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Organization of the Periodic Table 2.8 Formulas, Names of Compounds • Types of Chemical Formulas:  Empirical formula: shows the relative number of atoms of each element in the compound

GENERAL CHEMISTRY 1

(CHEMW2014)

28-Aug-16 1

Thanh M Le

Syllabus for 57NKN

WHAT IS THE TEXT BOOK?

28-Aug-16 2

Nature of Matter and Change (5th edition)

• A copy of the text book is on reserve at the TLU Library.

from Chapter 1 to Chapter 12

in the text book

WHAT DO STUDENTS NEED TO

COMPLETE?

28-Aug-16 3

You will not be allowed to attend to the final exam, IF:

lecturer’s permission.

80% of the prescribed time

THE COURSE GRADING

www.facebook.com/groups/generalchemistry1

Lecture PowerPoint

Chemistry The Molecular Nature of Matter and ChangeMartin S Silberberg

Chapter 1

Keys to the Study of

Chemistry

• Chemistry: is the study of matter, its properties, the

changes that matter undergoes, and the energy associated

with these changes

• Matter: anything that has both mass and volume - the “stuff”

of the universe: books, planets, trees, professors, students

• Composition: the types and amounts of simpler substances

that make up a sample of matter

• Properties: the characteristics that give each substance a

unique identity

1.1 Some Fundamental Definitions

• Physical Properties: properties a substance shows by itself

without interacting with another substance - color, melting point, boiling point, density

• Chemical Properties: properties a substance shows as it

interacts with, or transforms into, other substances flammability, corrosiveness

-• A physical change: occurs when a substance alters its

physical form, not its composition

• A chemical change, also called a chemical reaction: occurs

when a substance (or substances) is converted into a different substance (or substances)

Example 1: The scenes below represent an atomic-scale view of

substance A undergoing two different changes (B or C) Decide whether

each scene shows a physical or a chemical change?

Solution:

a) A  B: each particle of substance A is composed of one blue and two red spheres

Sample B is composed of two different types of particles, some have two red spheres

while some have one red and one blue As A changes to B, the chemical composition

has changed A  B is a chemical change.

b) A  C: each particle of C is still composed of one blue and two red spheres, but the

particles are closer together and are more organized The composition remains

Exercise 1: Decide whether each of the following processes is

primarily a physical or a chemical change, and explain briefly:(a) Frost forms as the temperature drops on a humid winter night.(b) A cornstalk grows from a seed that is watered and fertilized.(c) A match ignites to form ash and a mixture of gases

(d) Perspiration evaporates when you relax after jogging

(e) A silver fork tarnishes slowly in air

The States of Matter

• A solid has a fixed shape and volume

Solids may be hard or soft, rigid or flexible

• A liquid has a varying shape that conforms

to the shape of the container, but a fixed

volume A liquid has an upper surface.

• A gas has no fixed shape or volume and

therefore does not have a surface

Energy in Chemistry

• Energy is the ability to do work.

• Potential energy is energy due to the position of an object.

• Kinetic energy is energy due to the movement of an object Total Energy = Potential Energy + Kinetic Energy

• Lower energy states are more stable, and are favored over

higher energy states

• Energy is neither created nor destroyed, it is conserved, and can be converted from one form to another.

Example 2: A lower energy state is more stable.

• All measured quantities consist of a number and a unit.

• Units are used like numbers, in other words, units can be multiplied, divided, and canceled

• Try to make a habit of including units in all calculations whenever you practice

• A conversion factor is a ratio of equivalent quantities used to

express a quantity in different units

• Example 4: some conversions factors:

 Between mile and feet is: 1 mi = 5280 ft

 Between mile and kilometer is: 1 mi = 1.609 km

 Between joule and calorie is: 1 J = 0.000239 cal

Exercise 2: To wire your stereo equipment, you need 325

centimeters of speaker wire that sells for $0.15/ft What is the price of the wire? The conversion factor:

1 foot (ft) = 30.48 centimeters (cm)

upholster one chair Its Vietnamese supplier sends the fabric inbolts of exactly 200 m2 What is the maximum number of chairsthat can be upholstered by 3 bolts of fabric (1 m = 3.281 ft)?

28-Aug-16 13

1.5 Measurement in Scientific Study

• Our current system of measurement began in 1790, when the

newly formed National Assembly of France

• In 1960, another international committee met in France to

establish the International System of Units The units of this

system are called SI units, from the French Système

International d’Unités

Physical Quantity (Dimension)

Unit Name

Unit Abbreviation

SI Base Units (International System of Units)

Common Decimal Prefixes

used with SI Units

Quantity SI to English Equivalent English to SI Equivalent Length 1 km = 0.6214 mile

Some volume equivalents in SI

Exercise 4: A graduated cylinder contains 19.9

mL of water When a small piece of galena, an ore of lead, is added, it sinks and the volume increases to 24.5 mL What is the volume of the piece of galena in cm3and in L?

Exercise 5: Many international computer communications are

carried out by optical fibers in cables laid along the ocean floor

If one strand of optical fiber weighs 1.19 x 10-3lb/m, what is the mass (in kg) of a cable made of six strands of optical fiber, each long enough to link New York and Paris (8.94 x 103km)?

• At a given temperature and pressure, the density of a

substance is a characteristic physical property and has a

specific value

* At room temperature (20°C) and normal atmospheric pressure (1atm).

Substance Physical State Density (g/cm 3 )

Hydrogen gas 0.0000899 Oxygen gas 0.00133 Grain alcohol liquid 0.789 Water liquid 0.998 Table salt solid 2.16 Aluminum solid 2.70 Lead solid 11.3 Gold solid 19.3

Densities of Some Common Substances

Exercise 6: Lithium, a soft, gray solid with the lowest density of

any metal, is a key component of advanced batteries A slab of

lithium weighs 1.49x103mg and has sides that are 20.9 mm by

11.1 mm by 11.9 mm Find the density of lithium in g/cm3

Exercise 7: A cube of gold 3 cm on a side weighs 18.3 ounces.

What is the density of gold in g/cm3? The conversion factor:

1 ounce (oz) = 28.3495 grams (g)

Temperature

• Distinguish between temperature and heat:

 Temperature (T) is a measure of how hot or cold a substance is relative to another

substance

 Heat (Q) is the energy that flows between objects that are at different temperatures.

• Temperature Scales

 Kelvin (K): the “absolute temperature scale” begins at absolute zero and has only

positive values Note that the kelvin is not used with the degree sign (°).

 Celsius (oC): the Celsius scale is based on the freezing and boiling points of water

This is the temperature scale used most commonly around the world The Celsius and

Kelvin scales use the same size degree although their starting points differ.

 Fahrenheit ( o F): the Fahrenheit scale is commonly used in the US The Fahrenheit

scale has a different degree size and different zero points than both the Celsius and Kelvin scales.

T (K) = T ( o C) + 273.15; T ( o F) = T ( o C)*9/5 - 32

1.6 Uncertainty in Measurement:

Significant Figures

• Every measurement includes some uncertainty The

rightmost digit of any quantity is always estimated.

• The recorded digits, both certain and uncertain, are called

significant figures.

• The greater the number of significant figures in a quantity, the

greater its certainty

Example 5: the number of significant figures in a measurement.

Determining Which Digits are Significant

• All digits are significant, except zeros that are used only to

position the decimal point.

1 Make sure the measured quantity has a decimal point

 Start at the left and move right until you reach the first nonzero

digit.

 Count that digit and every digit to its right as significant.

2 If no decimal point is present, zeros at the end of the number

are not significant

Example 6: Determining which digits are significant.

Example 7: How many digits in these numbers are significant.

Solution:

a) 1.030 mL has 4 significant figures

b) 5300 L has 4 significant figures

c) 5300 L has only 2 significant figures

d) 0.00004715 m has 4 significant figures, = 4.715x10 -5 m

e) 0.0000007160 cm3has 4 significant figures, = 7.160x10 -7 cm 3

Exercise 8: For each of the following quantities, determine the

number of significant figures in each quantity

Exercise 9: Express the number in 8c), 8d), 8e), 8f) in

exponential notation

1 For multiplication and division The answer contains the

same number of significant figures as there are in the

measurement with the fewest significant figures

2 For addition and subtraction The answer has the same

number of decimal places as there are in the

measurement with the fewest decimal places

Rules for Significant Figures

Rules for Rounding Off Numbers

• If the digit removed is more than 5, the preceding number

increases by 1

• If the digit removed is less than 5, the preceding number is

unchanged

• If the digit removed is 5, followed by zeros, or with no

following digits, the preceding number increases by 1 if it is

odd and remains unchanged if it is even

• If the 5 is followed by other nonzero digits, rule 1 is

d) 17.6500 rounds to 17.6; but 17.6513 rounds to 17.7

•Be sure to carry two or more additional significant figures

through a multistep calculation and round off the final answer only.

Example 10: Significant figures of measuring devices.

1 g

1000 mg

Example 11: Perform the following calculations and round

each answer to the correct number of significant figures

1 g

1000 mg

= 48.0 g11.55 cm3 = 4.16 g/ cm 3

Precision, Accuracy, and Error

• Precision refers to how close the measurements in a series

are to each other

• Accuracy refers to how close each measurement is to the

actual value

• Systematic error produces values that are either all higher

or all lower than the actual value This error is part of the

experimental system

• Random error produces values that are both higher and

lower than the actual value

precise and accurate

precise but not accurate

Example 12a: Precision and accuracy in a laboratory calibration.

Chapter 2

The Components

of Matter

Thanh M Le

2.1 Elements, Compounds, and Mixtures

• Element - the simplest type of substance with unique physical and

chemical properties An element consists of only one type of atom

• Atom - the smallest constituent unit of ordinary matter that has the

properties of a chemical element

• Molecule - a structure that consists of two or more atoms which are

chemically bound together and thus behaves as an independent unit

• Compound - a substance composed of two or more elements which

are chemically combined

• Mixture - a group of two or more elements and/or compounds that

are physically intermingled

Exercise 1: The scenes below represent an atomic-scale view

of 7 samples of matter Describe each sample as an element, compound, or mixture

• Law of Conservation of Mass: “the total mass of substances

does not change during a chemical reaction”

=

• Example 2:

CaO + CO2 → CaCO3

(calcium oxide) (carbon dioxide) (calcium carbonate)

2.2 Mass Conservation & Mass fractions • Mass fractions (or mass percents) of any element in a

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2.5 The Atomic Theory today

Relative Mass (amu)

Relative charge

• The atomic mass (also called atomic weight) of an element is

the average mass of the masses of isotopes which are

weighted according to their abundances

• Atomic mass is calculated from its isotopes mass:

• Example 5: Silver (47Ag) has two occur naturally, 107Ag and

109Ag, with mass 106.90509 (amu) and 108.90476 (amu), with

abundance 51.84% and 48.16%, repectively So, the atomic

mass of silver is:

Exercise 2: Silicon(Si) is essential to the computer industry as

a major component of semiconductor chips It has three naturally occurring isoltopes: 28Si, 29Si, and 30Si Determine the number of protons, neutrons, and electrons in each silicon isotope, if atomic number of silicon is 14

Exercise 3: Boron (5B) has two naturally occurring isotopes Find the percent abundances of 10B and 11B given the atomic mass of B = 10.81 amu, the isotopic mass of 10B = 10.0129 amu, and the isotopic mass of 11B = 11.0093 amu

2.6 Periodic Table

• At the end of the 18thcentury,

Lavoisier compiled a list of the 23

elements known at that time

• By 1870, 65 were known

• In 1871, the Russian chemist

Dmitri Mendeleev published the

most successful of these

organizing schemes as a table of the elements

• By 1925, 88 were known Today, there are 116 and still counting!

Dmitri Mendeleev (1836–1907)

Mendeleev's 1871 periodic table

The Modern Periodic Table

• Each element has a box that contains its atomic number, atomic

symbol, and atomic mass

• The boxes lie in order of increasing atomic number (number of

protons) as you move from left to right

• The boxes are arranged into a grid of periods (horizontal rows) and groups (vertical columns).

• The elements are classified as metals, nonmetals, or metalloids

(semimetals)

Organization of the Periodic Table

2.8 Formulas, Names of Compounds

• Types of Chemical Formulas:

 Empirical formula: shows the relative number of atoms

of each element in the compound

 Molecular formula: shows the actual number of atoms of

each element in a molecule of the compound

 Structural formula: shows the number of atoms and the

bonds between them

• The cation name = the metal name

• The anion name = the nonmetal name + the suffix “ide”.

• Name of ionic compound = the cation name + the anion name.

• Example 6: Name the ionic compound formed from the

following pairs of elements: a) Mg and N; b) I and Cd;

Solution:

N3-= nitride;  Mg3N2= magnesium nitride

b) I = iodine → I-= iodide; Cd = cadmium → Cd2+= cadmium

 CdI2= cadmium iodide

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Names of multi cations of a family (4)

• With metal have more than one cation:

 The cation name with lower charge = the Latin metal root

+ the suffix -ous.

 The cation name with higher charge = the Latin metal root

+ the suffix -ic.

• Example 7: Name the following ions:

a) Fe2+and Fe3+ b) Cu+and Cu2+ c) Sn2+and Sn4+

Solution:

a) Fe = iron = ferrum (Latin name) → Fe 2+ = ferrous; Fe 3+ = ferric

b) Cu = copper = cuprum (Latin name) → Cu + = cuprous; Cu 2+ = cupric.

c) Sn = tin = stannum (stannum) → Sn 2+ = stannous; Sn 4+ = stannic

Exercise 4: Name the ionic compound formed from the following

pairs of elements:

a) Zn & O; b) Ag & Br; c) Li & Cl;

Exercise 5: Give the formulas for the following compounds:

a) tin(II) fluoride; b) ferric oxide; c) lead(IV) oxide; d) mercuric chloride; e) ferrous chloride; f) aluminum sulfide;g) calcium bromide h) chromium(III) iodide i) stannous fluoride

Names of multi anions of a family (5)

• With a family has 2 oxoanions (anions including oxygen atoms):

 The anion name with more O = the nonmetal root + suffix “ate”.

 The anion name with fewer O = the nonmetal root + suffix “ite”.

• With a family has 4 oxoanions:

 the anion name with most O atoms = the prefix “per” + the

nonmetal root + the suffix “ate”;

 the anion name with least (three fewer) O atoms = the prefix

“hypo” + the root + the suffix "ite”

• Example 8: Name the following ions:

• Example 9: HCl = hydrochloric acid HNO2= nitrous acid

HBrO4 = perbromic acid H2SO4= sulfuric acidHIO2= iodous acid HIO3 = iodic acid

Exercise 6: Give the names of the following formulas :

g) Ba(CH3COO)2 h) Na2SO3 i) Fe2(SO4)3

Exercise 7: Give the names of the following acids :

Names of Hydrates and Binary Covalent Compounds (7)

• Name of hydrates = name of compound + a Greek numerical prefix + “hydrate”

• Name of binary covalent compound (compound of 2 non-metal elements) = a Greek numerical prefix + name of element 1 + a

Greek numerical prefix + name of element 2 + suffix “ide”.

• The Greek numerical prefix = mono (1); di (2); tri (3); tetra (4); penta (5); hexa (6); hepta (7); octa (8); nona (9); deca (10)

• Example 10: CuSO4.5H2O = copper(II) sulfate pentahydrat

Ba(OH)2.8H2O = barium hydroxide octahydrate

CS2= carbon disulfidePCl5= phosphorus pentachloride

NO = dinitrogen tetraoxide

30-Aug-16 25

Exercise 8: Explain what is wrong with the following name or

formula If they are wrong, please correct it:

a) Ba(CH3COO)2is called barium diacetate

b) Sodium sulfide has the formula (Na)2SO3

c) Iron(II) sulfate has the formula Fe2(SO4)3

d) Cesium carbonate has the formula Cs2(CO3)

e) SF4is monosulfur pentafluoride

f) Dichlorine heptaoxide is Cl2O6

g) N2O3is dinitrotrioxide

2.9 Mixtures: Classification and Separation

• There are two broad classes of mixtures:

 A heterogeneous mixture: has one or more visible

boundaries between the components

 A homogeneous mixture has no visible boundaries

because the components are mixed as individual atoms, ions, and molecules A homogeneous mixture is also called

Chemistry is a practical science

• The mole (abbreviated mol) is the amount of a substance that

contains the same number of entities as there are atoms in exactly 0.012 kg of carbon-12

• One mole (1 mol) contains 6.022×1023entities (to four significant figures) This number is called Avogadro’s number

• The molar mass (M) of a substance is the mass per mole of its

entites (atoms, molecules)

• Molar mass has units of grams per mole (g/mol)

• Elements:

 For monatomic elements (Ne, Fe, Ba ), the molar mass is the

same as the atomic mass in gram per mole For example: the molar

mass of Ne = 20.18 g/mol.

 For molecular elements (O2, H 2 , S 8 ): you must know the molecular

formula to determine the molar mass For example: the molar mass of

O 2 = 2×M O = 2×16.00 g/mol = 32.00 g/mol.

• Compounds The molar mass of a compound is the sum of the

molar masses of the atoms of the elements in the formula

 For example: the molar mass of K2S = 2×M K + M S = (2×39.10 g/mol)+

a) How many grams of Ag are in 0.0342 mol of Ag?

b) How many Ga atoms are in 2.85 x 10-3mol of gallium?

c) How many Fe atoms are in 95.8 g of Fe?

Solution

a)

b)

c)

Mass Percent & Mass of an Element

• Assume with a compound AxBy

09-Oct-16 7

• Example 2: Glucose (C6H12O6) is a key nutrient for generating

chemical potential energy in biological systems

a) What is the mass percent of each element in glucose?

b) How many grams of carbon are in 16.55 g of glucose?

Solution

a)

b)

• Exercise 1: Nitrogen dioxide is a component of urban smog

that forms from the gases in car exhausts Assume we have a sample of 8.92 g of nitrogen dioxide

a) How many molecules are in there?

b) How many mol of N, mol of O are in the sample?

• Exercise 2: Ammonium carbonate, a white solid that

decomposes on warming, is an component of baking powder a) How many molecules are in 41.6 g of ammonium carbonate?b) How many O atoms are in this sample?

• Exercise 3: Tetraphosphorus decaoxide reacts with water to

formphosphoric acid, a major industrial acid In the laboratory,

the oxide is used as a drying agent

a) What is the mass (in g) of 4.65×1022molecules of

tetra-phosphorus decaoxide?

b) How many P atoms are present in this sample?

• Exercise 4: Ammonium nitrate is a common fertilizer

Agronomists base the effectiveness of fertilizers on their

nitrogen content

a) Calculate the mass percent of N in ammonium nitrate

b) How many grams of N are in 35.8 kg of ammonium nitrate?

• The empirical formula is the simplest formula for a compound

that agrees with the elemental analysis It shows the lowest whole number of moles and gives the relative number of atoms

of each element present For example, the empirical formula for glucose (C6H12O6) is CH2O

• The molecular formula shows the actual number of atoms of

each element in a molecule of the compound For example, the molecular formula for glucose is C6H12O6

3.2 Determining the Formula of an Unknown Compound

Steps to determine

empirical formula & molecular formula

• Step 1: Find the number of moles of each element and show

the preliminary formula

• Step 2: Divide by the lowest mol amount

• Step 3: If the results is not whole numbers, multiplying by the

smallest integer that gives whole numbers to get empirical

formula.

• Step 4: Divide Mmoleculefor Mempiricalto find a conversion factor

• Step 5: Multiply the empirical formula to the factor to get

molecular formula.

• Example 3: A sample of an unknown compound contains 0.21

mol of zinc, 0.14 mol of phosphorus, and 0.56 mol of oxygen What is its empirical formula?

09-Oct-16 13

• Example 4: Lactic acid (M = 90.08 g/mol) shows it contains

40.0 mass % C, 6.71 mass % H, and 53.3 mass % O

Determine the empirical formula and the molecular formula for

lactic acid?

Solution

• Assuming there are 100 g of lactic acid: CxHyOz

• The result give us the preliminary formula = C3.33P6.66O3.33

• Divide each fraction by the smallest mol, in this case 0.33, we

get empirical formula = CH2O

• Find the conversion factor:

• Multiply to the factor, the molecular formula = C3H6O3

14

Combustion Analysis of Organic Compounds

• In the combustion analysis of Organic Compounds (CxHyOz)

CxHyOz+ O2→ CO2+ H2O

 All the H in the Organic Compounds is converted to H2O

 All the C in the Organic Compounds is converted to CO2

 The mass of O in the Organic Compounds = the mass of organic compounds minus the sum of the C & H masses

t o

15

• Example 5: When a 1.000 g sample of vitamin C

(M= 176.12 g/mol) is placed in a combustion chamber and burned,

the following data are obtained: mass of CO2absorber after

combustion = 85.35 g; mass of CO2absorber before combustion =

83.85 g; mass of H2O absorber after combustion = 37.96 g; mass of

H2O absorber before combustion = 37.55 g What is the molecular

• Divide each fraction by the smallest mol, in this case 0.0341, then

multiply by the smallest integer, in this case 3, we get empirical

CH 2 O

Molecular Formula = CH2O = Formaldehyde

Molecular Formula = C2H4O2= acetic acidMolecular Formula = C3H6O3= lactic acid

Molecular Formula = C4H8O4= erythrose

Molecular Formula = C5H10O5= riboseMolecular Formula = C6H12O6= glucose

Chemical Formulas

• The types of Chemical Formulas:

 General formula For example: CxHyOz

 Preliminary formula For example: (C1.5H3O1.5)

 Empirical formula For example: (CH2O) or (CH2O)n

 Molecular formula For example: C6H12O6

 Condensed formula (semi-structural formula)

For example: CH2OH[CH(OH)]4CHO

 Structural formula For example:

09-Oct-16 19

• Exercise 5: Analysis of a sample of an ionic compound yields

2.82 g of Na, 4.35 g of Cl, and 7.83 g of O What is the

empirical formula and the name of the compound?

• Exercise 6: One of the most widespread environmental

carcinogens (cancer-causing agents) is benzo pyrene

(M=252.30 g/mol) It is found in coal dust, in cigarette smoke,

and even in charcoal-grilled meat Analysis of this hydrocarbon

shows 95.21 mass % C and 4.79 mass % H What is the

molecular formula of benzo pyrene?

• Exercise 7: A dry-cleaning solvent (M=146.99 g/mol) that

contains C, H, and Cl is suspected to be a cancer-causing agent When a 0.250-g sample was studied by combustion analysis, 0.451 g of CO2and 0.0617 g of H2O formed Find the

molecular formula?

• Exercise 8: Menthol (M=156.3 g/mol), a strong-smelling

substance used in cough drops, is a compound of carbon, hydrogen, and oxygen When 0.1595 g of menthol was subjected to combustion analysis, it produced 0.449 g of CO2and 0.184 g of H2O What is menthol’s molecular formula?

• A chemical equation uses formulas to express the identities

and quantities of substances involved in a physical or chemical

change

• The equation must be balanced; the same number and type of

each atom must appear on both sides

3.3 Writing and Balancing

Chemical Equations

Reactants are written on the left.

A yield arrow points from

reactants to products.

Products are written on the right.

Coefficients

Steps to Balance a Chemical Equation

• Step 1: Translate the chemical statements to its chemical signs Reactants are written on the left Products are written on the right

• Step 2: Put coefficients to balance the 2 sides of the equation;

do not change the formulas

• Step 3: Check that all atoms balance & adjust the coefficients if necessary

• Step 4: Write physical states of reactants and products

• Example 6: Within the cylinders of a car’s engine, the

hydrocarbon octane (C8H18), one of many components of

gasoline, mixes with oxygen from the air and burns to form

carbon dioxide and water vapor Write a balanced equation for

• 2C8H18(g) + 25O2(g) → 16CO2(g) + 18H2O(l)

• Example 7: The following molecular scenes depict an reaction The

blue spheres represent nitrogen while the red spheres represent oxygen Write a balanced equation for this reaction

Solution

• The reactant circle shows only one type of molecule, composed of 2

N and 5 O atoms The formula is thus N2O5 There are 4 N2O5molecules depicted

• The product circle shows two types of molecule; one has 1 N and 2 O atoms while the other has 2 O atoms The products are NO2and O2 There are 8 NO2molecules and 2 O2molecules shown

4 N2O5→ 8 NO2+ 2 O2.2N2O5→ 4NO2+ O2

09-Oct-16 25

• Exercise 9: Write a balanced equation for each of the following

chemical statements:

a) A characteristic reaction of Group 1A elements: chunks of

sodium react violently with water to form hydrogen gas and

sodium hydroxide solution

b) The destruction of marble statuary by acid rain: aqueous nitric

acid reacts with calcium carbonate to form carbon dioxide,

water, and aqueous calcium nitrate

c) Halogen compounds exchanging bonding partners:

phosphorus trifluoride is prepared by the reaction of

phosphorus trichloride and hydrogen fluoride; hydrogen

chloride is the other product The reaction involves gases only

d) Explosive decomposition of dynamite: liquid nitroglycerine

(C3H5N3O9) explodes to produce a mixture of gases-carbon

dioxide, water vapor, nitrogen, and oxygen

• 1 Write a balanced equation for the reaction

• 2 Convert the given mass (or number of entities) of the first substance to amount (mol)

• 3 Use the appropriate molar ratio from the balanced equation

to calculate the amount (mol) of the second substance

• 4 Convert the amount of the second substance to the desired mass (or number of entities)

3.4 Calculating Amounts of Reactant and Product

27

• Example 8: Copper is obtained from copper(I) sulfide by roasting it in

the presence of oxygen gas to form powdered copper(I) oxide and

gaseous sulfur dioxide

a) How many moles of oxygen are required to roast 10.0 mol of

c) nCu2O = mCu2O/MCu2O= 2.86×103/143.1 = 20.0 mol → nO2=

• Reactions often occur in sequence: the product of one reaction becomes a reactant in the next

• An overall reaction is written by combining the reactions;

any substance that forms in one reaction and reacts in the next can be eliminated

• Limiting reagent (limiting reactants) = is the substance that

is totally consumed when the chemical reaction is complete

The amount of product formed is limited by this reagent,

since the reaction cannot continue without it

If one or more other reagents are present in excess of the

quantities required to react with the limiting reagent, they are

described as excess reagents or excess reactants.

• There are 2 methods to find out which reactant is limiting reagent

Method 1: Comparison of reactant amounts

Method 2: Comparison of product amounts which can be

formed from each reactant

• Example 9: Chlorine reacts with fluorine to form chlorine trifluoride

The reaction is run with 0.750 mol of Cl2and 3.00 mol of F2 What mass of chlorine trifluoride will be produced?

Solution

Cl2(g) + 3F2(g) → 2ClF3(g)

• Compare: , the ratio is smaller than the other, that reactant will be limiting reactant: , so Cl2will be the limiting reactant

• Calculate the product always have to use the limiting reactant

• The theoretical yield is the amount of product calculated using

the molar ratios from the balanced equation

• The actual yield is the amount of product actually obtained

The actual yield is usually less than the theoretical yield

• Yields in Multistep Syntheses:

 Overall % yield = multiply every yields of all the steps

Overall % yield = %yield 1 × %yield 2 × %yield 3 × × %yield n

of SiC is formed from processing 100.0 kg of sand?

Solution

SiO2(s) + 3C(s) → SiC(s) + 2CO(g)

• Exercise 10: A fuel mixture used in the early days of rocketry

consisted of two liquids, hydrazine (N2H4) and dinitrogen

tetraoxide (N2O4), which ignite on contact to form nitrogen gas

and water vapor:

a) Write the balanced chemical equation for the reaction?

b) How many grams of nitrogen gas form when 1.00 x 102g of

N2H4and 2.00 x 102g of N2O4are mixed?

• Exercise 11: How many grams of solid aluminum sulfide can

be prepared by the reaction of 10.0 g of aluminum and 15.0 g of

sulfur? How much of the non-limiting reactant is in excess?

• Exercise 12: Marble (calcium carbonate) reacts with

hydrochloric acid solution to form calcium chloride solution, water, and carbon dioxide What is the percent yield of carbon dioxide if 3.65 g of the gas is collected when 10.0 g of marble reacts?

• Exercise 13: When 56.6 g of calcium and 30.5 g of nitrogen

gas undergo a reaction that has a 93.0% yield, what mass of calcium nitride forms?

3.5 Fundamentals of

Solution Stoichiometry

• A solution = a homogeneous mixture of two or more substances

A solution may exist in any phase

• A solvent = the component of a solution that is present in the

greatest amount It is the substance in which the solute is

dissolved For example: the solvent for seawater is water

• A solute = the substance that is dissolved in a solution.

• Concentration = a measurement of the amount of solute present

in a chemical solution, with respect to the amount of solvent

• Molarity = a concentration unit, defined to be the number of

moles of solute divided by the number of liters of solution

• Preparing a Dilute Solution from a Concentrated Solution: the moles of solvent is constant when the solution is diluted

n solute = C M1 ×V 1 = C M2 ×V 2

09-Oct-16 37

• Example 11: What is the molarity of an aqueous solution that

contains 0.715 mol of glycine (H2NCH2COOH) in 495 mL?

Solution

• Example 12: “Isotonic saline” is a 0.15 M aqueous solution of

NaCl How would you prepare 0.80 L of isotonic saline from a

6.0 M stock solution?

Solution

• nsolute= CM1×V1= CM2×V2 → 6.0 (M) × V1= 0.15 (M) × 0.8 (L)

→ V1= 0.020 (L) 6.0 M stock solution

• A 0.020 L portion of the concentrated solution must be diluted to

• Exercise 14: What mass of solute is in 1.75 L of 0.460 M

sodium monohydrogen phosphate buffer solution?

• Exercise 15: A 0.10 mol/L HCl solution is used to simulate the

acid concentration of the stomach What is the volume (L) of

“stomach acid” react with a tablet containing 0.10 g of magnesium hydroxide as the following reaction?

Mg(OH)2(s) + 2HCl (aq) → MgCl2(aq) + 2H2O (l)

• Exercise 16: In a simulation mercury removal from industrial

wastewater, 0.050 L of 0.010 mol/L mercury(II) nitrate reacts with 0.020 L of 0.10 mol/L sodium sulfide What is the mass of

mercury(II) sulfide that would form from the following reaction? Hg(NO3)2(aq) + Na2S (aq) → HgS (s) + 2NaNO3(aq)

 Acid-Base Reactions (Neutralization Reactions).

 Oxidation-Reduction Reactions (Redox Reactions)

 Metathesis Reactions (Displacement Reactions)

Types of Chemical Reactions

The whole H 2 O molecule is polar.

• Water is a polar molecule, since:

 it has uneven electron distribution

 it has a bent molecular shape

• Water readily dissolves a variety of substances

• Water interacts strongly with its solutes and often plays an active

role in aqueous reactions

4.1 The Role of Water as a Solvent

Each bond in H 2 O is polar.

Ionic Compounds in Water

• Many ionic compounds dissolve in water (NaCl, KNO3 ) →

“soluble” compound However, many others do not (AgCl,

BaSO4, ) → “insoluble” compound.

 In the latter cases, the electrostatic attraction among ions in

the compound remains greater than the attraction between

ions and water molecules, so the solid stays largely intact

• Example 1: Solubilities of NaCl & AgCl in water:

 Solubility of NaCl in H2O at 20oC = 365 g/L

→ NaCl = a “soluble” compound

 Solubility of AgCl in H2O at 20oC = 0.009 g/L

→ AgCl = a “insoluble” compound

• When an ionic compound dissolves, its aqueous solution

conduct an electric current → the ionic compound = an electrolyte For example, NaCl in water:

23-Oct-16 7

Covalent Compounds in Water

• Water can dissolves many covalent compounds, however,

many other covalent substances do not dissolve appreciably in

water

 In the former cases, even though these substances

dissolve, they do not dissociate into ions but remain as

intact molecules As a result, their aqueous solutions do not

conduct an electric current, so these substances are called

• Example 3: How many mol of each ion is in each solution?

a) 5.0 mol of ammonium sulfate dissolved in water

b) 78.5 g of cesium bromide dissolved in water

c) 7.42x1022molecules of copper(II) nitrate dissolved in water

Solution

a) The formula of ammonium sulfate is (NH4)2SO4so the equation

for dissociation is:

(NH4)2SO4(s) → 2NH4(aq) + SO42-(aq)

b) The formula of cesium bromide is CsBr, and 78.5 g CsBr =

0.369 mol CsBr, so the equation for dissociation is:

CsBr(s) → Cs+(aq) + Br-(aq)0.369mol 0.369mol 0.369mol

c) The formula of copper(II) nitrate is Cu(NO3)2, and 7.42x1022

molucules Cu(NO3)2 = 0.123 mol Cu(NO3)2, so the equationis:

Cu(NO3)2(s) → Cu2+(aq) + 2NO3-(aq)

• Exercise 1: How many moles of each ion are in each solution?

a) 2 mol of potassium perchlorate dissolved in waterb) 354 g of magnesium acetate dissolved in waterc) 1.88×1024molecules of ammonium chromate dissolved in waterd) 1.32 L of 0.55 M sodium bisulfate

• Exercise 2: Calculate the molarity of sodium sulfate & each ion

in a 23.55-mL solution which contains 28.24 mg of sodium sulfate (used in dyeing and printing textiles), M = 139.04 g/mol

• Three types of equations to represent aqueous ionic reactions:

 Molecular equation: shows all reactants and products as

if they were intact, undissociated compounds

 Total ionic equation: shows all soluble ionic substances

dissociated into ions.

 Net ionic equation: eliminates the spectator ions and

shows only the actual chemical change Spectator ions

are ions that are not involved in the actual chemical

change Spectator ions appear unchanged on both sides

of the total ionic equation

4.2 Equations for Aqueous Ionic Reactions

• Example 4: Write 3 types of equations to represent aqueous

ionic reaction between silver nitrate & sodium chromate

Solution

• Molecular equation:

2AgNO3(aq)+ Na2CrO4(aq)→ Ag2CrO4(s)+ 2NaNO3(aq)

• Total ionic equation:

2Ag+ (aq)+2NO3-

(aq)+2Na+ (aq)+CrO42-

(aq)→Ag2CrO4(s)+2Na+

(aq)+2NO3

-(aq)

• Net ionic equation:

2Ag+ (aq)+ CrO42-

(aq) → Ag2CrO4(s)

• Spectator ions = NO3-, Na+

23-Oct-16 13

• Example 5: Write 3 types of equations to represent aqueous

ionic reaction between barium hydroxide + sulfuric acid

Solution

• Molecular equation:

Ba(OH)2(aq)+ H2SO4(aq)→ BaSO4(s)+ 2H2O(l)

• Total ionic equation:

• The net ionic equation is the same as the total ionic equation

since there are no spectator ions.

• This reaction is both a neutralization reaction and a

precipitation reaction.

• Exercise 3: Write 3 types of equations to represent aqueous

ionic reactions between:

a) Sodium carbonate & Calcium nitrateb) Ammonium sulfate & Barium iodidec) Aluminum nitrate & Sodium phosphate

• Exercise 4: If 38.5 mL of lead(II) nitrate solution reacts

completely with excess sodium iodide solution to yield 0.628 g

of precipitate, what is the molarity of lead(II) ion in the original solution?

4.3 Precipitation Reactions

• In a precipitation reaction: two soluble ionic compounds react

to give an insoluble products, called a precipitate.

• The key event in a precipitation reaction is: the formation of an

insoluble product through the net removal of solvated ions from

solution

• It is possible for more than one precipitate to form in such a

reaction

• Most of the precipitation reactions are metathesis reactions

(or double displacement reactions)

Predicting whether a precipitate will form

1) Attention to the ions present in the reactants

2) Consider all possible cation-anion combinations

3) Use the solubility rules to decide whether any of the ion

b) FeSO4(aq)+ Sr(OH)2(aq)→ ?

Soluble Ionic Compounds

1 All common compounds of Group 1A ions (Li + , Na + , K + , etc.) and

ammonium ion (NH 4 ) are soluble.

2 All common nitrates (NO 3-), acetates (CH 3 COO - or C 2 H 3 O 2-) and most

perchlorates (ClO 4-) are soluble.

3 All common chlorides (Cl - ), bromides (Br - ) and iodides (I - ) are soluble,

except those of Ag+ , Pb 2+ , Cu + , and Hg 22+ All common fluorides (F - ) are

soluble except those of Pb2+ and Group 2A

4 All common sulfates (SO 22-) are soluble, except those of Ca2+ , Sr 2+ , Ba 2+ ,

Ag + , and Pb 2+

Insoluble Ionic Compounds

1 All common metal hydroxides are insoluble, except those of Group 1A

and the larger members of Group 2A(beginning with Ca 2+ ).

2 All common carbonates (CO 32-) and phosphates (PO 43-) are insoluble,

except those of Group 1A and NH4

3 All common sulfides are insoluble except those of Group 1A, Group 2A

23-Oct-16 19

• Exercise 5: Predict whether or not a reaction occurs when

each of the following pairs of solutions are mixed If a reaction

does occur, write balanced molecular, total ionic, and net ionic

equations, and identify the spectator ions

a) Potassium fluoride (aq) + strontium nitrate (aq) →

b) Ammonium perchlorate (aq) + sodium bromide (aq) →

c) Iron(III) chloride(aq) + cesium phosphate(aq) →

d) Sodium hydroxide(aq) + cadmium nitrate(aq) →

e) Magnesium bromide(aq) + potassium acetate(aq) →

f) Silver nitrate(aq) + barium chloride(aq) →

• Exercise 6: The following molecular views show reactant

solutions for a precipitation reaction (with H2O molecules omitted for clarity)

a) Which compound is dissolved in beaker A: KCl, Na2SO4, MgBr2, or Ag2SO4? Which compound is dissolved in beaker B:

NH4NO3, MgSO4, Ba(NO3)2, or CaF2?b) Name the precipitate and spectator ions when solutions A and

B are mixed, and write balanced molecular, total ionic, and net ionic equations for this process

• The Arrhenius Acid-Base Theory

 An acid = a substance that produces H+ions when

• The Bronsted-Lowry Acid-Base Theory

 Acids are substances that can donate H+ions to bases

 An acid is a "proton donor“.

 A base is a "proton acceptor"

• Since a hydrogen atom is a proton and one electron, technically

an H + ion is just a proton

• The acid-base reaction is essentially a proton transfer.

Strong Acid & Base

• perchloric acid, HClO4

• sodium hydroxide, NaOH

• potassium hydroxide, KOH

• calcium hydroxide, Ca(OH)2

• strontium hydroxide, Sr(OH)2

• barium hydroxide, Ba(OH)2

• Strong acids and strong bases dissociate completely into ions in aqueous solution They are strong electrolytes and

conduct well in solution For example:

• Weak acids and weak bases dissociate very little into ions

in aqueous solution They are weak electrolytes and conduct

poorly in solution For example:

• Weak acids and weak bases are kept unchanged in total ionic equations For example:

 Molecular equation:

CH3CH2COOH(aq)+ NH3(aq)→ CH3CH2COONH4(aq)

 Net ionic equation:

CH3CH2COOH(aq)+ NH3(aq)→ CH3CH2COO

-(aq) + NH4(aq)

23-Oct-16 25

Acid-Base Titrations

• Titration (titrimetry) = a common laboratory method of quantitative

chemical analysis, a known concentration of one reactant is used to

determine the concentration of the other.

• In a typical acid-base titration, a standardized solution (titrant) of base,

one whose concentration is known, is added slowly to an acid solution

(analyte) of unknown concentration.

• An acid-base indicator has different colors in acid and base, and is used to

monitor the reaction progress.

• At the equivalence point, the amount of H+ from the acid equals the amount

of OH - ion produced by the base.

• The end point occurs when there is a slight excess of base and the

indicator changes color permanently.

• Example 8: A 50.00 mL sample of HCl is titrated with 0.1524 M

NaOH The buret reads 0.55 mL at the start and 33.87 mL at

the end point Find the concentration of the HCl solution

Solution

• The balanced equation for the reaction

NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

• Volume of base NaOH = 33.87 mL – 0.55 mL = 33.32 mL

• Since 1 mol of HCl reacts with 1 mol NaOH, the amount of HCl

• Exercise 8: Write balanced molecular, total ionic, and net ionic

equations for the following acid-base reactions & identify the spectator ions

a) hydrochloric acid (aq) + potassium hydroxide (aq) → b) strontium hydroxide (aq) + perchloric acid (aq) → c) hydriodic acid (aq) + calcium hydroxide (aq) → d) potassium hydroxide (aq) + propionic acid (aq) →

• Exercise 9: What volume of 0.1292 M Ba(OH)2would

neutralize 50.00 mL of the HCl solution standardized above in

the sample problem?

• Exercise 10: In a titration 42.0 mL of 0.210 M H2SO4was

needed to neutralize 50.0 mL of a NaOH solution Determine

the molarity of the sodium hydroxide solution

4.5 Oxidation-Reduction Reactions

• Oxidation is the loss of electrons Reduction is the gain of

electrons Oxidation and reduction occur together

• The reducing agent loses electrons and is oxidized The oxidizing agent gains electrons and is reduced.

• A redox reaction involves electron transfer

• For example:

23-Oct-16 31 23-Oct-16 32

Oxidation Number (O.N.)

1 For an atom in its elemental form (Na, O2, Cl2, etc.): O.N = 0

2 For a monoatomic ion: O.N = ion charge

3 The sum of O.N values for the atoms in a compound equals zero The sum of O.N values for the atoms in a polyatomic ion equals the ion’s charge

4 Rules for specific atoms or periodic table groups

i For Group 1: O.N = +1 in all compounds

ii For Group 2: O.N = +2 in all compoundsiii For hydrogen: O.N = +1 in combination with nonmetals

iv For fluorine: O.N = -1 in combination with metals and boron

v For oxygen: O.N = -1 in peroxides, O.N = -2 in all other compounds (except with F)

vi For Group 7: O.N = -1 in combination with metals, nonmetals (except O), and other halogens lower in the group

• Example 9: Determine the oxidation number (O.N.) of each

element in these species:

a) zinc chloride b) sulfur trioxide c) nitric acid

Solution

a) ZnCl2→ the O.N for zinc is +2 (ion Zn2+) and that for

chloride is -1 (group 7A)

b) SO3→ each oxygen is an oxide with an O.N of -2 (oxygen

always is -2 in all most compounds) The O.N of sulfur must

therefore be +6 (calculate from the sum of O.N values in a

compound = zero)

c) HNO3→ element H has an O.N of +1 and each oxygen is

-2 The N must therefore have an O.N of +5

• Example 10: Use oxidation numbers to decide whether each of

the following is a redox reaction or not:

a) CaO(s) + CO2(g) → CaCO3(s)b) 4KNO3(s)→ 2K2O(s) + 2N2(g) + 5O2(g)c) NaHSO4(aq) + NaOH(aq)→ Na2SO4(aq) + H2O(l)

Solution

reaction, since no species change O.N

reaction, since N & O changes O.N

c) This is not a redox reaction, since no species change O.N

• Example 11: Identify the oxidizing agent and reducing agent in

each of the following reactions:

a) 2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3H2(g)

b) 2H2(g) + O2(g) → 2H2O(g)

Solution

a)

• Al changes O.N from 0 to +3 and is oxidized, Al is the reducing

agent; H changes O.N from +1 to 0 and is reduced, H2SO4is

the oxidizing agent

b)

• H2changes O.N from 0 to +1 and is oxidized, H2is the

reducing agent; O changes O.N from 0 to -2 and is reduced, O2

is the oxidizing agent

Balancing Redox Equations (oxidation number method)

1 Assign O.N.s to all atoms

2 Identify the reactants that are oxidized and reduced

3 Compute the numbers of electrons transferred, and draw lines from each reactant atom to the product atom to show the change

tie-4 Multiply the numbers of electrons by factor(s) that make the electrons lost equal to the electrons gained

5 Use the factor(s) as balancing coefficients

6 Complete the balancing by inspection and add states of matter

23-Oct-16 37

• Example 12: Use the oxidation number method to balance the

following Redox Equation:

Cu(s) + HNO3(aq)→ Cu(NO3)2(aq) + N2O(g) + H2O(l)

Solution

Redox Titrations

• In a redox titration, a known concentration of oxidizing agent is

used to find an unknown concentration of reducing agent

(or vice versa)

• A redox titration is a type of titration based on a redox reaction

between the analyte and titrant

• This application of stoichiometry is used in a wide range of situations, including measuring the iron content in drinking water and the vitamin C content in fruits and vegetables

c) KClO3(aq)+ HBr(aq)→ Br2(l)+ H2O(l)+ KCl(aq)

• Exercise 12: Identify each oxidizing agent and each reducing

agent:

a) 2Fe(s)+ 3Cl2(g)→ 2FeCl3(s)b) 2C2H6(g)+ 7O2(g)→ 4CO2(g)+ 6H2O(g)c) 5CO(g)+ I2O5(s)→ I2(s)+ 5CO2(g)

4.6 Elements in Redox Reactions

• Whenever atoms appear in the form of a free element on one

side of an equation and as part of a compound on the other,

there must have been a change in oxidation state and the

reaction is a redox process

• Redox reactions usually happen in 4 types:

 Reactions of Combining Two Elements, or Combining

Compound and Element

• Example 13: Classify each of the following redox reactions as

a combination, decomposition, or displacement reaction a) magnesium (s) + nitrogen (g) → magnesium nitride (aq)b) hydrogen peroxide (l) → water (l) + oxygen gasc) aluminum (s) + lead(II) nitrate (aq) → aluminum nitrate (aq) + lead (s)

Chapter 5 Gases and the Kinetic

Molecular Theory

5.1 Gaseous State

• Gas volume changes significantly with pressure & temperature:

 Gases expand when heated and shrink when cooled

 The volume change is 50 to 100 times greater for gases than for liquids and solids

• Gases flow very freely.

• Gases have relatively low densities.

• Gases form a solution in any proportions Gases are freely

miscible with each other

 The SI unit of pressure is the pascal (Pa).

 Standard atmosphere (atm), the average atmospheric

pressure measured at sea level and 0oC

 Millimeter of mercury (mmHg, torr).

pressure decreases with altitude

• Laboratory devices for measuring Gas Pressure:

 Barometer

 Manometer: closed end, open-end

• The barometer is a common

device used to measure

atmospheric pressure, basically

just a tube about 1 m long,

closed at one end, filled with

mercury, and inverted into a

dish containing more mercury

• The manometer:

 Closed end manometer

is a mercury-filled, curved tube, closed at one end and attached

to a flask at the other

 Open-end manometer

is also consists of a curved tube filled with mercury, but one end of the tube is open to the atmosphere and the other is connected to the gas sample

03-Nov-16 7

• Example 1: A geochemist heats a limestone (CaCO 3) sample

and collects the CO2released in an evacuated flask attached to

a closed-end manometer After the system comes to room

temperature, Δh = 291.4 mm Hg Calculate the CO2pressure in

torrs, atmospheres, and kilopascals

Solution

291.4 mmHg x 1torr

1 mmHg = 291.4 torr291.4 torr x 1 atm

 amount (number of moles, n)

• The variables are interdependent: any one of them can be determined by measuring the other three

• Three key relationships exist among the four gas variables:

Boyle’s, Charles’s, and Avogadro’s laws.

• Boyle’s Law: at constant temperature, the volume occupied by

a fixed amount of gas is inversely proportional to the external

pressure.

• Charles’s Law: at constant pressure, the volume occupied by

a fixed amount of gas is directly proportional to its absolute

(Kelvin) temperature.

• Avogadro’s Law: at fixed temperature and pressure, equal

volumes of any ideal gas contain equal numbers of particles

(or moles)

Gas Behavior at Standard Conditions

• STP (standard temperature and pressure) specifies:

 a pressure of 1 atm (760 torr)

 a temperature of 0°C (273.15 K).

• The standard molar volume is the volume of 1 mol of an ideal

gas at STP Standard molar volume = 22.4141 L or 22.4 L

The Ideal Gas Law

• The Ideal Gas Law combines the 3 gas laws into one relationship:

 R is the universal gas constant; the value of R depends

on the units used If P (atm), V (L), n (mol), T (K):

→ R = 0.082 atm L/mol K

03-Nov-16 13

• Example 2: Boyle’s apprentice finds that the air trapped in a J

tube occupies 24.8 cm3at 1.12 atm By adding mercury to the

tube, he increases the pressure on the trapped air to 2.64 atm

Assuming constant temperature, what is the new volume of air

• Example 3: A steel tank used for fuel delivery is fitted with a

safety valve that opens when the internal pressure exceeds 1.00x103torr It is filled with methane at 23°C and 0.991 atmand placed in boiling water at exactly 100°C Will the safety valve open?

• Exercise 1: A scale model of a blimp rises when it is filled with

helium to a volume of 55.0 dm3 When 1.10 mol of He is added

to the blimp, the volume is 26.2 dm3 How many more grams of

He must be added to make it rise? Assume constant T and P.

• Exercise 2: A steel tank has a volume of 438 L and is filled with

0.885 kg of O2 Calculate the pressure of O2at 21oC

• Exercise 3: A steel tank has a volume of 438 L and is filled with

0.885 kg of O2 The tank develops a slow leak that is discovered

and sealed The new measured pressure is 1.37 atm at 21oC

How many grams of O2was lost?

Which of the following balanced equations describes the reaction?1) A2(g) + B2(g) → 2AB(g) 2) 2AB(g) + B2(g) → 2AB2(g)3) A(g) + B2(g) → AB2(g) 4) 2AB2(g) → A2(g) + 2B2(g)

• Exercise 5: A rigid plastic container holds 35.0 g of ethylene

gas (C2H4) at a pressure of 793 torr What is the pressure if 5.0

g of ethylene is removed at constant temperature?

• Exercise 4: The piston-cylinders is

depicted before and after a gaseous reaction that is carried out at constant pressure The temperature

is 150 K before the reaction and 300

K after the reaction

5.4 Further Applications

of the Ideal Gas Law

• The density of a gas is directly proportional to its molar mass

and inversely proportional to its temperature

• Molar Mass:

• Partial pressure: the pressure exerted by each gas in a

mixture The partial pressure of a gas is proportional to its mole

03-Nov-16 19

• Example 5: An organic chemist isolates a colorless liquid from

a petroleum sample She places the liquid in a preweighed flask

and puts the flask in boiling water, causing the liquid to vaporize

and fill the flask with gas She closes the flask and reweighs it

She obtains the following data:

Volume (V) of flask = 213 mL; T = 100.0°C; P = 754 torr

mass of flask + gas = 78.416 g; mass of flask = 77.834 g

Calculate the molar mass of the liquid

Solution

• Example 6: In a study of O2uptake by muscle at high altitude,

a physiologist prepares an atmosphere consisting of 79 mole %

N2, 17 mole % 16O2, and 4.0 mole % 18O2 (The isotope 18O will

be measured to determine the O2uptake.) The pressure of the mixture is 0.75 atm to simulate high altitude Calculate the

mole fraction and partial pressure of 18O2in the mixture

Solution

• Exercise 6: What volume of H2gas at 765 torr and 225°C is

needed to reduce 35.5 g of copper(II) oxide to form pure copper

and water?

• Exercise 7: The alkali metals [Group 1A] react with the

halogens [Group 7A] to form ionic metal halides What mass of

potassium chloride forms when 5.25 L of chlorine gas at 0.950

atm and 293 K reacts with 17.0 g of potassium?

• Exercise 8: At 10.0oC and 102.5 kPa, the density of dry air is

1.26 g/L What is the average “molar mass” of dry air at these

conditions?

5.5 Gas Reaction Stoichiometry

• From the balanced equation:

 use stoichiometrically equivalent molar ratios to calculate the amounts (moles) of reactants and products

 converte these quantities into masses, numbers of molecules, or solution volumes

• Using the ideal gas law to convert between gas variables (P, T, and V) and amounts (moles) of gaseous reactants and products In effect, you combine a gas law problem with a stoichiometry problem; it is more realistic to measure the volume, pressure, and temperature of a gas than its mass

• Example 7: Acetylene (C2H2) is produced in the laboratory

when calcium carbide (CaC2) reacts with water:

CaC2(s) + 2H2O(l) → C2H2(g) + Ca(OH)2(aq)

A collected sample of acetylene has a total gas pressure of 738

torr and a volume of 523 mL At the temperature of the gas

(23oC), the vapor pressure of water is 21 torr How many

grams of acetylene are collected?

Solution

• Exercise 9: Copper dispersed in absorbent beds is used to

react with oxygen impurities in the ethylene used for producing polyethylene The beds are regenerated when hot H2reduces the metal oxide, forming the pure metal and H2O On a laboratory scale, what volume of H2at 765 torr and 225oC is needed to reduce 35.5 g of copper(II) oxide?

• Exercise 10: A small piece of zinc reacts with dilute HCl to form

H2, which is collected over water at 16oC into a large flask The total pressure is adjusted to barometric pressure (752 torr), and the volume is 1495 mL At this temperature of the gas (16oC), the vapor pressure of water is 13.6 torr Calculate the partial pressure and mass of H2

5.6 The Kinetic-Molecular Theory

• Postulate (assumption) 1 : Gas particles are tiny with large

spaces between them The volume of each particle is so small

compared to the total volume of the gas that it is assumed to be

zero

• Postulate (assumption) 2: Gas particles are in constant,

random, straight-line motion except when they collide with each

other or with the container walls

• Postulate (assumption) 3: Collisions are elastic, meaning that

colliding particles exchange energy but do not lose any energy

due to friction Their total kinetic energy is constant Between

collisions the particles do not influence each other by attractive

or repulsive forces

Kinetic Energy and Temperature

• At a given T, all gases in a sample have the same average kinetic energy (Ek) Kinetic energy depends on both the mass

and the speed of a particle

• The average speed (root-mean-square speed, v rms ) of a gas depends on both the molar mass and temperature At the same

T, a heavier gas particle moves more slowly than a lighter one

Graham’s Law of Effusion

• Effusion is the process by which a gas escapes through a

small hole in its container into an evacuated space

• Graham’s law of effusion: the rate of effusion of a gas is

inversely proportional to the square root of its molar mass.

 At the same T, a lighter gas moves more quickly and therefore has a

higher rate of effusion than a heavier gas

Solution

→ The effusion rate of helium is as twice as the one of methane

• Exercise 11: Rank the samples of gas described in the table

below in order of increasing average speed of the atoms in

them

I 2.1 mol argon (Ar) gas at 2.2 atm and 39°C

II 3.0 mol argon (Ar) gas at 1.7 atm and -5°C

III 1.7 mol xenon (Xe) gas at 2.6 atm and -5°C

IV 1.8 mol argon (Ar) gas at 1.2 atm and 17°C

• Exercise 12: Calculate the root-mean-square speed of the

following gases at 25°C?

C2H2 Kr NO CO2 Ar?

• Exercise 13: Rank the species in order of increasing average

kinetic energy per mole

I C5H12(g)at 35 oC; II C5H12(g)at 200 oC; III C8H18(g)at 200 oC;

Chapter 05: Suggested Practice Problems

6-12(even),18-30(even),39-45(odd),88,111,140,

52-60(even),61,68a,b,d,e,f,71

06-Nov-16 1

Chapter 6 Thermochemistry:

Energy Flow and Chemical Change

6.1 Forms of Energy and their Interconversion

• When energy is transferred from

one object to another, it appears as

work and heat.

• The system is the part of the

universe being studied, while the

surroundings are the rest of the

universe that interacts with the

system

• All energy is either potential or kinetic, and that these forms are

convertible from one to the other

Energy Flow to and from a System

• Each particle in a system has potential energy and kinetic

energy, and the sum of these energies for all the particles in

the system is the internal energy (E).

• The total energy of the universe remains constant

• In a system, changes from reactants to products and the

products return to the starting temperature, the internal energy

has changed

DE = Efinal– Einitial= Eproducts– Ereactants

(final state of the system minus the initial state)

• Energy diagrams for the transfer of internal

energy (E) between a

system and its surroundings

Heat and Work:

Two Forms of Energy Transfer

• Heat (or thermal energy, Q) = the energy transferred between

a system and its surroundings as a result of a difference in their

temperatures only

• Work (W) = the energy transferred when an object is moved by

 Unit of P (atm), V (L) → unit of W (atm∙L)

• The total change in a system’s internal energy (∆E)

∆E = Q + W

The Law of Energy Conservation

• The law of energy conservation (the first law of thermodynamics)

= the total energy of the universe is constant

 Energy is conserved, and is neither created nor destroyed

 Energy is transferred in the form of heat and/or work

DEuniverse= DEsystem+ DEsurroundings = 0

• The SI unit of energy is the joule (J).

1 J = 1 kg∙m2·s-2

• The calorie (cal) was once defined as the quantity of energy

needed to raise the temperature of 1 g of water by 1°C

• The British Thermal Unit (BTU) is often used to rate appliances

1 BTU = 1055 J

• The electron volt (eV) is the amount of energy gained (or lost)

by the charge of a single electron moving across an electric

potential difference of one volt

1 eV = 160 zJ (zeptojoules) = 1.6×10−21J

• The kilowatt hour (kWh): a measure of electrical energy

equivalent to a power consumption of 1,000 watts for 1 hour

• Example 1: When gasoline burns in a car engine, the heat

released causes the products CO2and H2O to expand, which pushes the pistons outward Excess heat is removed by the

car’s radiator If the expanding gases do 451 J of work on the pistons and the system releases 325 J of heat to the

surroundings, calculate the change in energy (∆E) in J, kJ, kcal.

Solution

• Heat is given out by the system, so Q = - 325 J

• The gases expand to push the pistons, so the system does

work on the surroundings and W = - 451 J

• DE = Q + W = -776 J = -0.776 kJ = -0.185 kcal

• Exercise 1: In a reaction, gaseous reactants form a liquid

product The heat absorbed by the surroundings is 26.0

kcal, and the work done on the system is 15.0 Btu

Calculate ∆E (in kJ)

• Exercise 2: A system receives 575 J of heat and delivers

425 J of work Calculate the change in the internal energy,

∆E, of the system

6.2 Enthalpy (H):

Heats of Reactions and Processes

• To determine ∆E, both heat and work must be measured The

most common chemical work is (P.∆V) work, the work done

when the volume of a system changes in the presence of an external pressure

• Enthalpy (H) is the change in heat for a system at constant

pressure, defined as (E + PV) so:

∆H = ∆E + P.∆V = Q p

• If a system remains at constant pressure and its volume does

not change much, then ∆H ≈ ∆E.

• The change in enthalpy equals the heat gained or lost at constant pressure With most changes occurring at constant

pressure, ∆H is more relevant and easier to find than ∆E.

Comparing ΔH & ΔE

• In fact, because many reactions involve little (if any) PV work, most

(or all) of the energy change occurs as a transfer of heat

 Reactions that do not involve gases: gases do not appear in

many reactions (precipitation, many acid-base, and many redox

reactions) → liquids and solids undergo verysmall volume

changes → ∆Vgas≈ 0 → ∆H ≈ ∆E

 Reactions in which the total amount (mol) of gas does not

change: the total amount of gaseous reactants equals the total

amount of gaseous products → ∆Vgas= 0 → ∆H = ∆E

 Reactions in which the amount (mol) of gas change → ∆Vgas

≠ 0 However, Qpis usually so much larger than P.∆V → ∆ H = Qp

= (∆E + P.∆V) ≈ ∆E.

• In conclusions: ∆H equals, or is very close to, ∆E, for many reactions

Exothermic and Endothermic Processes

• The enthalpy change of a reaction, also called the heat of reaction, ∆Hrxn, always refers to Hfinalminus Hinitial:

∆H rxn = H final - H initial

• An exothermic (“heat out”) process releases heat and results

in a decrease in the enthalpy of the system: ∆H < 0.

• An endothermic (“heat in”) process absorbs heat and results

in an increase in the enthalpy of the system: ∆H > 0.

• Example 2: In each of the following cases, determine the sign

of DH, state whether the reaction is exothermic or endothermic,

and draw and enthalpy diagram

a) H2(g) + ½O2(g) →H2O(l) + 285.8 kJ

b) 40.7 kJ + H2O(l) →H2O(g)

Solution

• Heat is a “product” for this reaction and is therefore given out,

so the reaction is exothermic → ∆H1= -285.8 kJ

• Heat is a “reactant” in this reaction and is therefore absorbed,

so the reaction is endothermic → ∆H2= +40.7 kJ

• Exercise 4: Choose the best answer for the following: “An

endothermic reaction causes the surroundings to ”

a) warm upb) become acidicc) condensed) decrease in temperaturee) become basic

6.3 Heat Capacity & Calorimetry

• The specific heat capacity (C) of a substance is the quantity of

heat required to change the temperature of 1 gram of the

substance by 1 K

Q = m•C•DT

 Q = heat (lost or gained) (J)

 C = specific heat capacity (J/g•K)

 m = mass (g); ∆T = (Tfinal– Tinitial) (K)

 Q = heat (lost or gained) (J)

 C = specific heat capacity (J/mol•K)

 n = moles (mol); ∆T = (Tfinal– Tinitial) (K)

• Example 3:

a) A layer of copper welded to the bottom of a skillet weighs 125g How much heat is needed to raise the temperature of the copper layer from 25°C to 300.°C? The specific heat

capacity (c) of Cu is 0.387 J/g∙K.

b) Find the heat transferred (in kJ) when 5.50 L of ethylene glycol (d = 1.11 g/mL; C = 2.42 J/g∙K) in a car radiator cools from 37.0 oC to 25.0 oC

Solution

a) Q1= m•C•∆T = 125 g • 0.387 J/g∙K • (300-25) K = 1.33×104 Jb) Q2= m•C•∆T = (550 • 1.11) g • 2.42 J/g∙K • (25-37) K = J

• The calorimeter is a device used to measure the heat released

(or absorbed) by a physical or chemical process

• Two common types are the constant-pressure calorimeters

(Coffee-cup calorimeters) and constant-volume calorimeters

(Bomb calorimeters).

 Qrxn+ Qwater= 0 (Coffee-cup calorimeter)

 Qrxn+ Qwater+ Qbomb= 0 (Bomb calorimeter)

 Qwater= Cwater•mwater•∆Twater

 Qbomb= Cbomb•∆Tbomb (Cbombin J/K)

• Example 4: A 22.05 g solid is heated in a test-tube to

100.00°C and added to 50.00 g of water in a coffee-cup

calorimeter The water temperature changes from 25.10°C to

28.49°C Find the specific heat capacity of the solid

Solution

• Qsolid+ Qwater= 0

• Qsolid= Csolid•msolid•∆Tsolid= Csolid•22.05 g•(28.49-100.00)K =

• Qwater= Cwater•mwater•∆Twater= 4.184 J/g∙K •50g•(28.49-25.10)K =

 Csolid= 0.450 J/g∙K

• Example 5: A manufacturer claims that its new dietetic dessert

has “fewer than 10 Calories per serving.” To test the claim, a chemist at the Department of Consumer Affairs places one serving in a bomb calorimeter and burns it in O2 The initial temperature is 21.862°C and the temperature rises to 26.799°C If the heat capacity of the calorimeter is 8.151 kJ/K,

Is the manufacturer’s claim correct?

• Exercise 5: You place 50.0 mL of 0.500 M NaOH in a

coffee-cup calorimeter at 25.00oC, and carefully add 25.0 mL of 0.500

M HCl, also at 25.00oC After stirring, the final temperature is

27.21oC Calculate Qsoln(in J) and the change in enthalpy,

∆Hrxn(in kJ/mol) Assume that the total volume is the sum of the

individual volumes, that D = 1.00 g/mL and C = 4.184 J/g∙K

• Exercise 6: A chemist burns 0.8650 g of graphite (a form of

carbon) in a new bomb calorimeter, and CO2forms If 393.5 kJ

of heat is released per mole of graphite and Tin creases 2.613

K, what is the heat capacity of the bomb calorimeter?

6.4 Stoichiometry of Thermochemical Equations

• A thermochemical equation is a balanced equation that