Download student version

1/9/2015
Chapter 7: Stoichiometry - Mass
Relations in Chemical Reactions
→
→
 How do we balance chemical equations?
 How can we used balanced chemical equations to
relate the quantities of substances consumed and
produced in chemical reactions?
 How can we determine a compound’s elemental
composition and chemical formula?
Chapter Outline








7.1 Chemical reactions and the Conservation of Mass
7.2 Balancing Chemical Equations
7.3 Combustion Reactions
7.4 Stoichiometric Calculations and the Carbon Cyle
7.5 Percent Composition and Empirical Formulas
7.6 Empirical and Molecular Formulas Compared
7.7 Combustion Analysis
7.8 Limiting Reactants and Percent Yield
1
1/9/2015
Law of Conservation of Mass
The law of conservation of mass states that the sum
of the masses of the reactants of a chemical equation
is equal to the sum of the masses of the products.
C(s)
+
O2(g)
→
CO2(g)
Law of Conservation of Mass
2 C(s)
+
O2(g)
→
2 CO(g)
2
1/9/2015
Chemical Equations
2 C(s)
+
O2(g)

→
2 CO(g)
The 2’s are called “stoichiometric coefficients”
The “” symbol means “reaction proceeds in this direction”
a “  “ symbol means the reaction is at equilibrium

(s) = solid phase
(l) = liquid phase
(g) = gas phase
(aq) = aqueous phase
 = heat
Stoichiometric Coefficients:
the ratios between reactants and/or products
4
2
3
1/9/2015
Chapter Outline








7.1 Chemical reactions and the Conservation of Mass
7.2 Balancing Chemical Equations
7.3 Combustion Reactions
7.4 Stoichiometric Calculations and the Carbon Cyle
7.5 Percent Composition and Empirical Formulas
7.6 Empirical and Molecular Formulas Compared
7.7 Combustion Analysis
7.8 Limiting Reactants and Percent Yield
Guidelines for Balancing
Chemical Equations
1. Write an expression using correct chemical formulas for the
reactants and products, separated by an arrow (). Include
symbols indicating physical states.
2. For each element, add up the numbers of atoms on each
side. Check whether the expression is already balanced. If
so, you’re done!
3. Otherwise - if present - choose an element that appears in
only one reactant and product to balance first. Insert the
appropriate coefficient(s) to balance this element.
4. Choose the element that appears in the next fewest total
reactants and products and balance it. Repeat the process
for additional elements if necessary.
4
1/9/2015
Example 1: balance the reaction that occurs between nitrogen
dioxide and water to form nitric acid and nitrogen monoxide
1. Write an expression using correct chemical formulas for the
reactants and products, separated by an arrow (). Include
symbols indicating physical states.
2. For each element, add up the numbers of atoms on each
side. Check whether the expression is already balanced. If
so, you’re done!
NO2(g) + H2O(l) → HNO3(aq) + NO(aq)
3. Otherwise - if present - choose an element that appears in
only one reactant and product to balance first. Insert the
appropriate coefficient(s) to balance this element.
= H, so try a 2 in front of HNO3
5
1/9/2015
NO2(g) + H2O(l) → 2 HNO3(aq) + NO(aq)
4. Choose the element that appears in the next fewest total
reactants and products and balance it. Repeat the process
for additional elements if necessary.
= N, try a 3 in front of NO2
Chapter Outline








7.1 Chemical reactions and the Conservation of Mass
7.2 Balancing Chemical Equations
7.3 Combustion Reactions
7.4 Stoichiometric Calculations and the Carbon Cyle
7.5 Percent Composition and Empirical Formulas
7.6 Empirical and Molecular Formulas Compared
7.7 Combustion Analysis
7.8 Limiting Reactants and Percent Yield
6
1/9/2015
Combustion Reactions
The reaction of an organic compound with oxygen
to produce CO2 + H2O, for example, balance CH4(g) + O2(g) → CO2(g) + H2O(g)
TIP FOR ALL COMBUSTION REACTIONS: Since oxygen
appears by itself, balance the other elements first, and then O 2
Combustion Reactions
Example: C2H6 (ethane) is combusted in oxygen
C2H6(g) + O2(g) → CO2(g) + H2O(g)
7
1/9/2015
Chapter Outline








7.1 Chemical reactions and the Conservation of Mass
7.2 Balancing Chemical Equations
7.3 Combustion Reactions
7.4 Stoichiometric Calculations and the Carbon Cycle
7.5 Percent Composition and Empirical Formulas
7.6 Empirical and Molecular Formulas Compared
7.7 Combustion Analysis
7.8 Limiting Reactants and Percent Yield
Stoichiometric Calculations and the Carbon Cycle
Photosynthesis: 6 CO2(g) + 6 H2O(g)  C6H12O6(aq) + 6 O2(g)
8
1/9/2015
Mauna Loa Observatory CO2 Concentrations since 1960 -
CO2 emissions over the last 800,000 years -
9
1/9/2015
Amounts of Reactants & Products
= STOICHIOMETRY
MM
ratio
MM
1. Write the balanced chemical equation
2. Convert the mass of the reactant into moles
3. Use coefficients in the balanced equation to calculate the
number of moles of product (stoichiometric ratio)
4. Convert moles of product into grams (or other desired
quantities)
Stoichiometry Example, p. 282
If the combustion of fossil fuels adds 8.2 x 10 12 kilograms of
carbon to the atmosphere each year as CO2, what is the mass
of added carbon dioxide?
Step 1: Write the balanced chemical equation
Step 2: Convert quantities of known substances (C) into moles
10
1/9/2015
Stoichiometry Example, p. 282
Step 3: Use coefficients in balanced equation to calculate the
number of moles of CO2 (stoichiometric ratio)
Step 4: Convert moles of CO2 into grams
Stoichiometry Example, p. 283
C7H6O3
MW =
138.12
C4H6O3
MW =
102.09
C9H8O4
MW =
180.16
Suppose we wish to prepare 1.00 kg of acetylsalicylic acid. How
many grams of salicylic acid and how many grams of acetic
anhydride are needed?
11
1/9/2015
Stoichiometry Example, p. 283
Step 1: Write the balanced chemical equation
Step 2: Convert quantities of known substances (ASA) into moles
Stoichiometry Example, p. 283
SA + AA  ASA + Ac
Step 3: Use coefficients in balanced equation to calculate the
number of moles of SA and AA (stoichiometric ratio)
12
1/9/2015
Stoichiometry Example, p. 283
Step 4: Convert moles of SA and AA into grams
Sample Exercise 7.3 – Calculating the Mass of a
Product from the Mass of a Reactant.
Each year, power plants in the U.S. consume about 1.1 x 10 11
kg of natural gas (CH4). How many kg of CO2 (MW = 44.01) are
released into atmosphere from these power plants. Given that
natural gas is mainly CH4 (MW = 16.04), base the calculation
on its combustion reaction –
Step 1: Write the balanced chemical equation
13
1/9/2015
Sample Exercise 7.3 – Calculating the Mass of a
Product from the Mass of a Reactant.
Step 2: Convert quantities of known substances (C) into moles
Step 3: Use coefficients in balanced equation to calculate the
number of moles of CO2 (stoichiometric ratio)
Step 4: Convert moles of CO2 into grams
Chapter Outline








7.1 Chemical reactions and the Conservation of Mass
7.2 Balancing Chemical Equations
7.3 Combustion Reactions
7.4 Stoichiometric Calculations and the Carbon Cycle
7.5 Percent Composition and Empirical Formulas
7.6 Empirical and Molecular Formulas Compared
7.7 Combustion Analysis
7.8 Limiting Reactants and Percent Yield
14
1/9/2015
Percent Composition: the composition of a compound in terms
of the percentage by mass of each element in the compound
n x molar mass of element
x 100%
molar mass of compound
n is the number of moles of the element in 1 mole
of the compound
3 x (1.008 g)
x 100% = 3.086%
97.99 g
1 x (30.97 g)
%P =
x 100% = 31.60%
97.99 g
4 x (16.00 g)
%O =
x 100% = 65.31%
97.99 g
%H =
H3PO4 MM = 97.99
g/mol
3.086% + 31.60% + 65.31% = 99.96%
= 100%
Sample Exercise 7.4 – Calculating Percent
Composition from a Chemical Formula
The mineral forsterite: Mg2SiO4,
MW = 140.71
15
1/9/2015
Empirical Formula from % Composition
A formula showing the smallest whole number
ratio of elements in a compound, e.g.
Benzene:
» Empirical = CH
» Molecular = C6H6
Glucose
» Empirical = CH2O
» Molecular = C6H12O6
Empirical Formula from % Composition
1. Assume 100 g
3. Divide by 4. Convert the mole
fewest
ratio from step 3
2. Convert to moles
number of
into small whole
moles
numbers if
necessary
16
1/9/2015
Sample Exercise 7.6
A sample of the carbonate mineral dolomite is 21.73% Ca,
13.18% Mg, 13.03% C, and the rest is oxygen. What is its
empirical formula? % O = 100 – 21.73 – 13.18 – 13.03 = 52.06 %
1. Assume 100 g
2. Convert to moles
Sample Exercise 7.6
3. Divide by fewest number of moles
4. Convert the mole ratio from step 3 into small whole
numbers if necessary – NOT REQUIRED HERE
17
1/9/2015
Example Illustrating Step 4
Vanillin is a common flavoring agent. It has a molar mass of 152
g/mol and is 63.15 %C and 5.30 %H; the rest is oxygen. What
are the empirical and molecular formulas?
% O = 100 – 63.15 – 5.30 = 31.55 %
1. Assume 100 g
2. Convert to moles
Example Illustrating Step 4
3. Divide by fewest number of moles
4. Convert the mole ratio from step 3 into small whole numbers if necessary
18
1/9/2015
Chapter Outline








7.1 Chemical reactions and the Conservation of Mass
7.2 Balancing Chemical Equations
7.3 Combustion Reactions
7.4 Stoichiometric Calculations and the Carbon Cycle
7.5 Percent Composition and Empirical Formulas
7.6 Empirical and Molecular Formulas Compared
7.7 Combustion Analysis
7.8 Limiting Reactants and Percent Yield
Empirical and Molecular
Formulas Compared
The molecular formula can be determined from the empirical
formula if the molecular weight of the compound is known.
empirical = CH2O
formula weight (FW)
= 30 g/mol
molecular = C2H4O2
Molecular weight (MW)
= 60 g/mol
Note that molecular formula = empirical x 2
glycolaldehyde
C2H4O2 = (CH2O) x 2 = C2H4O2
Or more generally –
molecular = (empirical) x n
19
1/9/2015
Empirical and Molecular
Formulas Compared
molecular = (empirical) x n
MW = (FW) x n
it follows that -
and therefore -
n = MW
FW
e.g. glucose – assume we know the
MW = 180.0
Empirical = CH2O
n=
180.0
30.0
FW = 30.0
=6
So molecular = (CH2O) x 6 = C6H12O6
Molecular Mass and Mass
Spectrometry
Acetylene
C2H2
Benzene
C6H6
20
1/9/2015
Sample Exercise 7.7 – Using Percent Composition
and Molecular weight to Derive a Molecular Fomula
Pheromones are chemical substances secreted by members of a species
to stimulate a response in other individuals of the same species. The
percent composition of eicosene, a compound similar to the Japanese
beetle mating pheromone, is 85.63% C and 14.37% H. Its molecular mass,
as determined by mass spectrometry, is 280 amu. What is the molecular
formula of eicosene?
1. Assume 100 g
2. Convert to moles
Sample Exercise 7.7 – Using Percent Composition
and Molecular weight to Derive a Molecular Fomula
3. Divide by fewest number of moles
4. Convert the mole ratio from step 3 into small whole numbers if necessary
– NOT NEEDED
21
1/9/2015
Chapter Outline








7.1 Chemical reactions and the Conservation of Mass
7.2 Balancing Chemical Equations
7.3 Combustion Reactions
7.4 Stoichiometric Calculations and the Carbon Cycle
7.5 Percent Composition and Empirical Formulas
7.6 Empirical and Molecular Formulas Compared
7.7 Combustion Analysis
7.8 Limiting Reactants and Percent Yield
Experimental Determination of Empirical
Formulas: Combustion Analysis
22
1/9/2015
Experimental Determination of Empirical
Formulas: Combustion Analysis
CxHyOz + O2(g)  x CO2(g) + y/2 H2O(g)
mass
sample
excess
Some of the oxygen comes
from the sample, some from
the excess O2
Calculation Outline:
g CO2
mol CO2
mol C
gC
g H2O
mol H2O
mol H
gH
g of O = g of sample – g of C - g of H
Sample Exercise 7.8: Deriving an Empirical
Formula from Combustion Analysis Data
Combustion of 1.000 grams of an organic compound known to
contain only C, H and O produces 2.360 g CO2 and 0.640 g
H2O. What is the empirical formula of the compound?
23
1/9/2015
Sample Exercise 7.8: Deriving an Empirical
Formula from Combustion Analysis Data
g of O = g of sample – g of C - g of H
and so the moles of O =
Sample Exercise 7.8: Deriving an Empirical
Formula from Combustion Analysis Data
Now divide by the smallest number of moles -
24
1/9/2015
Chapter Outline








7.1 Chemical reactions and the Conservation of Mass
7.2 Balancing Chemical Equations
7.3 Combustion Reactions
7.4 Stoichiometric Calculations and the Carbon Cycle
7.5 Percent Composition and Empirical Formulas
7.6 Empirical and Molecular Formulas Compared
7.7 Combustion Analysis
7.8 Limiting Reactants and Percent Yield
Limiting Reagents
Limiting Reagents - a reactant that is consumed completely
in a chemical reaction before the other reactant(s) run out.
The amount of product formed depends on the amount of the
limiting reagent available.
Each sandwich consists of 2 slices bread, 1 slice of cheese, and 1 slice of salami
8 slices
bread
4 slices
cheese
3 slices
salami
25
1/9/2015
Limiting Reagents
2 Br + 1 Ch + 1 Sal  1 sandwich
Given:
8
4
3
Here’s the logic – compare the given number of moles to
the stoichiometric ratio
Strategy for determining which
reactant is the Limiting Reagent:
aA + bB  C
1. Convert grams of each into moles
2. Calculate the stoichiometric ratio that’s the
largest, e.g. a/b or b/a
3. Calculate the given mole ratio in the same way
4. Compare to identify the LR
if
𝑚𝑜𝑙𝑒𝑠 𝐴
𝑚𝑜𝑙𝑒𝑠 𝐵
>
𝑔𝑖𝑣𝑒𝑛
𝑎
𝑏
𝑠𝑡𝑜𝑖𝑐ℎ𝑖𝑜𝑚𝑒𝑡𝑟𝑖𝑐
then A is in excess and B is the limiting reagent
26
1/9/2015
Sample Exercise 7.9: Identifying the
Limiting Reagent in a Reaction Mixture
The flame in an acetylene torch reaches temperatures as high as 3500 oC as
a result of the combustion of a mixture of acetylene (C2H2) and pure oxygen.
If these two gases flow from high-pressure tanks at the rates of 52.0 g C2H2
and 188 g O2 per minute, which reactant is the limiting reagent, or is the
mixture stoichiometric?
2 C2H2 + 5 O2  4 CO2 + 2 H2O
188 g
52.0 g
Given ratio:
Sample Exercise 7.9: Identifying the
Limiting Reagent in a Reaction Mixture
2 C2H2 + 5 O2  4 CO2 + 2 H2O
𝑖𝑠
𝑚𝑜𝑙𝑒𝑠 𝑂2
𝑚𝑜𝑙𝑒𝑠 𝐶2𝐻2
>
𝑔𝑖𝑣𝑒𝑛
Given ratio:
5.88 mol O2
= 2.94
2.00 mol C2H2
5 𝑚𝑜𝑙 𝑂2
2 𝑚𝑜𝑙 𝐶2𝐻2
???
𝑠𝑡𝑜𝑖𝑐ℎ𝑖𝑜𝑚𝑒𝑡𝑟𝑖𝑐
Stoichiometric ratio:
5 mol O2
= 2.5
2 mol C2H2
therefore O2 is in excess and so C2H2 is the LR
27
1/9/2015
Percent Yield
Theoretical Yield is the maximum amount of product
formed from given quantities of reactants (check if a
limiting reagent is present).
Actual Yield is the amount of product actually obtained
from a reaction.
% Yield =
Actual Yield
x 100
Theoretical Yield
Sample Exercise 7.10
The industrial process for making the ammonia used in fertilizer, explosives, and
many other products is based on the reaction between nitrogen and hydrogen at high
temperature and pressure.
If 18.2 kg of NH3 (MW = 17.03) is produced by a reaction mixture that initially
contains 6.00 kg of H2 (MW = 2.016) and an excess of N2, what is the percent yield of
the reaction?
Actual Yield
N2(g) + 3 H2(g) → 2 NH3(g)
x 100
% Yield =
Theoretical Yield
Excess
6.00 kg Actual yield =
(no LR)
18.2 kg
Calculate the theoretical yield of NH3 based on H2, and then calculate the %
yield -
28
1/9/2015
Sample Exercise 7.10
% Yield =
Actual Yield
x 100
Theoretical Yield
29