•The mole is a counting unit in chemistry.•Abbreviated mol.•Defined as the number equal to the number of carbon atoms in exactly 12 grams of pure 12C .•Has the value 6.022 x 1023.•Called Avogradro’s number to honor his contributions to chemistry.•One mole of anything consists of 6.022 x 1023 units of that substance.•Just like a dozen eggs is 12 eggs; a mole of eggs is 6.022 x 1023 eggs.

•How big is a mole?•1 mole of seconds represents a span of time 4 million times as long as the Earth as already existed.•1 mole of marbles is enough to cover the entire Earth to a depth of 50 miles.

•Remember that the mole is the number of atoms in exactly 12 grams of 12C therefore 12 grams of 12C contains exactly 6.022 x 1023 atoms.•Plus it tells us that a 12.01 gram sample of natural carbon contains 6.022 x 1023 atoms (natural carbon is a mixture of the isotopes of carbon).•Notice that the value 12.01 grams of natural carbon is the same as the atomic mass value (12.01 amu).•It also tells us that 26.98 grams of aluminum contains exactly 6.022 x 1023 atoms of aluminum.

•If a mole of natural carbon and aluminum have the same number of atoms why do they have different masses?•Consider a dozen doughnuts and a dozen tires.•Would they have the same mass?

•We can also define the mole such that a sample of a natural element with a mass equal to the element’s atomic mass expressed in grams contains 1 mole of atoms.

•Example: Determining Moles of Atoms•How many moles of copper are in an old copper penny that has a mass of 3.14 g? How many atoms are in this 3.14 g penny?(New pennies are made of zinc with only a thin surface coating of copper.)

•Solution: •The mass of 1 mole (6.022 x 1023 atoms) of copper is 63.55 g.•The sample we are considering has a mass of 3.14 g. •Since the mass is less than 63.55 g, this sample contains less than 1 mole of copper atoms.

1 mol Cu = 63.55 g Cu (gives two conversion factors)

Cug55.63Cumol1

orCumol1Cug55.63

Cug55.63Cumol1

orCumol1Cug55.63

•Which conversion factor is appropriate? We are converting from g to mol.

Cumol0494.0Cug55.63

Cumol1xCug14.3Cumol#

•How many atoms are in this 3.14 g penny?•If we have 1 mol Cu then we have 6.022 x 1023 atoms Cu.

1 mol Cu = 6.022 x 1023 atoms Cu

Cuatoms10x97.2Cumol1

Cuatoms10x022.6xCumol0494.0 22

23

•Example: Calculating Moles and Mass•Uranium is a silvery-white metallic element in the actinide series of the periodic table. Calculate both the number of moles in a sample of uranium containing 2.530 x 1015 atoms and the mass of the sample.

•Solution: •We know 1 mol U = 6.022 x 1023 atoms U.

Umol10x201.4

Uatoms10x022.6Umol1

xUatoms10x530.2

9

2315

•How many grams of U are contained in this sample?•We know if we have 1 mol U we have 238.03 g U.

1 mol U = 238.03 g U

Ug10x000.1Umol1

Ug03.238xUmol10x201.4 69

•Molar mass of a substance is the mass in grams of one mole of the compound.•Molar mass has units of g/mol.•Remember a chemical compound is a collection of bonded atoms.•If we want to know the mass in grams of one mole of methane, CH4 (the mass of 6.02 x 1023 atoms of CH4) we need to take into account that 1 mole of CH4 contains 1 mole of carbon atoms and 4 moles of hydrogen atoms.

•To determine the mass of 1 mole of CH4 we sum the atomic masses of carbon and hydrogen present:

Mass of 1 mol C = 12.02 gMass of 4 mol H = 4 x 1.008 gMass of 1 mol CH4 16.04 g

•Since 16.04 g represents the mass of 1 mole of methane molecules it is called the molar mass of methane.

•Example: Calculating Molar Mass of a Compound•What is the molar mass of calcium phosphate, Ca3(PO4)2, expressed to the second decimal place?

•Solution: Use atomic masses of 40.08 for Ca, 30.97 for P, and 16.00 for O.

Ca3(PO4)2 = 3 Ca + 2 P + 8 O

Molar mass of Ca3(PO4)2 = (3 x 40.08) + (2 x 30.97) + (8 x 16.00) =120.24 + 61.94 + 128.00 =310.18 g/mol or

1 mol Ca3(PO4)2 = 310.18 g Ca3(PO4)2

•Example: Calculating Grams from Moles for Compounds•A student has worked out the details for an experiment that will use 0.115 mol of Ca3(PO4)2, calcium phosphate, as a starting material. How many grams of Ca3(PO4)2 should be weighed out?

•Solution: For this problem we need an equality that relates the number of moles to the number of grams. We know that 1 mol Ca3(PO4)2 = 310.18 g Ca3(PO )2.

•Notice units of moles cancel and we are left with grams.

243243

243243 )PO(Cag7.35

)PO(Camol1)PO(Cag18.310

x)PO(Camol115.0

•Example: Calculating formula units (molecules if covalent) from grams.•How many formula units of Na2CO3 does 45.8 g of Na2CO3 contain?

•Solution: This problem requires 2 equalities. 1 mol Na2CO3 = ? g Na2CO3

1 mol Na2CO3 = 6.022 x 1023 formula unitsTo determine the number of grams in 1 mol of Na2CO3 sum the atomic masses of the atoms in the compound.(2 x 22.99 for Na) + (1 x 12.01 for C) + (3 x 16.00 for O) =

105.99 g/mol

3223

32

23

32

3232

CONaunitsformula10x60.2

CONamol1unitsformula10x022.6

xCONag99.105

CONamol1xCONag8.45

•How many sodium atoms were present in the 45.8 g of Na2CO3?

•Solution: •In the 45.8 g of Na2CO3 we calculated that we had 2.60 x 1023 formula units. Since each formula unit of Na2CO3 contains 2 Na atoms we can calculate as follows.

Naatoms10x20.5

CONaunitformula1Naatoms2

xCONaunitsformula10x60.2

2332

3223