© 2012 Pearson Education, Inc. Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations Ashley Warren Kings High School Lecture Presentation.

© 2012 Pearson Education, Inc.

Chapter 3

Stoichiometry:Calculations with

Chemical Formulas and Equations

Ashley Warren Kings High School

Lecture Presentation


Stoichiometry

• The area of study that examines the quantities of substances consumed and produced in chemical reactions.

• Stoicheion means “element” and metron means “measure.”

Stoichiometry


Law of Conservation of Mass

“We may lay it down as an incontestable axiom that, in all the operations of art and nature, nothing is created; an equal amount of matter exists both before and after the experiment. Upon this principle, the whole art of performing chemical experiments depends.”

--Antoine Lavoisier, 1789

Stoichiometry


Anatomy of a Chemical Equation

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

Reactants appear on the left side of the equation.

Products appear on the right side.

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

Coefficients are inserted to balance the equation.

Trumbull Rogers

AU: The ACS Style Guide and other standard reference works say these "states of matter" should be set on line--change OK thruout chapters?

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.

Stoichiometry

Interpreting and Balancing Chemical Equations

• The following diagram represents a chemical reaction in which the red spheres are oxygen atoms and the blue spheres are nitrogen atoms. – Write the chemical formulas for the reactants and products. – Write a balanced chemical equation for the reaction. – Is the diagram consistent with the Law of conservation of mass?


Stoichiometry

Interpreting and Balancing Chemical Equations

• In the diagram above, the white spheres represent hydrogen atoms and the blue spheres represent nitrogen atoms.

• To be consistent with the law of conservation of mass, how many NH3 molecules should be shown in the right (product) box?


Stoichiometry


Reaction Types

Stoichiometry


Combination Reactions

• Examples:– 2Mg(s) + O2(g) 2MgO(s)

– N2(g) + 3H2(g) 2NH3(g)

– C3H6(g) + Br2(l) C3H6Br2(l)

• In combination reactions two or more substances react to form one product.

Stoichiometry


• In a decomposition reaction one substance breaks down into two or more substances.

Decomposition Reactions

• Examples:– CaCO3(s) CaO(s) + CO2(g)

– 2KClO3(s) 2KCl(s) + O2(g)

– 2NaN3(s) 2Na(s) + 3N2(g)

Stoichiometry


Combustion Reactions

• Examples:– CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

– C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

• Combustion reactions are generally rapid reactions that produce a flame.

• Combustion reactions most often involve hydrocarbons reacting with oxygen in the air.

Stoichiometry

More Reactions


Stoichiometry

More Reactions


Stoichiometry

Final Reactions

Stoichiometry


Formula Weights

Stoichiometry


Formula Weight (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(40.08 amu)

+ Cl: 2(35.453 amu)

110.99 amu

• Formula weights are generally reported for ionic compounds.

• What is the formula weight for sulfuric acid?

Stoichiometry


Molecular Weight (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(12.011 amu)

30.070 amu+ H: 6(1.00794 amu)

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 weight)

(FW of the compound)x 100

Stoichiometry


Percent Composition

So the percentage of carbon in ethane (C2H6) is

%C =(2)(12.011 amu)

(30.070 amu)

24.022 amu

30.070 amu= x 100

= 79.887%

Stoichiometry


Moles

Stoichiometry


Avogadro’s Number

• 1 mole =

6.02 x 1023,

symbolized NA.

• 1 mole of 12C has a mass of 12.000 g.

Abbreviation = mol

Stoichiometry

Avogadro’s Number

• See exercise 3.7 and 3.8


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).

*”The atomic weight of an element in atomic mass units is numerically equal to the mass in grams of 1 mol of that element.”

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.

Stoichiometry

Molar Mass

• Which has more mass, a mole of water or a mole of glucose?

• Which contains more molecules, a mole of water or a mole of glucose?

• Calculate the molar mass of glucose.

• Calculate the molar mass of calcium nitrate.


Stoichiometry


Using Moles

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

Let’s think about units in all of these cases! If you know the units you will ALWAYS be able to figure out what to do!

Stoichiometry

Masses and Moles

• Calculate the number of moles of water in 4.389 grams of water.

• Calculate the mass, in grams, of 1.293 mol of potassium chlorite.

• How many molecules are in 10.33 grams of C3H8?


Stoichiometry


Finding Empirical Formulas

Stoichiometry


Calculating Empirical Formulas

One can calculate the empirical formula from the percent composition.

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 mol12.01 g

1 mol14.01 g

1 mol1.01 g

1 mol16.00 g

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 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

Molecular Formulas from Empirical

• Mesitylene, a hydrocarbon found in crude oil, has an empirical formula of C3H4, and an experimentally determined molecular weight of 121 amu. What is its molecular formula?

• The formula weight of C3H4 = 40.0 amu

• Whole number multiple = molecular weight

empirical formula weight

= 121 = 3.02

40

This means the molecular formula is triple the empirical.

Therefore the molecular formula is C9H12.


Stoichiometry


Combustion Analysis

• Compounds containing C, H, and O are routinely analyzed through combustion in a chamber like the one shown in Figure 3.14.– 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.

***See Exercise 3.15

Stoichiometry


Stoichiometric Calculations

The coefficients in the balanced equation give the ratio of moles of reactants and products as well as the relative numbers of molecules.

Stoichiometry



Starting with the mass of Substance A, you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant).

Stoichiometry



Starting with 1.00 g of C6H12O6… we calculate the moles of C6H12O6…(using molar mass)use the coefficients to find the moles of H2O…(using mole to mole ratio)and then turn the moles of water to grams…(using

molar mass again)

C6H12O6 + 6 O2 6 CO2 + 6 H2O

Stoichiometry


• If given the following reaction, how many grams of O2 can be prepared from 4.50 grams of KClO3?

• 2KClO3 (s) 2KCl (s) + 3 O2 (g)


Stoichiometry


Limiting Reactants

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).

– 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):

Percent yield = x 100actual yield

theoretical yield

Stoichiometry

Limiting Reactant/Theoretical Yield Calculations

• See Exercise 3.18, 3.19, and 3.20


Stoichiometry

Chapter 3 Homework Problems

• 1, 3, 5, 6, 8, 10, 12, 13, 15, 17, 19, 22, 23, 25, 27, 30, 31, 33, 35, 37, 40, 41, 43, 46, 47, 49, 52, 53, 55, 61, 63, 65, 67, 69, 71, 76, 77, 80, 81, 83, 87, 89a,b, 97, 101


© 2012 Pearson Education, Inc. Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations Ashley Warren Kings High School Lecture Presentation.

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© 2012 Pearson Education, Inc. Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations Ashley Warren Kings High School Lecture Presentation.