Chapter 4 Chemical Equations &

Stoichiometry

Chemical reactions are described by an equations that states which molecules have undergone a reaction (reactants) to produce molecules (products) under given reaction conditions

H2CO3

Heat CO2 + H2O

Reactant ProductsReaction Conditions

(aq) (g) (l)

Balancing Chemical Reactions

1) Recall the law of conservation of mass

Mass is always conserved

2) Recall Dalton’s Atomic Theory

Mass of Reactants = Mass of products

Reactions cause atoms to be recombined

Total # of each element in reactants = Total # of each element in products

The equality in mass and elemental composition between reactants and products is referred to as “Balance”

How to balance a chemical equation

1. Write the equation using one molecule of each type.

2. Check that each molecule has the right formula.

3. Balance one element at a time.

4. Balance polyatomic ions as a group.

5. Balance large molecules before small molecules.

6. Balance oxygen last, or any element common to several molecules

7. Ensure that the coefficients are the smallest possible whole numbers (“simplest whole number ratio”).

8. Add states of matter. - gas, liquid, aqueous, solid

Simple example: Formation of Rust

Fe + O2 Fe2O3

Balancing Fe:

Fe + Fe2O3O2

1 2+ Fe2O3O2Fe2

Balancing O:

+ Fe2O3O2Fe2

2 3

+

3/2

3 3Whole number factors:

Fe2O3O2Fe4 3 2

Fe2O3O2Fe2 +

2 2

+ Fe2O3 (s)O2 (g)Fe (s)4 3 2

Adding states of matter

Ex) Combustion of gasoline (Octane)

C8H18 + O2 CO2 + H2O

Balance the C’sC8H18 + O2 CO2 + H2O

8 1C8H18 + O2 CO2 + H2O8

8 8Balance the H’s

C8H18 + O2 CO2 + H2O8

18

2C8H18 + O2 CO2 + H2O8 9

18

18Balance O’s

C8H18 + O2 CO2 + H2O8 92 16 9

C8H18 + O2 CO2 + H2O8 925/2

25 25

C8H18 + O2 CO2 + H2O8 925/2

Ex) Combustion of gasoline (Octane)

Whole number factor

O2C8H18 + CO2 + H2O16 18252

States of matter

25C8H18(l) + CO2(g) + H2O(l)16 182 O2(g)

Double check the balance

Reactants

16 C

36 H

50 O

Products

16 C

36 H

50 O

Addition and Subtraction of Chemical Equations

Chemical equations can be added (or subtracted). Generally, this is done to describe the overall result of a multi-step reaction.

Fe2O3 + 3 C → 2 Fe + 3 CO

Fe2O3 + 3 CO → 2 Fe + 3 CO2

2 Fe2O3 + 3 C → 4 Fe + 3 CO2

2 Fe2O3 + 3 C + 3 CO → 4 Fe + 3 CO2 + 3 COXX

Multiplication of Chemical Equations by a Scalar

Chemical equations can be multiplied (or divided) by any number.

Every coefficient gets multiplied (or divided) by this number.

Fe2O3 + 3 C → 2 Fe + 3 CO6 ×

6 Fe2O3 + 18 C → 12 Fe + 18 CO

Fe2O3 + 3 C → 2 Fe + 3 CO1/2 ×

1/2 Fe2O3 + 3/2 C → Fe + 3/2 CO

StoichiometryIn chemistry it is often necessary to relate the quantity of molecules produced or consumed in a reaction.

Analysis using the molar ratios is called stoichiometry.

By the Law of Conservation of Mass, the Dalton Atomic Theory:

1) Chemical reactions recombine the elements but the total number of each remains the same.

2) If the total amount of each element is known, the proportion of all molecules can be predicted through the balanced chemical equation.

Ex) NO2 + CO → NO + CO2 Is balanced

If 10 NO2 are consumed:

10 CO are consumed 10 NO are produced

10 CO2 are produced

Ex) In a catalytic converter NO and CO are converted to N2, CO2 and O2

If 50.0 grams of NO is converted how many grams of N2 are produced?

NO + CO → N2 + CO2 + O2 Balance equation

4 NO (g) + 2 CO (g) → 2 N2 (g) + 2 CO2 (g) + O2(g)

4 molecules of NO produces 2 N2 molecule and 2 CO2 molecule

4 moles of NO produces 2 mole of N2 and 2 mole of CO2

Stoichiometry

2 NO + CO → 1 N2 + CO2 + O2

2 NO + 1 CO → 1 N2 + 1 CO2 + O2

N’s

C’s

2 NO + 1 CO → 1N2 + 1 CO2 + ½ O2 O’s

4 NO + 2 CO→ 2 N2 + 2 CO2 + O2(g) Whole number factors

States

Mass N2 = (moles N2)(molar mass N2)

= (moles N2)(1/2 mol NO/mol N2)(molar mass N2)

= 1/2 (moles NO) (molar mass N2)

= 1/2 (mass NO/molar mass NO) (molar mass N2)

= 1/2 (50.0 g/(14.01 + 15.99 g/mol)) (28.02 g/mol)

= 1/2 (50.0 g/(14.01 + 15.99 g/mol)) (28.02 g/mol)

Stoichoimetric ratio

= 23.4 g

Stoichiometry

4 NO (g) + 2 CO (g) → 2 N2 (g) + 2 CO2 (g) + O2(g)

4 moles NO produces 2 moles of N2

moles N2 = ½ moles NO ½ mol NO/mol N2 = 1

Limiting ReactantsReactants are almost never mixed in the correct ratios for a given reaction.

It is therefore desirable to compute how much product is expected based on a given amount of reactants that are combined

There will always be one reagent which runs out first limiting the amount of product produced, called the limiting reagent.

Ex) If 12.0 grams of NO and 16.0 grams of CO are combined.

Which is the limiting reagent? How much CO2 is produced?

4 NO (g) + 2 CO (g) → 2 N2 (g) + 2 CO2 (g) +1 O2 (g)

Recall the balanced equation:

2 moles of NO are consumed for every mole of CO

To determine which is the limiting reagent we compare (moles NO)/2 to moles CO

Moles NO = (mass NO)/(molar mass NO) = (12.0 g)/(14.01 + 15.99 g/mol)

= 0.400 mol NO

Therefore 0.400/2 moles = 0.200 moles of CO can be consumed

Limiting Regents

Moles CO = (mass CO)/(molar mass CO) = (16.0 g)/(12.01 + 15.99 g/mol)

= 0.571 mol

This implies that CO is in excess.

NO determines how much CO2 is produced.

For every 2 moles of NO consumed one mole of CO2 is produced

Since 0.400 mol of NO is consumed 0.200 mol CO2 is produced

Limiting Regents

How much CO is left?

For every 2 moles of NO consumed 1 mole CO is consumed.

0.40 mol NO was consumed. 0.20 mol CO was consumed.

Recall that there was 0.571 mol CO 0.371 mol CO remains.

Mass CO = (moles CO)(molar mass CO) = (0.371 moles)( 30.00 g/mol)

= 11.1 g CO remains

Mass CO2 = (moles CO2)(molar mass CO2)

= (0.200 mol)(12.01 + 2*15.99 g/mol)

= 8.80 g CO2

Percent YieldPercent yield is calculated by comparing the theoretical yield - the maximum possible amount of product - with the actual yield - how much product was actually obtained.

Theoretical yield can be computed using mass as the mass of product can also be predicted based on the mass of a reactant as a limiting reagent.

The comparison is made in moles of reactant to product using the correct stoichiometric relationship

% Yield = (100%)(Actual Yield/Theoretical Yield)

% Yield = (100%)(Actual moles/Theoretical moles)

% Yield = (100%)(Actual mass/Theoretical mass)

= (moles CO) (molar mass CO2)

Percent Yield

= (mass CO/molar mass CO) (molar mass CO2)

= (20.0 g/(12.01 + 15.99 g)) (12.01 + 2*15.99 g)

= (20.0 g/(28.00 g)) (43.99 g)

= 31.4 g CO2

% Yield CO2 = (100 %)(Actual Yield/Theoretical Yield)

= (100 %)(22.1 g/31.4 g)

= 70.4 %

Ex) If 20.0 grams of CO was used with NO in excess, and 22.1 grams of CO2 was produced. Compute the % Yield.

Mass CO2 = (moles CO2)(molar mass CO2)

= (moles CO2) (moles CO/moles CO2)(molar mass CO2)

= (moles CO) (molar mass CO2)

Quantitative AnalysisQuantitative analysis is the quantification of a substance through a chemical reaction where one chemical species involved is determined.

Ex) TitrationsA known quantity of a reagent, K, is added to an unknown quantity of another substance, U to be quantified.

By determining how much reagent is added the quantity of the unknown substance can be determined.

KU + Products

When (b/a)*(moles U) of K is added the reaction is complete - equivalence.

b Ka U +

Products Balanced

Completeness of reaction is usually indicated by some observation, i.e. a color change

Therefore when # moles of HCl added = # moles of NaOH present the equivalence condition is reached.

Mass of NaOH = (moles of NaOH)(molecular mass of NaOH)

= (moles of NaOH)(1 mole HCl/mole NaOH)(molecular mass of NaOH)

= (moles of HCl)(molecular mass of NaOH)

Mass of NaOH = (0.0123 mol)(22.99 + 15.99 + 1.01 g/mol)

Moles HCl = (volume HCl solution added)(concentration of HCl solution)

= (101.5 ml)(1/1000 l/ml)(0.121 mol/l)

= 0.0123 mol

= (0.0123 mol)(39.99 g/mol)

= 0.492 g

Ex) A unknown amount of sodium hydroxide is used to prepare a solution. By adding 101.5 ml of a 0.121 mol/l HCl solution the basic solution was neutralized. Compute the weight of NaOH present.

NaOH (aq) + HCl (aq) → H20 (l) + NaCl (aq)

1 Mole of HCl consumes 1 mole of NaOH

Example3.19 g of an organic compound with an unknown empirical formula upon complete combustion with excess oxygen gas, produced 8.79 g carbon dioxide and 1.44 g water. Determine the empirical formula.

Combustion AnalysisThe empirical formula of an organic compound can be determined by weighing the total amount of water and carbon dioxide produced

C xH yO z (s or l) + O2 (g) (excess) → x CO2 + y/2 H2O (l)

From the total mass and by determining the number of moles of CO2 and H2O we can relate then stoichiometrically to C and H to determine x, y and z.

Compute moles of CO2 and H2O

Moles CO2 = mass/molar mass Moles H2O = mass/molar mass

= (8.790 g)/(43.99 g/mol) = (1.440 g)/(18.01 g/mol)

= 0.200 mol = 0.0800 mol

Combustion AnalysisC xH yO z (s or l) + O2 (g) (excess) → x CO2 + y/2 H2O (l)

There a twice as many H’s in the molecule as water produced

The number of C’s in the molecules equals the number of CO2 produced

Therefore there are 0.200 mol of C’s in the molecule and 0.160 mol H’s

To determine he amount of O, we need to remove the contribution of C and H from the total mass.

Mass C = (0.200 mol)*(12.01 g/mol) = 2.40 g

Mass H = (0.160 mol)*(1.01 g/mol) = 0.162 g

Mass O = Total mass – mass C – mass H

Moles O = (0.63 g)/ (15.99 g/mol) = 0.039 mol

= 3.19 - 2.40- 0.162 g = 0.63 g

0.200 mol C : 0.160 mol H : 0.040 O

5 C : 4 H’s: 1 O 5.08 C’s :4.07 H’s : 1 0

Combustion AnalysisMolar ratio:

Closest whole numbers

Formula weight = 5*12.01 + 4*1.01 + 15.99 = 80.08 g/mol

Empirical formula = C5H4O

Moles of Empirical unit = (3.19 g)/ (80.08 g/mol) = 0.040 mol

Determine % Composition by weight

Recall that the molcule’s total weight can be broken down into 2.40 g from C, 0.162 g from H and 0.63 g from O

% C = (100 %)(2.40 g)/(3.19 g) = 75.2 %

% H = (100 %)(0.16 g)/(3.19 g) = 5.0%

% O = (100 %)(0.63 g)/(3.19 g) = 19.7 %

Concepts from Chapter 4

Balancing chemical equations

Stoichiometric calculations

Limiting reactants

Percent yield

Quantitative analysis