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Avogadro's number and the mole
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Counting atoms: Avogadro's number
Moles and their uses
Molar mass (molecular weight)
Molar volume
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, with 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
Amadeo Avogadro (1766-1856) never knew his own number; it was named in his honor by a French scientist in 1909. its value was first estimated by Josef Loschmidt, an Austrian chemistry teacher, in 1895.
Owing to their tiny size, atoms and molecules cannot be counted by direct observation. There are, however, a number of indirect methods that enable us to estimate the number of these particles in a sample of an element or compound. 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, 6.022137 × 1023, usually designated by NA.
What is the special significance of this huge number, 6.02 × 1023 ? To help you understand this extremely important point, take a moment to convince yourself of the reasoning embodied in the following sequence of problems:
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.
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.
Solution: The collection of N oxygen atoms would have a mass of 4/3 × 12 g = 16.0 g.
So what value must N have in order to make the weight of a pile of N atoms of any kind numerically equal to the atomic weight of the element? The answer is just Avogadro's number NA = 6.022137 × 1023.
Wikipedia has a good discussion of Avogadro's number
Things to understand about Avogadro's number
• It is a number, just as is "dozen", and thus is dimensionless; you can think of Avogadro's number as the "chemist's dozen".
• 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.
• Its value 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 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 the mole
National Mole Day site
Avogadro's number NA = 6.02 × 1023, like any pure number, is dimensionless. However, it also defines the mole, so we can also express NA as 6.02 × 1023 mol–1. 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–22 mol
Is this answer reasonable? Yes, because 80 is an extremely small fraction of NA.
Molar mass
The atomic-, molecular-, or formula weight of a substance is the ratio of its mass to 1/12 the mass of a C12 atom, and being a ratio, is dimensionless. But at the same time, this weight is also the mass of one mole (NA) of these entities, 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 to always bear in mind that the mole is a number and not a weight. But a mole of any specific substance corresponds to a certain weight of 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?
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
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 millilitre. Calculate the molar volume of methanol.
Solution: The molar volume will be the volume occpied by one molar mass (32 g) of the liquid. Expressing the density in litres instead of mL, we have
VM = (32 g mol–1) / (790 g L–1) = 0.0405 L mol–1
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.
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
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 xy 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? According to the WebElements page, 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 imortant 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.
