Tải bản đầy đủ - 0 (trang)
5: Naming Compounds That Contain Polyatomic Ions

5: Naming Compounds That Contain Polyatomic Ions

Tải bản đầy đủ - 0trang

Fresh

water

Membrane



Permission to redraw this illustration given by Poseidon Resources Corporation.



Seawater



2. The pretreated seawater

is forced through a dense,

semipermeable membrane

at extremely high pressure,

which separates salt and

minerals from fresh water.

The concentrated brine

residue is discharged back

into the sea at the end of

the cycle.



Discharge



1. Seawater enters the plant

through a pretreatment filtering

system that removes coarser

particles, sand, sediment, and dirt.



Fresh



3. Fresh water is stored in

a reservoir for later use by

the municipal water system.



A schematic diagram of the desalination plant in Carlsbad, California.



particle. This will occur only if the lost water–water interactions are replaced

by similar water–solute interactions. In the case of sodium chloride, strong

interactions occur between the polar water molecules and the Naϩ and ClϪ

ions. This allows the sodium chloride to dissolve. In the case of ethanol or

sucrose, hydrogen-bonding interactions can occur between the OOH groups

on these molecules and water molecules, making these substances soluble as

well. But oil molecules are not soluble in water, because the many water–

water interactions that would have to be broken to make “holes” for these

large molecules are not replaced by favorable water–solute interactions.

These considerations account for the observed behavior that “like dissolves like.” In other words, we observe that a given solvent usually dissolves

solutes that have polarities similar to its own. For example, water dissolves

most polar solutes, because the solute–solvent interactions formed in the solution are similar to the water–water interactions present in the pure solvent.

Likewise, nonpolar solvents dissolve nonpolar solutes. For example, drycleaning solvents used for removing grease stains from clothes are nonpolar

liquids. “Grease” is composed of nonpolar molecules, so a nonpolar solvent

is needed to remove a grease stain.



479



C H E M I S T R Y I N F OCUS

Green Chemistry



is not a hazard to the general public (little PERC

adheres to dry-cleaned garments), it represents a

major concern for employees in the dry-cleaning

industry. At high pressures CO2 is a liquid that,

when used with appropriate detergents, is a very

effective solvent for the soil found on dry-cleanonly fabrics. When the pressure is lowered, the

CO2 immediately changes to its gaseous form,

quickly drying the clothes without the need for

added heat. The gas can then be condensed and

reused for the next batch of clothes.

The good news is that green chemistry

makes sense economically. When all of the costs

are taken into account, green chemistry is usually

cheaper chemistry as well. Everybody wins.



culprits in the past for fouling the earth’s environment, that situation is rapidly changing. In fact, a

quiet revolution is sweeping through chemistry

from academic labs to Fortune 500 companies.

Chemistry is going green. Green chemistry means

minimizing hazardous wastes, substituting water

and other environmentally friendlier substances

for traditional organic solvents, and manufacturing products out of recyclable materials.

A good example of green chemistry is the

increasing use of carbon dioxide, one of the byproducts of the combustion of fossil fuels. For

example, the Dow Chemical Company is now using CO2 rather than chlorofluorocarbons (CFCs;

substances known to catalyze the decomposition of protective stratospheric ozone) to put

the “sponginess” into polystyrene egg cartons,

meat trays, and burger boxes. Dow does not

generate CO2 for this process but instead uses

waste gases captured from its various manufacturing processes.

Another very promising use of carbon dioxide is as a replacement for the solvent perchloroethylene (PERC; Cl

Cl), now used by

C C

Cl

Cl

about 80% of dry cleaners in the United States.

Chronic exposure to PERC has been linked to kidney and liver damage and cancer. Although PERC



Michael Newman/PhotoEdit



Although some chemical industries have been



The dry-cleaning agent PERC is a health concern for

workers in the dry-cleaning industry.



15.2 Solution Composition: An Introduction

OBJECTIVE:



To learn qualitative terms associated with the concentration of a solution.

Even for very soluble substances, there is a limit to how much solute can be

dissolved in a given amount of solvent. For example, when you add sugar to

a glass of water, the sugar rapidly disappears at first. However, as you continue to add more sugar, at some point the solid no longer dissolves but collects at the bottom of the glass. When a solution contains as much solute as

will dissolve at that temperature, we say it is saturated. If a solid solute is



480



15.3 Solution Composition: Mass Percent



481



added to a solution already saturated with that solute, the added solid does

not dissolve. A solution that has not reached the limit of solute that will dissolve in it is said to be unsaturated. When more solute is added to an unsaturated solution, it dissolves.

Although a chemical compound always has the same composition, a

solution is a mixture and the amounts of the substances present can vary in

different solutions. For example, coffee can be strong or weak. Strong coffee

has more coffee dissolved in a given amount of water than weak coffee. To

describe a solution completely, we must specify the amounts of solvent and

solute. We sometimes use the qualitative terms concentrated and dilute to describe a solution. A relatively large amount of solute is dissolved in a concentrated solution (strong coffee is concentrated). A relatively small

amount of solute is dissolved in a dilute solution (weak coffee is dilute).

Although these qualitative terms serve a useful purpose, we often need

to know the exact amount of solute present in a given amount of solution.

In the next several sections, we will consider various ways to describe the

composition of a solution.



15.3 Solution Composition: Mass Percent

OBJECTIVE:



To understand the concentration term mass percent and learn how to

calculate it.

Describing the composition of a solution means specifying the amount of

solute present in a given quantity of the solution. We typically give the

amount of solute in terms of mass (number of grams) or in terms of moles.

The quantity of solution is defined in terms of mass or volume.

One common way of describing a solution’s composition is mass percent (sometimes called weight percent), which expresses the mass of solute

present in a given mass of solution. The definition of mass percent is



The mass of the solution is the

sum of the masses of the solute

and the solvent.



mass of solute

ϫ 100%

mass of solution

grams of solute

ϭ

ϫ 100%

grams of solute ϩ grams of solvent



Mass percent ϭ



For example, suppose a solution is prepared by dissolving 1.0 g of

sodium chloride in 48 g of water. The solution has a mass of 49 g (48 g of

H2O plus 1.0 g of NaCl), and there is 1.0 g of solute (NaCl) present. The mass

percent of solute, then, is

1.0 g solute

ϫ 100% ϭ 0.020 ϫ 100% ϭ 2.0% NaCl

49 g solution



EXAMPLE 15.1



Solution Composition: Calculating Mass Percent

A solution is prepared by mixing 1.00 g of ethanol, C2H5OH, with 100.0 g of

water. Calculate the mass percent of ethanol in this solution.

SOLUTION

Where Are We Going?

We want to determine the mass percent of a given ethanol solution.



482 Chapter 15 Solutions

What Do We Know?

• We have 1.00 g ethanol (C2H5OH) in 100.0 g water (H2O).

• Mass percent ϭ



mass of solute

ϫ 100%.

mass of solution



How Do We Get There?

In this case we have 1.00 g of solute (ethanol) and 100.0 g of solvent (water).

We now apply the definition of mass percent.

Mass percent C2H5OH ϭ a



grams of C2H5OH

b ϫ 100%

grams of solution

1.00 g C2H5OH

ϭa

b ϫ 100%

100.0 g H2O ϩ 1.00 g C2H5OH

1.00 g

ϭ

ϫ 100%

101.0 g

ϭ 0.990% C2H5OH



R E A L I T Y C H E C K The percent mass is just under 1%, which makes sense

because we have 1.00 g of ethanol in a bit more than 100.0 g of solution.



Self-Check EXERCISE 15.1 A 135-g sample of seawater is evaporated to dryness, leaving 4.73 g of solid

residue (the salts formerly dissolved in the seawater). Calculate the mass percent of solute present in the original seawater.

See Problems 15.15 and 15.16. ■



EXAMPLE 15.2



Solution Composition: Determining Mass of Solute

Although milk is not a true solution (it is really a suspension of tiny globules

of fat, protein, and other substrates in water), it does contain a dissolved

sugar called lactose. Cow’s milk typically contains 4.5% by mass of lactose,

C12H22O11. Calculate the mass of lactose present in 175 g of milk.

SOLUTION

Where Are We Going?

We want to determine the mass of lactose present in 175 g of milk.

What Do We Know?

• We have 175 g of milk.

• Milk contains 4.5% by mass of lactose, C12H22O11.

• Mass percent ϭ



mass of solute

ϫ 100%.

mass of solution



How Do We Get There?

Using the definition of mass percent, we have

Mass percent ϭ



grams of solute

ϫ 100%

grams of solution



15.4 Solution Composition: Molarity



483



We now substitute the quantities we know:

Mass of lactose



Mass percent ϭ



Mass percent



grams of solute

ϫ 100% ϭ 4.5%

175 g

Mass of milk



We now solve for grams of solute by multiplying both sides by 175 g,

175 g ϫ



grams of solute

ϫ 100% ϭ 4.5% ϫ 175 g

175 g



and then dividing both sides by 100%,

Grams of solute ϫ



100%

4.5%

ϭ

ϫ 175 g

100%

100%



to give

Grams of solute ϭ 0.045 ϫ 175 g ϭ 7.9 g lactose



Self-Check EXERCISE 15.2 What mass of water must be added to 425 g of formaldehyde to prepare a

40.0% (by mass) solution of formaldehyde? This solution, called formalin, is

used to preserve biological specimens.

HINT: Substitute the known quantities into the definition for mass percent,

and then solve for the unknown quantity (mass of solvent).

See Problems 15.17 and 15.18. ■



15.4 Solution Composition: Molarity

OBJECTIVES:



To understand molarity. • To learn to use molarity to calculate the

number of moles of solute present.

When a solution is described in terms of mass percent, the amount of solution is given in terms of its mass. However, it is often more convenient to

measure the volume of a solution than to measure its mass. Because of this,

chemists often describe a solution in terms of concentration. We define the

concentration of a solution as the amount of solute in a given volume of solution. The most commonly used expression of concentration is molarity

(M). Molarity describes the amount of solute in moles and the volume of the

solution in liters. Molarity is the number of moles of solute per volume of solution in liters. That is

M ϭ molarity ϭ



moles of solute

mol

ϭ

liters of solution

L



A solution that is 1.0 molar (written as 1.0 M) contains 1.0 mole of solute per

liter of solution.



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



Tài liệu bạn tìm kiếm đã sẵn sàng tải về

5: Naming Compounds That Contain Polyatomic Ions

Tải bản đầy đủ ngay(0 tr)

×