the mole standards standards the mole 1 dozen = 1 gross = 1 ream = 1 mole = 12 144 500 6.02 x 10 23...

The The MoleMole

StandardsStandards

The MoleThe Mole

1 dozen =1 gross =

1 ream =

1 mole =

12

144

500

6.02 x 1023

There are There are exactlyexactly 12 grams of 12 grams of carbon-12 in one mole of carbon-12 in one mole of carbon-12.carbon-12.

Avogadro’s NumberAvogadro’s Number6.02 x 1023 is called “Avogadro’s Number” in honor of the Italian chemist Amadeo Avogadro (1776-1855).

Amadeo Avogadro

I didn’t discover it. Its just named after

me!

I didn’t discover it. Its just named after

me!

Calculations with Moles:Calculations with Moles:Converting moles to gramsConverting moles to grams

How many grams of lithium are in 3.50 moles of lithium?

3.50 mol Li= g Li

1 mol Li

6.9 g Li 45.145.1

Calculations with Moles:Calculations with Moles:Converting grams to molesConverting grams to moles

How many moles of lithium are in 18.2 grams of lithium?

18.2 g Li= mol Li

6.94 g Li

1 mol Li2.622.62

Calculations with Moles:Calculations with Moles:Using Avogadro’s NumberUsing Avogadro’s Number

How many atoms of lithium are in 3.50 moles of lithium?

3.50 mol

= atoms1 mol

6.02 x 1023 atoms 2.07 x 102.07 x 102424

Calculations with Moles:Calculations with Moles:Using Avogadro’s NumberUsing Avogadro’s Number

How many atoms of lithium are in 18.2 g of lithium?

18.2 g Li

= atoms Li

1 mol Li 6.02 x 1023 atoms Li

1.58 x 1024

6.9 g Li 1 mol Li

(18.2)(6.02 x 1023)/6.9

Calculating Formula MassCalculating Formula MassCalculate the formula mass of carbon Calculate the formula mass of carbon dioxide, COdioxide, CO22..

12.0 g + 2(16.0 g) =12.0 g + 2(16.0 g) = 44.0 g44.0 g

One mole of CO2 (6.02 x 1023 molecules) has a mass of 44.0 grams

Standard Molar VolumeStandard Molar Volume

Equal volumes of all gases at the same temperature and pressure contain the same number of molecules.

- Amedeo Avogadro

Equal volumes of all gases at the same temperature and pressure contain the same number of molecules.

- Amedeo Avogadro

At STP (Standard Temperature and Pressure):

1 mole of a gas occupies 22.4 liters of volume

At STP (Standard Temperature and Pressure):

1 mole of a gas occupies 22.4 liters of volume

The Mole HighwayThe Mole Highway

Mass (g)

Rep Part

Vol (L)

Form Mass

1 mol

1 mol

Form Mass

22.4 L

1 mol

1 mol

22.4 L

6.02x1023

Rep Part

1 mol

1 mol

6.02x1023

Rep Part

xx xx x x

Calculations with Moles:Calculations with Moles:

Two-Step ProblemTwo-Step Problem

How many atoms of lithium are in 18.2 g of lithium?

18.2 g Li

= atoms Li

1 mol Li 6.02 x 1023 atoms Li

1.58 x 1024

6.9 g Li 1 mol Li

(18.2)(6.02 x 1023)/6.9

Calculating Percentage Calculating Percentage CompositionComposition

Calculate the percentage composition of Calculate the percentage composition of magnesium carbonate, MgCOmagnesium carbonate, MgCO33..

From previous slide: From previous slide: 24.3 g + 12.0 g + 3(16.0 g) = 84.3 g24.3 g + 12.0 g + 3(16.0 g) = 84.3 g

24.31100 28.83%

84.32Mg

12.01

100 14.24%84.32

C 48.00

100 56.93%84.32

O

100.00

FormulasFormulas

molecular formula = (empirical molecular formula = (empirical formula)formula)nn [ [nn = integer] = integer]

molecular formula = Cmolecular formula = C66HH66 = =

(CH)(CH)6 6

empirical formula = CHempirical formula = CH

Empirical formula: the lowest whole number ratio of atoms in a compound.

Molecular formula: the true number of atoms of each element in the formula of a compound.

FormulasFormulas (continued) (continued)

Formulas for Formulas for ionic compoundsionic compounds are are ALWAYSALWAYS empirical (lowest whole empirical (lowest whole number ratio).number ratio).

Examples:Examples:

NaCl MgCl2 Al2(SO4)3 K2CO3

FormulasFormulas (continued) (continued)

Formulas for Formulas for molecular compoundsmolecular compounds MIGHTMIGHT be empirical (lowest whole be empirical (lowest whole number ratio).number ratio).

Molecular:Molecular:

H2O

C6H12O6 C12H22O11

Empirical:

H2O

CH2O C12H22O11

Empirical Formula Empirical Formula DeterminationDetermination

RulesRules

1.1. Base calculation on 100 grams of Base calculation on 100 grams of compound. compound.

2.2. Determine moles of each element in 100 Determine moles of each element in 100 grams of compound. grams of compound.

3.3. Divide each value of moles by the Divide each value of moles by the smallest of the values. smallest of the values.

4.4. Multiply each number by an integer to Multiply each number by an integer to obtain all whole numbers.obtain all whole numbers.

Empirical Formula Empirical Formula DeterminationDetermination

ExampleExampleAdipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid?

49.32 14.107

12.01

g C mol Cmol C

g C

6.85 16.78

1.01

g H mol Hmol H

g H

43.84 12.74

16.00

g O mol Omol O

g O

Empirical Formula Empirical Formula DeterminationDetermination

(part 2)(part 2)

4.1071.50

2.74

mol C

mol O

6.782.47

2.74

mol H

mol O

2.741.00

2.74

mol O

mol O

Divide each value of moles by the smallest Divide each value of moles by the smallest of the values.of the values.

Carbon:Carbon:

Hydrogen:Hydrogen:

Oxygen:Oxygen:

Empirical Formula Empirical Formula DeterminationDetermination

(part 3)(part 3)

Multiply each number by an integer to Multiply each number by an integer to obtain all whole numbers.obtain all whole numbers.

Carbon: 1.50Carbon: 1.50 Hydrogen: 2.50Hydrogen: 2.50 Oxygen: 1.00Oxygen: 1.00x 2 x 2 x 2

33 55 22

Empirical formula:C3H5O

2

Finding the Molecular Finding the Molecular FormulaFormula

The empirical formula for adipic acid The empirical formula for adipic acid is Cis C33HH55OO22. The molecular mass of . The molecular mass of adipic acid is 146 g/mol. What is the adipic acid is 146 g/mol. What is the molecular formula of adipic acid?molecular formula of adipic acid?

1. Find the formula mass of 1. Find the formula mass of CC33HH55OO22

3(12.0 g) + 5(1.0) + 2(16.0) = 73.0 g3(12.0 g) + 5(1.0) + 2(16.0) = 73.0 g

Finding the Molecular Finding the Molecular FormulaFormula

(continued)(continued)

3(12.0 g) + 5(1.0) + 2(16.0) = 73.0 g3(12.0 g) + 5(1.0) + 2(16.0) = 73.0 g

2. Divide the molecular mass by 2. Divide the molecular mass by the mass given by the empirical the mass given by the empirical formula.formula.

1462

73

3. Multiply the empirical formula by this3. Multiply the empirical formula by this number to get the molecular formula.number to get the molecular formula.

(C(C33HH55OO22) x 2 =) x 2 = CC66HH1010OO44