Warmup – Mass Percent
Find the percent by mass:
1. What is the mass % of silver in silver
nitrate? (63.5%)
2. What is the mass % of N in ammonium
carbonate? (29.2%)
Stoichiometry
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Finding
Empirical
Formulas
Stoichiometry
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Calculating Empirical Formulas
One can calculate the empirical formula from
the percent composition.
Stoichiometry
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Rhyme
•
•
•
•
Rhyme to remember order of steps to
convert % composition  empirical
formula:
Percent to mass
Mass to mole
Divide by small
Times till whole
Stoichiometry
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Calculating Empirical Formulas
Ascorbic acid (vitamin C) contains 40.92% C, 4.58% H,
and 54.50% O by mass.
What is the empirical formula of vitamin C?
(C3H4O3)
Stoichiometry
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Calculating Empirical Formulas
Assuming 100.00 g of ascorbic acid,
C:
H:
O:
1 mol = 3.407 mol C
12.01 g
1 mol
4.58 g x
= 4.53 mol H
1.01 g
1 mol
54.50 g x
= 3.406 mol O
16.00 g
40.92 g x
Stoichiometry
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Calculating Empirical Formulas
Calculate the mole ratio by dividing by the smallest number
of moles:
C:
3.407 mol
3.406 mol
= 1.000
H:
4.53 mol
3.406 mol
= 1.33
O:
3.406 mol
3.406 mol
= 1.000
Multiply all by 3: 3 C:4 H:3 O
Stoichiometry
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Calculating Empirical Formulas
These are the subscripts for the empirical formula:
C3H4O3
Stoichiometry
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Practice Exercise 3.13
A 5.325-g sample of methyl benzoate, a
compound used in the manufacture of
perfumes, is found to contain 3.758 g of
carbon, 0.316 g of hydrogen, and 1.251 g of
oxygen.
What is the empirical formula of this
substance?
(C4H4O)
Stoichiometry
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Molecular Formula from Empirical
Formula
The empirical formula (relative ratio of
elements in the molecule) may not be the
molecular formula (actual ratio of elements in
the molecule).
e.g. ascorbic acid (vitamin C) has empirical
formula C3H4O3.
The molecular formula is C6H8O6.
Stoichiometry
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Molecular Formula from Empirical
Formula
To get the molecular formula from the
empirical formula, we need to know the
molecular weight, MW.
(“molecular molar mass”)
The ratio of molecular weight (MW) to formula
weight (FW) of the empirical formula must be
a whole number.
(“formula molar mass”)
Stoichiometry
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Sample Exercise 13.14 (p. 96)
Mesitylene, a hydrocarbon that occurs in
small amounts in crude oil, has an empirical
formula of C3H4. The experimentally
determined molecular weight of this
substance is 121 amu.
What is the molecular formula of mesitylene?
(Hint: you do not need to figure out the
empirical formula first)
(C9H12)
Stoichiometry
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Practice Exercise 13.14
Ethylene glycol, the substance used in
automobile antifreeze, is composed of 38.7%
C, 9.7% H, and 51.6% O by mass. Its
(molecular) molar mass is 62.1 g/mol.
a) What is the empirical formula of ethylene
glycol? (CH3O)
b) What is its molecular formula? (C2H6O2)
Stoichiometry
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Hydrates
• Anhydrous = dry, does not contain
water
• Hydrate = a compound formed by the
combination of water with some other
substance in which the water retains its
molecular state as H2O
Stoichiometry
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Determining Hydrate Formulas
1. Find the formula of a hydrate from experimental
mass data:
a) Calculate the number of moles of the anhydrous
salt (mass  mole conversion)
b) Calculate the mass of H2O by subtracting mass of
anhydrous salt from mass of hydrated salt
c) Calculate the moles of H2O
(mass  mole conversion)
d) Calculate ratio: # mol H2O:# mol anhydrous salt
e)  correct formula = salt · x H2O (x = mole ratio)
Stoichiometry
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Determining Hydrate Formulas
2. Find the formula from mass % data:
a) Use the method of using 100 g (from 100 %) to
give mass in grams of the salt and water
Use mass  mole conversion to calculate the
# moles of the anhydrous salt and the # moles of
water.
b) Calculate the mole ratio of H2O:anhydrous salt
Stoichiometry
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Determining Hydrate Formulas
3. Find the % H2O in a hydrate from
its formula:
Use mass percent.
Stoichiometry
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Hydrate practice problems
1. A mass of 2.50 g of blue hydrated
copper sulfate (CuSO4 • x H2O) is
placed in a crucible and heated. After
heating, 1.59 g white anhydrous
copper sulfate (CuSO4) remains.
What is the formula for the hydrate?
Name the hydrate.
(CuSO4 • 5 H2O)
Stoichiometry
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Hydrate practice problems
2. A hydrate is found to have the
following percent composition:
48.8% MgSO4 and 51.2% H2O.
What is the formula and the name for
this hydrate?
(MgSO4 • 7 H2O)
Stoichiometry
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Hydrate practice problems
3. The formula for a hydrate is SnCl2 • 2
H2O.
What is the mass percent of water in
this hydrate?
(15.97% H2O)
Stoichiometry
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Combustion Analysis
• Compounds containing C, H and O are routinely
analyzed through combustion in a chamber like this.
– 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
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Sample Exercise 3.15 (p. 97)
Isopropyl alcohol, a substance sold as
rubbing alcohol, is composed of C, H,
and O. Combustion of 0.255 g of
isopropyl alcohol produces 0.561 g CO2
and 0.306 g H2O. Determine the
empirical formula of isopropyl alcohol.
(C3H8O)
Stoichiometry
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Practice Exercise 3.15
a)
Caproic acid, which is responsible for the foul odor
of dirty socks, is composed of C, H, and O atoms.
Combustion of a 0.225 g sample of this compound
produces 0.512 g CO2 and 0.209 g H2O. What is
the empirical formula of caproic acid?
(C3H6O)
b)
Caproic acid has a molar mass of 116 g/mol. What
is its molecular formula?
(C6H12O2)
Stoichiometry
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Elemental Analyses
Compounds
containing other
elements are
analyzed using
methods analogous
to those used for C,
H and O.
Stoichiometry
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Stoichiometric Calculations
The coefficients in the balanced equation give
the ratio of moles of reactants and products.
Stoichiometry
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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
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Stoichiometric Calculations
C6H12O6 + 6 O2  6 CO2 + 6 H2O
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.
Stoichiometry
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Limiting
Reactants
Stoichiometry
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How Many Cookies Can I Make?
• You can make cookies
until you run out of one
of the ingredients.
• Once this family runs
out of sugar, they will
stop making cookies
(at least any cookies
you would want to eat).
Stoichiometry
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How Many Cookies Can I Make?
• In this example the
sugar would be the
limiting reactant,
because it will limit the
amount of cookies you
can make.
Stoichiometry
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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
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Limiting Reactants
In the example below, the O2 would be the
excess reagent.
Stoichiometry
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Sample Exercise 3.18 (p. 104)
The most important commercial process for
converting N2 from the air into nitrogencontaining compounds is based on the
reaction of N2 and H2 to form ammonia (NH3):
N2(g) + 3 H2(g)  2 NH3(g)
How many moles of NH3 can be formed from
3.0 mol of N2, and 6.0 mol of H2?
(4.0 mol NH3)
Stoichiometry
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Practice Exercise 3.18
Consider the reaction
2 Al(s) + 3 Cl2(g)  2 AlCl3(s).
A mixture of 1.50 mol of Al and 3.00 mol of
Cl2 are allowed to react.
a) What is the limiting reactant? (Al)
b) How many moles of AlCl3 are formed?
(1.50 mol)
c) How many moles of the excess reactant
remain at the end of the reaction?
(0.75 mol Cl2)
Stoichiometry
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Sample Exercise 3.19 (p. 104)
Consider the following reaction that occurs in a fuel
cell:
2 H2(g) + O2(g)  2 H2O(l)
This reaction, properly done, produces energy in the
form of electricity and water. Suppose a fuel cell is
set up with 150 g of hydrogen gas and 1500 grams of
oxygen gas (each measurement is given with two
significant figures). How many grams of water can
be formed?
(1400 g H2O)
Stoichiometry
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Practice Exercise 3.19
A strip of zinc metal weighing 2.00 g is placed in an
aqueous solution containing 2.50 g of silver nitrate,
causing the following reaction to occur:
Zn(s) + 2 AgNO3(aq)  2 Ag(s) + Zn(NO3)2(aq)
a) Which reactant is limiting? (AgNO3)
b) How many grams of Ag will form? (1.59 g)
c) How many grams of Zn(NO3)2 will form? (1.39 g)
d) How many grams of the excess reactant will be
left at the end of the reaction? (1.52 g Zn)
Stoichiometry
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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.
Stoichiometry
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Percent Yield
One finds the percent yield by
comparing the amount actually obtained
(actual yield) to the amount it was
possible to make (theoretical yield).
Actual Yield
Percent Yield =
Theoretical Yield
x 100
Stoichiometry
© 2009, Prentice-Hall, Inc.
Sample Exercise 3.20 (p. 106)
Adipic acid, H2C6H8O4, is used to produce nylon. It is made
commercially by a controlled reaction between cyclohexane
(C6H12) and O2:
2 C6H12(l) + 5 O2(g)  2 H2C6H8O4(l) + 2 H2O(g)
a)
Assume that you carry out this reaction starting with 25.0 g
of cyclohexane, and that cyclohexane is the limiting
reactant. What is the theoretical yield of adipic acid?
(43.5 g H2C6H8O4)
b)
If you obtain 33.5 g of adipic acid from your reaction, what
is the percent yield of adipic acid?
(77.0%)
Stoichiometry
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Practice Exercise 3.20
Imagine that you are working on ways to improve the process by
which iron ore containing Fe2O3 is converted into iron. In your
tests you carry out the following reaction on a small scale:
Fe2O3(s) + 3 CO(g)  2 Fe(s) + 3 CO2(g)
a)
If you start with 150 g of Fe2O3 as the limiting
reagent (reactant), what is the theoretical yield of
Fe?
(105 g Fe)
b)
If the actual yield of Fe in your test was 87.9 g,
what was the percent yield?
(83.7%)
Stoichiometry
© 2009, Prentice-Hall, Inc.