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7: Writing Formulas from Names

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



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



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





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. ■



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