Stoichiometry © 2009, Prentice-Hall, Inc. Chapter 3 Stoichiometry Definition: Mathematical...

Stoichiometry

© 2009, Prentice-Hall, Inc.

Chapter 3Stoichiometry

Definition: Mathematical calculations for chemical formulas & equations.

Mrs. Deborah Amuso

Camden High School

Camden, NY 13316

Chemistry, The Central Science, 11th editionTheodore L. Brown, H. Eugene LeMay, Jr.,

and Bruce E. Bursten

Stoichiometry


Chemical EquationsChemical equations are concise representations of chemical reactions.

(example)

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

Stoichiometry


Anatomy of a Chemical Equation

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

Reactants: Products: appear on the appear on the left side of the right side of the equation equation

The states are written in parentheses.

Coefficients are inserted to balance the equation.

Stoichiometry


Subscripts vs Coefficients

• Subscripts tell the number of atoms of each element in a molecule.

H2O (The 2 means 2 atoms of H)

• Coefficients tell the number of molecules.

3 H2O (The 3 means 3 molecules of water)

Stoichiometry

Counting Atoms

Total # atoms = Coefficient x Subscript

Practice:

3 H2SO4

6-H 3-S 12-O

or 3 – SO4

2 Ca3(PO4)2

6-Ca 4-P 16-O

or 4 – PO4© 2009, Prentice-Hall, Inc.

Stoichiometry

Balancing Equations

When balancing equations use coefficients to get the # of each type of atom on both sides of the equation the same.

Consult your teacher for balancing tips.


Stoichiometry

ReactionTypes

1. Combination

2. Decomposition

3. Combustion

4. Single Replacement

5. Double Replacement


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 a combination reaction two or more substances react to form one product.

Stoichiometry


• In a decomposition reaction one reactant 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)

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)

• In a combustion reaction a carbon compound reacts with oxygen in the air to produce carbon dioxide & water.

• These are generally rapid reactions that produce a flame.

Stoichiometry

Single Replacement Reactions

In a single replacement reaction, an element & compound react to form a different element & compound.


Examples: Zn (s) + CuSO4 (aq) Cu(s) + H2SO4 (aq)

2 HCl (aq) + Mg(s) MgCl2 (aq) + H2(g)

Stoichiometry

Double Replacement Reactions

In a double replacement reaction, two compounds in solution react by exchanging ions to form two different compounds.


Examples:

ZnCl2 (aq) + 2 AgNO3 (aq) 2AgCl(s) + Zn(NO3)2(aq)

Ba(NO3)2(aq) + Na2SO4(aq) BaSO4(s) + 2 NaNO3 (aq)

Stoichiometry


Formula Mass (FM)

• A formula mass is the sum of the atomic masses for the atoms in a chemical formula.

• So, the formula mass of calcium chloride, CaCl2, would be

Ca: 1(40.1 amu) = 40.1

+ Cl: 2(35.5 amu) = 71.0

111.1 amu

• Formula masses are generally reported for ionic compounds.

Stoichiometry


Molecular Mass (MM)

• A molecular mass is the sum of the atomic masses for the atoms in a molecule.

• For the molecule ethane, C2H6, the molecular mass would be

• FM & MM are found the same way!

C: 2(12.0 amu) = 24.0

30.0 amu+ H: 6(1.0 amu) = 6.0

Stoichiometry


Mass Percent or Percent Composition

The mass percent of each element in a compound is found by using this equation:

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

(FM of the compound)x 100

Or simply: % = Part/Whole x 100

Stoichiometry


Percent Composition

So the percentage of carbon in ethane, C2H6, is…

%C =(2)(12.0 amu)

(30.0 amu)

24.0 amu

30.0 amu= x 100

= 80.0%

Stoichiometry


Moles & Avogadro’s Number

• 1 mole = 6.02 x 1023

• The mass of 1 molecule of water is 18.0 amu.• The mass of 1 mole of water is 18.0 grams.• Why? 1 gram = 6.02 x 1023 amu.

Stoichiometry


Molar Mass or GFM

• Molar mass is the mass of 1 mole of a substance in g/mol.– The molar mass of an element is the mass

number that we find on the periodic table.

example: Cu = 63.5 O2 = 2(16.0) = 32

– The molar mass of a compound is the same number as its formula mass or molecular mass.

example: C2H6 = 30.0

Stoichiometry


Using Moles

Moles provide a bridge from the molecular scale to the real-world scale.

# moles = given mass

GFM

GFM is the same as Molar Mass (It’s the mass in grams of 1 mole of the substance)

Stoichiometry

Three Helpful EquationsUse these 3 equations to convert between masses,

moles, molecules (or units) , & atoms (or ions):

#moles = mass/GFM

#molecules = (#moles) (6.02 x 1023)

(or # units)

#atoms = (#molecules) (#atoms/molecule)

(or # ions)


Stoichiometry

Mole Relationships:Check Your Understanding

How many atoms of oxygen are in 10.0 grams of CH3COOH?

Think: Mass Moles Molecules Atoms

#moles = mass/GFM =

10.0g / 60.0 g/mol = 0.167 moles

#molecules = (#moles)((6.02 x 1023) =

(0.167 moles) (6.02 x 1023) =

1.00 x 1023 molecules

#atoms = (#molecules) (#atoms/molecule) =

(1.00 x 1023 molecules)(2 oxygen atoms/molecule) =

2.00 x 1023 oxygen atoms


Stoichiometry

Mole Relationships:Check Your UnderstandingHow many hydroxide ions are in 10.0 grams of Ca(OH)2

Think: Mass Moles Units Ions

#moles = mass/GFM =

10.0g / 74.1g/mol = 0.135 moles Ca(OH)2

#units = (#moles) (6.02 x 1023) =

(0.135 mol) (6.02 x 1023) =

8.12 x 1022

#ions = (#units) (#ions/unit) =

(8.12 x 1022) (2 OH-/unit) =

1.62 x 1023 OH- ions


Stoichiometry


Calculating Empirical Formulas

One can calculate the empirical formula from the percent composition of each element in the formula.

Steps:

1) Assume percent of each element = its mass

2) Divide mass of each element by that element’s atomic mass to get # moles

3) Calculate the mole ratio by dividing by the smallest # of moles

4) These are the #’s in the empirical formula

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.

C: 61.31 g / 12.00 g/mol = 5.105 mol C

H: 5.14 g / 1.01 g/mol = 5.09 mol H

N: 10.21 g / 14.01 g/mol = 0.7288 mol N

O: 23.33 g / 16.00 g/mol = 1.456 mol O

Stoichiometry


Calculating Empirical FormulasCalculate 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 mol0.7288 mol

5.09 mol0.7288 mol

0.7288 mol0.7288 mol

1.458 mol0.7288 mol

Stoichiometry



These are the subscripts for the empirical formula:

C7H7NO2

Stoichiometry



Find the empirical formula of a compound whose mass percentage is 50.0% sulfur and 50.0% oxygen.

Step 1S: 50.0 g / 32.1 g/mol = 1.56 mol S

O: 50.0 g / 16.0 g/mol = 3.13 mol O

Step 2

S: 1.56 / 1.56 = 1

O: 3.13 / 1.56 = 2

Step 3 Empirical Formula is SO2

Stoichiometry

Calculating Molecular FormulasFrom Empirical Formulas & Molar Mass

What is the MF if the EF is C4H5N2O and the MM is 194 g/mol?

Step 1: Determine molar mass of empirical formula

C4H5N2O = 97 g/mol

Step 2: Divide:

Molar Mass given/Molar Mass of empirical formula

194 / 97 = 2

Step 3: Multiply:

Answer from Step 2 * Empirical Formula

(2)(C4H5N2O) = C8H10N4O2 © 2009, Prentice-Hall, Inc.

Stoichiometry

Calculating Molecular FormulasFrom Empirical Formulas & Molar Mass

What is the MF if the EF is CH2 and the molecular mass is

84.16 amu?

Step 1: Determine molecular mass of empirical formula

CH2 = 14.03 amu

Step 2: Divide:

Molecular mass given/molecular mass of emp formula

84.16 amu /14.03 amu = 6

Step 3: Multiply:

Answer from Step 2 * Empirical Formula

(6)(CH2) = C6H12 © 2009, Prentice-Hall, Inc.

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.

– Then you can determine the Empirical Formula of the compound

Stoichiometry

Combustion AnalysisWhen 0.255 g of caproic acid is burned, you get 0.512 g CO2 and 0.209 g

H2O. What is the emp formula of the acid?

#mole CO2 = mass/GFM = 0.512g / 44.0g/mol = 0.0116 mol

0.0116 mol CO2 * (1 mole C/1 mol CO2) = 0.0116 mol C

0.0116 mol C * (12.0 g/mol) = 0.140 g C

#mole H2O = mass/GFM = 0.209g / 18.0g/mol = 0.0116 mol

0.0116 mol H2O * (2 mole H/1 mol H2O)= 0.0232 mol H

0.0232 mol H * (1.01 g/mol) = 0.0235 g H

Mass O = Mass caproic acid – (mass C + mass H)

Mass O = 0.225 g – (0.140 g + 0.0235 g) = .0615 g


Stoichiometry

2nd Part of Problem

For carbon: 0.140 g C / 12.0 g/mol = 0.0117 mol

For hydrogen: 0.0235 g H / 1.01 g/mol = 0.0233 mol

For oxygen: 0.0615 g O / 16.0 g/mol = 0.00384 mol

For carbon: 0.0117 / 0.00384 = 3.05

For hydrogen: 0.0233 / 0.00384 = 6.07

For oxygen: 0 0.00384 / 0.00384 = 1

Empirical Formula = C3H6O


Stoichiometry


Stoichiometric Calculations

The coefficients in the balanced equation represent the # of molecules or # of moles of reactants and products.

The coefficients tell the ‘recipe’ for the rxn.

Stoichiometry



Stoichiometry



Starting with 1.00 g of C6H12O6… we calculate the moles of C6H12O6…use the coefficients to find the moles of H2O…and then turn the moles of water to grams.

C6H12O6 + 6 O2 6 CO2 + 6 H2O

Stoichiometry

Stoichiometric Calculation Steps1) Complete & balance equation using coefficients.

2) Identify Given & Unknown.

3) Convert Given mass into moles:

#molesgiven = massgiven/GFMgiven

4) Use mole ratio to find moles of Unknown.

#molesgiven = #molesunk

Coeffgiven Coeffunk

5) Convert moles of ‘unknown’ to answer.

massunknown = (#molesunk) (GFMunk)


Stoichiometry

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

How many grams of water are produced when 32.0 grams of methane completely react?

1) Given = 32.0 g CH4 Unk = ? g H2O

2) Find # moles CH4

#moles CH4 = mass/GFM = 32.0 g / 16.0 g/mol = 2.00 moles CH4

3) Find # moles H2O

2.00 moles CH4 = ? moles H2O

1 mole 2 mole Ans. = 4.00 moles H2O

4) Find mass of H2O

mass H2O = (#moles)(GFM) = (4.00 mol)(18.0 g/mol) =

Ans. = 72.0 g H2O


Stoichiometry

2 H2 (g) + O2 (g) 2 H2O (g)

How many grams of water are produced when 6.0 grams of oxygen gas completely react?

1) Given = 6.0 g O2 Unk = ? g H2O

2) Find # moles O2

#moles O2 = mass/GFM = 6.0 g / 32.0 g/mol = 0.19 moles O2

3) Find # moles H2O

0.19 moles O2 = ? moles H2O

1 mole 2 mole Ans. = 0.38 moles H2O

4) Find mass of H2O

mass H2O = (#moles)(GFM) = (0.38 mol)(18.0 g/mol) =

Ans. = 6.8 g H2O


Stoichiometry


Limiting Reactants:How Many Cookies Can I Make?

• You can make cookies until you run out of one of the ingredients.

• Once you run out of sugar, you can’t make any more cookies.

• 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 (or limiting reagent) is the reactant you’ll run out of first

• What is the LR in this reaction?

H2

Stoichiometry

How to Determine Which Reactant is the Limiting Reactant

Step 1: Convert the masses given to moles.

Step 2: Use these equations:

# moles reactant A / coefficient A = x

# moles reactant B / coefficent B = y

Step 3: The smaller # (x or y) identifies the

Limiting Reagent.

Step 4: Use the # of moles of the Limiting

Reagent to solve the rest of the

problem.


Stoichiometry

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

Given 50.0 g of CH4 and 60.0 g of O2

Which is the Limiting Reagent?

Step 1:

#moles CH4 = mass/GFM = 50.0 g/16.0 g/mol = 3.13 moles

#moles O2 = mass/GFM = 60.0 g/32.0 g/mol = 1.88 moles


# moles reactant CH4 / coefficient CH4= 3.13 / 1 = 3.12

# moles reactant O2 / coefficent O2 = 1.88 / 2 = 0.94

Step 3: The smaller # (x or y) identifies the Limiting Reagent

0.94 is smaller, so O2 is the L.R.


Stoichiometry

2 CHCl3 (g) + 2 Cl2 (g) 2 CClO4 (g) + 2 HCl (g)

Given 15.9 g of CHCl3 and 12.6 g of Cl2

Which is the Limiting Reagent?

Step 1:

#moles CHCl3 = mass/GFM = 15.9 g/119.5 g/mol = 0.133 mol

#moles Cl2 = mass/GFM = 12.6 g/71.0 g/mol = 0.177 mol


# mol reactant CHCl3 / coefficient CHCl3= 0.133/ 2 = 0.0665

# mol reactant Cl2 / coefficent Cl2 = 0.177/ 2 = 0.0885

Step 3: The smaller # (x or y) identifies the Limiting Reagent

0.0665 is smaller, so CHCl3 is the L.R.


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 YieldTheoretical YieldPercent Yield = x 100

Stoichiometry

Percent Yield Problem

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

Previously, you determined 72.0 g of water should be produced from 32.0 g of methane. Assume only 68.0 g of water are produced when this experiment is actually conducted. What is the percent yield?

% yield = [actual yield/theoretical yield] x 100

% yield = [68.0g / 72.0 g] x 100

% yield = 94.4%


Stoichiometry © 2009, Prentice-Hall, Inc. Chapter 3 Stoichiometry Definition: Mathematical...

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