Mole Concept Analysis

The 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 kg.

Alternatively: (12 g/mol) ÷ (6.022 × 10²³ 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.

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 × 10²³, 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'.

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

Solution: (80 atoms) / (6.02E23 atoms mol^–1) = 1.33E–22mol

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 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.

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, Na₂B₄O₇. 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 Na₂B₄O₇ 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?

Avogadro's Number And Mole Concept Chart

Mole Avogadro's Number Worksheet

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.

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.

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

Avogadro's Number And Mole Concept Pdf

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 cm³ mol^–1

Avogadro's Number And Mole Concept Book

The volume of each 'box' is (33.7 cm³ mol^–1) / (6.02E23 mol^–1) = 5.48E–23 cm³
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 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, 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.

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 10²⁵ 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 10²³ particles per mole.

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).