1. 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
Heat
CO2
(g)
+ H2O
(l)
H2CO3
(aq)
Reaction
Conditions
Reactant
Products

2. Balancing Chemical Reactions
1) Recall the law of conservation of mass
Mass is always conserved
Mass of Reactants
= Mass of products
2) Recall Dalton’s Atomic Theory
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”

3. 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

4. Simple example: Formation of RustFe + O2
Fe2O3
Balancing Fe: Fe +
Fe2O3 O2 1 2 2 +
Fe2O3 O2 Fe 2 2
Balancing O: +
Fe2O3 O2 Fe 2 2 3
Fe2O3 O2 Fe 2 +
3/2 3 3
Whole number factors: 4 Fe + 3 O2 2
Fe2O3
Adding states of matter 4
Fe (s) + 3
O2 (g) 2
Fe2O3 (s)

5. Ex) Combustion of gasoline (Octane)
C8H18 + O2
CO2 +
H2O
Balance the C’s
C8H18 + O2
CO2
H2O 8 1
C8H18 + O2
CO2
H2O 8 8 8
Balance the H’s
C8H18 + O2
CO2
H2O 8 18 2
C8H18 + O2
CO2
H2O 8 9 18 18
Balance O’s
C8H18 + O2
CO2
H2O 8 9 2 16 9
C8H18 + O2
CO2
H2O 8 9
25/2 25 25

6. Ex) Combustion of gasoline (Octane)
C8H18 + O2
CO2
H2O 8 9
25/2
Whole number factor O2
C8H18 +
CO2
H2O 16 18 25 2
States of matter 25
C8H18(l) +
CO2(g)
H2O(l) 16 18 2
O2(g)
Double check the balance
Reactants
Products
16 C
16 C
36 H
36 H
50 O
50 O

7. X X 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.
Fe2O C → 2 Fe CO
Fe2O CO → 2 Fe CO2 X X
2 Fe2O C + 3 CO → 4 Fe CO CO
2 Fe2O C → 4 Fe CO2

8. 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.
6 ×
Fe2O C → 2 Fe CO
6 Fe2O C → 12 Fe CO
1/2 ×
Fe2O C → 2 Fe CO
1/2 Fe2O /2 C → Fe /2 CO

9. Stoichiometry
In 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 CO2 are produced
10 CO are consumed
10 NO are produced

10. Stoichiometry NO + CO → N2 + CO2 + O2 2 NO + CO → 1 N2 + CO2 + O2 N’s
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
2 NO + CO → 1 N2 + CO2 + O2
N’s
2 NO CO → 1 N CO2 + O2
C’s
O’s
2 NO CO → 1N CO2 + ½ O2
4 NO + 2 CO→ 2 N CO2 + O2(g)
Whole number factors
4 NO (g) + 2 CO (g) → 2 N2 (g) + 2 CO2 (g) + O2(g)
States
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

11. Stoichiometry 4 NO (g) + 2 CO (g) → 2 N2 (g) + 2 CO2 (g) + O2(g)
Mass N2
= (moles N2)(molar mass N2)
Stoichoimetric ratio
4 moles NO produces 2 moles of N2
moles N2 = ½ moles NO
½ mol NO/mol N2 = 1
= (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/( g/mol)) (28.02 g/mol)
= 1/2 (50.0 g/( g/mol)) (28.02 g/mol)
= 23.4 g

12. Limiting Reactants
Reactants 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?
Recall the balanced equation:
4 NO (g) + 2 CO (g) → 2 N2 (g) + 2 CO2 (g) +1 O2 (g)
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

13. Limiting Regents Moles NO = (mass NO)/(molar mass NO)
= (12.0 g)/( g/mol)
= mol NO
Therefore 0.400/2 moles = moles of CO can be consumed
Moles CO = (mass CO)/(molar mass CO)
= (16.0 g)/( g/mol)
= 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 mol of NO is consumed mol CO2 is produced

14. Limiting Regents How much CO is left?
Mass CO2 = (moles CO2)(molar mass CO2)
= (0.200 mol)( * g/mol)
= 8.80 g CO2
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 mol CO
0.371 mol CO remains.
Mass CO = (moles CO)(molar mass CO)
= (0.371 moles)( g/mol)
= 11.1 g CO remains

15. Percent Yield
Percent yield is calculated by comparing the theoretical yield - the maximum possible amount of product - with the actual yield - how much product was actually obtained.
% Yield = (100%)(Actual Yield/Theoretical Yield)
The comparison is made in moles of reactant to product using the correct stoichiometric relationship
% Yield = (100%)(Actual moles/Theoretical moles)
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.
% Yield = (100%)(Actual mass/Theoretical mass)

16. Percent Yield
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)
= (moles CO) (molar mass CO2)
= (mass CO/molar mass CO) (molar mass CO2)
= (20.0 g/( g)) ( *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 %

17. Quantitative Analysis
Quantitative analysis is the quantification of a substance through a chemical reaction where one chemical species involved is determined.
Ex) Titrations
A known quantity of a reagent, K, is added to an unknown quantity of another substance, U to be quantified. U + K
Products
a U +
b K
Products
Balanced
When (b/a)*(moles U) of K is added the reaction is complete - equivalence.
Completeness of reaction is usually indicated by some observation, i.e. a color change
By determining how much reagent is added the quantity of the unknown substance can be determined.

18. NaOH (aq) + HCl (aq) → H20 (l) + NaCl (aq)
Ex) A unknown amount of sodium hydroxide is used to prepare a solution. By adding ml of a 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
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)
Moles HCl = (volume HCl solution added)(concentration of HCl solution)
= (101.5 ml)(1/1000 l/ml)(0.121 mol/l)
= mol
Mass of NaOH = ( mol)( g/mol)
= ( mol)(39.99 g/mol)
= g

19. Combustion Analysis
The 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.
Example
3.19 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.
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)
= mol
= mol

20. Combustion Analysis
C 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 mol of C’s in the molecule and 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) = g
Mass O = Total mass – mass C – mass H
= g = 0.63 g
Moles O = (0.63 g)/ (15.99 g/mol) = mol

21. Combustion Analysis Determine % Composition by weight Molar ratio:
0.200 mol C : mol H : O
5.08 C’s :4.07 H’s : 1 0
Closest whole numbers
5 C : 4 H’s: 1 O
Empirical formula = C5H4O
Formula weight = 5* * = g/mol
Moles of Empirical unit = (3.19 g)/ (80.08 g/mol) = mol
Determine % Composition by weight
Recall that the molcule’s total weight can be broken down into 2.40 g from C, 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 %

22. Concepts from Chapter 4 Balancing chemical equations
Stoichiometric calculations
Limiting reactants
Percent yield
Quantitative analysis