chem 1410 moles and eqs

7/28/2019 Chem 1410 Moles and Eqs

1/39

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 10-24g H

Avogadros Number


2/39


3/39

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 Avogadros number of C atoms?

g

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


4/39


5/39

C. Definition of a mole

1. Amole 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. Amole 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?


6/39

How Large a Quantity is Avogadros Number??


7/39


8/39

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)


9/39


10/39

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


11/39

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 Ca

X 6.02 x 1023

atoms

1.00 mol Ca= 1.81 x 1024 atoms

of Ca


12/39

1.0 x 106 molecules CO

6.02 x 1023 molecules CO

1.00 mole CO

X

=1.66 x 10-18 mol CO

How many moles are represented by

1,000,000 molecules of carbon monoxide (CO) ?

MOLES

OFSUBSTANCE

PARTICLES

OFSUBSTANCE

Avogadros Number


13/39


14/39


15/39

ANOTHER PAIR OF

"CONVERSION FACTORS"

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

1 mole of CO2or

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

44.01 g of CO2

How many grams are contained in4.25 moles of carbon dioxide (CO2) ?

4.25 moles CO 2 X

44.01 g of CO2

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

= 187 g CO2


16/39

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

MOLES

OF

SUBSTANCE

GRAMS

OF

SUBSTANCE

Molar Mass


17/39


18/39


19/39

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

7.67 g Co X6.02 x 1023 atoms

1.00 mol Co

X1 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 CO2X = 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


20/39

MOLES

OF

SUBSTANCE

GRAMS

OF

SUBSTANCE

PARTICLES

OF

SUBSTANCE


21/39


22/39


23/39


24/39

Information Available from A Balanced Equation


25/39

Conversion Factors Relevant to Stoichiometry

Use Avogadros

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


26/39

Solution Map for Stoichiometry

Moles of

A

Moles of

B

Grams ofA

Grams ofB

Particles of

A

Particles of

B

Avogadros Number Avogadros Number

Molar MassMolar Mass

Coefficients


27/39


28/39

Limiting Reactant, Theoretical Yield, and Percent Yield

Limiting Reactant

The 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 Yield

Percent Yield = --------------------------------- x 100%Theoretical Yield


29/39

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, whatis 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 1.5

Percent Yield = --------------------------------- x 100% = -------- x 100% = 94%

Theoretical Yield 1.6

(theoretical yield)


30/39


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

If we begin the reaction with 0.552 mol of aluminum and 0.887 mol ofchlorine, 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 .448

Percent Yield = --------------------------------- x 100% = -------- x 100% = 81.1%

Theoretical Yield .552


31/39


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 16.4

Percent Yield = --------------------------------- x 100% = -------- x 100% = 63.1%



32/39


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 thelimiting 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 86.4

Percent Yield = --------------------------------- x 100% = -------- x 100% = 80.0%

Theoretical Yield 108



33/39


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 14.2

Percent Yield = --------------------------------- x 100% = -------- x 100% = 73.2%


(theoretical yield)

id h i


34/39


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 Cuare obtained. Find the limiting reactant, theoretical yield and percent yield?

(theoretical yield)

Actual Yield 87.4

Percent Yield = --------------------------------- x 100% = -------- x 100% = 85.9%



35/39

EMPIRICAL FORMULA

A chemical formula that indicates the relative

proportions of the elements in a molecule rather thanthe actual number of atoms of the elements.

(An empirical formula may be obtained from percentagecomposition of elements in a compound.)

MOLECULAR FORMULA

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

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


36/39

Molecular FormulasMay Differ from Empirical Formulas

Benzene

Empirical Formula, CH

Molecular Formula, C6H6

Acetylene

Empirical Formula, CH

Molecular Formula, C2H2


37/39

Molecular FormulasMay Differ from Empirical Formulas

Glucose

Empirical Formula, CH2OMolecular Formula, C6H12O6

Fructose

Empirical Formula, CH2OMolecular Formula, C6H12O6


38/39


39/39

Solutions of Emprical Formula Problems

chem 1410 moles and eqs

Documents