# Chapter 6 worked problems.

```Worked Example 6.1
Molar Mass and Avogadro’s Number: Number of Molecules
Pseudoephedrine hydrochloride
is a nasal decongestant commonly found in cold medication.
(a) What is the molar mass of pseudoephedrine hydrochloride? (b) How many molecules of pseudoephedrine
hydrochloride are in a tablet that contains 30.0 mg of this decongestant?
Analysis
We are given a mass and need to convert to a number of molecules. This is most easily
accomplished by using the molar mass of pseudoephedrine hydrochloride calculated in part (a)
as the conversion factor from mass to moles and realizing that this mass (in grams) contains
.
Ballpark Estimate The formula for pseudoephedrine contains 10 carbon atoms (each one of atomic weight 12.0
amu), so the molecular weight is greater than 120 amu, probably near 200 amu. Thus, the
molecular weight should be near 200 g/mol. The mass of 30 mg of pseudoepinephrine HCl is
less than the mass of 1 mol of this compound by a factor of roughly
(0.03 g versus 200 g),
which means that the number of molecules should also be smaller by a factor of
(on
the order of
in the tablet versus
in 1 mol).
Solution
(a) The molecular weight of pseudoephedrine is found by summing the atomic weights of all atoms in the molecule:
_____________________________________________
Remember that atomic mass in amu converts directly to molar mass in g/mol. Also, following the rules for
significant figures from Sections 1.9 and 1.11, our final answer is rounded to the second decimal place.
Fundamentals of General, Organic, and Biological Chemistry, 7e
John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson
(b) Since this problem involves unit conversions, we can use the step-wise solution introduced in Chapter 1.
STEP 1: Identify known information. We are given the mass of
pseudoephedrine hydrochloride (in mg).
30.0 mg pseudoephedrine hydrochloride
STEP 2: Identify answer and units. We are looking for the number
of molecules of pseudoephedrine hydrochloride in a 30 mg tablet.
STEP 3: Identify conversion factors. Since the molecular weight of
pseudoephedrine hydrochloride is 201.70 amu, 201.70 g contains
molecules. We can use this ratio as a conversion factor
to convert from mass to molecules. We will also need to convert 30
mg to g.
STEP 4: Solve. Set up an equation
so that unwanted units cancel.
Ballpark Estimate Our estimate for the number of molecules was on the
order of 1019, which is consistent with the calculated
Fundamentals of General, Organic, and Biological Chemistry, 7e
John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson
Worked Example 6.2 Avogadro’s Number: Atom to Mass Conversions
A tiny pencil mark just visible to the naked eye contains about
pencil mark in grams?
atoms of carbon. What is the mass of this
Analysis We are given a number of atoms and need to convert to mass. The conversion factor can be
obtained by realizing that the atomic weight of carbon in grams contains Avogadro’s number of
atoms
.
Ballpark Estimate Since we are given a number of atoms that is six orders of magnitude less than Avogadro’s
number, we should get a corresponding mass that is six orders of magnitude less than the
molar mass of carbon, which means a mass for the pencil mark of about
g.
Solution
STEP 1: Identify known information. We know the
number of carbon atoms in the pencil mark.
atoms of carbon
STEP 2: Identify answer and units.
STEP 3: Identify conversion factors. The atomic
weight of carbon is 12.01 amu, so 12.01 g of carbon
contains
atoms.
STEP 4: Solve. Set up an equation using the conversion
factors so that unwanted units cancel.
Ballpark Check The answer is of the same magnitude as
our estimate and makes physical sense.
Fundamentals of General, Organic, and Biological Chemistry, 7e
John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson
Worked Example 6.3 Molar Mass: Mole to Gram Conversion
The nonprescription pain relievers Advil and Nuprin contain ibuprofen
, whose molecular weight is
206.3 amu (Problem 6.1a). If all the tablets in a bottle of pain reliever together contain 0.082 mol of ibuprofen,
what is the number of grams of ibuprofen in the bottle?
Analysis We are given a number of moles and asked to find the mass. Molar mass is the conversion factor
between the two.
Ballpark Estimate
Since 1 mol of ibuprofen has a mass of about 200 g, 0.08 mol has a mass of about
0.08 × 200 g = 16 g.
Solution
STEP 1: Identify known information.
STEP 2: Identify answer and units.
STEP 3: Identify conversion factors. We use the
molecular weight of ibuprofen to convert from moles
to grams.
STEP 4: Solve. Set up an equation using the known
information and conversion factor so that unwanted
units cancel.
Ballpark Check The calculated answer is consistent
with our estimate of 16 g.
Fundamentals of General, Organic, and Biological Chemistry, 7e
John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson
Worked Example 6.4 Molar Mass: Gram to Mole Conversion
The maximum dose of sodium hydrogen phosphate
that should be taken
in one day for use as a laxative is 3.8 g. How many moles of sodium hydrogen phosphate, how many moles of
ions, and how many total moles of ions are in this dose?
Analysis Molar mass is the conversion factor between mass and number of moles. The chemical formula
shows that each formula unit contains 2
ions and
ion.
Ballpark Estimate The maximum dose is about two orders of magnitude smaller than the molecular
weight (approximately 4 g compared to 142 g). Thus, the number of moles of
sodium hydrogen phosphate in 3.8 g should be about two orders of magnitude less
than one mole. The number of moles of
and total moles of ions, then,
should be on the order of 10-2.
Solution
STEP 1: Identify known information. We are given
the mass and molecular weight of
.
STEP 2: Identify answer and units. We need to find
the number of moles of
, and the total
number of moles of ions.
STEP 3: Identify conversion factors. We can use the
molecular weight of
to convert from grams
to moles.
Fundamentals of General, Organic, and Biological Chemistry, 7e
John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson
STEP 4: Solve. We use the known information and
conversion factor to obtain moles of
; since
1 mol of
contains 2 mol of
ions and 1
mol of
ions, we multiply these values by the
number of moles in the sample.
Ballpark Check The calculated answers (0.027 mol
order of
Fundamentals of General, Organic, and Biological Chemistry, 7e
John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson
, 0.081 mol ions) are on the
consistent with our estimate.
10-2,
Worked Example 6.5 Balanced Chemical Equations: Mole Ratios
Rusting involves the reaction of iron with oxygen to form iron (III) oxide,
:
(a) What are the mole ratios of the product to each reactant and of the reactants to each other?
(b) How many moles of iron (III) oxide are formed by the complete oxidation of 6.2 mol of iron?
Analysis and Solution
(a) The coefficients of a balanced equation represent the mole ratios:
(b) To find how many moles of
are formed, write down the known information—6.2 mol of iron—and select
the mole ratio that allows the quantities to cancel, leaving the desired quantity:
Note that mole ratios are exact numbers and therefore do not limit the number of significant figures in the result of a
calculation.
Fundamentals of General, Organic, and Biological Chemistry, 7e
John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson
Worked Example 6.6 Mole Ratios: Mole to Mass Conversions
In the atmosphere, nitrogen dioxide reacts with water to produce NO and nitric acid, which contributes to pollution
by acid rain:
How many grams of
is 63.0 amu.
are produced for every 1.0 mol of NO2 that reacts? The molecular weight of
Analysis We are given the number of moles of a reactant and are asked to find the mass of a product.
Problems of this sort always require working in moles and then converting to mass, as outlined in
Figure 6.2 .
Ballpark Estimate The molar mass of nitric acid is approximately 60 g/mol, and the coefficients
in the balanced equation say that 2 mol of
are formed for each 3 mol
of
that undergo reaction. Thus, 1 mol of
Solution
mol
, or 2/3 mol × 60 g/mol = 40 g.
STEP 1: Write balanced equation.
STEP 2: Identify conversion factors. We need a mole to mole conversion
to find the number of moles of product, and then a mole to mass conversion
to find the mass of product. For the first conversion we use the mole ratio of
to
as a conversion factor, and for the mole to mass calculation,
we use the molar mass of
(63.0 g/mol) as a conversion factor.
Fundamentals of General, Organic, and Biological Chemistry, 7e
John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson
STEP 3: Set up factor labels. Identify appropriate mole
ratio factor labels to convert moles
to moles
, and moles
to grams.
STEP 4: Solve.
Ballpark Check Our estimate was 40 g!
Fundamentals of General, Organic, and Biological Chemistry, 7e
John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson
Worked Example 6.7 Mole Ratios: Mass to Mole/Mole to Mass Conversions
The following reaction produced 0.022 g of calcium oxalate
. What mass of calcium chloride was used
as reactant? (The molar mass of
is 128.1 g/mol, and the molar mass of
is 111.0 g/mol.)
Analysis Both the known information and that to be found are masses, so this is a mass to mass conversion
problem. The mass of
is first converted into moles, a mole ratio is used to find moles of
, and the number of moles of
is converted into mass.
is formed for each mole
Ballpark Estimate The balanced equation says that 1 mol of
of
that reacts. Because the formula weights of the two substances are
similar, it should take about 0.02 g of
to form 0.02 g of
.
Solution
STEP 1: Write balanced equation.
STEP 2: Identify conversion factors. Convert the mass of
into moles, use a mole ratio to find moles of
,
and convert the number of moles of
to mass. We will
need three conversion factors.
Fundamentals of General, Organic, and Biological Chemistry, 7e
John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson
STEP 3: Set up factor labels. We will need to perform
gram to mole and mole to mole conversions to get from
grams
to grams
.
STEP 4: Solve.
Ballpark Check The calculated answer (0.019 g) is
consistent with our estimate (0.02 g).
Fundamentals of General, Organic, and Biological Chemistry, 7e
John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson
Worked Example 6.8 Percent Yield
The combustion of acetylene gas
produces carbon dioxide and water, as indicated in the following reaction:
When 26.0 g of acetylene is burned in sufficient oxygen for complete reaction, the theoretical yield of
Calculate the percent yield for this reaction if the actual yield is only 72.4 g
.
is 88.0 g.
Analysis The percent yield is calculated by dividing the actual yield by the theoretical yield and multiplying
by 100.
Ballpark Estimate The theoretical yield (88.0 g) is close to 100 g. The actual yield (72.4 g) is
about 15 g less than the theoretical yield. The actual yield is thus about 15%
less than the theoretical yield, so the percent yield is about 85%.
Solution
Ballpark Estimate The calculated percent yield agrees very well with our estimate of 85%.
Fundamentals of General, Organic, and Biological Chemistry, 7e
John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson
Worked Example 6.9
Mass to Mole Conversions: Limiting Reagent and Theoretical Yield
The element boron is produced commercially by the reaction of boric oxide with magnesium at high temperature:
What is the theoretical yield of boron when 2350 g of boric oxide is reacted with 3580 g of magnesium? The molar
masses of boric oxide and magnesium are 69.6 g/mol and 24.3 g/mol, respectively.
Analysis To calculate theoretical yield, we first have to identify the limiting reagent. The theoretical yield in
grams is then calculated from the amount of limiting reagent used in the reaction. The calculation
involves the mass to mole and mole to mass conversions discussed in the preceding section.
Solution
STEP 1: Identify known information. We have
the masses and molar masses of the reagents.
STEP 2: Identify answer and units. We are solving
for the theoretical yield of boron.
STEP 3: Identify conversion factors. We can use the molar
masses to convert from masses to moles of reactants (
,
Mg). From moles of reactants, we can use mole ratios from
the balanced chemical equation to find the number of moles of B
produced, assuming complete conversion of a given reactant.
is the limiting reagent, since complete conversion of this
reagent yields less product (67.6 mol B formed) than does
complete conversion of Mg (98.0 mol B formed).
Fundamentals of General, Organic, and Biological Chemistry, 7e
John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson
STEP 4: Solve. Once the limiting reagent has been
Identified (
), the theoretical amount of B that
should be formed can be calculated using a mole to
mass conversion.
Fundamentals of General, Organic, and Biological Chemistry, 7e
John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson
Worked Example 6.10 Mass to Mole Conversion: Percent Yield
The reaction of ethylene with water to give ethyl alcohol
occurs with 78.5% actual yield. How many
grams of ethyl alcohol are formed by reaction of 25.0 g of ethylene? (For ethylene, MW = 28.0 amu; for ethyl
alcohol, MW = 46.0 amu. )
Analysis Treat this as a typical mass relationship problem to find the amount of ethyl alcohol that can
theoretically be formed from 25.0 g of ethylene, and then multiply the answer by 0.785 (the fraction
of the theoretical yield actually obtained) to find the amount actually formed.
Ballpark Estimate
The 25.0 g of ethylene is a bit less than 1 mol; since the percent yield is about 78%, a bit
less than 0.78 mol of ethyl alcohol will form—perhaps about 3/4 mol, or 3/4 × 46 g = 34
g.
Solution
The theoretical yield of ethyl alcohol is:
and so the actual yield is:
41.1 g ethyl alc. × 0.785 = 32.3 g ethyl alcohol
Ballpark Check
The calculated result (32.3 g) is close to our estimate (34 g).
Fundamentals of General, Organic, and Biological Chemistry, 7e
John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson
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