1 Chapter 1-5 Review Brady & Senese 5th Ed. 1.5 Measurements are essential to describe properties2 Machines or instruments are used to take measurements

1

Chapter 1-5 Review

Brady & Senese 5th Ed

1.5 Measurements are essential to describe properties 2

• Machines or instruments are used to take measurements

• Generally require units

• Involve numbers that are inexact (estimated). This uncertainty is due to the limitations of the observer and the instruments used

Measurements:

1.6. Measurements always contain some uncertainty 3

• Because each measurement involves an estimate, measurements always have error.

• Record all measured numbers, including the first estimated digit

• These digits are called significant digits or significant figures

• Exact numbers have no error and do not affect sig figs in calculations

Measurement Error

4

Measured Numbers

Measurements are subject to error

For digital instruments record all digits displayed

For analog instruments use the markings to record as many digits as possible and estimate one digit beyond.


Significant Digits In A Measurement Are Limited By Instrument Precision

• Using the first thermometer, the temperature is 21.3 ºC (3 significant digits)

• Using the more precise (second) thermometer, the temperature is 21.32 ºC (4 significant digits)


Reducing Error:

• Errors can often be detected by making repeated measurements

• Error can be reduced by calibrating equipment

• The average or mean reduces data variations: it helps find a central value


• An accurate measurement is close to the true or correct value, a “hole-in-one”

• A precise measurement is close to the average of a series of repeated measurements

• When calibrated instruments are used properly, the greater the number of significant figures, the greater is the degree of precision for a given measurement

Accuracy vs. Precision


• All non-zero digits are significant

• Zeroes are only significant when: Between non-zero digits

To the right of non-zero digits in a number that contains a decimal point are significant (Trailing with a decimal point)

• Zeros to the left of the first nonzero digit are never counted as significant (Leading)

• Zeros at the end of a number without a decimal point are assumed not to be significant (Trailing without a decimal place)

Rules For Significant Figures (Sig Figs)


Learning Check: How Many Significant Figures Are There In The Following?

2.33 3

500.0 4

1000 1

.0500 3


Rules for combining measurements depend on the type of operation performed:

• Multiplication and division The number of sig. figs in the answer should not be

greater than the number of sig. figs in the factor with the fewest sig. figs

figs.) sig. (2

13

figs.) sig. (2figs.) sig. (4 figs.) sig. (3

0.642.751 3.14

Measurements Limit The Precision Of Calculated Results


The answer should have the same number of decimal places as the quantity with the fewest number of decimal places (least precise). You may also use columns if decimal places aren’t present.

3.247 ← 3 decimal places 41.36 ← 2 decimal places +125.2 ← 1 decimal place 169.8 ← answer rounded to 1 decimal place

Addition and Subtraction


Exact Numbers

• Numbers that come from definitions are exact and have no uncertainty

• Typically describe relationships between numbers within the same system (ie: 1m = 100cm)

• They do not affect sig figs in our calculations

13

Unit Conversions - Equalities on Food Labels

The contents of packaged foods

In the U.S. are listed as both metric and U.S. units.

Indicate the same amount of a substance in two different units.

14

Write the initial and final units.

Write a unit plan to convert the initial unit to the final unit.

Write equalities and conversion factors.

Use conversion factors to cancel the initial unit and provide the final unit.

Unit 1 x Unit 2 = Unit 2Unit 1

Initial x Conversion = Final unit factor unit

Problem Setup

15

Setting up a Problem

How many minutes are 2.5 hours?

Initial unit = 2.5 hr

Final unit = ? min

Plan = hr min

Setup problem to cancel hours (hr).

Initial Conversion Finalunit factor unit

2.5 hr x 60 min = 150 min (2 SF)

1 hr

16

Often, two or more conversion factors are required to obtain the unit needed for the answer.Unit 1 Unit 2 Unit 3

Additional conversion factors are placed in the setup to cancel each preceding unit Initial unit x factor 1 x factor 2 = Final unitUnit 1 x Unit 2 x Unit 3 = Unit 3

Unit 1 Unit 2

Using Two or More Factors

17

How many minutes are in 1.4 days? Initial unit: 1.4 days

Factor 1 Factor 2 Plan: days hr min

Set up problem: 1.4 days x 24 hr x 60 min = 2.0 x 103 min

1 day 1 hr

2 SF Exact Exact = 2 SF

Example: Problem Solving

18

Be sure to check your unit cancellation in the setup. The units in the conversion factors must cancel to

give the correct unit for the answer.

What is wrong with the following setup?1.4 day x 1 day x 1 hr

24 hr 60 min

Units = day2/min is not the unit needed

Units don’t cancel properly.

Check the Unit Cancellation

1.8. Density is a useful intensive property 19

Density (d)

• intensive property defined as the ratio of an object’s mass (m) to volume (v), d = m/v

• characteristic of pure substances at a specified temperature

• Since most substances expand when heated, densities decrease when heated.

• units : g/L for gases and g/mL for solids and liquids.

20

Densities of Common Substances

21

Volume by Displacement

A solid completely submerged in water displaces its own volume of water.

The volume of the solid is calculated from the volume difference.

45.0 mL - 35.5 mL

= 9.5 mL = 9.5 cm3

22

Density Using Volume Displacement

The density of the zinc object isthen calculated from its massand volume.

mass = 68.60 g = 7.2 g/cm3 volume 9.5 cm3

23

Sink or Float

In water,

Ice floats because the density of ice is less than the density of water.

Aluminum sinks because its density is greater than the density of water.

2.2 Atoms are composed of subatomic particles 24

Subatomic Particles

Particle Symbol Mass (u) Location Charge

electron 5.48579903(10-4) orbital 1-

proton 1.007276470 nucleus 1+

neutron 1.008664904 nucleus 0

-01 eor e

Hor 11

11 p

010 nor n


• Dalton’s atomic theory states that atoms of an element have a characteristic atomic mass or atomic weight measured in amu (u)

• Atomic masses units are based on a standard mass: 1/12 of an atom of Carbon-12 (the most common type of carbon atom)

Atomic Mass Unit


• All elements in nature are a mixtures of two or more forms with slightly different masses

• Atoms of the same element with different masses are called isotopes For example: there are naturally occurring 3 isotopes of

hydrogen and 4 isotopes of iron

• Isotopes have virtually identical properties other than mass, density and sometimes radioactivity

• Isotopes have the same number of protons but differing number of neutrons

Isotopes


Atomic Notationor Nuclear Symbol

• An element is a substance whose atoms all contain the identical number of protons, called the atomic number (Z)

• Isotopes are distinguished by mass number (A): Atomic number, Z = number of protons Mass number, A = (number of protons) + (number of

neutrons) Note that for atoms, A is greater than Z: the symbol is

top-heavy

• The term atom indicates neutral charge overall so the number of electrons = the number of protons

SyAZ


This information can be summarized: Number of protons = 92 ( = number of electrons) Number of neutrons = 143 Atomic number (Z) = 92 Mass number (A) = 92 + 143 = 235 Chemical symbol = U

Mass number, A (protons + neutrons) Chemical Symbol Atomic number, Z (number of protons)

235 U 92

Example: uranium-235

2.6 Molecular compounds contain neutral particles called molecules 29

Alkanes

• Alkanes are hydrocarbons (contain only C and H)

• Always have a ratio of atoms CnH2n+2

• Named using a prefix designating the number of C

• All have –ane suffix.

C

1

2

3

4

5

6

7

8

9

10

Prefix

Meth-

Eth-

Prop-

But-

Pent-

Hex-

Hept-

Oct-

Non-

Dec-

Suffix

-ane

Name

Methane

Ethane

Propane


Learning Check: Name that alkane

• ethane

• butane

• octane

Your turn!

Which of the following is heptane?

• C6H12

• C7H14

• C6H14

• C7H16

Your turn!

Which is the correct name for C4H10?

A. methane

B. ethane

C. propane

D. pentane

E. none of these butane


Other Organic Compounds

Alkenes- hydrocarbons with fewer H than the alkanes. CnH2n. Use the same prefixes, but have the suffix -ene. C2H4 : ___________ C3H6: _____________propene

ethene


Other Organic Compounds (Cont.)

Alcohols- Replace one H in an alkane with an -OH group

• Same prefixes, suffix becomes –anol CH3OH is ____________ C2H5OH is _______________

methanol

ethanol

Your Turn!

What is the name of CH3CH2CH2CH2OH?

A. butanol

B. propanol

C. pentanol

D. tetranol

E. none of these

Your Turn!

What is a formula for heptene?A. C6 H12

B. C7H14

C. C6H14

D. C7H16

E. none of these

2.7 Ionic compounds are composed of charged particles called ions 37

Ionic Compounds

• Positively charged ions are called cations

• Negatively charged ions are called anions

• subscripts in the formula always specify the smallest whole-number ratio of the ions needed to make a neutral combination (formula unit)

Fe O O2-Fe3+3232


What About Ions?

• Number of protons = number of e- if neutral• Number of protons < number of e- if negative• Number of protons > number of e- if positive• The number of protons never changes

How does Ca form Ca2+? Ca loses 2 electrons

How is N3- formed? N gains 3 electrons


Learning Check:

Fill in the blanks:

Symbol neutrons protons electrons60Co3+

81Br-

36 29 27

33 27 24

46 35 36265

29Cu


The Charges On Many Representative Elements Can Be Predicted

• Noble gases are especially stable • Main group elements will often gain or lose

electrons to have the same number of electrons as the nearest noble gas

• Metals form cations by losing electrons What is the expected charge on:Ca? Na?

• Nonmetals form anions by gaining electrons What is the expected charge on:N? O?

2+ +

3- 2-

2.8 The formulas of many ionic compounds can be predicted 41

Rules For Writing Formulas Of Ionic Compounds

• The cation (metal) is listed first in the formula• The subscripts in the formula must produce an

electrically neutral formula unit• The subscripts should be the set of smallest whole

numbers possible• The charges on the ions are not included in the

finished formula of the substance


Determining The Formula Of An Ionic Compound

• The “Criss-cross” rule often works as a shortcut for determining formula subscripts from charges of ions.

• If you choose this approach, make sure that the subscripts are reduced to the lowest whole number.

Al3+ O2- Mg2+ O2-

NH4+ PO4

3-

(NH4)+ (PO4)3- (NH4)3PO4

Mg2+ O2- MgO

Al2O3Al3+ O2-

Your Turn!

Which of the following is the correct formula for the formula unit composed of potassium and oxygen ions?

a) KO b) KO2 c) K2O d) none of these

Which of the following is the correct formula for the formula unit composed of Fe3+ and sulfide ions?

a) FeO b) Fe3O2 c) Fe2O3 d) none of these


Transition Metals:

Chromium Cr2+, Cr3+ Copper Cu+, Cu2+

Manganese Mn2+, Mn3+ Iron Fe2+, Fe3+

Cobalt Co2+, Co3+ Gold Au+, Au3+

Mercury Hg22+, Hg2+ etc…

Tin and Lead:

Tin Sn2+, Sn4+ Lead Pb2+, Pb4+

Exceptions:

Zinc Zn2+ Silver Ag+ Cadmium Cd2+

Transition Metals as well as Tin (Sn) and Lead (Pb) Generally Multiple Charges


Some Polyatomic Ions (Ions With Two Or More Atoms):

NH4+ Ammonium ion CO3

2- carbonate ion

OH- hydroxide ion H3O+ hydronium ion

NO2- nitrite ion SO3

2- sulfite ion

NO3- nitrate ion SO4

2- sulfate ion

ClO2- chlorite ion CrO4

2- chromate ion

ClO3- chlorate ion Cr2O7

2- dichromate ion

PO43- phosphate ion

2.9 Molecular and ionic comounds are named following a system 46

• Cations: If the metal forms only one positive ion, the cation

name is the English name for the metal If the metal forms more than one positive ion, the

cation name is the English name followed, without a space, by the numerical value of the charge written as a Roman numeral in parentheses

• Anions: monatomic anions are named by adding the “–ide”

suffix to the stem name for the element polyatomic ions use guide on next slide

The Stock System Of Naming Ionic Compounds

Naming Oxoanions

- Many nonmetals form two different oxoanions:- The ion with more oxygen atoms has the –ate suffix

- The ion with fewer oxygen atoms has the -ite suffix

- NO3- = Nitrate and NO2

- = Nitrite

- SO42- = Sulfate and SO3

2- = Sulfite

Some elements (generally halogens) form more than two oxoanions:

- The ion with the greatest oxygens has the per- prefix

- The ion with the least oxygens has the hypo- prefix

- ClO4- = Perchlorate, ClO3

- = Chlorate, ClO2- = Chlorite

and ClO- = Hypochlorite


Chemical Name as Name asSymbol Stem First Element Second Element O ox- oxygen oxide N nitr- nitrogen nitride P phosph- phosphorus phosphide Cl chlor- chlorine chloride I iod- iodine iodide

Naming Binary Molecules

The first element in the formula is identified by its English name, the second by appending the suffix –ide to its stem


Naming Binary Covalent Molecules

• Format:number prefix + 1st element name number prefix + stem_ide for 2nd element.

• Greek prefixes mono- = 1 (omitted on 1st atom) hexa- = 6

di- = 2 hepta- = 7

tri- = 3 octa- = 8

tetra- = 4 nona- = 9

penta- = 5 deca- = 10

EXAMPLES OF THE MOLE CONCEPT

1 mole H2O = 18.02 g H2O = 6.022x1023 H2O molecules

1 mole CO2 = 44.01 g CO2 = 6.022x1023 CO2 molecules

1 mole NH3 = 17.03 g NH3 = 6.022x1023 NH3 molecules

THE MOLE AND CHEMICAL CALCULATIONS

The mole concept can be used to obtain factors that are useful in chemical calculations involving both elements and compounds.

3.1 The mole conveniently links mass to number of atoms or molecules 51

Molar Mass

• One mole contains the same number of particles as the number of atoms in exactly 12 g of carbon-12

• The molar mass of a substance has the same numeric value as the formula mass

• The value is different because the units are different Thus if the formula mass of Ba3(PO4)2 is 610.332 u, the

molar mass of Ba3(PO4)2 is 610.332 g

Percentage Composition

Percentage

% A = 100

The percentage composition of a compound is the

percentage by mass of each element in the compound.

mass total Aof mass


Example:

Determine the percentage composition of table sugar, C12H22O11.

Solution:

In one mole of C12H22O11:

12(12.01 g C) + 22(1.008 g H) + 11(16.00 g O) =

342.30 g C12H22O11

(Solution continued on the next slide)


12(12.01 g C) + 22(1.008 g H) + 11(16.00 g O) =342.30 g

C12H22O11

100 = 42.10% C

100 = 6.479% H

100 = 51.42% O

Check: 42.10% + 6.479% + 51.42% = 100.00%

12(12.01 g C)342.30 g C

12H

22O

11

22(1.008 g H)342.30 g C

12H

22O

11

11(16.00 g O)342.30 g C

12H

22O

11

Empirical Formula

Empirical Formula

The lowest whole-number ratio of atoms of the elements in a compound.

The empirical formula of C2H4 is CH2. Likewise, the empirical formula of C6H12O6 is CH2O

All compounds with the general formula CnH2n have the same empirical formula and therefore the same percentage composition.

Compounds with Empirical Formula CH2O(Composition by Mass: 40.0% C, 6.71% H, 53.3% O)

NameMolecular Formula

Whole-Number Multiple

M(g/mol) Use or Function

formaldehyde

acetic acid

lactic acid

erythrose

ribose

glucose

CH2O

C2H4O2

C3H6O3

C4H8O4

C5H10O5

C6H12O6

1

2

3

4

5

6

30.03

60.05

90.09

120.10

150.13

180.16

disinfectant; biological preservative

acetate polymers; vinegar(5% soln)

part of sugar metabolism

sour milk; forms in exercising muscle

component of nucleic acids and B2

major energy source of the cell

CH2O C2H4O2 C3H6O3 C4H8O4 C5H10O5 C6H12O6

Empirical FormulaHow to Find an Empirical Formula

1. Find the masses of different elements in a sample of the compound.

2. Convert the masses into moles of atoms of the different elements.

3. Determine the ratio of moles of atoms by dividing through by the smallest number of moles.

4. Express the moles of the atoms as the smallest possible ratio of integers.

5. Write the empirical formula, using the number for each atom in the integer ratio as the subscript in the formula.

Empirical FormulaExample:

What is the empirical formula of a compound that analyzes as 79.95% carbon, 9.40% hydrogen, and 10.65% oxygen?

Solution:

It is usually helpful in an empirical formula problem to organize the calculations in a table with the following headings:

Mole FormulaEmpirical

Element Grams Moles Ratio Ratio Formula

Empirical Formula

Mole Formula EmpiricalElement Grams Moles Ratio Ratio Formula

C 79.95 10 6.657 10.00

H 9.40 14 C10H14O 9.33 14.0

O 10.65 1 0.6656 1.000

9.40 g H1.008 g/mol H

=

79.95 g C12.01 g/mol C

10.65 g O16.00 g/mol O

=

6.6570.6656

=

9.330.6656

=

0.66560.6656

=

mass(g) of each element

Determining the Empirical Formula

PROBLEM:

PLAN:

SOLUTION:

amount(mol) of each element

empirical formula

Elemental analysis of a sample of an ionic compound showed 2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. What are the empirical formula and name of the compound?

preliminary formula

change to integer subscripts

use # of moles as subscripts

divide by M(g/mol)

Once we find the relative number of moles of each element, we can divide by the lowest mol amount to find the relative mol ratios (empirical formula).

2.82 g Namol Na

22.99 g Na= 0.123 mol Na

4.35 g Clmol Cl

35.45 g Cl= 0.123 mol Cl

7.83 g Omol O

16.00 g O= 0.489 mol O

Na1 Cl1 O3.98 NaClO4Na1 Cl1 O3.98 NaClO4

NaClO4 is sodium perchlorate.

Molecular Formula

The molecular formula of a compound can be found by determination of the number of empirical formula units in the molecule.

For example, if the empirical formula of a compound is CH2, its molecular formula must be: CH2, C2H4, C3H6, C4H8, etc.

Empirical formula units in 1 molecule =

molar mass of compoundmolar mass of empirical formula

Molecular Formula

Example:What is the molecular formula of a compound with the empirical

formula C5H10O and a molar mass of 258 g/mol?

Solution:

The molar mass of the empirical formula unit, C5H10O, is5(12.01 g/mol C) + 10(1.008 g/mol H) + 1(16.00 g/mol O) =

86.13 g/mol C5H10OThe number of empirical formula units per molecule is

(C5H10O)3 = C15H30O3

258 g/mol molecule86.13 g/mol empirical formula unit

= 3

assume 100g lactic acid and find the mass of each element

Molecular Formula

PROBLEM:

PLAN:

amount(mol) of each element

During physical activity. lactic acid (M=90.08 g/mol) forms in muscle tissue and is responsible for muscle soreness. Elemental anaylsis shows that this compound contains 40.0 mass% C, 6.71 mass% H, and 53.3 mass% O.

(a) Determine the empirical formula of lactic acid.

(b) Determine the molecular formula.

preliminary formula

empirical formula

divide each mass by mol mass(M)

molecular formulause # mols as subscripts

convert to integer subscripts

divide mol mass by mass of empirical formula to get a

multiplier

Molecular Formula

continued

SOLUTION: Assuming there are 100. g of lactic acid, the constituents are

40.0 g C 6.71 g H 53.3 g Omol C

12.01g C

mol H

1.008 g H

mol O

16.00 g O

3.33 mol C 6.66 mol H 3.33 mol O

C3.33 H6.66 O3.33

3.33 3.33 3.33CH2O empirical formula

mass of CH2O

molar mass of lactate 90.08 g

30.03 g3 C3H6O3 is the

molecular formula

3.3 Chemical formulas can be determined from experimental mass measurements 65

Combustion Analysis:

• Empirical formulas may also be calculated indirectly

• When a compound made only from carbon, hydrogen, and oxygen burns completely in pure oxygen, only carbon dioxide and water are produced


Combustion Analysis:

Empirical formulas may be calculated from the analysis of combustion information grams of C can be derived from amount of CO2

grams of H can be derived from amount of H2O

the mass of oxygen is obtained by difference: g O = g sample – ( g C + g H )


Learning Check:

The combustion of a 5.217 g sample of a compound of C, H, and O gave 7.406 g CO2 and 4.512 g of H2O. Calculate the empirical formula of the compound.

2

2CO g 44.00956C g 12.0107

CO g .4067

H: 1.00794; C:12.0107; O: 15.99943

OH g 18.01531

H g 2.01588OH g 512.4

22 0.5048 g H

2.021 g C

5.217g - 2.021 g C - 0.5048 g H = 2.690 g O


Learning Check (con.):

Calculate the empirical formula of the compound.

H: 1.00794; C:12.0107; O: 15.99943

C H O

mass 2.02118 0.504884 2.69094

MM 12.0107 1.00794 15.99943

mol

low ratio

integer ratio 3 11

.500907 .16819.16828

2.97 11

CH3O

THE MOLE AND CHEMICAL EQUATIONS

• The mole concept can be applied to balanced chemical equations and used to calculate mass relationships in chemical reactions.

• Balanced equations can be interpreted in terms of the mole concept and the results used to provide factors for use in factor-unit solutions to numerical problems.

Calculating the Mass of Product

When 18.6 g ethane gas C2H6 burns in oxygen, how

many grams of CO2 are produced?

2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(g) 18.6 g ? g

The plan and factors would be

g C2H6 mole C2H6 mole CO2 g CO2

molar mole-mole molar

mass C2H6 factor mass CO2

Calculating the Mass of Product

The setup would be

18.6 g C2H6 x 1 mole C2H6 x 4 moles CO2 x 44.0 g CO2

30.1 g C2H6 2 moles C2H6 1 mole CO2

molar mole-mole molar

mass C2H6 factor mass CO2

= 54.4 g CO2

Theoretical, Actual, and Percent Yield

Theoretical yield

The maximum amount of product, which is calculated using the balanced equation.

Actual yield

The amount of product obtained when the reaction takes place.

Percent yield

The ratio of actual yield to theoretical yield.

percent yield = actual yield (g) x 100 theoretical yield (g)

You prepared cookie dough to make 5 dozen cookies.

The phone rings and you answer. While you talk, a sheet

of 12 cookies burn and you throw them out. The rest

of the cookies are okay. What is the percent yield of

edible cookies?

Theoretical yield 60 cookies possible

Actual yield 48 cookies to eat

Percent yield 48 cookies x 100 = 80% yield 60 cookies

Calculating Percent Yield

Limiting Reactant

A limiting reactant in a chemical reaction is the

substance that is used up first.

Limits the amount of product that can form and stops the reaction.

Example of An Everyday Limiting Reactant

• How many peanut butter sandwiches could be made from 8 slices of bread and 1 jar of peanut butter?

• With 8 slices of bread, only 4 sandwiches could be made. The bread is the limiting item.

Example of An Everyday Limiting Reactant

• How many peanut butter sandwiches could be made from 8 slices bread and 1 tablespoon of peanut butter?

• With 1 tablespoon of peanut butter, only 1 sandwich could be made. The peanut butter is the limiting item.

Limiting Reactants Using Mass

Calculate the mass of water produced when 8.00 g H2

and 24.0 g O2 react?

2H2(g) + O2(g) 2H2O(l)

Limiting Reactants Using Mass

Moles H2O from H2:

8.00 g H2 x 1 mole H2 x 2 moles H2O = 3.97 moles H2O

2.02 g H2 2 moles H2 (not possible)

Moles H2O from O2:

24.0 g O2 x 1 mole O2 x 2 moles H2O = 1.50 moles H2O

32.0 g O2 1 mole O2 O2 is limiting

The maximum amount of product is 1.50 moles H2O,which is converted to grams.

1.50 moles H2O x 18.0 g H2O = 27.0 g H2O

1 mole H2O

4.1. Special terminology applies to solutions 79

Solutions

• solution –a homogeneous mixture in which the two or more components mix freely

• solvent - the component present in the largest amount

• solute – the substance dissolved in the solvent. The solution is named by the solute.

• concentration - a solute-to-solvent or solute-to-solution ratio describing the composition of the mixture

Examples of Solutions

4.1. Special terminology applies to solutions 81

• saturated –no more solute can be dissolved at the current temperature in the given amount of solvent

• solubility - the amount of solute that can dissolve in the specified amount of solvent at a given temperature (usually g solute/ 100 g solvent or moles solute/L solution)

• unsaturated - contains less solute than the solubility allows

• supersaturated- contains more solute than solubility predicts

Solubility

Effect of Temperature on Solubility

•Solubility:•Depends on temperature.

•Of most solids increases as temperature increases.

•Of gases decreases as temperature increases.

Solubility and Pressure

•Henry’s Law states •The solubility of a gas in a liquid is directly related to the pressure of that gas above the liquid.

•At higher pressures, more gas molecules dissolve in the liquid.

Water as a polar solvent

•Polar molecules have an asymmetrical shape•Have an electron rich area (often containing lone pairs)•Have an electron poor area•Water is polar

O is electron richH is electron poor

Formation of a Solution

•Na+ and Cl- ions•On the surface of a NaCl crystal are attracted to polar water molecules.•Are hydrated in solution with many H2O molecules surrounding each ion.

• In water, • Strong electrolytes produce ions and conduct

an electric current. • Weak electrolytes produce a few ions. • Nonelectrolytes do not produce ions.

Solutes and Ionic Charge

4.2. Ionic compounds conduct electricity when dissolved in water 87

Ionic equations show dissociated ions

• hydrated ions, with the symbol (aq), are written separately

• Na2SO4(s) → 2Na+(aq) + SO4

2-(aq)

• you might encounter the equation as:

• Na2SO4(s) → 2Na+ + SO42-

Accepted because only 2 states allow for dissociated ions (plasma and aqueous). Aqueous is far more common

It is vague and not preferred


Writing chemical equations

• Molecular equation: Balanced, shows states, all substances electrically neutral AgNO3(aq) + KCl(aq) →AgCl(s) + KNO3(aq)

• Ionic equation: Balanced, shows states, shows strong electrolytes as

dissociated ions, net charges balance Ag+

(aq) + NO3-(aq) + K+

(aq) + Cl-(aq) →AgCl(s) + K+

(aq) + NO3-(aq)

• Net ionic equation: Balanced, shows states, eliminates spectator ions from the

ionic equation, net charges balance Ag+

(aq) + Cl-(aq) →AgCl(s)


Learning check:

• Write the ionic equations for each:

• BaCl2(aq) + Pb(NO3)2(aq)→PbCl2(s) + Ba(NO3)2(aq)

• Ba2+(aq) + 2Cl-

(aq) + Pb2+(aq) + 2NO3

-(aq) →PbCl2(s) +

Ba2+(aq) + 2NO3

-(aq)

• Na2CO3(aq) +CaCl2(aq) →CaCO3(s) +2NaCl(aq)

• 2Na+(aq) + CO3

2-(aq) + Ca2+

(aq) + 2Cl-(aq) → CaCO3(s) +

2Na+(aq) + 2Cl-

(aq)

15.1. Brønsted-Lowry acids and bases exchange protons90

Acid/Base Theories

• Arrhenius Theory Acids produce hydronium when dissolved in water Bases produce hydroxide when dissolved in water

• Bronsted-Lowry Theory Acids are proton (H+) donors. Bases are proton (H+) acceptors.

4.5. Ionic reactions can often be predicted 91

Predicting acid-base reactions

• Neutralization: metathesis reaction in which acid + metal hydroxide or metal oxide forms water and salt NaOH(aq) + HCl(aq) →H2O(l) + NaCl(aq)

• Acid-base reaction: reaction of weak base and acid transferring a H+ ion, driven by the formation of a weaker acid. HCl(aq) + NH3(aq) →NH4Cl(aq)


Learning check

Determine the molecular, total ionic and net ionic equations

• Molecular Equation

• Total Ionic Equation (TIE)

• Net Ionic Equation (NIE)

2HCl(aq) + Ca(OH)2(aq) → 2H2O(l) + CaCl2(aq)

H+(aq) + OH-

(aq) → H2O(l)

2H+(aq)+2Cl-

(aq)+ Ca2+

(aq) +2OH-

(aq) 2H2O(l)→ + Ca 2+(aq)+ 2Cl-

(aq)


Solubility rules: soluble compoundsA general idea as to whether a fair amount of solid

will dissolve is achieved using solubility rules

1. All compounds of the alkali metals (Group IA)

2. All salts containing NH4+, NO3

−, ClO4−, ClO3

−, and C2H3O2

−

3. All chlorides, bromides, and iodides (salts containing Cl−, Br−, or I−) except when combined with Ag+, Pb2+, and Hg2

2+

4. All sulfates (salts containing SO42−) except those of Pb2+,

Ca2+, Sr2+, Hg22+, and Ba2+


Solubility rules: insoluble compounds

5. All metal hydroxides (ionic compounds containing OH−) and all metal oxides (ionic compounds containing O2−) are insoluble except those of Group IA and Group IIA

When metal oxides dissolve, they react with water to form hydroxides. The oxide ion, O2−, does not exist in water. For example, Na2O(s) +H2O(l)

→ 2NaOH(aq)

6. All salts that contain PO43−, CO3

2−, SO32−, and S2− are

insoluble, except those of Group IA and NH4+.


Learning check:

Which of the following compounds are expected to be soluble in water?

Ca(C2H3O2)2

FeCO3

AgCl

Yes

No

No


Learning Check:

• Pb(NO3)2(aq) + Ca(OH)2(aq) →

• BaCl2(aq) + Na2CO3(aq) →

• Na3PO4(aq) + Hg2(NO3)2(aq) →

• NaCl(aq) + Ca(NO3)2(aq) →

Pb(OH)2(s) + Ca(NO3)2(aq)

BaCO3(s) + NaCl(aq)

NaNO3(aq) + (Hg2)3(PO4) 2(s)

CaCl2(aq) + NaNO3(aq)

NR (No reaction)

Predict the products of the following:

4.6. The composition of a solution is described by its concentration 97

Molar concentrations

• In solutions, solutes are dispersed in a larger volume

• Molarity expresses the relationship between the moles of solute and the volume of the solution

• Molarity (M)=moles solute/L solution Hence, a 6.0M solution of HCl contains 6.0 mole HCl

in a liter of solution

4.6. The composition of a solution is described by its concentration 98

Dilution

• Adding solvent to a solution creates a less concentrated solution

• moles of solute do not change, hence CstockVstock= CnewVnew

C=concentration V=volume

• Using volumetric glassware ensures that the volumes are known precisely

4.7. Molarity is used for problems in solution stoichiometry 99

• What volume of 2.00M HCl is needed to react 25.2 g Na2CO3 (MM=105.9887) completely?

• How many moles of BaSO4 will form if 20.0 mL of 0.600 M BaCl2 is mixed with 30.0 mL of 0.500 M MgSO4?

• BaCl2(aq) + MgSO4(aq) →BaSO4(s) + MgCl2(aq)

Solution stoichiometry

L 0.238HCl mol 2.00

L

CONa mol 1

HCl mol 2

105.9887g

CONa mol 1

1

CONa g 25.2

32

3232

44

44

42

42

BaSO mol 0150.0MgSO mol 1

BaSO mol 1

L

MgSO mol 0.500

1

L 0.0300

BaSO mol 0.0120BaCl mol 1

BaSO mol 1

L

BaCl mol 0.600

1

L 0.0200

0.0120 mol

4.8. Chemical analysis and titration are applications of solution stoichiometry 100

.

Titration

• Is the controlled addition of one reactant (titrant) to a known quantity of another (titrate) until the reaction is complete

• Often, an indicator is used to signal the reaction completion

• Endpoint: the volume of titrant required to complete the reaction


Titration in practice:


Learning Check:

• 25.00 mL of HCl are titrated with 75.00 mL of 1.30M Ca(OH)2. What is the concentration of HCl?

• 2HCl(aq) + Ca(OH)2(aq)→CaCl2(aq) + 2H2O(l)

0.02500L

1

Ca(OH) mol 1

HCl mol 2

L

Ca(OH) mol 1.30

1

L 0.07500

2

2

7.80 M HCl

Documents

1 Chapter 1-5 Review Brady & Senese 5th Ed. 1.5 Measurements are essential to describe properties2 Machines or instruments are used to take measurements