Ratios 10/27/2011
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Stoichiometry: of Combination
8.1 Chemical Equations
8.2 Combustion Analysis
8.3 Calculations with Balanced Chemical Equations
8.4 Limiting Reactants
8.5 Periodic Trends in Reactivity of the Main Group Elements
Chemical Equations
A chemical equation uses chemical symbols to denote what occurs in a
chemical reaction.
Ammonia and hydrogen chloride react to produce ammonium chloride.
Balancing Chemical Equations
Chemical equations must be balanced so that the law of conservation of
mass is obeyed.
NH3 + HCl NH4Cl
reactants
products
Chemical formulas indicate elements or compounds that are reacting or
being formed
Labels in parentheses indicate the physical state
(s)
(l)
(g)
(aq)
Balancing is achieved by writing stoichiometric coefficients to the
left of the chemical formulas.
solid
liquid
gas
aqueous (dissolved in water)
NH3 (g) + HCl (g) NH4Cl (s)
SO3 (g) + H2O (l) H2SO4 (aq)
Balancing Chemical Equations
1. Change the coefficients of compounds before changing
the coefficients of elements.
2. Treat polyatomic ions that appear on both sides of the
equation as units.
3. Count atoms and/or polyatomic ions carefully, and track
their numbers each time you change a coefficient.
Example
Write the balanced chemical equation that represents the
combustion of propane.
Step 1: Write the unbalanced equation:
C3H8(g) + O2(g) CO2(g) + H2O(l)
Step 2: Leaving O2 until the end, balance each of the atoms:
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)
Step 3: Double check to make sure there are equal numbers of each type on
atom on both sides of the equation.
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Examples
O2(g) + 2 H(g) H2H2O(g)
O (g) + H2 2(g) 2 O(g)
SiO2(s) + C(s) iCiC(s) + O(g) (g)
3 C(s) S S (s) + C 2 CO
C8H18(l) + O2(g) CO2(g) + H2O(g)
C8H18+ O2 CO2 + H2O
C8H18+ O2 8 CO2 + H2O
C8H18+ O2 8 CO2 + 9 H2O
C8H18+ 25/2 O2 8 CO2 + 9 H2O
2 C8H18+ 25 O2 16 CO2 + 18 H2O
Examples
O2(g) + 2 H2(g) 2 H2O(g)
SiO2(s) + 3 C(s) SiC(s) + 2 CO(g)
2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g)
2 (s) + 3 H2 4(aq) A Al2(SO (aq) + 3 (g)
AlAl(s) +H2SOSO4(aq) l2(SO4)34)3(aq) + H2H2(g)
2 Al + 6 HCl A 2 AlCl3( + + (g)
Al(s)(s) +HCl(aq()aq) lCl3(aq) aq) H23 H2(g)
H |C |O
18 8
2
21
3
18 8
2
2 8 17
18 8
2
18 8 25
18 8 25
18 8 25
36 16 50
36 16 50
Patterns of Chemical Reactivity
Three of the most commonly encountered reaction types
are combination, decomposition, and combustion.
Combination two or more reactants combine to form a single product
NH3(g) + HCl(g) NH4Cl(s)
Decomposition two or more products form from a single reactant
CaCO3 CaO + CO2
Combustion a substance burns in the presence of oxygen.
Combustion of a compound that contains C and H (or C, H, and O)
produces carbon dioxide gas and water.
CH2O(l) + O2(g) CO2(g) + H2O(l)
Combustion Analysis
Combustion Analysis
The experimental determination of an empirical formula is
carried out by combustion analysis.
In the combustion of 18.8 g of glucose, 27.6 g of CO2 and
11.3 g of H2O are produced. Calculate the empirical
formula of glucose.
It is possible to determine the mass of carbon and hydrogen in the
original sample as follows:
mass of C = 27.6 CO2
mass of H = 11.3 H2 O
1 CO2
1 C
12.01 C
= 7.53 C
44.01 CO2 1 CO2
1 C
1 H2 O
2 H
1.008 H
= 1.26 H
18.01 H2 O 1 H2 O
1 H
The remaining mass is oxygen:
18.8 glucose 7.53 C + 1.26 H = 10.0 O
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Combustion Analysis
Combustion Analysis
It is now possible to calculate the empirical formula.
The molar mass of glucose is 180 g/mol. What is its
molecular formula?
Step 1: Determine the number of moles of each element.
C = 7.53 C
1 C
= 0.627 C
12.01 C
1 H
= 1.25 H
1.008 H
Empirical formula of glucose: CH2O
1 O
= 0.626 O
16.00 H
Empirical formula mass = 12.01
H = 1.26 H
O = 7.53 O
Step 2: Write the empirical formula and divide by the smallest
subscript to find the whole number ratio.
C0.627H1.25O0.626 simplifies to CH2O
0.626
To determine the molecular formula, divide the molar mass by the
empirical formula mass.
+ 2 1.008
+ 16.00
30
Molar mass: 180 g/mol
Molar mass/Empirical mass: 180/30 = 6
Molecular formula = [CH2O] x 6 = C6H12O6
0.626 0.626
Calculations with Balanced Chemical
Equations
Calculations with Balanced Chemical
Equations
Making Pancakes: Relationships
Making Pancakes: Relationships
Similar to a recipe
5 Pancakes
Numerical relationship
1 cup of flour
2 eggs
tsp baking powder
2 Eggs 5 Pancakes
Two eggs are required to
produce 5 pancakes.
8 eggs are required to produce
20 pancakes
1 cup Flour 5 Pancakes
tsp Baking Powder 5 Pancakes
1 cup Flour tsp Baking Powder
If any of your ingredients runs out in proportion to the rest
you cant make any more pancakes
Calculations with Balanced Chemical
Equations
A balanced chemical equation is like a recipe: atoms must combine in
precise mole ratios.
The recipe (mole ratios) are given by the coefficients of the balanced
chemical equation. The ratio of coefficients is called stoichiometry
Calculations with Balanced Chemical
Equations
If we have 6 moles of O2 and more than enough CO,
how much CO2 can we make?
2 CO (g) + O2(g) 2 CO2 (g)
Given: 6 mol O2
Find: mol CO2
Conversion Factors: 1 mol O2 = 2 mol CO2
6 O2
2 moles of CO combine with 1 mole of O2 to produce 2 moles of CO2.
2 CO2
= 12 CO2
O2
2 moles of CO is stoichiometrically equivalent to 2 moles of CO2.
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Calculations with Balanced Chemical
Equations
Calculations with Balanced Chemical
Equations
Consider the complete reaction of 3.82 moles of CO to form
CO2. Calculate the number of moles of CO2 produced.
Mass of Reactant to Mass of Product
Mass A Moles A Moles B Mass B
Moles A
Mass A
Equivalence from
balanced
chemical equation
CO2 produced = 3.82 CO
2 CO2
= 3.82 CO2
2 CO
Making Molecules: Mass to Mass
Calculate the mass of CO2 emitted upon the combustion
of 5.0 x 102 grams of octane (C8H18).
mass
C8H18
moles
C8H18
moles
CO2
mass
CO2
2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g)
5.0 102 C8 H18
Molar mass B
Moles B
More Pancakes: Limiting Reactant,
Theoretical Yield, and Percent Yield
Our recipe:
1 cup of Flour + 2 Eggs + tsp Baking Powder 5 pancakes
3 cups of flour, 10 eggs, and 4 tsp baking powder
3 cups of flour 15 pancakes
10 eggs 25 pancakes
4 tsp baking powder 40 pancakes
You can only make 15 pancakes
1 C8 H18
= 4.4 C8 H18
114.2 C8 H18
4.4 C8 H18
35 CO2
Mass B
16 CO2
= 35 CO2
2 C8 H18
44.01 CO2
= 1.5 103 CO2
1 CO2
Note: Do not actually round until the final step
More Pancakes: Limiting Reactant, Theoretical
Yield, and Percent Yield
1 cup of Flour + 2 Eggs + tsp Baking Powder 5 pancakes
More Pancakes: Limiting Reactant, Theoretical
Yield, and Percent Yield
Summary
Theoretical Yield
Limiting Reactant
Actual Yield
Excess reactants
15 pancakes Most that can be produced
Reactant Completely consumed in Reaction
What was actually produced 11 pancakes
Burned a pancake
Dropped a pancake
Spilled some batter
Dog ate one
Reactants present in quantities greater than necessary to react with the
quantity of the limiting reactant.
Theoretical Yield
Amount of product that can be produced based on Limiting Reactant
Actual Yield
Percent Yield
Percent of the theoretical yield actually attained
11
= 100 = 15 100 = 73%
Amount of product actually produced in the reaction
Percent Yield
Actual Yield
x 100 = % Yield
Theoretical Yield
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Limiting Reactants
Example
Consider the reaction between 5 moles of CO and 8 moles of H2 to
produce methanol.
CO(g) + 2H2(g) CH3OH(l)
How many moles of H2 are necessary in order for all the CO to react?
moles of H2 = 5 mol CO
2 mol H2
1 mol CO
= 10 mol H 2
If we begin with 1.8 moles of titanium and 3.2 moles of
chlorine, what is the limiting reactant in moles of TiCl4?
Ti(s) + 2 Cl2(g) TiCl4(s)
Given: 1.8 moles Ti and 3.2 moles Cl2
Find: Limiting reactant
Conversions: 1 mole Ti 1 mole TiCl4
2 mole Cl2 1 mole TiCl4
1 TiCl4
= 1.8 TiCl4
1 Ti
1 TiCl4
3.2 Cl2
= 1.6 TiCl4
2 Cl2
How many moles of CO are necessary in order for all of the H2 to react?
moles CO of = 8 mol H 2
1 mol CO
2 mol H2
= 4 mol CO
10 moles of H2 required; 8 moles of H2 available; limiting reactant.
1.8 Ti
Limiting
Reactant
Cl2
Least Amount
of Product
4 moles of CO required; 5 moles of CO available; excess reactant.
The 3.2 mol of Cl2 will run out before the 1.8 mol of Ti is used up, and hence
Cl2 is the limiting reactant, and will produce 1.6 mol of TiCl4. Therefore, the
theoretical yield is 1.6 mol of TiCl4.
Example
Example
In the following synthesis reaction:
2 Na(s) + Cl2(g) 2 NaCl(s)
If we have 53.2 g of Na and 65.8 g of Cl2, what is the limiting reactant?
In the following synthesis reaction:
2 Na(s) + Cl2(g) 2 NaCl(s)
If we have 53.2 g of Na and 65.8 g of Cl2, what is the limiting reactant?
Given: 53.2 g Na (atomic mass 22.99 g/mol)
65.8 g Cl2 (molar mass 70.91 g/mol)
Find: Limiting Reactant
mass Na
moles NaCl
mass NaCl
1 mol Na
2 mol NaCl
58 .44 g NaCl
22 .99 g Na
2 moles Na 2 moles NaCl
1 mole Cl2 2 moles NaCl
Conversion:
moles Na
2 mol Na
1 mol NaCl
mass Cl2
moles Cl2
moles NaCl
mass NaCl
1 mol Cl 2
58 .44 g NaCl
1 mol Cl 2
1 mol NaCl
Example
In the following synthesis reaction:
2 Na(s) + Cl2(g) 2 NaCl(s)
If we have 53.2 g of Na and 65.8 g of Cl2, what is the limiting reactant?
53.2 Na
= 2.31 Na
22.99 /
2 NaCl
= 2.31 NaCl
2 Na
65.8 Cl2
= 0.928 Cl2
70.91 /
2.31 Na
Limiting
Reactant
Cl2
2 mol NaCl
70 .90 g Cl 2
Example
Smallest
amount
determine the
limiting reactant
0.928 Cl2
Least Amount
2 NaCl
= 1.86 NaCl
of Product
1 Cl2
Add Another Step Percent Yield
In the following synthesis reaction:
2 Na(s) + Cl2(g) 2 NaCl(s)
If we have 53.2 g of Na and 65.8 g of Cl2, what is the limiting reactant?
What is the Percent Yield if the actual yield of NaCl was 86.4 grams?
Given: 53.2 grams Na
65.8 grams Cl2
Find: Limiting Reactant, Theoretical Yield, and Percent Yield (you need
theoretical yield plus the actual yield to get the percent yield.)
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Example
Example
10.4 g iron metal reacts with 11.8 g liquid sulfur to produce 14.2 g iron(III)
sulfide. Find the limiting reactant, theoretical yield and percent yield.
Iron metal: Fe(s), 10.4 g
Liquid sulfur: S(l), 11.8 g
Iron(III) sulfide: Fe2S3(s), 14.2 g
Theoretical Yield is 108 g of NaCl (least amount of product)
Percent Yield =
Actual Yield
86.4
100 =
100 = 80.0%
Theoretical Yield
108
Step 1: W rite the balanced chemical equation:
2 Fe(s) + 3 S(l) Fe2S3(s)
Step 2: Convert to moles:
Atomic mass Fe = 55.85 g/mol
Atomic mass S = 32.06 g/mol
Molar mass Fe2S3 = (2 x 55.85 g/mol + 3 x 32.06 g/mol) = 207.9 g/mol
Example
10.4 g iron metal reacts with 11.8 g liquid sulfur to produce 14.2 g iron(III)
sulfide. Find the limiting reactant, theoretical yield and percent yield.
The limiting reactant is Fe(s)
The theoretical yield is 19.4 g Fe2S3(s)
The actual yield is 14.2 g Fe2S3(s)
14.2
= 100 = 19.4 100 = 73.2 %
Example
Example
Lakes that have been acidified by acid rain can be neutralized by limestone
(CaCO3). How much limestone in kg is required to neutralize a lake with
volume 5.2x109 L that contains 5.0x10-3 g of H2SO4 per L?
Aspirin can be synthesized by reacting acetic anhydride (C4H6O3) with salicylic
acid (C7H6O3) to form aspirin (C9H8O4) and acetic acid (C2H4O2) according to
the reaction:
C4H6O3 + C7H6O3 C9H8O4 + C2H4O2
Molar mass CaCO3 = 100.09 g/mol
Molar mass H2SO4 = 98.09 g/mol
If 5.00 mL of acetic anhydride (density=1.08 g/mL) is reacted with 2.08 g of
salicylic acid, 2.01 g of aspirin is produced. What is the limiting reactant,
theoretical yield and percent yield for the reaction?
Balanced chemical equation:
H2SO4(aq) + CaCO3(s) CaSO4(aq) + H2O(l) + CO2(g)
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mol H 2 SO4 5.2 109 L
5.0 10 g H 2 SO4
L
1 mol H 2 SO4
98.09 g H 2 SO4
Molar mass C7H6O3 = 138.13 g/mol
2.65 105 mol H 2 SO4
1 mol CaCO3 100.09 g CaCO3 1 kg
kg CaCO3 2.65 10 mol H 2 SO4
1 mol H 2 SO4
1 mol CaCO3
1000 g
5
2.7 10 4 kg CaCO3
Molar mass C4H6O3 = 102.10 g/mol
ANS: 2.7104 kg
Molar mass C9H8O4 = 180.17 g/mol
5.00 mL C4 H 6O3
1.08 g C4 H 6O3
mL C4 H 6O3
2.08 g C7 H 6O3
1 mol C4 H 6O3
1 mol C9 H 8O4
102.10 g C4 H 6O3 1 mol C4 H 6O3
1 mol C7 H 6O3
1 mol C9 H 8O4
138.13 g C7 H 6O3 1 mol C7 H 6O3
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Example
Practice
Aspirin can be synthesized by reacting acetic anhydride (C4H6O3) with salicylic
acid (C7H6O3) to form aspirin (C9H8O4) and acetic acid (C2H4O2) according to
the reaction:
How much salicylic acid, in an excess of acetic anhydride, would be required to
produce 10.0 g aspirin if the percent yield was 74.1%?
C4H6O3 + C7H6O3 C9H8O4 + C2H4O2
C4H6O3 + C7H6O3 C9H8O4 + C2H4O2
If 5.00 mL of acetic anhydride (density=1.08 g/mL) is reacted with 2.08 g of
salicylic acid, 2.01 g of aspirin is produced. What is the limiting reactant,
theoretical yield and percent yield for the reaction?
Molar mass C4H6O3 = 102.10 g/mol
Molar mass C7H6O3 = 138.13 g/mol
Thus, if we want to produce 10.0 g of aspirin, it will require more salicylic acid
(1/0.741 = 1.350 times more).
Molar mass C9H8O4 = 180.17 g/mol
mol C9 H 8O4
180.17 g C9 H 8O4
1 mol C9 H 8O4
Percent Yield
According to the equation, 1 mol of acetic anhydride is required to produce 1
mol of aspirin (the theoretical yield). But if the percent yield is only 74.1%, then
the amount of aspirin that is actually produced is only 0.741 times the
theoretical yield.
2.08 g C9 H 8O4
2.71 g C9 H 8O4
2.71 g C9 H 8O4
ANS: 10.3 g of salicylic acid
100 74.1%
ANS: salicylic acid, 2.71 g, 74.1 %
Periodic Trends in Chemical Properties of the
Main Group Elements
Ionization Energy and Electron Affinity enable us to understand types
of reactions that elements undergo and the types of compounds
formed.
General Trends in Reactivity
Hydrogen (1s1)
Grouped by itself
Forms a cation with a +1 charge (H+)
Forms an anion with a -1 charge (H-)
Hydrides react with water to produce hydrogen gas and a base.
CaH2(s) + H2O(l) Ca(OH)2(aq) + H2(g)
Reactions of the Active Metals
Group 1A Elements
(ns1,
Reactions of the Active Metals
n 2)
Group 2A Elements (ns2, n 2)
Low IE
Less reactive than 1A
Never found in nature in pure elemental state
Some react with H2O to produce H2
React with oxygen to form metal oxides
Some react with acid to produce H2
sodium
Alkali metals
reacting with
water
potassium
cesium
Barium reacting
with water
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Reactions of Other Main Group Elements
Group 3A elements (ns2np1, n 2)
Reactions of Other Main Group Elements
Group 4A elements (ns2np2, n 2)
Metalloid (B) and metals (all others)
Nonmetal (C); metalloids (Si, Ge) and others metals
Al forms Al2O3 with oxygen
Form +2 and +4 oxidation states
Al forms +3 ions in acid
Sn, Pb react with acid to produce H2
Others form +1 and +3
Finely-divided aluminum sprinkled
into a flame to form Al2O3
Reactions of Other Main Group Elements
Group 5A elements
(ns2np3,
n 2)
Nonmetal (N2, P), metalloid (As, Sb), and metal (Bi)
Reactions of Other Main Group Elements
Group 6A elements (ns2np4, n 2)
Nonmetals (O, S, Se)
Nitrogen, N2, forms variety of oxides
Oxygen, O2
Phosphorus, P4
Sulfur, S8
As, Sb, Bi (crystalline)
HNO3 and H3PO4 important industrially
Selenium, Se8
Metalloids (Te, Po)
Te, Po (crystalline)
SO2, SO3, H2S, H2SO4
Forest damaged by acid rain
Nonmetal oxides added to water
produce an acid
Reactions of Other Main Group Elements
Reactions of Other Main Group Elements
Group 7A elements (ns2np5, n 2)
Group 8A elements (ns2np6, n 2)
All diatomic
All monatomic
Do not exist in elemental form in nature
Filled valence shells
Form ionic salts
Considered inert until 1963 when Xe and Kr were used to
form compounds
Form molecular compounds
with each other
No major commercial use
React with hydrogen to form
hydrogen halides
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Comparison of Group 1A and Group 1B
Elements
Have single valence electron
Properties differ
Group 1B much less reactive than 1A
High IE of 1B - incomplete shielding of nucleus by inner d outer s electron of 1B strongly attracted to nucleus
1B metals often found elemental in nature (coinage metals)
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