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The MoleThe Mole

6.02 X 6.02 X 10102323

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What is What is ““The Mole??The Mole??””

• A counting unit (similar to a dozen)

• 6.02 X 1023 (in scientific notation) =

602 billion trillion = 602,000,000,000,000,000,000,000

• This number is named in honor of Amedeo Avogadro (1776 – 1856)Amedeo Avogadro (1776 – 1856)

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A mole is like a dozen.

1. How many paper clips in 1 dozen?

a) 1 b) 4 c) 12

2. How many oranges in 2.0 dozen?

a) 4 b) 12 c) 24

3. How many dozen contain 36 donuts?

a) 3 b) 12 c) 36

Learning CheckLearning Check

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= 6.02 x 1023 C atoms

= 6.02 x 1023 Na atoms

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

1 mole C

1 mole Na

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The MoleThe Mole

• 1 dozen cookies = 12 cookies• 1 mole of cookies = 6.02 X 1023 cookies

• 1 dozen cars = 12 cars• 1 mole of cars = 6.02 X 1023 cars

• 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!

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Avogadro’s number

• The number of particles in 1 mole is called Avogadro’s Number

(6.02 x 1023)

• This is the magic number that allows us to turn atomic masses (amu) into grams

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Atomic Masses• Unit is the amu.

– atomic mass unit– 1 amu = 1.66 x 10-24g

• We define the masses of atoms and molecules in terms of atomic mass units. Amu’s can be found on the Periodic Table.– 1 Carbon atom = 12.01 amu– 1 Zinc atom = 65.39 amu

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Molecular massesMolecular mass of a molecule = atomic masses of all the atoms of that molecule

–1 O2 molecule = 2(16.00 amu) = 32.00 amu–1 MgCl2 = 24.31 + 2(35.45) = 95.21 amu

• Atomic mass, molecular mass, and formula mass are ALL measured in amu.

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Practice• Calculate the molecular mass (in amu) of the

following compounds:• Sodium chloride (NaCl):• (22.99 amu + 35.45 amu =) 58.44 amu

• Sulfur dioxide (SO2):

• (32.07 amu + 2 x 16.00 amu =) 64.07 amu

• Iron(III) hydroxide (Fe(OH)3):

• (55.85 amu + 3 x (16.00 amu + 1.01 amu)) = 106.87 amu

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Molar Mass• The molar mass is the mass in grams

of one mole of a compound• The molecular mass can be calculated

from atomic masses:

water = H2O = 2(1.01 amu) + 16.00 amu = 18.02 amu

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• 1 molecule has a molecular mass of 18.02 amu

• 1 mole of H2O will weigh 18.02 g, therefore the molar mass of H2O is 18.02 g

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• 1 mole of C atoms (= 12.01 amu) = 12.01 g• 1 mole of Zn atoms (= 65.39 amu) = 65.39 g• 1 mole of O2 molecules (= 32.00 amu) =

32.00 g• 1 mole of MgCl2 formula unit (= 95.21 amu)

= 95.21 g

• Notice:Notice: When you just have a SINGLE element, the molar mass = atomic mass

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IMPORTANT!!IMPORTANT!!

The masses of individual atoms or molecules

are measured in

AMU

The mass of a mole of (6.02 x 1023) atoms or

(6.02 x 1023) molecules are measured in

GRAMS

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6.02 x 1023 particle

1 mole

or

1 mole

6.02 x 1023 particle

Note that a particle can be an atom or a molecule

AvogadroAvogadro’’s Number as s Number as Conversion FactorConversion Factor

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1. Number of atoms in 0.500 mole of Ala) 500 Al atoms

b) 6.02 x 1023 Al atomsc) 3.01 x 1023 Al atoms

2.Number of moles of S in 1.8 x 1024 S atomsa) 1.0 mole S atomsb) 3.0 mole S atomsc) 1.1 x 1048 mole S atoms

Learning CheckLearning Check

0.500 mol x 6.02 x 1023 atoms = 1 mol

1.8 x 1024 atoms x 1 mol_____ = 6.02 x 1023 atoms

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molar mass (g) of atom

1 mole

or

1 mole

molar mass (g) of atom

Molar Mass as Conversion Molar Mass as Conversion FactorFactor

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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

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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

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

REMEMBER: Use the conversion factor that has the given unit ON THE BOTTOM!

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÷ molar mass ÷ Avogadro’s number

Grams Moles Particles

x molar mass x Avogadro’s number

Everything must go through Moles!!!

CalculationsCalculations

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Atoms/Molecules and GramsAtoms/Molecules and Grams

How many atoms of Cu are present in 35.4 g of Cu?

= 3.36 X 1023 atoms Cu

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Learning Check!Learning Check!

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

= 1.21 x 10 24 atoms K

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Practice

• How many grams are 4.25 x 1023 molecules of CH4?

• How many moles are there in 27.2 g of potassium oxide?

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Molar Volume

• At STP, 1 mol (6.02 x 1023 particles) of ANY gas occupies a volume of 22.4 L.– The quantity 22.4 L is called the molar

volume of a gas.– STP = standard temperature and pressure

• 0o C and 1 atm

• What volume will 0.375 mol of O2 gas occupy at STP?

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Mass Percent• Measures the mass (by percent) of a single

atom in a compound.

Mass Percent = Element mass x 100 %

Total mass

• Ex. – A 35.0 g sample contains 11.2 g of potassium. What is the mass percent of potassium?

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Percent Composition

• Percentage of each element in a compound– By mass

• Can be determined from:

-the formula of the compound or

-the experimental mass analysis of the

compound• The percentages may not always total to

100% due to rounding

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100%wholepart

Percentage

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Examples

1. A 23.9 g sample contains 14.3 g C, 3.2 g H, and 6.4 g O. What is the % composition?

2. What is the % composition of calcium chloride?

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Formulas

• Empirical formula: formula in its simplest form; gives the lowest ratio of the atoms in a compound.

• Molecular formula: actual formula of a compound.

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Examples

• C2H6:

• Empirical formula = CH3

• N2O4:

• Empirical formula = NO2

• C6H12O6:

• Empirical formula = CH2O

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Example

• An unknown compound contains 0.426 grams of carbon and 0.107 grams of hydrogen. Determine the empirical formula.

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Calculations

• 1. Determine the amount in grams of all the elements in the compound

• 2. Convert those grams into moles for each element

• 3. Divide the number of moles by the smallest and round up or down

• 4. Turn those rounded numbers into whole numbers when necessary

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• C: 0.426 g / 12.01 g/mol = 0.0355 mol• H: 0.107 g / 1.008 g/mol = 0.106 mol

• C: 0.0355 / 0.0355 = 1.00; round to 1• H: 0.106 / 0.0355 = 2.99; round to 3

• Empirical formula: CH3

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Percentages

• Convert percentages in grams assuming 100 % = 100 grams

• Treat as grams problems

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Practice

• A compound is made out of 30.4 % of nitrogen and 69.6 % of oxygen.

• Determine its empirical formula.

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• N: 30.4 % = 30.4 g; 30.4 g / 14.01 g/mol = 2.17 mol

• O: 69.6 % = 69.6 g; 69.6 g / 16.00 g/mol = 4.35 mol

• N: 2.17 / 2.17 = 1.00• O: 4.35 / 2.17 = 2.00

• Empirical formula: NO2

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Formulas

• Empirical formula: formula in its simplest form.

• Molecular formula: actual formula of a compound.

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Examples

• C2H6: Empirical formula = CH3

• N2O4: Empirical formula = NO2

• C6H12O6: Empirical formula = CH2O

• Molecular formula = (empirical formula)n where n is whole number

• C2H6= (CH3)2 N2O4= (NO2)2

• C6H12O6= (CH2O)6

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Example problem

• The empirical formula of a compound is C2H5 and its molar mass is 58.12 g/mol. What is the molecular formula of this compound?

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Calculations

• 1. Convert empirical formula to empirical formula molar mass by using the molar mass

• 2. Determine n by dividing the actual molar mass by the empirical formula molar mass

• 3. Molecular formula = n x empirical formula

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• C2H5: empirical molar mass = 2 x 12.01 + 5 x 1.008 = 29.06 g/mol

• n = actual molar mass / empirical molar mass =58.12 g/mol / 29.06 g/mol = 2

Molecular formula = 2 x empirical formula = C4H10

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Percentages

• When percentages are given, first calculate the empirical formula.

• Use this empirical formula to calculate the molecular formula.

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Practice

• A compound is made out of 30.4 % of nitrogen and 69.6 % of oxygen.

• Determine its empirical formula.

• The molar mass of the compound is: 92.02 g/mol

• Determine the actual formula.

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• Molar mass of empirical formula: (1 x 14.01 g/mol + 2 x 16.00 g/mol) = 46.01 g/mol

• n = 92.02 g/mol / 46.01 g/mol = 2

• The actual formula is: (NO2) x 2 = N2O4