7: Writing Formulas from Names
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484 Chapter 15 Solutions
EXAMPLE 15.3
Solution Composition: Calculating Molarity, I
Calculate the molarity of a solution prepared by dissolving 11.5 g of solid
NaOH in enough water to make 1.50 L of solution.
SOLUTION
Where Are We Going?
We want to determine the concentration (M) of a solution of NaOH.
What Do We Know?
• 11.5 g of NaOH is dissolved in 1.50 L of solution.
• Mϭ
moles of solute
.
liters of solution
What Information Do We Need?
• We need to know the number of moles of NaOH in 11.5 g NaOH.
How Do We Get There?
We have the mass (in grams) of solute, so we need to convert the mass of
solute to moles (using the molar mass of NaOH). Then we can divide the
number of moles by the volume in liters.
Mass of
solute
Use molar
mass
Moles of
solute
Molarity
Moles
Liters
We compute the number of moles of solute, using the molar mass of NaOH
(40.0 g).
11.5 g NaOH ϫ
1 mol NaOH
ϭ 0.288 mol NaOH
40.0 g NaOH
Then we divide by the volume of the solution in liters.
Molarity ϭ
EXAMPLE 15.4
moles of solute
0.288 mol NaOH
ϭ
ϭ 0.192 M NaOH ■
liters of solution
1.50 L solution
Solution Composition: Calculating Molarity, II
Calculate the molarity of a solution prepared by dissolving 1.56 g of gaseous
HCl into enough water to make 26.8 mL of solution.
SOLUTION
Where Are We Going?
We want to determine the concentration (M) of a solution of HCl.
What Do We Know?
• 1.56 g of HCl is dissolved in 26.8 mL of solution.
• Mϭ
moles of solute
.
liters of solution
15.4 Solution Composition: Molarity
485
What Information Do We Need?
• We need to know the number of moles of HCl in 1.56 g.
• We need to know the volume of the solution in liters.
How Do We Get There?
We must change 1.56 g of HCl to moles of HCl, and then we must change
26.8 mL to liters (because molarity is defined in terms of liters). First we calculate the number of moles of HCl (molar mass ϭ 36.5 g).
1.56 g HCl ϫ
1 mol HCl
ϭ 0.0427 mol HCl
36.5 g HCl
ϭ 4.27 ϫ 10 Ϫ2 mol HCl
Next we change the volume of the solution from milliliters to liters, using
the equivalence statement 1 L ϭ 1000 mL, which gives the appropriate conversion factor.
26.8 mL ϫ
1L
ϭ 0.0268 L
1000 mL
ϭ 2.68 ϫ 10 Ϫ2 L
Finally, we divide the moles of solute by the liters of solution.
Molarity ϭ
4.27 ϫ 10 Ϫ2 mol HCl
2.68 ϫ 10 Ϫ2 L solution
ϭ 1.59 M HCl
Self-Check EXERCISE 15.3 Calculate the molarity of a solution prepared by dissolving 1.00 g of ethanol,
C2H5OH, in enough water to give a final volume of 101 mL.
See Problems 15.37 through 15.42. ■
It is important to realize that the description of a solution’s composition may not accurately reflect the true chemical nature of the solute as it
is present in the dissolved state. Solute concentration is always written in
terms of the form of the solute before it dissolves. For example, describing a
solution as 1.0 M NaCl means that the solution was prepared by dissolving
1.0 mole of solid NaCl in enough water to make 1.0 L of solution; it does
not mean that the solution contains 1.0 mole of NaCl units. Actually the solution contains 1.0 mole of Naϩ ions and 1.0 mole of ClϪ ions. That is, it contains 1.0 M Naϩ and 1.0 M ClϪ.
EXAMPLE 15.5
Remember, ionic compounds
separate into the component
ions when they dissolve in
water.
Co(NO3)2
FeCl3
Co2ϩ
Ϫ
NO3
NO3Ϫ
Fe3ϩ
Ϫ
Cl
ClϪ
Solution Composition: Calculating Ion
Concentration from Molarity
Give the concentrations of all the ions in each of the following solutions:
a. 0.50 M Co(NO3)2
b. 1 M FeCl3
SOLUTION
a. When solid Co(NO3)2 dissolves, it produces ions as follows:
ClϪ
H2O(l)
Co(NO3)2(s) — Co2ϩ(aq) ϩ 2NO3Ϫ(aq)
486 Chapter 15 Solutions
which we can represent as
H2O(l)
1 mol Co(NO3)2(s) — 1 mol Co2ϩ(aq) ϩ 2 mol NO3Ϫ(aq)
Therefore, a solution that is 0.50 M Co(NO3)2 contains 0.50 M Co2ϩ
and (2 ϫ 0.50) M NO3Ϫ, or 1.0 M NO3Ϫ.
b. When solid FeCl3 dissolves, it produces ions as follows:
H2O(l)
FeCl3(s) — Fe3ϩ(aq) ϩ 3ClϪ(aq)
or
H2O(l)
1 mol FeCl3(s) — 1 mol Fe3ϩ(aq) ϩ 3 mol ClϪ(aq)
A solution that is 1 M FeCl3 contains 1 M Fe3ϩ ions and 3 M ClϪ
ions.
Self-Check EXERCISE 15.4 Give the concentrations of the ions in each of the following solutions:
a. 0.10 M Na2CO3
b. 0.010 M Al2(SO4)3
See Problems 15.49 and 15.50. ■
MATH SKILL BUILDER
moles of solute
Mϭ
liters of solution
Liters ϫ M
Moles of solute
Often we need to determine the number of moles of solute present in a
given volume of a solution of known molarity. To do this, we use the definition of molarity. When we multiply the molarity of a solution by the volume (in liters), we get the moles of solute present in that sample:
Liters of solution ϫ molarity ϭ liters of solution ϫ
ϭ moles of solute
EXAMPLE 15.6
moles of solute
liters of solution
Solution Composition: Calculating Number
of Moles from Molarity
How many moles of Agϩ ions are present in 25 mL of a 0.75 M AgNO3
solution?
SOLUTION
Where Are We Going?
We want to determine the number of moles of Agϩ in a solution.
What Do We Know?
• We have 25 mL of 0.75 M AgNO3.
Tom Pantages
• Mϭ
A solution of cobalt(II) nitrate.
moles of solute
.
liters of solution
How Do We Get There?
A 0.75 M AgNO3 solution contains 0.75 M Agϩ ions and 0.75 M NO3Ϫ ions.
Next we must express the volume in liters. That is, we must convert from mL
to L.
15.4 Solution Composition: Molarity
25 mL ϫ
487
1L
ϭ 0.025 L ϭ 2.5 ϫ 10 Ϫ2 L
1000 mL
Now we multiply the volume times the molarity.
2.5 ϫ 10 Ϫ2 L solution ϫ
0.75 mol Ag ϩ
ϭ 1.9 ϫ 10 Ϫ2 mol Ag ϩ
L solution
Self-Check EXERCISE 15.5 Calculate the number of moles of ClϪ ions in 1.75 L of 1.0 ϫ 10Ϫ3 M AlCl3.
See Problems 15.49 and 15.50. ■
A standard solution is a solution whose concentration is accurately
known. When the appropriate solute is available in pure form, a standard solution can be prepared by weighing out a sample of solute, transferring it
completely to a volumetric flask (a flask of accurately known volume), and
adding enough solvent to bring the volume up to the mark on the neck of
the flask. This procedure is illustrated in Figure 15.7.
EXAMPLE 15.7
Solution Composition: Calculating Mass from Molarity
To analyze the alcohol content of a certain wine, a chemist needs 1.00 L of
an aqueous 0.200 M K2Cr2O7 (potassium dichromate) solution. How much
solid K2Cr2O7 (molar mass ϭ 294.2 g) must be weighed out to make this
solution?
SOLUTION
Where Are We Going?
We want to determine the mass of K2Cr2O7 needed to make a given solution.
What Do We Know?
• We want 1.00 L of 0.200 M K2Cr2O7.
Volume
marker
(calibration
mark)
Wash
bottle
Weighed
amount
of solute
a
Put a weighed amount of a
substance (the solute) into the
volumetric flask, and add a small
quantity of water.
b
Dissolve the solid in the water by
gently swirling the flask (with the
stopper in place).
Figure 15.7
Steps involved in the preparation of a standard aqueous solution.
c
Add more water (with gentle swirling)
until the level of the solution just reaches
the mark etched on the neck of the flask.
Then mix the solution thoroughly by
inverting the flask several times.
488 Chapter 15 Solutions
• The molar mass of K2Cr2O7 is 294.2 g/mol.
• Mϭ
moles of solute
.
liters of solution
How Do We Get There?
We need to calculate the number of grams of solute (K2Cr2O7) present (and thus
the mass needed to make the solution). First we determine the number of moles
of K2Cr2O7 present by multiplying the volume (in liters) by the molarity.
Liters ϫ M
Moles of solute
1.00 L solution ϫ
0.200 mol K2Cr2O7
ϭ 0.200 mol K2Cr2O7
L solution
Then we convert the moles of K2Cr2O7 to grams, using the molar mass of
K2Cr2O7 (294.2 g).
0.200 mol K2Cr2O7 ϫ
294.2 g K2Cr2O7
ϭ 58.8 g K2Cr2O7
mol K2Cr2O7
Therefore, to make 1.00 L of 0.200 M K2Cr2O7, the chemist must weigh out
58.8 g of K2Cr2O7 and dissolve it in enough water to make 1.00 L of solution.
This is most easily done by using a 1.00-L volumetric flask (see Figure 15.7).
Self-Check EXERCISE 15.6 Formalin is an aqueous solution of formaldehyde, HCHO, used as a preservative for biologic specimens. How many grams of formaldehyde must be
used to prepare 2.5 L of 12.3 M formalin?
See Problems 15.51 and 15.52. ■
15.5 Dilution
OBJECTIVE:
The molarities of stock solutions
of the common concentrated
acids are:
Sulfuric (H2SO4)
Nitric (HNO3)
Hydrochloric (HCl)
18 M
16 M
12 M
Dilution with water doesn’t alter
the number of moles of solute
present.
To learn to calculate the concentration of a solution made by diluting a
stock solution.
To save time and space in the laboratory, solutions that are routinely used are
often purchased or prepared in concentrated form (called stock solutions).
Water (or another solvent) is then added to achieve the molarity desired for
a particular solution. The process of adding more solvent to a solution is
called dilution. For example, the common laboratory acids are purchased
as concentrated solutions and diluted with water as they are needed. A typical dilution calculation involves determining how much water must be
added to an amount of stock solution to achieve a solution of the desired
concentration. The key to doing these calculations is to remember that only
water is added in the dilution. The amount of solute in the final, more dilute
solution is the same as the amount of solute in the original, concentrated
stock solution. That is,
Moles of solute after dilution ϭ moles of solute before dilution
The number of moles of solute stays the same but more water is added,
increasing the volume, so the molarity decreases.
Remains constant
moles of solute
Mϭ
volume (L)
Decreases
Increases
(water added)
15.5 Dilution
489
For example, suppose we want to prepare 500. mL of 1.00 M acetic acid,
HC2H3O2, from a 17.5 M stock solution of acetic acid. What volume of the
stock solution is required?
The first step is to determine the number of moles of acetic acid needed
in the final solution. We do this by multiplying the volume of the solution
by its molarity.
Volume of dilute molarity of
moles of solute
ϫ
ϭ
solution (liters)
dilute solution present
The number of moles of solute present in the more dilute solution
equals the number of moles of solute that must be present in the more concentrated (stock) solution, because this is the only source of acetic acid.
Because molarity is defined in terms of liters, we must first change
500. mL to liters and then multiply the volume (in liters) by the molarity.
500. mL solution ϫ
MATH SKILL BUILDER
Liters ϫ M
Moles of solute
1 L solution
ϭ 0.500 L solution
1000 mL solution
Vdilute solution
(in mL)
Convert mL to L
0.500 L solution ϫ
1.00 mol HC2H3O2
ϭ 0.500 mol HC2H3O2
L solution
Mdilute solution
Now we need to find the volume of 17.5 M acetic acid that contains
0.500 mole of HC2H3O2. We will call this unknown volume V. Because
volume ϫ molarity ϭ moles, we have
V (in liters) ϫ
17.5 mol HC2H3O2
ϭ 0.500 mol HC2H3O2
L solution
Solving for V aby dividing both sides by
Vϭ
17.5 mol
b gives
L solution
0.500 mol HC2H3O2
ϭ 0.0286 L, or 28.6 mL, of solution
17.5 mol HC2H3O2
L solution
Therefore, to make 500. mL of a 1.00 M acetic acid solution, we take 28.6 mL
of 17.5 M acetic acid and dilute it to a total volume of 500. mL. This process
is illustrated in Figure 15.8. Because the moles of solute remain the same before and after dilution, we can write
Initial Conditions
M1
Molarity
before
dilution
ϫ
V1
Final Conditions
ϭ moles of solute ϭ
Volume
before
dilution
M2
ϫ
Molarity
after
dilution
V2
Volume
after
dilution
We can check our calculations on acetic acid by showing that M1 ϫ V1 ϭ
M2 ϫ V2. In the above example, M1 ϭ 17.5 M, V1 ϭ 0.0286 L, V2 ϭ
0.500 L, and M2 ϭ 1.00 M, so
mol
ϫ 0.0286 L ϭ 0.500 mol
L
mol
M2 ϫ V2 ϭ 1.00
ϫ 0.500 L ϭ 0.500 mol
L
M1 ϫ V1 ϭ 17.5
and therefore
M1 ϫ V1 ϭ M2 ϫ V2
This shows that the volume (V2) we calculated is correct.
490 Chapter 15 Solutions
500 mL
a
Figure 15.8
EXAMPLE 15.8
c
b
28.6 mL of 17.5 M
acetic acid solution
is transferred to a
volumetric flask that
already contains some
water.
Water is added to the
flask (with swirling) to
bring the volume to the
calibration mark, and the
solution is mixed by
inverting the flask
several times.
The resulting solution is
1.00 M acetic acid.
Calculating Concentrations of Diluted Solutions
What volume of 16 M sulfuric acid must be used to prepare 1.5 L of a
0.10 M H2SO4 solution?
SOLUTION
Where Are We Going?
We want to determine the volume of sulfuric acid needed to prepare a given
volume of a more dilute solution.
What Do We Know?
Initial Conditions (concentrated)
M1 ϭ 16
mol
L
M2 ϭ 0.10
V1 ϭ ?
Tom Pantages
Final Conditions (dilute)
mol
L
V2 ϭ 1.5 L
Moles of solute ϭ M1 ϫ V1 ϭ M2 ϫ V2
Approximate dilutions can be
carried out using a calibrated
beaker. Here, concentrated
sulfuric acid is being added to
water to make a dilute solution.
How Do We Get There?
We can solve the equation
M1 ϫ V1 ϭ M2 ϫ V2
15.6 Stoichiometry of Solution Reactions
491
for V1 by dividing both sides by M1,
M1 ϫ V1
M2 ϫ V2
ϭ
M1
M1
to give
V1 ϭ
M2 ϫ V2
M1
Now we substitute the known values of M2, V2, and M1.
V1 ϭ
It is always best to add
concentrated acid to water,
not water to the acid. That
way, if any splashing occurs
accidentally, it is dilute acid
that splashes.
a0.10
mol
b (1.5 L)
L
ϭ 9.4 ϫ 10Ϫ3 L
mol
16
L
9.4 ϫ 10 Ϫ3 L ϫ
1000 mL
ϭ 9.4 mL
1L
Therefore, V1 ϭ 9.4 ϫ 10Ϫ3 L, or 9.4 mL. To make 1.5 L of 0.10 M H2SO4 using 16 M H2SO4, we must take 9.4 mL of the concentrated acid and dilute it
with water to a final volume of 1.5 L. The correct way to do this is to add the
9.4 mL of acid to about 1 L of water and then dilute to 1.5 L by adding more
water.
Self-Check EXERCISE 15.7 What volume of 12 M HCl must be taken to prepare 0.75 L of 0.25 M HCl?
See Problems 15.57 and 15.58. ■
15.6 Stoichiometry of Solution Reactions
OBJECTIVE:
To understand the strategy for solving stoichiometric problems for solution
reactions.
Because so many important reactions occur in solution, it is important to be
able to do stoichiometric calculations for solution reactions. The principles
needed to perform these calculations are very similar to those developed in
Chapter 9. It is helpful to think in terms of the following steps:
Steps for Solving Stoichiometric Problems Involving Solutions
See Section 7.3 for a discussion
of net ionic equations.
Step 1 Write the balanced equation for the reaction. For reactions involving
ions, it is best to write the net ionic equation.
Step 2 Calculate the moles of reactants.
Step 3 Determine which reactant is limiting.
Step 4 Calculate the moles of other reactants or products, as required.
Step 5 Convert to grams or other units, if required.
492 Chapter 15 Solutions
EXAMPLE 15.9
Solution Stoichiometry: Calculating Mass
of Reactants and Products
Calculate the mass of solid NaCl that must be added to 1.50 L of a 0.100 M
AgNO3 solution to precipitate all of the Agϩ ions in the form of AgCl. Calculate the mass of AgCl formed.
SOLUTION
Where Are We Going?
We want to determine the mass of NaCl.
What Do We Know?
• We have 1.50 L of a 0.100 M AgNO3 solution.
What Information Do We Need?
• We need the balanced equation between AgNO3 and NaCl.
• We need the molar mass of NaCl.
How Do We Get There?
This reaction was discussed in
Section 7.2.
MATH SKILL BUILDER
Liters ϫ M
Moles of solute
Step 1 Write the balanced equation for the reaction.
When added to the AgNO3 solution (which contains Agϩ and NO3Ϫ ions),
the solid NaCl dissolves to yield Naϩ and ClϪ ions. Solid AgCl forms according to the following balanced net ionic reaction:
Agϩ(aq) ϩ ClϪ(aq) S AgCl(s)
Step 2 Calculate the moles of reactants.
In this case we must add enough ClϪ ions to just react with all the Agϩ ions
present, so we must calculate the moles of Agϩ ions present in 1.50 L of a
0.100 M AgNO3 solution. (Remember that a 0.100 M AgNO3 solution contains 0.100 M Agϩ ions and 0.100 M NO3Ϫ ions.)
1.50 L ϫ
0.100 mol Ag ϩ
ϭ 0.150 mol Ag ϩ
L
Moles of Agϩ present
in 1.50 L of 0.100 M AgNO3
Richard Megna/Fundamental Photographs
Step 3 Determine which reactant is limiting.
In this situation we want to add just enough ClϪ to react with the Agϩ present. That is, we want to precipitate all the Agϩ in the solution. Thus the Agϩ
present determines the amount of ClϪ needed.
Step 4 Calculate the moles of ClϪ required.
We have 0.150 mole of Agϩ ions and, because one Agϩ ion reacts with one
ClϪ ion, we need 0.150 mole of ClϪ,
0.150 mol Agϩ ϫ
When aqueous sodium chloride is
added to a solution of silver
nitrate, a white silver chloride
precipitate forms.
1 mol ClϪ
1 mol Agϩ
ϭ 0.150 mol ClϪ
so 0.150 mole of AgCl will be formed.
0.150 mol Agϩ ϩ 0.150 mol ClϪ S 0.150 mol AgCl
15.6 Stoichiometry of Solution Reactions
493
Step 5 Convert to grams of NaCl required.
To produce 0.150 mol ClϪ, we need 0.150 mol NaCl. We calculate the mass
of NaCl required as follows:
0.150 mol NaCl ϫ
Moles
58.4 g NaCl
ϭ 8.76 g NaCl
mol NaCl
times molar mass
Mass
The mass of AgCl formed is
0.150 mol AgCl ϫ
EXAMPLE 15.10
See Section 7.2 for a discussion
of this reaction.
143.3 g AgCl
ϭ 21.5 g AgCl ■
mol AgCl
Solution Stoichiometry: Determining Limiting
Reactants and Calculating Mass of Products
When Ba(NO3)2 and K2CrO4 react in aqueous solution, the yellow solid BaCrO4
is formed. Calculate the mass of BaCrO4 that forms when 3.50 ϫ 10Ϫ3 mole of
solid Ba(NO3)2 is dissolved in 265 mL of 0.0100 M K2CrO4 solution.
SOLUTION
Where Are We Going?
We want to determine the mass of BaCrO4 that forms in a reaction of known
amounts of solutions.
What Do We Know?
• We react 3.50 ϫ 10Ϫ3 mol BaNO3 with 265 mL of 0.0100 M K2CrO4.
What Information Do We Need?
• We will need the balanced equation between BaNO3 and K2CrO4.
• We will need the molar mass of BaCrO4.
How Do We Get There?
Step 1 The original K2CrO4 solution contains the ions Kϩ and CrO42Ϫ.
When the Ba(NO3)2 is dissolved in this solution, Ba2ϩ and NO3Ϫ ions are
added. The Ba2ϩ and CrO42Ϫ ions react to form solid BaCrO4. The balanced
net ionic equation is
Ba2ϩ(aq) ϩ CrO42Ϫ(aq) S BaCrO4(s)
Step 2 Next we determine the moles of reactants. We are told that
3.50 ϫ 10Ϫ3 mole of Ba(NO3)2 is added to the K2CrO4 solution. Each formula unit of Ba(NO3)2 contains one Ba2ϩ ion, so 3.50 ϫ 10Ϫ3 mole of
Ba(NO3)2 gives 3.50 ϫ 10Ϫ3 mole of Ba2ϩ ions in solution.
Richard Megna/Fundamental Photographs
3.50 ϫ 10Ϫ3
mol Ba(NO3)2
dissolves to give
3.50 ϫ 10Ϫ3
mol Ba2ϩ
Because V ϫ M ϭ moles of solute, we can compute the moles of K2CrO4 in
the solution from the volume and molarity of the original solution. First we
must convert the volume of the solution (265 mL) to liters.
Barium chromate precipitating.
265 mL ϫ
1L
ϭ 0.265 L
1000 mL
494 Chapter 15 Solutions
Next we determine the number of moles of K2CrO4, using the molarity of the
K2CrO4 solution (0.0100 M).
0.265 L ϫ
0.0100 mol K2CrO4
ϭ 2.65 ϫ 10 Ϫ3 mol K2CrO4
L
We know that
2.65 ϫ 10Ϫ3
mol K2CrO4
2.65 ϫ 10Ϫ3
mol CrO42Ϫ
dissolves to give
so the solution contains 2.65 ϫ 10Ϫ3 mole of CrO42Ϫ ions.
Step 3 The balanced equation tells us that one Ba2ϩ ion reacts with one
CrO42Ϫ ion. Because the number of moles of CrO42Ϫ ions (2.65 ϫ 10Ϫ3) is
smaller than the number of moles of Ba2ϩ ions (3.50 ϫ 10Ϫ3), the CrO42Ϫ will
run out first.
Ba2ϩ(aq)
3.50 ϫ 10Ϫ3
mol
ϩ CrO42Ϫ(aq)
BaCrO4(s)
2.65 ϫ 10Ϫ3
mol
Smaller (runs
out first)
Therefore, the CrO42Ϫ is limiting.
Moles of
CrO42Ϫ
limits
Moles of
BaCrO4
Step 4 The 2.65 ϫ 10Ϫ3 mole of CrO42Ϫ ions will react with 2.65 ϫ 10Ϫ3 mole
of Ba2ϩ ions to form 2.65 ϫ 10Ϫ3 mole of BaCrO4.
2.65 ϫ 10Ϫ3
mol Ba2ϩ
ϩ
2.65 ϫ 10Ϫ3
mol CrO42Ϫ
2.65 ϫ 10Ϫ3
mol BaCrO4(s)
Step 5 The mass of BaCrO4 formed is obtained from its molar mass
(253.3 g) as follows:
2.65 ϫ 10 Ϫ3 mol BaCrO4 ϫ
253.3 g BaCrO4
ϭ 0.671 g BaCrO4
mol BaCrO4
Self-Check EXERCISE 15.8 When aqueous solutions of Na2SO4 and Pb(NO3)2 are mixed, PbSO4 precipitates. Calculate the mass of PbSO4 formed when 1.25 L of 0.0500 M Pb(NO3)2
and 2.00 L of 0.0250 M Na2SO4 are mixed.
HINT: Calculate the moles of Pb2ϩ and SO42Ϫ in the mixed solution, decide
which ion is limiting, and calculate the moles of PbSO4 formed.
See Problems 15.65 through 15.68. ■