The MoleThe Mole

6.02 X 6.02 X 10102323

The MoleThe Mole• 1 pair = 2 representative particles1 pair = 2 representative particles• 1 dozen = 12 representative particles1 dozen = 12 representative particles• 1 mole = 6.02 x 101 mole = 6.02 x 102323 representative particles representative particles

The Mole (mol): A unit to count numbers of particles– 1 dozen cookies = 12 cookies– 1 mole of cookies = 6.02 X 1023 cookies

– 1 dozen Al atoms = 12 Al atoms– 1 mole of Al atoms = 6.02 X 1023 atoms

Note that the NUMBER is always the same, but the MASS is very different!

A Mole of ParticlesA Mole of Particles Contains 6.02 x 1023 particles

• Representative ParticleRepresentative Particle::– Refers to the species present.Refers to the species present.– Usually atoms, molecules, or formula units.Usually atoms, molecules, or formula units.– Example:Example:

• Atomic Nitrogen = NAtomic Nitrogen = N

• Nitrogen gas = NNitrogen gas = N22

• Water = HWater = H22OO

• Calcium ion = CaCalcium ion = Ca2+2+

• Glucose = CGlucose = C66HH1212OO66

1 mole of all of these =6.02 x 1023 rep. particles

6.02 x 1023 particles

1 mole

or

1 mole

6.02 x 1023 particles

Note that a particle could be an atom OR a molecule!

Avogadro’s Number as Avogadro’s Number as Conversion FactorConversion Factor

1. Number of atoms in 0.500 mole of Ala) 0.500 mole of Al

b) 6.02 x 1023 Al atoms in 1 molec) So, in 0.5 mole, 3.01 x 1023 Al atoms

2.Number of moles of S in 1.8 x 1024 S atomsa) 6.02 x 1023 S atoms in 1.0 mole of S atom

b) So, 3 x 6.02 x 1023 = 1.8 x 1024 S atoms in 3 x 1 mole= 3.0 mole S atoms

c) 3.0 mole S atoms

Learning CheckLearning Check

Gram Atomic MassGram Atomic Mass• Gram Atomic MassGram Atomic Mass – –

– Atomic mass of an element expressed in grams.Atomic mass of an element expressed in grams.– The gram atomic mass is the mass of 1 mol of atoms The gram atomic mass is the mass of 1 mol of atoms

of any element.of any element.

1 mole of C atoms = 12.0 g

1 mole of Mg atoms = 24.3 g

1 mole of Cu atoms = 63.5 g

# of C atoms# of C atoms in in12.0 12.0 g of C atoms = g of C atoms = # of Mg atoms# of Mg atoms in in24.324.3 g of Mg atoms = g of Mg atoms = # of O atoms# of O atoms in in 16.016.0 g g• How many is that?How many is that?

6.02 x 106.02 x 102323!!!!

Other NamesOther Names

• Gram Molecular Mass/Molecular Weight

• Gram Formula Mass/Formula Weight

• What is the gram atomic mass (GAM) of 1 What is the gram atomic mass (GAM) of 1 mol of the following compounds?mol of the following compounds?– Gram Molecular Mass of SOGram Molecular Mass of SO33 and H and H22OO = ? = ?

• SOSO33 = 80.1g = 80.1g HH22O = 18 gO = 18 g

– Gram Formula Mass of NaCl and CaIGram Formula Mass of NaCl and CaI22 = ? = ?

• NaCl = 58.5 g NaCl = 58.5 g CaICaI22 = 293.9 g = 293.9 g

Molar Mass( atoms, molecules and Molar Mass( atoms, molecules and compounds)compounds)

A.1 mole of Br atomsB.1 mole of Sn atoms

= 79.9 g/mole

= 118.7 g/mole

Molar MassMolar Mass: : mass of one mol of any substancemass of one mol of any substance((usually we round to the tenths place).

Mass in grams of 1 mole equal numerically to the sum of the atomic masses

1 mole of CaCl2 = 111.1 g/mol 1 mole Ca x 40.1 g/mol + 2 moles Cl x 35.5 g/mol = 111.1 g/mol CaCl2

1 mole of N2O4 = 92.0 g/mol

A. Molar Mass of K2O = ? Grams/mole

B. Molar Mass of antacid Al(OH)3 = ? Grams/mole

Example:Example:1.1. How many grams are in 9.45 mol dinitrogen How many grams are in 9.45 mol dinitrogen

trioxide? trioxide? 718.20 g N718.20 g N22OO33

2. How many moles are in 92.2 g iron2. How many moles are in 92.2 g iron (III) oxide? (III) oxide?•

0.58 mol Fe0.58 mol Fe22OO33

Learning Check!Learning Check!

10

What is the formula mass of Ca3(PO4)2 ?

1 formula unit of Ca3(PO4)2

3 Ca 3 x 40.08

2 P 2 x 30.97

8 O + 8 x 16.00310.18 amu

Volume of a MolVolume of a Mol• Standard temperature and pressure:Standard temperature and pressure:

– STPSTP– 0°C and 101.3 kPa or 1 atm0°C and 101.3 kPa or 1 atm– At these conditions, 1 mol of any gas = 22.4 LAt these conditions, 1 mol of any gas = 22.4 L

• 1. What is the volume, in L, at 0.60 mol SO1. What is the volume, in L, at 0.60 mol SO22 gas at gas at STP?STP?13.44 L SO13.44 L SO22 (g) (g)

• 2. The density of a gaseous compound containing 2. The density of a gaseous compound containing carbon and oxygen is 1.964 g/L at STP. What is carbon and oxygen is 1.964 g/L at STP. What is the molar mass?the molar mass?43.99 g/mol43.99 g/mol

Problems, cont.Problems, cont.

• 3. A gaseous compound composed of 3. A gaseous compound composed of sulfur and oxygen is linked in the formation sulfur and oxygen is linked in the formation of acid rain and has a density of 3.58 g/L of acid rain and has a density of 3.58 g/L at STP. What is the molar mass?at STP. What is the molar mass?

80.19 g/mol80.19 g/mol

• 4. What is the density of Kr gas at STP?4. What is the density of Kr gas at STP?

3.74 g/L3.74 g/L

Big HelperBig Helper……

moles Representativeparticles

Liters

GramsMolar Mass

(Molecules,Atoms,Ions,Formula units)

NA=6.02 x 1023

22.4 liters(STP)

Molar volume

Gram atomic mass/Gram molecular mass/formula mass

Aluminum is often used for the structure of light-weight bicycle frames. How many grams of Al are in 3.00 moles of Al?

3.00 moles Al ? g Al

Converting Moles and GramsConverting Moles and Grams

1. Molar mass of Al 1 mole Al = 27.0 g Al

2. Conversion factors for Al

27.0g Al or 1 mol Al

1 mol Al 27.0 g Al

3. Setup 3.00 moles Al x 27.0 g Al

1 mole Al

Answer = 81.0 g Al

Atoms/Molecules and GramsAtoms/Molecules and Grams

• Since 6.02 X 1023 particles = 1 mole AND

1 mole = molar mass (grams)

• You can convert atoms/molecules to moles and then moles to grams! (Two step process)

• You can’t go directly from atoms to grams!!!! You MUST go thru MOLES.

molar mass Avogadro’s number Grams Moles particles

Everything must go through Moles!!!

CalculationsCalculations

Atoms/Molecules and GramsAtoms/Molecules and GramsHow many atoms of Cu are present

in 35.4 g of Cu?

35.4 g Cu 1 mol Cu 6.02 X 1023 atoms Cu 63.5 g Cu 1 mol Cu

= 3.4 X 1023 atoms Cu

Learning Check!Learning Check!

How many atoms of K are present in 78.4 g of K?

Learning Check!Learning Check!

How many atoms of O are present in 78.1 g of oxygen?

78.1 g O2 1 mol O2 6.02 X 1023 molecules O2 2 atoms O 32.0 g O2 1 mol O2 1 molecule O2

Percent CompositionPercent Composition

• Percent compositionPercent composition::– The percent by mass of each element in a The percent by mass of each element in a

compound.compound.

% mass of element = % mass of element = grams of elementgrams of element grams of compound grams of compound

• 1. Calculate the percent composition of each 1. Calculate the percent composition of each element in propane, Celement in propane, C33HH88..

• C = 81.82%C = 81.82%• H = 18.18%H = 18.18%

X 100%

Types of FormulasTypes of Formulas

• Empirical FormulaEmpirical Formula

The formula of a compound that The formula of a compound that expresses the expresses the smallest whole number smallest whole number ratioratio of the atoms present. of the atoms present.

Ionic formula are always empirical formulaIonic formula are always empirical formula

• Molecular FormulaMolecular Formula

The formula that states the The formula that states the actualactual number of each kind of atom found in number of each kind of atom found in one one moleculemolecule of the compound. of the compound.

Empirical FormulaEmpirical Formula

• Empirical FormulasEmpirical Formulas::• The lowest whole number ratio of atoms of the elements The lowest whole number ratio of atoms of the elements

in a compound.in a compound.

– Example: CO (could be CExample: CO (could be C22OO22, C, C44OO44, but not CO, but not CO22))

• *Empirical Formula may or may not be the molecular *Empirical Formula may or may not be the molecular formula.formula.

• Examples: Examples:

–Hydrogen Peroxide = HHydrogen Peroxide = H22OO22 but the empirical but the empirical

formula is HOformula is HO

–Carbon dioxide = COCarbon dioxide = CO22 and it is the empirical and it is the empirical

formulaformula

To obtain an To obtain an Empirical Empirical FormulaFormula

1.1. Determine the mass in grams of each Determine the mass in grams of each element present, if necessary.element present, if necessary.

2.2. Calculate the number of Calculate the number of molesmoles of of each each element.element.

3.3. Divide each by the smallest number of moles Divide each by the smallest number of moles to obtain the to obtain the simplest whole number ratio.simplest whole number ratio.

4.4. If whole numbers are not obtainedIf whole numbers are not obtained** in step in step 3), multiply through by the smallest number 3), multiply through by the smallest number that will give all whole numbersthat will give all whole numbers

** Be careful! Do not round off numbers prematurelyBe careful! Do not round off numbers prematurely

Empirical ProblemsEmpirical Problems

• 1. What is the empirical formula of a 1. What is the empirical formula of a compound that is 25.9% nitrogen and compound that is 25.9% nitrogen and 74.1% oxygen?74.1% oxygen?

• Looks hard, but follow the three steps…Looks hard, but follow the three steps…

Problem, cont.Problem, cont.

• Step 1:Step 1:– Convert % to moles.Convert % to moles.

• 1.85 mol N1.85 mol N• 4.63 mol O4.63 mol O

• Step 2:Step 2:– Divide both answers by the lowest one.Divide both answers by the lowest one.

• 1.85 mol N is lowest.1.85 mol N is lowest.1 mol N1 mol N–2.5 mol O2.5 mol O

Problem, cont.Problem, cont.

• Step 3:Step 3:– If one answer is a decimal, multiply both by a If one answer is a decimal, multiply both by a

number so it is converted to a whole number.number so it is converted to a whole number.• 2.5 mol O becomes 5 mol2.5 mol O becomes 5 mol• 1 mol N becomes 2 mol1 mol N becomes 2 mol

• Answer: NAnswer: N22OO55

Chemical Formulas of Chemical Formulas of CompoundsCompounds

• Formulas give the relative numbers of atoms or Formulas give the relative numbers of atoms or moles of each element in a formula unit - always moles of each element in a formula unit - always a whole number ratio (the law of definite a whole number ratio (the law of definite proportions).proportions).

NONO22 2 atoms of O for every 1 atom of N 2 atoms of O for every 1 atom of N

1 mole of NO1 mole of NO22 : 2 moles of O atoms to : 2 moles of O atoms to

every 1 mole of N atomsevery 1 mole of N atoms

• If we know or can determine the relative number If we know or can determine the relative number of moles of each element in a compound, we can of moles of each element in a compound, we can determine a formula for the compound.determine a formula for the compound.

A sample of a brown gas, a major air pollutant, is A sample of a brown gas, a major air pollutant, is found to contain 2.34 g N and 5.34g O. Determine a found to contain 2.34 g N and 5.34g O. Determine a formula for this substance.formula for this substance.

require require molemole ratios so convert grams to moles ratios so convert grams to moles

moles of N = moles of N = 2.34g of N 2.34g of N = 0.167 moles of N= 0.167 moles of N

14.01 g/mole14.01 g/mole

moles of O = moles of O = 5.34 g5.34 g = 0.334 moles of O = 0.334 moles of O

16.00 g/mole16.00 g/mole

Formula:Formula:

0.334 0.167ON 0.167 0.334 2

0.167 0.167

N O NO

Determine Empirical Formula

1. 75% carbon, 25% hydrogen

2. 52.7% potassium, 47.3% chlorine

3. 22.1% aluminium, 25.4% phosphorous, 52.5% oxygen

4. 13% magnesium, 87% bromine

5. 32.4% sodium, 22.5% sulfur, 45.1% oxygen

Molecular FormulasMolecular Formulas

• Molecular FormulasMolecular Formulas::– Either the same as the empirical formula or a Either the same as the empirical formula or a

simple whole number multiple of it.simple whole number multiple of it.

Known molar mass of moleculeKnown molar mass of molecule = =

Molar mass of empirical form.Molar mass of empirical form.

the number of empirical formula units needed (EFU).

Multiply empirical formula with EFU to get the molecular formula

Calculation of the Molecular FormulaCalculation of the Molecular Formula

A compound has an empirical formula A compound has an empirical formula of NOof NO22. The colourless liquid, used in . The colourless liquid, used in

rocket engines has a molar mass of rocket engines has a molar mass of 92.0 g/mole. What is the 92.0 g/mole. What is the molecular molecular formula formula of this substance?of this substance?