Chapter 3

Stoichiometry:Calculations with

Chemical Formulas and Equations

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

3.1 Chemical Equations

Chemical equations are concise representations of chemical reactions.

2 molecule of hydrogen + 1 molecule of oxygen 2 molecule of water

2 moles of hydrogen + 1 mole of oxygen 2moles of water

Stoichiometry

Chemical Equations

Stoichiometry

Anatomy of a Chemical Equation

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

Coefficients are inserted to balance the equation.

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

3.2 Some Simple Patterns of Chemical Reactivity

• 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.

Combination Reactions

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 reactionsare generally rapid reactions that produce a flame.

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

Stoichiometry

3.3 Formula Weights (FW)

• A formula weight is the sum of the atomic weights for the atoms in a chemical formula.

(often called formula mass or molecular mass)

• 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.

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

So the percentage of carbon in ethane is

%C =(2)(12.011 amu)

(30.070 amu)

24.022 amu

30.070 amu= x 100

= 79.887%

Percent Composition

Stoichiometry

3.4 Avogadro’s Number & The Mole

• 6.02 x 1023

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

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).

Stoichiometry

Using Moles

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

Stoichiometry

• 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

• Sample Problem

How many molecules of carbon dioxide

are in a 3.61g sample?

232

22

222

6.022 10 molecules CO3.61 g CO

44.01 g CO

4.94 10 molecules CO

How many moles of HCl are in 102g?

1 mol HCl102 g HCl 2.80 mol HCl

36.46mol HCl

Stoichiometry

Sample Problem

How many moles of gold are there

in 8.62×1020 atoms of gold?

2023

3

1mol Au8.62 10 atomsAu

6.022 10 atoms Au

1.43 10 mol Au

Stoichiometry

One can calculate the empirical formula from the percent composition.

3.5 Empirical Formulas from Analyses

Ex) 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

Calculating Empirical Formulas

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

These are the subscripts for the empirical formula:

C7H7NO2

Stoichiometry

From Empirical formula to Molecular Formula

The empirical formula of a carbohydrate is CH2O. If the molar mass is 180 g/mol, what is the molecular formula?

2 6 12 6n

The empirical formula mass is:

12.0 2 1.0 16.0 30.0 g/mol

Dividing the empirical formula mass by

the molar mass:

180 g/moln = 6

30.0 g/mol

CH O C H O

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.

Stoichiometry

3.6 Quantitative Information from Balanced Equations : Stoichiometric Calculations

The coefficients in the balanced equation give the ratio of moles of reactants and products.

Stoichiometry

1. Write the formulas for the reactants on the left side and the formulas for the products on the right side of the equation.

Propane reacts with oxygen to form carbon dioxide and water

2. Adjust the numbers in front of the compounds until the number of atoms of each element is the same on both sides of the equation.

C3H8 + O2 CO2 + H2O

Balancing Equations

Stoichiometry

Balancing Equations

C3H8 + O2 CO2 + H2O3 C 2 O 1 C 2 H

8 H 2 O 1 O

1) Balance C:C3H8 + O2 3CO2 + H2O

2) Balance H:C3H8 + O2 3CO2 + 4H2O

Stoichiometry

Balancing Equations

3) Balance O:

C3H8 + 5O2 3CO2 + 4H2O

4) Check if equation is balanced:

3C 10O 3C 8H

8H 6O 4O

Stoichiometry

Stoichiometric Calculations

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

Working with Equations

How many grams of water will be produced by the combustion of 122g of propane with excess oxygen?

C3H8(g) + 2O2(g) 3CO2(g) + 4H2O(l)

3 8 2 2

3 8 23 8 3 8 2

1 mol C H 4 mol H O 18.0g H O 122g C H 199g H O

44.1g C H 1 mol C H 1 mol H O

1 mol 4 mol122 g

X mol

?? g

4X mol

Stoichiometry

Stoichiometric Calculations

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

v

3.7 Limiting Reactants

Stoichiometry

• The limiting reactant is the reactant used up first in a reaction.

• The excess reactant is the reactant present is greater amount than is necessary for the reaction.

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).

Stoichiometry

Limiting Reactants

In the example below, the O2 would be the excess reagent.

Stoichiometry

Sample Problem

If 12.0g of NO are reactant with 10.0g of O2, how much NO2 will be formed? Which is the limiting reactant?

2NO(g) + O2(g) 2NO2(g)

22

2

2 22 2

2 2

2 mol NO1 mol NO 46.0g12.0g NO 18.4g NO

30.0g 2 mol NO 1 mol NO

1 mol O 2 mol NO 46.0g10.0g O 28.8g NO

32.0g 1 mol O 1 mol NO

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.

Percent yield = X 100actual yield

theoretical yield

Stoichiometry

Sample Problem

The theoretical yield of NO2 based on NO was 18.4g. If 13.7g were obtained from the reaction, what is the percentage yield?

actual yield%yield = 100%

theoretical yield

13.7g= 100%

18.4g

= 74.5%

Stoichiometry

Homework

• 필수숙제: 14, 38, 54, 66, 82, 104, 112

• 추가연습

12, 20, 34, 50, 52, 56, 62, 68, 76, 88