2 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 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:

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
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 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; 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'.
Problem Example 3: number of moles in N particles
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 NA.
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 (NA) 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.
Problem Example 4: Boron content of borax
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.
Problem Example 5: 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: (.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 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

VM = (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
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.

Number Of Moles Formula

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.

Color By Number The Mole