The The MoleMole

CA StandardsCA Standards

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.

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

12.01 g + 2(16.00 g) =12.01 g + 2(16.00 g) =44.01 44.01 gg

One mole of CO2 (6.02 x 1023 molecules) has a mass of 44.01 grams

Mole RelationshipsMole Relationships

Mole Mole

Atomsor

molecules

Atomsor

molecules

MassMass

6.02 X 1023 Atoms1 Mole

1 Mole

6.02 X 1023 Atoms

1 Mole

Molar Mass (g)

Molar Mass (g)

1 Mole

How would you convert,Atoms Moles Moles Atoms

Moles Mass (g) Mass (g) Moles

Mass (g) Atoms Atoms Mass (g)

atoms

atoms

mole

6.02 x 1023

1 moles atoms

mole

6.02 x 1023

1moles= = atoms

moles

mole1

gg=

g

g

mole1moles=

g

g

1 mole

1 mole

atoms6.02 x 1023

=atoms

atoms

atoms6.02 x 1023

1 mole

1 mole

gg=

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

6.94 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 6.022 x 1023 atoms Li

1.58 x 1024

6.94 g Li 1 mol Li

(18.2)(6.022 x 1023)/6.94

1 mol NaOH

40.00 g NaOH

• How many moles is 5.69 g of NaOH?

Na = 22.99 g/mol

O = 16.00 g/mol

H = 1.01 g/ mol 40.00 g/mol

5.69 g NaOH = 0.142 mol NaOH

• How many grams are in 9.45 mol of dinitrogen trioxide?

N2O3 = 2(14.01) + 3(16.00) = 76.02 g/mol

9.45 mol N2O3 76.02 g N2O3

1 mol N2O3

= 718.2 =718 g N2O3

Percent Composition, Empirical and Percent Composition, Empirical and Molecular FormulasMolecular Formulas

Percent Composition, Empirical and Percent Composition, Empirical and Molecular FormulasMolecular Formulas

Courtesy www.lab-initio.com

7.3: Calculating Percent Composition of a Compound

• Like all percent problems: Part

whole

• Find the mass of each component,

• then divide by the total mass (assume one mole).

x 100

100cmpd of mass

element of masselement of mass %

• What is the % composition of water.– Find MM: 2(1.01) + 16.00 = 18.02 g/mol

• Double check: %’s should add up to 100.

100 =

Percent Composition

%H =2.02 g H

18.02 g H2O 100 =

%O =16.00 g O

18.02 g H2O

11.2 % H

88.8 % O

What is the percent carbon in C5H8NO4

(MSG monosodium glutamate), a

compound used to flavor foods and

tenderize meats?

1. Find mass of C

1. 5 x 12.0 g = 60.0 g C

2. Find mass of MSG

1. 5(12.0) + 8(1.0) +14.0 +4(16.0)=146.0 g

3. Mass of element/ mass of cmpd x 100

1. 60.0 g C/146.0 g x100 = 41.1% C

Practice Problem

%Fe =28 g

36 g 100 =78% Fe

%O =8.0 g

36 g 100 =22% O

• Find the percentage composition of a sample that is 28 g Fe and 8.0 g O.– Determine total mass:

Percent Composition

28 g + 8.0 g = 36 g

Calculating Percentage CompositionCalculating Percentage CompositionCalculate the percentage Calculate the percentage composition of magnesium composition of magnesium carbonate, MgCOcarbonate, MgCO33..Formula mass of magnesium carbonate:Formula mass of magnesium carbonate:24.31 g + 12.01 g + 3(16.00 g) = 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g84.32 g

100.00 100.00

%83.2810032.84

31.24

Mg

%24.1410032.84

01.12

C

%93.5610032.84

00.48

O

• How many grams of copper are in a 38.0-gram sample of Cu2S?

– Multiply mass of cmpd with % element

(38.0 g Cu2S)x(0.79852) = 30.3 g Cu

Using Percent as a Conversion Factor

Cu2S is 79.8% Cu (Sample 7-10 p 227)

EF vs MF

• Empirical Formula (EF)

• Def’n: lowest whole # ratio

• H2O

• HO• CH

• CH2O

• Molecular Formula (MF)

• Def’n: actual ratio

• H2O

• H2O2

• C6H6

• C6H12O6

Waterperoxid

ebenzenesugar

B. Empirical Formula

C2H6

CH3

reduce subscripts

• Smallest whole number ratio of atoms in a compound

Calculating Empirical Formula

• It is not just the ratio of atoms, it is also the ratio of moles of atoms.

• In 1 mole of CO2 there is 1 mole of carbon and 2 moles of oxygen.

Steps for Calculating the Empirical Formula

We get a ratio from the percent composition.

Assume you have a 100 g sample.1. % = g

2. Convert grams to moles (÷ atomic mass)

3. Find lowest whole number ratio.

(÷ by smallest # of moles)

(if this step gives a decimal, multiply by 2, 3, or 4 to get whole #’s)

4. Use ratio to write the EF

• Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N.

Part 1 Calculate the number of moles

Assume 100 g• 38.67 g C x 1mol C = 3.220 mole C

12.01 g C • 16.22 g H x 1mol H = 16.09 mole H

1.01 g H• 45.11 g N x 1mol N = 3.219 mole N

14.01 g N

Part 2 calculate the formula!

3.220 mol C

16.09 mol H

3.219 mol N

3.219

3.219

3.219

=

=

=

1 mol C

5 mol H

1 mol N

Divide by the smallest number of moles

CH5N

Empirical Formula• Find the empirical formula for a

sample of 25.9% N and 74.1% O.

25.9 g 1 mol

14.01 g = 1.85 mol N

74.1 g 1 mol

16.00 g = 4.63 mol O

1.85 mol

1.85 mol

= 1 N

= 2.5 O

Empirical Formula

N1O2.5

Need to make the subscripts whole numbers multiply by 2

N2O5

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

carbonmolcarbong

carbonmolcarbong107.4

01.12

132.49

hydrogenmolhydrogeng

hydrogenmolhydrogeng78.6

01.1

185.6

oxygenmoloxygeng

oxygenmoloxygeng74.2

00.16

184.43

1. Convert to grams, calculate # of moles

2. Divide each value of moles by the 2. Divide each value of moles by the smallest of the values.smallest of the values.

Carbon:Carbon:

Hydrogen:Hydrogen:

Oxygen:Oxygen:

50.174.2

107.4

mol

carbonmol

47.274.2

78.6

mol

hydrogenmol

50.174.2

74.2

mol

oxygenmol

3. Multiply each number by an 3. Multiply each number by an integer to obtain all whole numbers.integer to 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:C3H5O2

Molecular Formula• “True Formula” - the actual number

of atoms in a compound

CH3

C2H6

empiricalformula

molecularformula

?

Steps to calculating Molecular Formula1. Find the empirical formula.2. Find the empirical formula mass.3. Divide the molecular mass by the empirical mass.4. Multiply each subscript by the answer from step 3.

nmass EF

mass MF nEF

Molecular Formula• The empirical formula for ethylene is CH2. Find the

molecular formula if the molecular mass is 28.1 g/mol?

28.1 g/mol

14.03 g/mol = 2.00

empirical mass = 14.03 g/mol

(CH2)2 C2H4

1. Find the empirical formula and its mass

2. Divide the molecular mass by the empirical mass.

3. Multiply each subscript by the answer from step 3.

Finding the Molecular FormulaFinding the Molecular FormulaThe empirical formula for adipic The empirical formula for adipic acid is Cacid is C33HH55OO22. The molecular . The molecular mass of adipic acid is 146 g/mol. mass of adipic acid is 146 g/mol. What is the molecular formula of What is the molecular formula of adipic acid?adipic acid?

1. Find the formula mass of C1. Find the formula mass of C33HH55OO22

3(12.01 g) + 5(1.01) + 2(16.00) = 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g73.08 g

Finding the Molecular FormulaFinding the Molecular Formula

3(12.01 g) + 5(1.01) + 2(16.00) = 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g73.08 g

273

146

2. Divide the molecular mass by the empirical mass.

3. Multiply each subscript by the answer from step 3.

(C(C33HH55OO22) x ) x 22 = = CC66HH1010OO44

Example

• A compound is known to be composed of 71.65 % Cl, 24.27% C and 4.07% H. Its molar mass is known (from gas density) to be 98.96 g. What is its molecular formula?–EF: CClH2

–MF: C2Cl2H4