Stoichiometry: Calculations with Chemical Formulas …holcombslab.yolasite.com/resources/WCU...

6/10/2012

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Stoichiometry

© 2009, Prentice-Hall, Inc.

Chapter 3

Stoichiometry:

Calculations with Chemical

Formulas and Equations

John D. Bookstaver St. Charles Community College

Cottleville, MO

Chemistry, The Central Science, 11th edition

Theodore L. Brown, H. Eugene LeMay, Jr.,

and Bruce E. Bursten

Stoichiometry


Chapter 3 Problems

• Problems 11, 13, 19, 21, 23, 33, 45, 51,

59, 73

Stoichiometry


Chemical Equations

Depict the kind of reactants and

products and their relative amounts

in a reaction.

4 Al(s) + 3 O2(g) ---> 2 Al2O3(s)

The numbers in the front are called

stoichiometric coefficients The letters (s), (g), and (l) are the

physical states of compounds.

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Stoichiometry


Reaction of Iron with Cl2

Notice the stoichiometric coefficients and the

physical states of the reactants and products.

Stoichiometry


Chemical Equations 4 Al(s) + 3 O2(g)

---> 2 Al2O3(s)

This equation means

4 Al atoms + 3 O2 molecules

---give--->

2 “molecules” of Al2O3

4 moles of Al + 3 moles of O2

---give--->

2 moles of Al2O3

Stoichiometry


Chemical Equations • Because the same atoms are present in a

reaction at the beginning and at the end, the

amount of matter in a system does not

change.

• The Law of the Conservation of Matter

2HgO(s) ---> 2 Hg(liq) + O2(g)

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Stoichiometry


Because of the principle of the

conservation of matter,

an equation must be balanced.

It must have the same

number of atoms of the

same kind on both sides.

Chemical Equations

Lavoisier, 1788

Stoichiometry


Because of the principle of the

conservation of matter,

an equation must be balanced.

It must have the same

number of atoms of the

same kind on both sides.

Chemical Equations

Lavoisier, 1788

Stoichiometry


Anatomy of a Chemical Equation

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

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Stoichiometry



Reactants appear on the left

side of the equation.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Stoichiometry



Products appear on the

right side of the equation.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Stoichiometry



The states of the reactants and products are written in parentheses to the right of each compound.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

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Stoichiometry



Coefficients are inserted

to balance the equation.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Stoichiometry


Subscripts and Coefficients Give

Different Information

• Subscripts tell the number of atoms of

each element in a molecule.

Stoichiometry


Subscripts and Coefficients Give

Different Information

• Subscripts tell the number of atoms of

each element in a molecule

• Coefficients tell the number of

molecules.

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Stoichiometry


Chemical Reaction

Nitrogen monoxide + oxygen → nitrogen dioxide

Step 1: Write the reaction using chemical symbols.

NO + O2 → NO2

Step 2: Balance the chemical equation.

2 1 2

Stoichiometry


Writing and Balancing

Chemical Equations

• Write a word equation.

• Convert word equation into formula

equation.

• Balance the formula equation by the

use of prefixes (coefficients) to balance

the number of each type of atom on the

reactant and product sides of the

equation.

Stoichiometry


Example

Hydrogen gas reacts with oxygen gas to

produce water.

Step 1.

hydrogen + oxygen -----> water

Step 2.

H2 + O2 -----> H2O

Step 3.

2 H2 + O2 -----> 2 H2O

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Stoichiometry


Balancing Equation Strategy

• Balance elements that occur in only one

compound on each side first.

• Balance free elements last.

• Balance unchanged polyatomics as

groups.

• Fractional coefficients are acceptable

and can be cleared at the end by

multiplication.

Stoichiometry


Example

Iron(III) oxide reacts with carbon monoxide to produce the iron oxide (Fe3O4) and carbon dioxide.

iron(III) oxide + carbon monoxide -----> Fe3O4 + carbon

dioxide

Fe2O3 + CO -----> Fe3O4 + CO2

3 Fe2O3 + CO -----> 2 Fe3O4 + CO2

Stoichiometry


Balancing Equations

____C3H8(g) + _____ O2(g) ---->

_____CO2(g) + _____ H2O(g)

____B4H10(g) + _____ O2(g) ---->

___ B2O3(g) + _____ H2O(g)

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Stoichiometry


Reaction

Types

Stoichiometry


Combination Reactions

• Examples:

– 2 Mg (s) + O2 (g) 2 MgO (s)

– N2 (g) + 3 H2 (g) 2 NH3 (g)

– C3H6 (g) + Br2 (l) C3H6Br2 (l)

• In this type of

reaction two

or more

substances

react to form

one product.

Stoichiometry


• In a decomposition

one substance breaks

down into two or more

substances.

Decomposition Reactions

• Examples:

– CaCO3 (s) CaO (s) + CO2 (g)

– 2 KClO3 (s) 2 KCl (s) + O2 (g)

– 2 NaN3 (s) 2 Na (s) + 3 N2 (g)

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Stoichiometry


Combustion Reactions

• Examples:

– CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

– C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (g)

• These are generally

rapid reactions that

produce a flame.

• Most often involve

hydrocarbons

reacting with oxygen

in the air.

Stoichiometry


Writing and Balancing an Equation: The Combustion of a

Carbon-Hydrogen-Oxygen Compound.

Liquid triethylene glycol, C6H14O4, is used a a solvent and

plasticizer for vinyl and polyurethane plastics. Write a

balanced chemical equation for its complete combustion.

Stoichiometry


15 2

6 7 C6H14O4 + O2 → CO2 + H2O 6

2. Balance H.

2 C6H14O4 + 15 O2 → 12 CO2 + 14 H2O

4. Multiply by two

Example 4-2

3. Balance O.

and check all elements.

Chemical Equation:

1. Balance C.

6 7

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Stoichiometry


Formula Weight/Molecular

Weight

Stoichiometry


Formula Weight (Mass) (FW)

• A formula weight is the sum of the

atomic weights for the atoms in a

chemical formula.

• So, the formula weight of calcium

chloride, CaCl2, would be Ca: 1(40.1 amu)

+ Cl: 2(35.5 amu)

111.1 amu

• Formula weights are generally reported

for ionic compounds.

Stoichiometry


Molecular Weight (Mass) (MW)

• A molecular weight is the sum of the

atomic weights of the atoms in a

molecule.

• For the molecule ethane, C2H6, the

molecular weight would be

C: 2(12.0 amu)

30.0 amu

+ H: 6(1.0 amu)

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Stoichiometry


Molecular Weight (Mass)

HOO

H

H

HO

H

OH

OHHH

OH

Molecular formula C6H12O6

Empirical formula CH2O

Glucose

6 x 12.01 + 12 x 1.01 + 6 x 16.00

Molecular Mass: Use the naturally occurring mixture of isotopes,

= 180.18

Stoichiometry


Percent Composition

One can find the percentage of the mass

of a compound that comes from each of

the elements in the compound by using

this equation:

% element = (number of atoms)(atomic mass)

(FM of the compound) x 100

Stoichiometry


Percent Composition

So the percentage of carbon in ethane

is…

%C = (2)(12.0 amu)

(30.0 amu)

24.0 amu

30.0 amu = x 100

= 80.0%

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Stoichiometry


Moles

Stoichiometry


Counting Atoms

Mg burns in air (O2) to

produce white magnesium

oxide, MgO.

How can we figure out how

much oxide is produced

from a given mass of Mg?

Stoichiometry


Counting Atoms

Chemistry is a quantitative

science—we need a

“counting unit.”

1 mole is the amount of

substance that contains as

many particles (atoms,

molecules) as there are in

12.0 g of 12C.

MOLE

518 g of Pb, 2.50 mol

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Stoichiometry


Particles in a Mole

6.02214199 x 1023

Avogadro’s Number

There is Avogadro’s number of

particles in a mole of any substance.

Amedeo Avogadro

1776-1856

Stoichiometry


Molar Mass 1 mol of 12C

= 12.00 g of C = 6.022 x 1023 atoms of C

12.00 g of 12C is its

MOLAR MASS

Taking into account all of

the isotopes of C, the

molar mass of C is

12.011 g/mol

Stoichiometry


Molar Mass

• By definition, a molar mass is the mass

of 1 mol of a substance (i.e., g/mol).

– The molar mass of an element is the mass

number for the element that we find on the

periodic table.

– The formula weight (in amu’s) will be the

same number as the molar mass (in

g/mol).

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Stoichiometry


One-mole Amounts

Stoichiometry


Using Moles

Moles provide a bridge from the molecular

scale to the real-world scale.

Stoichiometry


Mole Relationships

• One mole of atoms, ions, or molecules contains

Avogadro’s number of those particles.

• One mole of molecules or formula units contains

Avogadro’s number times the number of atoms or

ions of each element in the compound.

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Stoichiometry


PROBLEM: What amount of

Mg is represented by 0.200 g?

How many atoms?

Mg has a molar mass of 24.3050 g/mol.

0.200 g • 1 mol

24.31 g = 8.23 x 10-3 mol

8.23 x 10-3 mol • 6.022 x 1023 atoms

1 mol

= 4.95 x 1021 atoms Mg

How many atoms in this piece of Mg?

Stoichiometry


Finding

Empirical

Formulas

Stoichiometry


Calculating Empirical Formulas

One can calculate the empirical formula from

the percent composition.

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Stoichiometry


Empirical formula

1. Choose an arbitrary sample size (100g).

2. Convert masses to amounts in moles.

3. Write a formula.

4. Convert formula to small whole numbers.

5. Multiply all subscripts by a small whole

number to make the subscripts integral.

5 Step approach:

Stoichiometry



The compound para-aminobenzoic acid (you may have

seen it listed as PABA on your bottle of sunscreen) is

composed of carbon (61.31%), hydrogen (5.14%),

nitrogen (10.21%), and oxygen (23.33%). Find the

empirical formula of PABA.

Stoichiometry



Assuming 100.00 g of para-aminobenzoic acid,

C: 61.31 g x = 5.105 mol C

H: 5.14 g x = 5.09 mol H

N: 10.21 g x = 0.7288 mol N

O: 23.33 g x = 1.456 mol O

1 mol

12.01 g

1 mol

14.01 g

1 mol

1.01 g

1 mol

16.00 g

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Stoichiometry



Calculate the mole ratio by dividing by the smallest number

of moles:

C: = 7.005 7

H: = 6.984 7

N: = 1.000

O: = 2.001 2

5.105 mol

0.7288 mol

5.09 mol

0.7288 mol

0.7288 mol

0.7288 mol

1.458 mol

0.7288 mol

Stoichiometry



These are the subscripts for the empirical formula:

C7H7NO2

Stoichiometry


Determining the Empirical and Molecular Formulas of a

Compound from Its Mass Percent Composition.

Dibutyl succinate is an insect repellent used against household

ants and roaches. Its composition is 62.58% C, 9.63% H and

27.79% O. Its experimentally determined molecular mass is

230 u. What are the empirical and molecular formulas of

dibutyl succinate?

Step 1: Determine the mass of each element in a 100g sample.

C 62.58 g H 9.63 g O 27.79 g

Dibutyl succinate is an insect repellent used against household

ants and roaches. Its composition is 62.58% C, 9.63% H and

27.79% O. Its experimentally determined molecular mass is

230 u. What are the empirical and molecular formulas of

dibutyl succinate?

Example

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Stoichiometry


Step 2: Convert masses to amounts in moles.

OmolOg

OmolOgn

HmolHg

HmolHgn

CmolCg

CmolCgn

O

H

C

737.1999.15

179.27

55.9008.1

163.9

210.5011.12

158.62

Step 3: Write a tentative formula.

Step 4: Convert to small whole numbers.

C5.21H9.55O1.74

C2.99H5.49O

Example

Stoichiometry


Step 5: Convert to a small whole number ratio.

Multiply 2 to get C5.98H10.98O2

The empirical formula is C6H11O2

Step 6: Determine the molecular formula.

Empirical formula mass is 115 u.

Molecular formula mass is 230 u.

The molecular formula is C12H22O4

Example

Stoichiometry


Combustion Analysis

• Compounds containing C, H and O are routinely

analyzed through combustion in a chamber like this.

– C is determined from the mass of CO2 produced.

– H is determined from the mass of H2O produced.

– O is determined by difference after the C and H have been

determined.

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Stoichiometry


Combustion analysis

Stoichiometry


Menthol, the substance we can smell in mentholated cough drops,

is composed of C, H and O. A 0.1005g sample of menthol is

combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What

is the empirical formula for menthol?

Mass of Carbon

mass CO2 moles CO2 moles C grams C

Empirical Formulas from Analyses

Combustion Analysis

Stoichiometry






Mass of Carbon



Cmol

Cg

COmol

Cmol

COg

COmolg

1

011.12

1

1

011.44

12829.0

22

2

Combustion Analysis

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Stoichiometry






Mass of Carbon



CgCmol

Cg

COmol

Cmol

COg

COmolg 07721.0

1

011.12

1

1

011.44

12829.0

22

2

Combustion Analysis

Stoichiometry






Mass of Hydrogen

mass H2O moles H2O moles H grams H


Combustion Analysis

Stoichiometry






Mass of Hydrogen



OHg

OHmolg

2

2

02.18

11159.0

Combustion Analysis

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Stoichiometry






Mass of Hydrogen



OHmol

Hmol

OHg

OHmolg

22

2

1

2

02.18

11159.0

Combustion Analysis

Stoichiometry






Mass of Hydrogen



Hmol

Hg

OHmol

Hmol

OHg

OHmolg

1

01.1

1

2

02.18

11159.0

22

2

Combustion Analysis

Stoichiometry






Mass of Hydrogen



gHmol

Hg

OHmol

Hmol

OHg

OHmolg 01299.0

1

01.1

1

2

02.18

11159.0

22

2

Combustion Analysis

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Stoichiometry






Mass of Oxygen

mass O = mass of sample – (mass C +mass H)


gggOxygenmass 01299.007721.01005.0

Combustion Analysis

Stoichiometry






Mass of Oxygen

mass O = mass of sample – (mass C +mass H)


g

gggOxygenmass

01030.0

01299.007721.01005.0

Combustion Analysis

Stoichiometry






Now we can determine the empirical formula

Mass of elements:

C 0.07721g

H 0.01299g

O 0.01030g


Combustion Analysis

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Stoichiometry







Moles of elements:

C 0.07721g/12.011g/mol =

H 0.01299g/1.01g/mol =

O 0.01030g/16.00g/mol =


Combustion Analysis

Stoichiometry







Moles of elements:

C 0.07721g/12.011g/mol = 0.006428 mol

H 0.01299g/1.01g/mol = 0.01286 mol

O 0.01030g/16.00g/mol = 0.0006438 mol


Combustion Analysis

Stoichiometry







Moles of elements:

C 0.07721g/12.011g/mol = 0.006428 mol

H 0.01299g/1.01g/mol = 0.01286 mol

O 0.01030g/16.00g/mol = 0.0006438mol


0006438.001286.0006428.0 OHC

Combustion Analysis

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Stoichiometry







Moles of elements:

C 0.07721g/12.011g/mol = 0.006428 mol

H 0.01299g/1.01g/mol = 0.01286 mol

O 0.01030g/16.00g/mol = 0.0006438 mol


0006438.0

0006438.0

0006438.0

01286.0

0006438.0

006428.0 OHC

Combustion Analysis

Stoichiometry







Moles of elements:

C 0.07721g/12.011g/mol = 0.006428 mol

H 0.01299g/1.01g/mol = 0.01286 mol

O 0.01030g/16.00g/mol = 0.0006438 mol


12010 OHC

Combustion Analysis

Stoichiometry


STOICHIOMETRY

- the study of the quantitative aspects of chemical reactions.

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Stoichiometry


Chemical Equations and Stoichiometry

• Stoichiometry includes all the

quantitative relationships involving:

– atomic and formula masses

– chemical formulas.

• Mole ratio is a central conversion factor.

Stoichiometry


Stoichiometric Calculations

The coefficients in the balanced equation give

the ratio of moles of reactants and products.

Stoichiometry


GENERAL PLAN FOR STOICHIOMETRY CALCULATIONS

Mass

reactant

Stoichiometric

factor Moles

reactant

Moles

product

Mass

product

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Stoichiometry


PROBLEM: If 454 g of NH4NO3 decomposes, how much N2O and H2O are formed? What is the theoretical yield of products?

STEP 1

Write the balanced chemical equation

NH4NO3 ---> N2O + 2 H2O

Stoichiometry


454 g of NH4NO3 --> N2O + 2 H2O

STEP 2 Convert mass reactant (454 g) --> moles

454 g • 1 mol

80.04 g = 5.68 mol NH4NO3

STEP 3 Convert moles reactant

(5.68 mol) --> moles product

Stoichiometry


454 g of NH4NO3 --> N2O + 2 H2O

STEP 3 Convert moles reactant --> moles

product

Relate moles NH4NO3 to moles product

expected.

1 mol NH4NO3 --> 2 mol H2O

Express this relation as the

STOICHIOMETRIC FACTOR.

2 mol H2O produced

1 mol NH4NO3 used

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Stoichiometry


454 g of NH4NO3 --> N2O + 2 H2O

= 11.4 mol H2O produced

5.68 mol NH4NO3 • 2 mol H2O produced

1 mol NH4NO3 used

STEP 3 Convert moles reactant (5.68 mol) --> moles product

Stoichiometry


454 g of NH4NO3 --> N2O + 2 H2O

11.4 mol H2O • 18.02 g

1 mol = 204 g H2O

STEP 4 Convert moles product (11.4 mol) --> mass product

Called the THEORETICAL YIELD

ALWAYS FOLLOW THESE STEPS IN SOLVING STOICHIOMETRY PROBLEMS!

Stoichiometry


Limiting

Reactants

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Stoichiometry


How Many Cookies Can I Make?

• You can make cookies

until you run out of one

of the ingredients.

• Once this family runs

out of sugar, they will

stop making cookies

(at least any cookies

you would want to eat).

Stoichiometry


How Many Cookies Can I Make?

• In this example the

sugar would be the

limiting reactant,

because it will limit the

amount of cookies you

can make.

Stoichiometry


Limiting Reactants

• The limiting reactant is the reactant present in

the smallest stoichiometric amount.

– In other words, it’s the reactant you’ll run out of first (in

this case, the H2).

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Stoichiometry


Limiting Reactants

In the example below, the O2 would be the

excess reagent.

Stoichiometry


Theoretical Yield

• The theoretical yield is the maximum

amount of product that can be made.

– In other words it’s the amount of product

possible as calculated through the

stoichiometry problem.

• This is different from the actual yield,

which is the amount one actually

produces and measures.

Stoichiometry


Percent Yield

One finds the percent yield by

comparing the amount actually obtained

(actual yield) to the amount it was

possible to make (theoretical yield).

Actual Yield

Theoretical Yield Percent Yield = x 100

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Stoichiometry


454 g of NH4NO3 --> N2O + 2 H2O

STEP 5 How much N2O is formed?

Total mass of reactants =

total mass of products

454 g NH4NO3 = ___ g N2O + 204 g

H2O

mass of N2O = 250. g

Stoichiometry


454 g of NH4NO3 --> N2O + 2 H2O

STEP 6

Calculate the percent yield

If you isolated only 131 g of N2O, what

is the percent yield?

This compares the theoretical (250.

g) and actual (131 g) yields.

Stoichiometry


454 g of NH4NO3 --> N2O + 2 H2O

% yield = actual yield

theoretical yield • 100%

STEP 6 Calculate the percent yield

% yield = 131 g

250. g • 100% = 52.4%

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Stoichiometry


Determining Limiting Reagent:

Another Example

• The reactant that is completely consumed

determines the quantities of the products

formed.

Stoichiometry


Determining the Limiting Reactant in a Reaction.

Phosphorus trichloride , PCl3, is a commercially important

compound used in the manufacture of pesticides, gasoline

additives, and a number of other products. It is made by the

direct combination of phosphorus and chlorine

P4 (s) + 6 Cl2 (g) → 4 PCl3 (l)

What mass of PCl3 forms in the reaction of 125 g P4 with

323 g Cl2?

Example

Strategy: Compare the actual mole ratio

to the required mole ratio.

Stoichiometry


Example

nCl2 = 323 g Cl2 × = 4.56 mol Cl2

1 mol Cl2

70.91 g Cl2

nP4 = 125 g P4 × = 1.01 mol P4

1 mol P4

123.9 g P4

4.56 mol Cl2 x 4PCl3/6 Cl2 = 3.04 mole PCl3

1.01 mol P4 x 4PCl3/1P4 = 4.04 mole PCl3

Chlorine gas is the limiting reagent.

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