MOLES AND CALCULATIONS USING THE MOLE CONCEPT

INTRODUCTORY TERMS

A. What is an amu ? 1.66 x 10-24 g

B. We need a conversion to the macroscopic world.

1. How many hydrogen atoms are in 1.00 g of hydrogen?

1 H atom 1.00 g H x = 6.02 x 1023 H atoms 1.66 x 10-24 g H

Avogadro’s Number

2. Consider carbon-12 (the most abundant isotope of C) What is the mass of one carbon-12 atom ?

amu g g 12 x 1.66 x 10-24 = 1.99 x 10-23

C atom amu C atom

3. What is the mass of Avogadro’s number of C atoms?

g 6.02 x 1023 C atoms x 1.99 x 10-23 = 12.00 g C atom

C. Definition of a mole

1. A mole is the amount of any substance that contains as many elementary entities as there are atoms in exactly 1.00 g of hydrogen-1.

2. A mole is the amount ... in exactly 12.00 g of carbon-12.

3. 6.02 x 1023 of anything

4. It is important to state the entities involved:

atoms, molecules, ions, electrons, etc.

5. How large a number is this?

How Large a Quantity is Avogadro’s Number??

D. Definition of a molar mass (g/mol)

1. Molar mass of an element - The mass in grams that is numerically equal to the atomic weight of the element.

C, 12.011 g/mol Na, 22.99 g/mol Xe, 131.30 g/mol

2. Molar mass of a molecular form of an element - May be different

N2, 28.02 g/mol Cl2, 70.906 g/mol (O2, I2, F2, Br2)

3. Molar mass of a compound - The mass in grams that is numerically equal to the formula weight of the compound.(The sum of the atomic weights involved)

H2O, 18.0152 g/mol (1.01 + 1.01 + 15.99)

BaCl2, 208.23 g/mol (137.33 + 35.45 + 35.45)

CO2, 44.01 g/mol (12.01 + 15.99 + 15.99)

4. Sample Calculations of Molar Mass

a. Na2HPO4 Na2 2 x 22.99 = 45.98 H 1 x 1.008 = 1.008 P 1 x 30.97 = 30.97 O4 4 x 16.00 = 64.00

141.96 g/mol

b. Ca3(PO4)2 Ca3 3 x 40.08 = 120.24 (P )2 2 x 30.97 = 61.94 ( O4)2 8 x 16.00 = 128.00

310.18 g/mol

c. C15H22ClNO2 C15 15 x 12.01 = 180.15 (Demerol) H22 22 x 1.008 = 22.22 Cl 1 x 35.45 = 35.45 N 1 x 14.04 = 14.01 O2 2 x 15.99 = 31.98

283.81 g/mol

THE ABOVE CONCEPTS LEAD TO NEW "CONVERSION FACTORS"

1 mole--------------------------

6.02 x 1023 objectsor

6.02 x 1023 objects--------------------------

1 mole

How many atoms are represented by 3.00 moles of calcium (Ca) ?

3.00 mol CaX

6.02 x 1023 atoms1.00 mol Ca

= 1.81 x 1024 atoms of Ca

1.0 x 106 molecules CO 6.02 x 1023 molecules CO

1.00 mole COX

=1.66 x 10-18 mol CO

How many moles are represented by 1,000,000 molecules of carbon monoxide (CO) ?

MOLESOF

SUBSTANCE

PARTICLESOF

SUBSTANCEAvogadro’s Number

ANOTHER PAIR OF "CONVERSION FACTORS"

44.01 g of CO2-----------------------

1 mole of CO2 or

1 mole of CO2--------------------

44.01 g of CO2

How many grams are contained in 4.25 moles of carbon dioxide (CO2) ?

4.25 moles CO 2 X44.01 g of CO2

-----------------------1 mole of CO2

= 187 g CO2

3.77 g of CO2 is equal to how many moles of CO2 ?

3.77 g CO2 X1.00 mole of CO2

--------------------44.01 g of CO2

= .0856 mol CO2

MOLESOF

SUBSTANCE

GRAMSOF

SUBSTANCE

Molar Mass

How many atoms are in 7.67 g of cobalt (Co) ?

7.67 g Co X 6.02 x 1023 atoms1.00 mol CoX

1 mol of Co--------------------58.93 g of Co

= 7.84 x 1022 atoms of Co

What is the mass of 1.77 x 1030 molecules of CO2 ?

44.01 g of CO2-----------------------

1 mole of CO2 X = 1.29 x 108 g CO2

1.77 x 1030 molecules of CO 2 X 6.02 x 1023 molecules CO 2

1.00 mol CO 2

MOLESOF

SUBSTANCE

GRAMSOF

SUBSTANCE

PARTICLESOF

SUBSTANCE

Information Available from A Balanced Equation

Conversion Factors Relevant to Stoichiometry

Use Avogadro’s number as a conversion factor.

Moles of A Particles of A

Use molar mass as a conversion factor.

Moles of A Grams of A

Use mole ratio as a conversion factor.

Moles of A Moles of B

Solution Map for Stoichiometry

Moles of A

Moles of B

Grams of A

Grams of B

Particles of A

Particles of B

Avogadro’s Number Avogadro’s Number

Molar MassMolar Mass

Coefficients

Limiting Reactant, Theoretical Yield, and Percent Yield

Limiting ReactantThe reactant that is completely consumed in a chemical reaction

Theoretical YieldThe amount of product that can be made in a chemical reaction based on the amount of limiting

reactant

Actual YieldThe amount of product actually produced by a chemical reaction

Actual YieldPercent Yield = --------------------------------- x 100%

Theoretical Yield

Consider the Reaction:

Ti (s) + 2 Cl2 (g) -------> TiCl4 (s)

If we begin the reaction with 1.8 mol of titanium and 3.2 mol of chlorine, what is the limiting reactant and theoretical yield of TiCl4 in moles?

If only 1.5 mol of TiCl4 is actually isolated from the reaction, what is the percent yield for the process?

Actual Yield 1.5Percent Yield = --------------------------------- x 100% = -------- x 100% = 94% Theoretical Yield 1.6

(theoretical yield)

Consider the Reaction:

2 Al (s) + 3 Cl2 (g) -------> 2AlCl3 (s)

If we begin the reaction with 0.552 mol of aluminum and 0.887 mol of chlorine, what is the limiting reactant and theoretical yield of AlCl3 in moles?

(theoretical yield)

If only 0.448 mol of AlCl3 is actually isolated from the reaction, what is the percent yield for the process?

Actual Yield .448Percent Yield = --------------------------------- x 100% = -------- x 100% = 81.1% Theoretical Yield .552

Consider the Reaction:

2 NO (g) + 5 H2 (g) -------> 2 NH3 (g) + 2 H2O (g)

What is the maximum amount of ammonia in grams that can be synthesized from 45.8 g of NO and 12.4 g of H2?

(theoretical yield)

If only 16.4 g of NH3 is actually isolated from the reaction, what is the percent yield for the process?

Actual Yield 16.4Percent Yield = --------------------------------- x 100% = -------- x 100% = 63.1% Theoretical Yield 26.0

Consider the Reaction:

2 Na (s) + Cl2 (g) -------> 2 NaCl (s)

If we begin the reaction with 53.2 g of sodium and 65.8 g of chlorine, what is the limiting reactant and theoretical yield of NaCl in grams?

(theoretical yield)

If only 86.4 g of NaCl is actually isolated from the reaction, what is the percent yield for the process?

Actual Yield 86.4Percent Yield = --------------------------------- x 100% = -------- x 100% = 80.0% Theoretical Yield 108

Consider the Reaction:

2 Fe (s) + 3 S (l) -------> Fe2S3 (s)

When 10.4 g of Fe are allowed to react with 11.8 g of S, 14.2 g of Fe2S3 are obtained. Find the limiting reactant, theoretical yield and percent yield?

Actual Yield 14.2Percent Yield = --------------------------------- x 100% = -------- x 100% = 73.2% Theoretical Yield 19.4

(theoretical yield)

Consider the Reaction:

Cu2O (s) + C (g) -------> 2 Cu (s) + CO (g)

When 11.5 g of C are allowed to react with 114.5 g of Cu2O, 87.4 g of Cu are obtained. Find the limiting reactant, theoretical yield and percent yield?

(theoretical yield)

Actual Yield 87.4Percent Yield = --------------------------------- x 100% = -------- x 100% = 85.9% Theoretical Yield 101.7

EMPIRICAL FORMULA

A chemical formula that indicates the relative proportions of the elements in a molecule rather than

the actual number of atoms of the elements.(An empirical formula may be obtained from percentage

composition of elements in a compound.)

MOLECULAR FORMULA

A chemical formula that indicates the actual number of atoms of the elements in a molecule.

(Information in addition to percentage composition of elements is needed to determine a molecular formula.)

Molecular Formulas May Differ from Empirical Formulas

BenzeneEmpirical Formula, CH

Molecular Formula, C6H6

AcetyleneEmpirical Formula, CH

Molecular Formula, C2H2

Molecular Formulas May Differ from Empirical Formulas

GlucoseEmpirical Formula, CH2O

Molecular Formula, C6H12O6

FructoseEmpirical Formula, CH2O

Molecular Formula, C6H12O6

Solutions of Emprical Formula Problems