Solutions to Stoichiometry Problems
Problem 1 (warm-up)
A solution of lead (II) nitrate is mixed with a solution of potassium bromide.
a) Write balanced molecular, complete ionic, and net ionic equations for this reaction.
The first step is to determine the formulas of all substances from the names. We notice that both
compounds contain a metal and a nonmetal (or polyatomic ion), and hence are ionic compounds.
The charges of the ions are determined as follows:
Pb2+ (given by numeral II in name), NO3¯ (memorized), K+ (periodic table), Br¯ (periodic table)
Finally, by balancing charges, we can determine the correct formulas of each ionic compound:
Pb(NO3)2 (aq), KBr (aq)
(note that we know both are aq, not only because we are told they are
solutions in the problem statement, but also due to solubility rules: all nitrates are soluble, and all
alkali metal salts are soluble)
To predict the outcome of the reaction, we recognize first that this is a double displacement reaction
(two ionic compounds reacting), and then determine the correct formulas of the “new partners” from
charge balancing: PbBr2, KNO3
In order for there to be an actual reaction, something must change from reactants to products that
provides a “driving force” for the reaction to move forward. This could include formation of a solid, of
a weak or nonelectrolyte (e.g., water or weak acid), or of a gas, or alternatively, the disappearance of
any of the above if they existed as one of the reactants.
In this case, we know that KNO3 will still be soluble (aq) due to both of the solubility rules mentioned
earlier. PbBr2 will form a precipitate, however, because although most halide salts (salts containing
halogen elements such as Br) are soluble, the exception is with Ag+, Hg22+, or Pb2+.
At last, we can write the entire reaction equation and balance it. To write the complete ionic equation,
we simply separate any dissolved (aq) ionic compounds (or other strong electrolytes such as strong
acids) into their ions. Note that “insoluble” solids (s), gases (g), or soluble (aq) nonelectrolytes or
weak electrolytes (e.g., water or weak acids) are not separated into ions. For the net ionic equation,
we simply remove the “spectator ions”, which appear identical on both sides of the reaction arrow.
Pb(NO3)2 (aq) + 2 KBr (aq) Æ PbBr2 (s) + 2KNO3 (aq)
Pb2+ + 2 NO3¯ + 2 K+ + 2 Br¯ Æ PbBr2 (s) + 2K+ + 2NO3¯
Pb2+ + 2 Br¯ Æ PbBr2 (s)
Chem 131
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b) How many moles of lead (II) bromide could be produced from 6 moles of potassium
bromide?
From the balanced molecular eqn, Pb(NO3)2 (aq) + 2 KBr (aq) Æ PbBr2 (s) + 2KNO3 (aq), we see
that 1 PbBr2 is formed from every 2 KBr reacted (i.e., for some number of KBr, we get half as many
PbBr2). Thus, from 6 KBr, we should get 3 PbBr2, or more explicitly:
⎛ 1 mol PbBr2 ⎞
⎟⎟ = 3 mol PbBr2
6 mol KBr ⎜⎜
⎝ 2 mol KBr ⎠
c) How many moles of lead (II) bromide could be produced from 0.846 moles of potassium
bromide?
Applying the same logic as above:
⎛ 1 mol PbBr2
0.846 mol KBr ⎜⎜
⎝ 2 mol KBr
⎞
⎟⎟ = 0.423 mol PbBr2
⎠
Problem 2 (intermediate warmup)
In reality, gasoline is a mixture of hydrocarbons and other organic compounds. But for this
problem, assume that gasoline can be accurately represented as pure octane, C8H18.
a) Write a balanced reaction equation for the complete combustion of gasoline.
Complete combustion is the reaction of an organic compound (containing C, H, and possibly other
elements) with oxygen gas (O2) to form carbon dioxide and water (and possibly other oxidized
compounds if the organic compound contains elements other than C, H, or O).
Note that the water can be produced in either liquid (l) or vapor (g) form, depending upon the
conditions (temperature and pressure) of the reaction. In this case, we will assume vapor is formed.
To balance combustion reactions, it is easiest to first balance C (by putting coefficient in front of CO2),
then H (by putting coefficient in front of H2O), then O (by putting coefficient in front of O2). If an even
number of O’s are needed the coefficient in front of O2 will be an integer; if an odd number are
needed, this coefficient will be fractional (something over 2). For most reactions, it would be
conventional to double all of the coefficients so as to remove the fraction. In the case of combustion
reactions, however, it is commonplace to leave the fractional coefficient as is, since this leaves a
coefficient of “1” in front of the organic compound of interest.
C8H18 (l) +
Chem 131
25
O2 (g) Æ 8 CO2 (g) + 9 H2O (g)
2
Solutions to Stoichiometry+ Problems
Page 2
b) Which substance is oxidized and which is reduced in this reaction? (Verify by determining
oxidation numbers for all of the elements in the reaction.)
First, note that all combustion reactions are redox (oxidation-reduction) reactions, where the organic
compound gets oxidized by the oxygen gas (which in turn gets reduced). Side note: In fact, oxidation
of substances by atmospheric O2 is a principal component of most of the important biogeochemical
cycles that we observe on the earth (the other “half” of the cycle typically involving the reduction of
those oxidized compounds by plants, microorganisms, or human society). The importance of O in
earth chemistry will be a continuous theme throughout the course.
Oxidation numbers are a useful tool for identifying oxidation and reduction in all types of redox
reactions (we will get into a more thorough discussion of the physical significance of oxidation
numbers later in the semester). The following rules apply when determining oxidation numbers:
•
All elemental substances are assigned an oxidation number of zero (e.g., 0 for O2).
•
Ions are assigned an oxidation number equal to their ionic charge (e.g., +1 for Na+ in NaCl).
•
In covalently bonded substances (molecular compounds, polyatomic ions, covalent networks):
o
Assign H an oxidation number of +1
note that when H is in a compound with metals, it is NOT covalently bonded; it exists as
the hydride ion, H−, which has a charge of −1
o
Assign O an oxidation number of −2
there are some rare exceptions to this, but we will ignore them for now
o
Assign other elements whatever is needed to give the correct net charge.
for molecules the net charge is zero, while for polyatomic ions the net charge is the ionic
charge (e.g., for SO42− the net charge is −2.)
Applying these rules to the combustion reaction above, we have (in order of difficulty):
O2
O is zero, since elemental
H2O
H = +1, O = −2, (2)(+1) + (1)(−2) = 0, automatically results in correct net charge.
CO2
C = x, O = −2, (1)(x) + (2)(−2) = 0 (net charge), so x = +4 on C
C8H18
C = x, H = +1, (8)(x) + (18)(+1) = 0 (net charge), so x = −18/8 = −2.25
−2.25 x 8 +1 x 18
Summarizing:
C8H18 (l)
0
+
25
O2 (g)
2
+4 x 1 −2 x 2
Æ
8 CO2 (g)
+1 x 2 −2 x 1
+
9 H2O (g)
Remember that oxidation is loss of electrons, while reduction is gain of electrons (OIL RIG). Since
electrons have a -1 charge, loss (oxidation) results in something become more positively charged,
while gain (reduction) results in becoming more negatively charged.
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In this reaction, we see that C has gone from −2.25 to +4 (oxidation), O has gone from 0 to −2
(reduction), and H has not changed (+1 on both sides).
When ionic substances are involved in redox reactions, it is customary to say that the element gets
oxidized or reduced. When molecular substances are involved in redox reactions, it is customary to
say that the substance gets oxidized or reduced.
Thus, for this reaction, we would say:
C8H18 gets oxidized and O2 gets reduced
c) What masses of CO2 and of H2O would be produced from the combustion of 1 gallon of
gasoline? (The density of gasoline is 6.07 lb/gal. 1 lb = 454 g.)
First we need to convert the volume of gasoline into customary mass units (g):
⎛ 6.07 lb C8 H 18 ⎞⎛ 454 g C8 H 18 ⎞
⎟⎟⎜⎜
⎟⎟ = 2756 g C8 H 18
1 gal C8 H 18 ⎜⎜
1
gal
C
H
1
lb
C
H
8
18 ⎠⎝
8 18 ⎠
⎝
I’m assuming the calculation is for exactly 1 gallon of gasoline, hence 3 sig figs due to conversion factor.
In all stoichiometric calculations, we must first move from the macroscopic measurable perspective
(e.g., mass) to the atomic-level perspective (number of moles). We do this using the molar mass of
the substance, derived from the atomic molar masses on the periodic table (8 x 12.011 + 18 x
1.0079):
⎛ 1 mol C8 H 18 ⎞
⎟⎟ = 24.127 mol C8 H 18
2756 g C8 H 18 ⎜⎜
⎝ 114.23 g C8 H 18 ⎠
Next, we determine the number of moles of each of the product substances, knowing the
stoichiometric ratios that are given in the balanced reaction equation:
⎛ 8 mol CO2 ⎞
⎟⎟ = 193.0 mol CO2
24.127 mol C8 H 18 ⎜⎜
⎝ 1 mol C8 H 18 ⎠
⎛ 9 mol H 2 O ⎞
24.127 mol C8 H 18 ⎜⎜
⎟⎟ = 217.1 mol H 2 O
⎝ 1 mol C8 H 18 ⎠
Finally, we convert from the atomic-level perspective (moles) back to the macro perspective (mass):
⎛ 44.01 g CO2
193.0 mol CO2 ⎜⎜
⎝ 1 mol CO2
⎞
⎟⎟ = 8490 g CO2
⎠
⎛ 18.015 g H 2 O ⎞
217.1 mol H 2 O⎜⎜
⎟⎟ = 3910 g H 2 O
⎝ 1 mol H 2 O ⎠
Chem 131
(= 8.49 kg CO2 or 18.7 lb CO2 )
(= 3.91 kg H 2 O or 8.62 lb H 2 O )
Solutions to Stoichiometry+ Problems
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Once you are accustomed to doing stoichiometric calculations, it is reasonable to compress all of the
steps above into a single multi-conversion calculation for each substance:
mass A
Æ moles A
Æ moles B
Æ mass B
⎛ 1 mol C8 H 18 ⎞⎛ 8 mol CO2 ⎞⎛ 44.01 g CO2 ⎞
⎟⎟⎜⎜
⎟⎟⎜⎜
⎟⎟ = 8490 g CO2
2756 g C8 H 18 ⎜⎜
⎝ 114.23 g C8 H 18 ⎠⎝ 1 mol C8 H 18 ⎠⎝ 1 mol CO2 ⎠
⎛ 1 mol C8 H 18 ⎞⎛ 9 mol H 2 O ⎞⎛ 18.015 g H 2 O ⎞
2756 g C8 H 18 ⎜⎜
⎟⎟⎜⎜ 1 mol C H ⎟⎟⎜⎜ 1 mol H O ⎟⎟ = 3910 g H 2 O
114
.
23
g
C
H
8
18
8
18 ⎠⎝
2
⎠
⎝
⎠⎝
***A little conceptual check: note that we started out with about 6 lb of gasoline and ended up making
almost 19 lb of carbon dioxide and 9 lb of water. Where did the extra mass come from?
Problem 3 (standard easy stoich problem)
Iron metal reacts with chlorine gas to produce iron (III) chloride.
a) Write a balanced reaction equation for this reaction.
As with Problem 1a, the first step is to determine the formulas of all substances: iron metal is Fe (s)
and chlorine gas is Cl2 (g) (remember that the diatomic elements are H2, N2, O2, F2, Cl2, Br2, I2; note
their locations on the periodic table to help you remember). Iron (III) chloride is FeCl3 (s), following
the same approach as in 1a above, and noting that virtually all pure ionic substances are solids at
room temperature. After balancing, we have:
2 Fe (s) + 3 Cl2 (g) Æ 2 FeCl3 (s)
b) What type of reaction is this?
We know that this is not a double displacement reaction (which includes precipitation, acid-base, and
gas-forming reactions), so one might suspect that it is a redox (oxidation-reduction) reaction. We can
verify this by determining oxidation numbers:
0
0
2 Fe (s) + 3 Cl2 (g)
+3 x 1 −1 x 3
Æ 2 FeCl3 (s)
Hence, it is a redox reaction, where iron gets oxidized and chlorine gets reduced.
c) If 18.0g of iron chloride were obtained from the reaction of 10.0g of iron with excess
chlorine, what is the percentage yield of the reaction?
The excess chlorine indicates that it will not limit us from consuming all 10.0g of iron, and we can
determine the mass of iron chloride that should theoretically result:
⎛ 1 mol Fe ⎞⎛ 2 mol FeCl3 ⎞⎛ 162.20 g FeCl3 ⎞
⎟⎟ = 29.04 g FeCl3
⎟⎟⎜⎜
⎟⎟⎜⎜
10.0 g Fe⎜⎜
⎝ 55.845 g Fe ⎠⎝ 2 mol Fe ⎠⎝ 1 mol FeCl3 ⎠
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Since we actually obtained only 18.0g of iron chloride from the reaction, our percentage yield was:
⎛ 18.0 g FeCl 3 ⎞
⎛ Actual Yield ⎞
⎟ × 100 % = 62.0%
⎟⎟ × 100 % = ⎜
% Yield = ⎜⎜
⎜
⎟
Theoretica
l
Yield
29
.
0
4
g
FeCl
⎝
⎠
3 ⎠
⎝
d) If 10.0g each of iron metal and chlorine gas are reacted, what is the theoretical yield of iron
chloride?
Since amounts of both reactants are specified, it is possible that we could run out of one of the
reactants before we consume all of the other. In other words, one of the two substances may be a
limiting reactant, which prevents further creation of product. From the calculation earlier, we know
that if we use up all of the iron, we will produce 29.0g FeCl3. On the other hand, if we use up all
10.0g of chlorine, we will produce:
⎛ 1 mol Cl 2 ⎞⎛ 2 mol FeCl3 ⎞⎛ 162.20 g FeCl3 ⎞
⎟⎟ = 15.25 g FeCl3
⎟⎟⎜⎜
⎟⎟⎜⎜
10.0 g Cl 2 ⎜⎜
70
.
905
g
Cl
3
mol
Cl
1
mol
FeCl
2
2
3
⎝
⎠⎝
⎠⎝
⎠
Hence, Cl2 is the limiting reactant. Once we produce 15.3g FeCl3, we will run out of Cl2 and will be
unable to make any more. Thus, under these conditions, the theoretical yield of FeCl3 is 15.3g.
Problem 4 (‘kitchen sink’ problem)
In order to identify an unknown alkaline earth metal “M”, an initial experiment is performed, in
which a 0.248g sample of the metal is oxidized completely in excess oxygen. The metal oxide
product has a mass of 0.413g.
a) Determine the mass % oxygen in the metal oxide.
Conceptually, we understand that the solid metal sample has gained mass during the reaction, as
oxygen atoms are incorporated into the solid crystal structure of the metal oxide product.
Quantitatively, the gain in mass is due entirely to the oxygen that was added:
Mass oxygen = Mass metal oxide – mass metal = 0.413g – 0.248g = 0.165g
The percentage by mass that the oxygen represents of the metal oxide is thus:
0.165 g oxygen
× 100 % = 40.0 % O
0.413 g metal oxide
b) Given that the metal is an alkaline earth metal, what must be the empirical formula of the
compound?
MO, since all alkaline earth metals form ions with +2 charge, while oxygen forms ions with -2 charge.
Chem 131
Solutions to Stoichiometry+ Problems
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c) Determine the molar mass and the identity of the metal.
The molar mass of any pure substance is the ratio of the mass to the number of moles of a sample of
the substance. In this case, we know the mass of the original metal sample. We can determine the
number of moles of metal in that sample by first determining the number of moles of oxygen in the
sample, and then recognizing the mol-to-mol ratio between the metal and oxygen given by the
empirical formula above.
⎛ 1 mol O ⎞⎛ 1 mol M
⎟⎟⎜⎜
moles metal = 0.165 g O⎜⎜
⎝ 15.9994 g O ⎠⎝ 1 mol O
⎞
⎟⎟ = 0.0103 mol M
⎠
The molar mass of the metal is therefore:
⎛ 0.248 g M
⎜⎜
⎝ 0.0103 mol M
⎞
⎟⎟ = 24.0 g mol
⎠
The only alkaline earth metal with a molar mass in this vicinity is magnesium, Mg.
d) How many protons and electrons are in an atom of this metal?
All neutral magnesium atoms have 12 protons and 12 electrons.
e) If the common isotopes of the metal have mass numbers of 24, 25, and 26, how many
neutrons are in the most common atom of this metal? Justify your answer.
The average atomic mass of a naturally occurring sample of Mg is about 24.3 amu (from periodic
table). This can only be possible if the Mg-24 isotope is more common than the two heavier isotopes;
otherwise, the average atomic mass would be greater than 24.5 amu.
The Mg-24 isotope has a total of 24 protons and neutrons. Since 12 of these are protons (see part d
above), then there must be 24 – 12 = 12 neutrons.
In a second experiment, a 0.127g piece of this metal is placed in a test tube containing 1.25 x 10-2
moles of hydrochloric acid.
f) Write the molecular, complete ionic, and net ionic equations for the reaction that occurs
between the metal and hydrochloric acid.
The formulas of the reactants are Mg (s) and HCl (aq). We know that this cannot be a double
displacement reaction since Mg (s) is not a compound. We might suspect that it is a redox reaction,
especially if we are aware that (1) alkali and alkaline earth metals are fairly reactive and tend to form
positively charged ions (cations), and (2) most metals dissolve in acids to produce salt solutions of
the metal ions, as well as hydrogen gas (H2). (We could verify the hypothesis via experiment.)
Mg (s) + 2 HCl (aq) Æ MgCl2 (aq) + H2 (g)
Mg (s) + 2 H+ + 2 Cl– Æ Mg2+ + 2 Cl– + H2 (g)
Mg (s) + 2 H+ Æ Mg2+ + H2 (g)
Chem 131
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Remember that in writing the complete ionic equation, we separate aqueous (aq) ionic compounds
(such as MgCl2) and strong electrolytes (such as HCl) into the appropriate ions.
g) Which substance is oxidized and which reduced in this reaction?
By assigning oxidation numbers in the net ionic equation, this answer becomes evident:
0
+1
Mg (s) + 2 H+
Æ
+2
0
2+
Mg
+ H2 (g)
Mg goes from 0 to +2, indicating a loss of electrons and hence oxidation.
H goes from +1 to 0, indicating a gain of electrons and hence reduction.
h) Does the metal dissolve completely? Show calculations to support your answer.
In the problem statement earlier, we were given amounts of both reactants. This indicates that we
may have a limiting reactant problem, as in Problem 3d above. If the Mg metal is the limiting
reactant, then it will be completely converted to MgCl2, and hence will dissolve completely. If, on the
other hand, HCl is the limiting reactant, then there will still be some Mg metal left at the end of
reaction. The quickest way to answer the question in this case is to determine how many moles of
HCl are needed to react all of the Mg metal:
⎛ 1 mol Mg ⎞⎛ 2 mol HCl ⎞
⎟⎟⎜⎜
⎟⎟ = 1.05 × 10 − 2 mol HCl required
0.127 g Mg ⎜⎜
24
.
305
g
Mg
1
mol
Mg
⎝
⎠⎝
⎠
Since 1.25 x 10-2 moles of HCl are present, there is more than enough HCl to react all of the Mg.
Therefore, the metal dissolves completely.
Problem 5 (combustion analysis)
Limonene is a mildly sweet-smelling organic compound in lemon peel. In this problem, you will
determine the molecular formula of limonene.
a) 0.500 g of limonene was burned in excess oxygen. 1.615 g of carbon dioxide and 0.529 g of
water (and nothing else) were produced. What is the empirical formula of limonene?*
See the Step-by-Step Combustion Analysis below for a thorough explanation of the approach.
⎛ 1 mol CO2 ⎞⎛ 1 mol C ⎞
⎟⎟⎜⎜
⎟⎟ = 3.670 × 10 − 2 mol C
1.615 g CO2 ⎜⎜
⎝ 44.01 g CO2 ⎠⎝ 1 mol CO2 ⎠
⎛ 1 mol H 2 O ⎞⎛ 2 mol H ⎞
⎟⎟⎜⎜
⎟⎟ = 5.873× 10 − 2 mol H
0.529 g H 2 O⎜⎜
18
.
015
g
H
O
1
mol
H
O
2
2
⎝
⎠⎝
⎠
These quantities of C and H correspond to the following masses:
Chem 131
Solutions to Stoichiometry+ Problems
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⎛ 12.011 g C ⎞
⎟⎟ = 0.4408 g C
3.670 × 10 − 2 mol C ⎜⎜
⎝ 1 mol C ⎠
⎛ 1.0079 g H ⎞
⎟⎟ = 0.0592 g H
5.873 × 10 − 2 mol H ⎜⎜
⎝ 1 mol H ⎠
Since the sum of these masses is 0.5000g, equal to the starting mass of limonene, we know that no
other elements (e.g., O) are present in limonene. The mol-mol ratio of H to C is:
5.873× 10 −2 mol H
3.670 × 10 − 2 mol C
=
1.600 mol H ⎛ 5 ⎞ 8.00 mol H
, so the empirical formula is C5 H8.
⎜ ⎟=
1 mol C ⎝ 5 ⎠
5 mol C
b) If the molar mass of the compound is known to be 136 g/mol, what is the molecular formula
of limonene?
By definition, the molecular formula is an integer multiple of the empirical formula, so the molar mass
of the compound must be an integer multiple of the formula mass of C5H8, which is 68.1 g/mol:
136 g mol
= 2.00 , so the molecular formula must be twice the empirical formula, or C10 H16.
68.1 g mol
c) Can you say with certainty that the condensed structural formula shown below is or is not
that of limonene? Explain.
CH3-CH=CH-CH2-CH=CH-CH=CH-CH2-CH3
The condensed structural formula shown could be that of limonene, since it does contain 10 C atoms
and 16 H atoms. But we cannot be certain, since there are other structures that also match this
molecular formula.
*if you’re struggling with this problem, try the step-by-step problem on the following page first.
Chem 131
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Page 9
Step-by-Step Combustion Analysis Problem
After reading the appropriate section of the textbook (p. 171-173), try to solve each step of this
problem without using the hints. If, after thinking about the question, you’re still lost, read the
hint, and see if that helps. Note that parts of this question are similar to Example 4.4.
After you’ve successfully completed this problem, make sure you can do Study Questions 33-34,
which are similar, but don’t lead you step-by-step through the problem.
Complete combustion of 50.00g of an organic compound produced 151.44g of carbon dioxide
and 77.51g of water (and nothing else).
First, remember that the combustion involves the reaction of the organic compound with oxygen
(O2) to produce carbon dioxide (CO2) and water (H2O).
a) What are all of the possible elements that could be found in the organic compound?
C, H, and O, since these are the only elements contained in the products.
b) What mass of oxygen was consumed during the combustion reaction?
Note that the masses of all other substances have been given. From conservation of
mass, we know that total mass of reactants must equal the total mass of products, so:
Mass of oxygen = 151.44g + 77.51g – 50.00g = 178.95g
c) How many moles of oxygen atoms is this?
⎛ 1 mol O ⎞
⎟⎟ = 11.185mol O
178.95 g O ⎜⎜
15
.
9994
g
O
⎝
⎠
d) How many moles of oxygen and of carbon atoms are contained in the 151.44g of carbon
dioxide?
⎛ 1 mol CO2 ⎞⎛ 2 mol O ⎞
⎟⎟⎜⎜
⎟⎟ = 6.8821mol O
151.44 g CO2 ⎜⎜
⎝ 44.010 g CO2 ⎠⎝ 1 mol CO2 ⎠
⎛ 1 mol CO2 ⎞⎛ 1 mol C ⎞
⎟⎟⎜⎜
⎟⎟ = 3.4411mol C
151.44 g CO2 ⎜⎜
⎝ 44.010 g CO2 ⎠⎝ 1 mol CO2 ⎠
e) How many moles of oxygen and of hydrogen atoms are contained in the 77.51g of water?
⎛ 1 mol H 2O ⎞⎛ 1 mol O ⎞
⎟⎟⎜⎜
⎟⎟ = 4.302mol O
77.51g H 2O ⎜⎜
⎝ 18.015 g H 2O ⎠⎝ 1 mol H 2O ⎠
⎛ 1 mol H 2O ⎞⎛ 2 mol H ⎞
⎟⎟⎜⎜
⎟⎟ = 8.605mol H
77.51g H 2O ⎜⎜
18
.
015
1
g
H
O
mol
H
O
2
2
⎝
⎠⎝
⎠
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Page 10
f) Is there any oxygen in the organic compound?
All of the O atoms in CO2 and in H2O must have come from either the O2 or the organic
compound. We know the quantity of O atoms in all but the organic compound, so we
can compute this by difference:
4.302 + 6.8821 – 11.185 = 0.000 mol O
No, all O accounted for by H2O, CO2, and O2
g) How many moles of carbon and of hydrogen atoms are contained in the organic
compound?
All of the C and H atoms in the products (CO2 and H2O) must have come from the
organic compound, since the other reactant (O2) does not contain these atoms.
3.4411 mol C and 8.605 mol H
h) What is the empirical formula of the organic compound?
The empirical formula is just the simplest whole number ratio of the elements contained
in the compound, which in this case is just C and H. To find the ratio, divide the larger
by the smaller number, and then multiply by whatever common factor is needed to just
convert both to integers (or very nearly so):
⎛ 8.605 mol H ⎞ 2.500 mol H
⎜⎜
⎟⎟ =
1 mol C
⎝ 3.4411mol C ⎠
⎛ 2 ⎞ 5.000 mol H
, so empirical formula is C2H5
⎜ ⎟=
2 mol C
⎝2⎠
i) If it is known that the molar mass of the compound is approximately 58g/mol, what is the
compound’s molecular formula?
By definition, the molecular formula is an integer multiple of the empirical formula, so the
molar mass of the compound must be an integer multiple of the formula mass of C2H5,
which is 29.06 g/mol:
⎛ 58 g mol ⎞
⎜⎜
⎟⎟ = 2.0 , so the molecular formula must be twice the empirical formula, or C4 H10.
g
mol
29
⎝
⎠
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