AP Chemistry Summer Assignment · AP Chemistry Summer Assignment: Briar Woods High School, Ashburn...

AP Chemistry Summer Assignment: Briar Woods High School, Ashburn VA

We are looking forward to teaching you during the 2019-2020 school year! Attached is the summer assignment. The first section is a summary of the information you should have learned in first-year chemistry. After the summary notes, you’ll find several worksheets. These are your homework assignments for the first unit of AP Chemistry. You are expected to have these completed by the third day of class (August 28th or 29th). Your first assessment will evaluate your understanding of the information from the summer assignment. It is important to come to class well prepared.

You are also expected to join the AP Chemistry Summer 2019 Google Classroom. The code is 44ek8r. A copy of the summer assignment packet is posted on Classroom. We will also post EdPuzzle videos during the summer to help you review the first-year chemistry material.

It is recommended that you do not wait until the last minute to start the assignment, do not procrastinate. Remember AP Chemistry is an equivalent course to an Introductory Chemistry college course. Taking a college level course in high school is difficult and it requires commitment, hard work and time.

If you are having difficulty with a specific unit there are many online resources available to you to assist you. You can also post to the forum on Classroom. Please feel free to contact us if you are having difficulty or have any questions.

Have a great summer and enjoy chemistry!

Mrs. Matthews Mr. Smith

[email protected] [email protected]

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mailto:[email protected]

© Adrian Dingle’s Chemistry Pages and ChemEducator LLC 2014 – All rights reserved.

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UNIT 00: AP Chemistry Preamble

TOPIC 0A: Chemistry, Scientific Method and Chemical & Physical Change

What is chemistry?

Chemistry can be described as the science that deals with matter, and the changes that matter

undergoes. It is sometimes called the central science because so many naturally occurring

phenomena involve chemistry and chemical change.

Scientific problem solving

Scientific (logical) problem solving involves three steps;

1. State the problem and make observations. Observations can be quantitative (those

involving numbers or measurement) or qualitative (those not involving numbers).

2. Formulate a possible explanation (this is known as a hypothesis).

3. Perform experiments to test the hypothesis. The results and observations from these

experiments lead to the modification of the hypothesis and therefore further

experiments.

Eventually, after several experiments, the hypothesis may graduate to become a theory. A theory

gives a universally accepted explanation of the problem. Of course, theories should be constantly

challenged and may be refined as and when new data and new scientific evidence comes to light.

Theories are different to laws. Laws state what general behavior is observed to occur naturally.

For example, the law of conservation of mass exists since it has been consistently observed that

during all chemical changes mass remains unchanged (i.e., it is neither created nor destroyed).


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States of matter and particle representations

All matter has two distinct characteristics. It has mass and it occupies space. Properties

associated with the three states of matter, and the behaviors of the particles that make up each,

are summarized below.

SOLIDS LIQUIDS GASES

Have a definite shape and

definite volume.

The particles in a solid are

packed tightly together and

only vibrate relatively gently

around fixed positions.

Have no shape of their own

but take the shape of their

container. A liquid has a

definite volume.

The particles in a liquid are

free to move around one

another.

Have neither a definite shape

nor a definite volume.

The particles in a gas spread

apart filling all the space of the

container available to them

and interactions between the

particles are considered to be

negligible.

The circles in the diagrams below represent the relative positions and movements of the particles

in the three states of matter. Expect to see many such particulate representations during the AP

course.


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Physical and chemical changes and properties

All matter exhibits physical and chemical properties by which it can be classified. Examples of

physical properties are color, odor, density, hardness, solubility, melting point, and boiling point.

Chemical properties are those exhibited when a substance reacts with other substances.

Examples of chemical properties are reactions with acids and bases, oxidation and reduction

(REDOX) and a huge number of other chemical reactions. Changes in which the physical or

chemical properties of a substance are altered are considered physical or chemical changes,

respectively.

Physical change

If some aspect of the physical state of matter is altered, but the chemical composition remains the

same, then the change is considered to be a physical change. The most common physical

changes are changes of state. These are summarized below.

SOLID → LIQUID Melting

LIQUID → GAS Boiling

GAS → LIQUID Condensing SOLID → GAS Sublimation

GAS → SOLID Reverse sublimation or deposition

LIQUID → SOLID Freezing

In solids, the particles have relatively little energy and vibrate around fixed positions. If a solid is

heated, the particles gain energy, move around move, and eventually gain enough energy to

break away from their fixed positions and form a liquid. Continued heating leads to the liquid

particles gaining sufficient energy to break away from one another and form a gas. In a gas the

particles move freely and with relatively large amounts of energy.

Chemical change

In a chemical change, which is often called a chemical reaction, the atoms of a substance are

rearranged to form new substances. A chemical change requires that the new substance or

substances formed have a different chemical composition to the original substance or

substances. Chemical changes are often accompanied by observable changes such as color

changes and energy changes.


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There is a very important distinction to be made between these two types of change that you will

encounter in UNIT 2. More on this later, but for now, note the following.

• During physical changes, the intermolecular forces (the forces between particles) are

disrupted, e.g., boiling water separates one water molecule (H2O) from another water

molecule but does not break any, individual water molecule apart.

• During chemical changes the intra forces (the forces within substances) are disrupted,

e.g., during the electrolysis of water, one water molecule (H2O) splits up to form O and H

atoms. Individual water molecules do break apart.


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TOPIC 0B: Measurement

Measurements

Measurements, and subsequently calculations applied to those measurements, allow the

determination of some of the quantitative properties of a substance; for example, mass and

density.

Scientific notation

Measurements and calculations in chemistry often require the use of very large or very small

numbers. In order to make handling them easier, such numbers can be expressed using scientific

notation. All numbers expressed in this manner are represented by a number between 1 and 10

which is then multiplied by 10, raised to a particular power.

The number of places the decimal point has moved determines the power of 10. If the decimal

point has moved to the left then the power is positive, if it has moved to the right then it is

negative.

For example, the number 42000.0 is converted to scientific notation by using the number 4.2. In

the process the decimal point has moved four places to the left, so the power of 10 used is +4.

42000.0 = 4.2 x 104

The number 0.00012 is converted to scientific notation by using the number 1.2. In the process

the decimal point has moved four places to the right, so the power of 10 used is -4.

0.00012 = 1.2 x 10-4

Task 0Bi

1 Convert the following numbers to scientific notation.

(a) 24500

(b) 356

(c) 0.000985

(d) 0.222

(e) 12200


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2. Convert the following scientific notation numbers to non-scientific notation numbers.

(a) 4.2 x 103 (b) 2.15 x 10-4

(c) 3.14 x 10-6

(d) 9.22 x 105

(e) 9.57 x 102

SI units

Units tell us the scale that is being used for measurement. Prefixes are used to make writing very

large or small numbers easier. Common SI (System International) units and prefixes are given

below.

Base quantity Name of unit Symbol

Mass Kilogram kg Length Meter m Time Second s

Amount of substance Mole mol Temperature Kelvin K

Prefix Symbol Meaning Giga G 109 Mega M 106 Kilo k 103 Deci d 10-1 Centi c 10-2 Milli m 10-3

Micro µ 10-6 Nano n 10-9 Pico p 10-12


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Converting units and dimensional analysis (the factor label method)

One unit can be converted to another unit by using a conversion factor. Application of the simple

formula below will allow the conversion of one unit to another. This method of converting between

units is called dimensional analysis or the factor-label method.

(unit a) (conversion factor) = unit b

The conversion factor is derived from the equivalence statement of the two units. For example, in

the equivalence of 1.00 inch = 2.54 cm, the conversion factor will either be,

2.54 cm1.00 inch

or 1.00 inch2.54 cm

The correct choice is the one that allows the cancellation of the unwanted units. For example, to

convert 9.00 inches to cm, perform the following calculation

9.00 inch x

2.54 cm1.00 inch

⎛⎝⎜

⎞⎠⎟

= 22.86 cm

To convert 5.00 cm into inches, perform the following calculation

5.00 cm

x 1.00 inch2.54 cm

⎛⎝⎜

⎞⎠⎟

= 1.97 inches

Task 0Bii

1. Convert the following quantities from one unit to another, using the following

equivalence statements; 1.000 m = 1.094 yd, 1.000 mile = 1760 yd, 1.000 kg = 2.205 lbs

(a) 30 m to miles

(b) 1500 yd to miles (c) 206 miles to m

(d) 34 kg to lbs

(e) 34 lb to kg

2. In each case below, which is the larger quantity?

(a) A distance of 3.00 miles or 3000. m.

(b) A mass of 10.0 kg or 25 lbs.


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Temperature

There are three scales of temperature that you may come across in your study of chemistry. They

are Celsius (oC), Fahrenheit (oF) and Kelvin (K). The following conversion factors will be useful.

Temperature Conversion factors

Celsius to Kelvin

oT in K = T in C + 273

Kelvin to Celsius

oT in C = T in K - 273

Celsius to Fahrenheit

o oT in F = (1.8 (T in C)) + 32

Fahrenheit to Celsius

o

o (T in F - 32)T in C = 1.8

Task 0Biii

1. Convert the following temperatures from one unit to the other.

(a) 263 K to oF

(b) 38 K to oF

(c) 13 oF to oC

(d) 1390 oC to K

(e) 3000 oC to oF

2. When discussing a change in temperature, why will it not matter if the change is

recorded in Celsius or Kelvin?


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Derived units

All other units can be derived from base quantities. One such unit that is very important in

chemistry is volume. Volume has the unit, length3. Common units for volume are liters (L) or

milliliters (mL).

1.000 mL = 1.000 cm3

and

1.000 L = 1000. mL = 1000. cm3 = 1.000 dm3

Density is the ratio of the mass to volume.

volumemassdensity =

This relationship is particularly useful when dealing with liquids in chemistry. Liquids are most

conveniently measured by pouring them into, say, a graduated cylinder. The graduated cylinder

records a volume, not a mass. In order to calculate the mass of a known volume of a liquid

(assuming the density is known) the relationship below can be applied.

mass = (density) (volume)

Assuming that density has the units of g/L, volume has units of L, and by using dimensional

analysis, it can be seen that the resultant unit for mass in this case is g.

( )⎛ ⎞

⎜ ⎟⎝ ⎠

Lg = gL


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Uncertainty, significant figures and rounding

When reading the scale on a piece of laboratory equipment such as a graduated cylinder or a

buret, there is always a degree of uncertainty in the recorded measurement. The reading will

often fall between two divisions on the scale and an estimate must be made in order to record the

final digit. This estimated final digit is said to be uncertain and is reflected in the recording of the

numbers by using +/-. All of the digits that can be recorded with certainty are said to be certain.

The certain and the uncertain numbers taken together are called significant figures.

Determining the number of significant figures present in a number

1. Any non-zero integers are always counted as significant figures.

2. Leading zeros are those that precede all of the non-zero digits and are never counted as

significant figures.

3. Captive zeros are those that fall between non-zero digits and are always counted as

significant figures.

4. Trailing zeros are those at the end of a number and are only significant if the number is

written with a decimal point.

5. Exact numbers have an unlimited number of significant figures. (Exact numbers are those

which are as a result of counting e.g., 3 apples or by definition e.g., 1.000 kg = 2.205 lb).

6. In scientific notation the 10x part of the number is never counted as significant.

Determining the correct number of significant figures to be shown as the result of a

calculation

1. When multiplying or dividing. Limit the answer to the same number of significant figures

that appear in the original data with the fewest number of significant figures.

2. When adding or subtracting. Limit the answer to the same number of decimal places that

appear in the original data with the fewest number of decimal places.

i.e., don’t record a greater degree of significant figures or decimal places in the calculated answer

than the weakest data will allow.


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Rounding

Calculators will often present answers to calculations with many more figures than the significant

ones. As a result many of the figures shown are meaningless, and the answer, before it is

presented, needs to be rounded.

In a multi-step calculation it is possible to leave the rounding until the end i.e., leave all numbers

on the calculator in the intermediate steps, or round to the correct number of figures in each step,

or round to an extra figure in each intermediate step and then round to the correct number of

significant figures at the end of the calculation. In most cases in the AP chemistry course you will

leave numbers on the calculator and round at the end.

Whichever method is being employed, use the simple rule that if the digit directly to the right of

the final significant figure is less that 5 then the preceding digit stays the same, if it is equal to or

greater than 5 then the preceding digit should be increased by one.

Task 0Biv

1. Determine the number of significant figures in the following numbers.

(a) 250.7 (b) 0.00077

(c) 1024

(d) 4.7 x 10-5

(e) 34000000

(f) 1003.

2. Use a calculator to carry out the following calculations and record the answer to the correct number of significant figures.

(a) (34.5) (23.46)

(b) 123 / 3

(c) (2.61 x 10-1) (356)

(d) 21.78 + 45.86

(e) 23.888897 - 11.2

(f) 6 - 3.0


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Accuracy and precision

Accuracy relates to how close the measured value is to the actual value of the quantity. Precision

refers to how close two or more measurements of the same quantity are to one another.

Task 0Bv

1. Consider three sets of data that have been recorded after measuring a piece of wood

that is exactly 6.000 m long.

SET X SET Y SET Z

5.864 m 6.002 m 5.872 m

5.878 m 6.004 m 5.868 m

Average Length 5.871 m 6.003 m 5.870 m

(a) Which set of data is the most accurate?

(b) Which set of data is the most precise?

Percentage error

The data that are derived in experiments will often differ from the accepted, published, actual

value. When this occurs, a common way of expressing accuracy is percentage error.

( )Actual Value - Calculated ValuePercentage Error = x 100

Actual Value


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TOPIC 0C: Atomic Theory

Brief history of atomic theory

Circa. 400-5 BC. Greek philosopher Democritus proposes the idea of matter being made up of

small, indivisible particles (atomos).

Late 18th Century. Lavoisier proposes the Law of conservation of mass and Proust proposes the

Law of constant composition.

Early 19th Century. Using the previously unconnected ideas above, John Dalton formulates his

Atomic Theory.

Dalton’s atomic theory

1. Elements are made from tiny particles called atoms.

2. All atoms of a given element are identical (N.B., see isotopes).

3. The atoms of a given element are different to those of any other element.

4. Atoms of different elements combine to form compounds. A given compound always has

the same relative numbers and types of atoms. (Law of constant composition).

5. Atoms cannot be created or destroyed in a chemical reaction they are simply rearranged

to form new compounds. (Law of conservation of mass).

Structure of the atom and the periodic table

Several experiments were being carried out in the 19th and 20th centuries that began to identify

the sub-atomic particles that make up the atom. A summary of those experiments is given below.

Scientist Experiment Knowledge gained Relating to

Crookes Cathode Ray Tube Negative particles of some kind exist Electron

J. J. Thomson Cathode Ray Deflection

Mass/charge ratio of the electron determined

Electron

Millikan Oil Drop Experiment Charge on the electron Electron

Rutherford, Marsden and Geiger Gold Foil Experiment Nucleus present in

atom The nucleus of an

atom and the proton

In the first part of the 20th Century, Bohr built upon Rutherford’s idea by introducing quantum

theory to the Solar System Model, and proposed the idea that the atom was made up of a

LO 1.13


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nucleus containing protons, that was being orbited by electrons, but only in specific, allowed

orbits. Schrödinger subsequently expanded upon Bohr’s model, in order to incorporate the wave

nature of the electrons. Once Chadwick’s discovered the neutron in 1932, the modern picture of

the atom in its simplest form was complete.

Particle Charge Mass in atomic mass units (amu) Position in atom

PROTON +1 1 Nucleus

NEUTRON 0 1 Nucleus

ELECTRON -1 11836

Outside of the nucleus

The atomic numbers (in the periodic table below shown above the element symbol and

sometimes referred to as Z) and mass numbers (in the periodic table below shown below the

symbol and sometimes referred to as A) have specific meanings.

Atomic number = the number of protons in the nucleus of one atom of the element

Since all atoms are neutral it also tells us the number of electrons surrounding the nucleus.

N.B., when atoms lose or gain electrons the proton and electron numbers become unbalanced

and the atoms become charged particles, i.e., they are no longer neutral. These charged particles

are called ions. A negative ion is formed when an atom gains electrons to possess a greater

number of electrons than protons, and is called an anion. A positive ion is formed when an atom

loses electrons to possess a fewer number of electrons than protons, and is called a cation.

Mass number = the number of protons + the number of neutrons in one atom of the element


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Period GROUP 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

1 1 H 1

2 He 4

2 3 Li 7

4 Be 9

T R A N S I T I O N M E T A L S

5 B 11

6 C 12

7 N 14

8 O 16

9 F 19

10 Ne 20

3 11 Na 23

12 Mg 24

13 Al 27

14 Si 28

15 P 31

16 S 32

17 Cl

35.5

18 Ar 36

4 19 K 39

20 Ca 40

21 Sc 45

22 Ti 48

23 V 51

24 Cr 52

25 Mn 55

26 Fe 56

27 Co 59

28 Ni 59

29 Cu 64

30 Zn 65

31 Ga 70

32 Ge 73

33 As 75

34 Se 79

35 Br 80

36 Kr 84

5 37 Rb 86

38 Sr 88

39 Y 89

40 Zr 91

41 Nb 93

42 Mo 96

43 Tc 99

44 Ru 101

45 Rh 103

46 Pd 106

47 Ag 108

48 Cd 112

49 In

115

50 Sn 119

51 Sb 122

52 Te 128

53 I

127

54 Xe 131

6 55 Cs 133

56 Ba 137

57 La* 139

72 Hf

178

73 Ta 181

74 W

184

75 Re 186

76 Os 190

77 Ir

192

78 Pt

195

79 Au 197

80 Hg 201

81 Tl

204

82 Pb 207

83 Bi

209

84 Po 210

85 At

210

86 Rn 222

7 87 Fr

223

88 Ra 226

89 Ac† 226

104 Rf

105 Db

106 Sg

107 Bh

108 Hs

109 Mt

110 Ds

111 Rg

112 Cn

113 Nh

114 Fl

115 Mc

116 Lv

117 Ts

118 Og

*Lanthanides 58 Ce 140

59 Pr

141

60 Nd 144

61 Pm 147

62 Sm 150

63 Eu 152

64 Gd 157

65 Tb 159

66 Dy 163

67 Ho 165

68 Er

167

69 Tm 169

70 Yb 173

71 Lu

175

†Actinides 90 Th 232

91 Pa 231

92 U

238

93 Np 237

94 Pu 242

95 Am 243

96 Cm 247

97 Bk 251

98 Cf

251

99 Es 254

100 Fm 253

101 Md 256

102 No 254

103 Lr

257

KEY:

Metal Semi Metal Non-metal

13 Al 27

14 Si 28

15 P 31

In this example Al is a metal, Si is a semi-metal (metalloid) and P is a non-metal.

Task 0Ci

1. Determine the number of protons, electrons and neutrons in,

(a) 210Pb82

(b) 34S16

2. Using only the periodic table above, determine how many elements within the first 20,

have atoms with;

(a) The same numbers of protons and electrons (b) The same numbers of protons and neutrons


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TOPIC 0D: Nomenclature

Nomenclature

Nomenclature is the language of chemistry, and a grasp of it is essential to studying the subject.

Symbols

Each element has a symbol displayed on the periodic table. Some elements have a symbol that

is a single letter while others have a symbol made up of two letters. It is important when writing

the two letter symbols to ensure that you use a lower case letter for the second letter. This may

sound trivial but is very important, for example, Co (cobalt), a metal element, is not the same as

CO (carbon monoxide), a gaseous compound made from carbon (C) and oxygen (O).

Binary compounds of metals and non-metals (ionic compounds)

Binary compounds are those formed between only two elements. In compounds where one is a

metal and one a non-metal an ionic compound is formed. An ion is a charged particle and ionic

formulae and names can be determined by considering the charge on the ions. To find the

formula of an ionic compound the positive and negative charges must be balanced, i.e., there

must be no net charge.

To name a binary compound of a metal and a non-metal, the unmodified name of the positive ion

is written first followed by the root of the negative ion with the ending modified to -ide. For

example, NaCl is sodium chloride.

A few common ions, their charges and formulae are listed below.


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Negative ions (ANIONS) Positive ions (CATIONS)

Name Charge Symbol Name Charge Symbol

Bromide 1- Br- Aluminum 3+ Al3+

Chloride 1- Cl- Barium 2+ Ba2+

Fluoride 1- F- Calcium 2+ Ca2+

Hydride 1- H- Copper (I) 1+ Cu+

Iodide 1- I- Copper (II) 2+ Cu2+

Nitride 3- N3- Hydrogen 1+ H+

Oxide 2- O2- Iron (II) 2+ Fe2+

Phosphide 3- P3- Iron (III) 3+ Fe3+

Sulfide 2- S2- Lead (II) 2+ Pb2+

Lead (IV) 4+ Pb4+

Lithium 1+ Li+

Magnesium 2+ Mg2+

Manganese (II) 2+ Mn2+

Nickel (II) 2+ Ni2+

Potassium 1+ K+

Silver 1+ Ag+

Sodium 1+ Na+

Strontium 2+ Sr2+

Tin (II) 2+ Sn2+

Tin (IV) 4+ Sn4+

Zinc 2+ Zn2+

Most transition metal ions (and a few other metal ions) include a Roman numeral after the name,

for example, copper (II). These metals form ions with varying charges, and the Roman numeral

identifies the charge in each case. Elements that commonly form an ion with only a single charge

for example, sodium, do not have Roman numerals associated with them.


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Task 0Di

1. Name these binary compounds.

(a) NaCl

(b) SrO

(c) AlN

(d) BaCl2(e) K2O

(f) CuO

(g) Cu2O

2. Convert these names to formulae.

(a) Magnesium nitride

(b) Barium bromide

(c) Aluminum phosphide

(d) Potassium iodide

(e) Lithium chloride

(f) Sodium fluoride(g) Tin (IV) bromide

Binary acids

Acids will be discussed at great length later in the course, but for the purposes of nomenclature,

an acid can be defined as a compound that produces hydrogen ions (H+) when it is dissolved in

water, and the formulae of acids start with ‘H’. Binary acids are formed when hydrogen ions

combine with monatomic anions.

To name a binary acid use the prefix ‘hydro’ followed by the other non-metal name modified to an

–ic ending. Then add the word ‘acid’. For example, HCl is hydrochloric acid.


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Polyatomic ions

Polyatomic ions are those where more than one element are combined together to create a

species with a charge. Some of these ions can be named systematically, others names must be

learned. Some common polyatomic ions, their charges and formulae are listed below.

Common Polyatomic ions Name Charge Formula

Ammonium 1+ NH4+

Carbonate 2- CO32-

Chromate (VI) 2- CrO42-

Dichromate (VI) 2- Cr2O72-

Ethanedioate 2- C2O42-

Hydrogen carbonate 1- HCO3-

Hydrogen sulfate 1- HSO4-

Hydroxide 1- OH- Manganate (VII) (permanganate) 1- MnO4

- Nitrate 1- NO3

- Nitrite 1- NO2

- Phosphate 3- PO4

3- Sulfate 2- SO4

2- Sulfite 2- SO3

2-

Polyatomic anions where oxygen is combined with another non-metal are called oxoanions and

can be named systematically. In these oxoanions certain non-metals (Cl, N, P and S) form a

series of oxoanions containing different numbers of oxygen atoms. Their names are related to the

number of oxygen atoms present, and are based upon the system below.

Name Number of oxygen atoms

Hypo(element)ite q

Increase in number of oxygen atoms

q

(element)ite

(element)ate

Per(element)ate


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Where there are only two members in such a series the endings are –ite and –ate. For example,

sulfite (SO32-) and sulfate (SO4

2-). When there are four members in the series the hypo- and per-

prefixes are used additionally.

Some oxoanions contain hydrogen and are named accordingly, for example, HPO42-, hydrogen

phosphate. The prefix thio- means that a sulfur atom has replaced an atom of oxygen in an anion.

To name an ionic compound that contains a polyatomic ion, the unmodified name of the positive

ion is written first followed by unmodified name of the negative ion. For example, K2CO3 is

potassium carbonate.

Oxoacids

Oxoacids are formed when hydrogen ions combine with polyatomic oxoanions. This gives a

combination of hydrogen, oxygen and another non-metal.

To name an oxoacid use the name of the oxoanion and replace the -ite ending with –ous or the -

ate ending with -ic. Then add the word ‘acid’. For example, H2SO4 is sulfuric acid.

To illustrate the names of these oxoanions and oxoacids consider the following example

using chlorine as the non-metal.

Formula and name of oxoacid Formula and name of corresponding oxoanion HClO Hypochlorous acid ClO- Hypochlorite HClO2 Chlorous acid ClO2

- Chlorite HClO3 Chloric acid ClO3

- Chlorate HClO4 Perchloric acid ClO4

- Perchlorate


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Task 0Dii

1. What are the formulae for the following ionic compounds?

(a) Ammonium nitrate

(b) Copper (II) bromide

(c) Copper (I) bromide

(d) Zinc hydrogen sulfate

(e) Aluminum sulfate

(f) Sodium perchlorate

(g) Copper (II) iodite

2. Convert the following formulae to names.

(a) NaNO3

(b) KMnO4

(c) CaC2O4

(d) CuSO4

(e) Cu2SO4

(f) KNO2

(g) LiClO4


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Binary compounds of two non-metals (molecular compounds)

If the two elements in a binary compound are non-metals, then the compound is molecular.

To name a molecular compound of two non-metals, the unmodified name of the first element is

followed by the root of the second element with ending modified to -ide. In order to distinguish

between several different compounds with the same elements present use the prefixes mono, di,

tri, tetra, penta and hexa to represent one, two, three, four, five and six atoms of the element

respectively. For example, SO2 is sulfur dioxide.

Some other examples are given below.

Formula Name BCl3 Boron trichloride CCl4 Carbon tetrachloride CO Carbon monoxide CO2 Carbon dioxide NO Nitrogen monoxide NO2 Nitrogen dioxide

Note that the prefix mono is only applied to the second element present in such compounds, if the

prefix ends with ‘a’ or ‘o’, and the element name begins with ‘a’ or ‘o’, then the final vowel of the

prefix is often omitted.

Some compounds have trivial names that have come to supersede their systematic names, for

example, H2O is usually ‘water’, not dihydrogen monoxide!


7/31/2016 5:03 PM of 24

Task 0Diii

1. Write formula or names for the following molecular compounds.

(a) Dinitrogen tetroxide

(b) Phosphorous pentachloride

(c) Iodine trifluoride

(d) Nitrogen dioxide

(e) Dihydrogen monoxide

2. Convert the following formulae to names.

(a) N2O5

(b) PCl3(c) SF6

(d) H2O

(e) Cl2O

Hydrates

Hydrates are ionic formula units with water molecules associated with them. The water molecules

are incorporated into the solid structure of the ions. Strong heating can generally drive off the

water in these salts. Once the water has been removed the salts are said to be anhydrous

(without water).

To name a hydrate use the normal name of the ionic compound followed by the term ‘hydrate’

with an appropriate prefix to show the number of water molecules per ionic formula unit. For

example, CuSO4.5H2O is copper (II) sulfate pentahydrate.


8/9/2016 7:40 PM Page 3 of 14

Formulae of compounds

The chemical formula of a compound shows the exact ratio of the different elements that are

present. The numbers of each element are recorded using a subscript to the right of the elements

symbol. When only one atom is present, the subscript one is assumed (understood), and not

written.

Percentage composition in chemical formula

To determine the % composition of an individual element within a compound, simply express the

mass of each element as a % of the mass of the compound.

Empirical formulae

The empirical formula is the simplest whole number ratio of the atoms of each element in that

compound and can be calculated from mass data. There is a simple method to follow.

Take the percentage of each element present, assume a sample of 100 g, and divide that

mass by the appropriate mass number. (This gives the number of moles - see later).

Find the smallest number calculated in and divide all the results of the calculations in by

that number. (This gives the molar ratio). N.B. Avoid rounding up or down too much at this stage

and be lenient with significant figures.

The results from should be in a convenient ratio and give the empirical formula. N.B. It may

be that the ratio includes a decimal (fraction) such as .500, .250 or .333 etc. If so then multiply all

numbers by 2, 4 or 3 as appropriate to remove the decimal.


8/9/2016 7:40 PM Page 4 of 14

For example; calculate the empirical formula for the compound containing 40.1% carbon, 6.60%

hydrogen and 53.3% oxygen.

C H O Assuming 100g

sample, the % by mass 40.1 6.60 53.3

RAM (from periodic table)

12.011 1.0079 16.00

% by mass ÷ RAM 3.34 6.55 3.33

divide by smallest 1 2 1

Empirical formula C1H2O1 or CH2O

Task 04a

1. Calculate the empirical formulae of the three oxides of iron shown below.

(a) 77.78% Fe, 22.22% O

(b) 70.00% Fe, 30.00% O

(c) 72.40% Fe, 27.60% O

2. A hydrocarbon (a compound containing only hydrogen & carbon) is found to be

7.690% H and 92.31% C by mass. Calculate its empirical formula.

Formulae of Molecules - Molecular formulae

Once the empirical formula has been established, and given further appropriate data, the

molecular formula can be calculated. The molecular formula tells us exactly how many atoms of

each element are present in the compound rather than just the simplest whole number ratio. It is

a simple multiple of the empirical formula. Hence, in the example of an empirical formula of

CH2O, the molecular formula could be C2H4O2 or C3H6O3 etc. To find the molecular formula it is

necessary to know the Molar mass or Relative Molecular Mass (RMM). Given the RMM to be 60

g mol-1

it is clear the molecular formula is C2H4O2, i.e., twice the empirical.

Task 04b

The same compound as in question #2 in Task 4a has a molar mass of 78.00 g mol-1

. What

is the molecular formula of the compound?


8/9/2016 7:40 PM Page 5 of 14

Chemical equations

Chemical equations are a shorthand method used to illustrate what happens during a chemical

reaction. There are a number of steps to writing an equation.

1. Write down the equation in words.

2. Fill in the correct formulae for all the substances.

3. Balance the equation. Balancing the equation can be tricky and requires practice. It involves

the following steps.

I. Ensure the correct formulae are being used for all the reactants and products

II. Balance each element in turn remembering to multiply brackets out carefully. This

process in rather unscientific and is essentially a process of trial and error but can be

helped by the following tips;

If an element appears in only one compound on each side of the equation, try balancing that first.

Secondly, if one of the reactants or products appears as the free element, try balancing that last.

III. When balancing, only place numbers in front of whole formulae. Do not change the

(correct) formulae of any of the reactants or products, or add any extra formulae. The

numbers that appear in front of each formula are called the stoichiometric coefficients.

They have an extremely important role to play in calculations since they give the

reacting ratio (i.e. the number of moles of one substance that react with, or are produced

from, others).

4. Add state symbols, (s) for solid, (l) for liquid, (g) for gas and (aq) for aqueous meaning in

solution with water.


8/9/2016 7:40 PM Page 6 of 14

Task 04c

1. Write balanced equations for the following reactions.

(a) Hydrogen + Copper (II) oxide Copper + Water

(b) Carbon + Oxygen Carbon monoxide

(c) Magnesium + Sulfuric acid Hydrogen + Magnesium Sulfate

2. Balance the following equations.

(a) Ca + H2O H2 + Ca(OH)2

(b) Cu + O2 CuO

(c) Na + O2 Na2O

(d) Fe + HCl FeCl2 + H2

(e) Fe + Br2 FeBr3

(f) C4H8 + O2 CO2 + H2O

(g) Na2CO3 + HI NaI + CO2 + H2O

(h) CuCO3 CuO + CO2

(i) Pb(NO3)2 PbO + NO2 + O2

(j) H2SO4 + KOH K2SO4 + H2O

(k) NaHCO3 Na2CO3 + H2O + CO2

(l) Al + O2 Al2O3


8/9/2016 7:40 PM Page 7 of 14

The mole concept and calculations from equations

Atomic Mass Units

We have seen previously how atoms are comprised of protons, neutrons and electrons and how

the protons and neutrons have masses of approx. 1 atomic mass unit (amu) respectively and that

electrons do not contribute significantly to the mass. If we define the amu as having a mass of

1.66 x 10-24

g we can see that atoms are extremely small and have an extremely small mass.

For example, 1 atom of Cl35

contains 17 protons and 18 neutrons. This is a total of 35 amu

(ignoring the electrons) and has a mass of 5.81 x 10-23

g. This is a very small number so we use

the concept of the mole (see below) to overcome the problem of handling such small quantities.

As you will see 1 mole contains 6.022 x 1023

particles. So if we take 1 mole of Cl35

atoms they will

have a mass of (5.81 x 10-23

g) (6.022 x 1023

) = 35.0 g.

Relative Atomic Mass (RAM) or Molar Mass

RAM is defined as the weighted average of the masses of all the atoms in a normal isotopic

sample of the element based upon the scale where 1 mole of atoms of the C12

isotope has a

mass of exactly 12.00 g.

We have seen previously how elements occur in nature as a number of different isotopes an as a

result they have a RAM that is not an integer. For example, chlorine occurs in nature as approx.

75% Cl 35

and 25% Cl 37

Average Relative Atomic Mass of 1 mole of chlorine atoms =

100

25 x 3775 x 35 = 35.5 g

Average relative atomic masses are the mass numbers recorded on the periodic table. The

relative masses of atoms shown on the periodic table can be used to determine the relative

masses of molecules and ions by simple summation.

Relative Molecular Mass (RMM) or Molar Mass - Found by adding all of the individual RAM’s

together in one molecule of a compound.

Relative Formula Mass (RFM) or Molar Mass - Found by adding all of the individual RAM’s

together in one formula unit of an ionic compound.


8/9/2016 7:40 PM Page 8 of 14

Avogadro’s number and the mole concept

In chemistry, amounts of substances are measured in moles (mols). The mole is a standard

number of particles (atoms, ions or molecules) and can be defined as, the amount of any

substance that contains the same number of particles, as there are C12

atoms in 12.00g of the C12

isotope. The actual number of particles in a mole, known as the Avogadro constant or number is

found to be 6.022 x 1023

particles per mole and has the unit, mol-1

. For example, one mole of

atoms = 6.02 x 1023

atoms.

The average Relative Atomic Mass (RAM) of each element is given on the periodic table. The

figure shows the average mass of one mole of atoms of that particular element. This leads to the

relationships below.

mass of sample mass of sampleNumber of Moles of an Element

RAM Molar Mass

mass of sample mass of sampleNumber of Moles of a Molecular compound =

RMM Molar Mass

mass of sample mass of sampleNumber of Moles of an Ionic compound

RFM Molar Mass

Task 04d

1. What is the mass of one mole of sodium chloride, NaCI

2. How many moles of Ca atoms are there in 140. g of calcium?

3. How many moles of CuBr2 are there in 0.522 g of copper(II) bromide?

4. How many moles of CO2 molecules are there in 23.0 g of carbon dioxide?

5. How many ‘particles’ are present in each of the chemicals in questions #1-4 above?


8/9/2016 7:40 PM Page 9 of 14

If we know the number of moles of a substance that is present in a reaction and we know a

balanced chemical equation, (i.e. we know the reacting ratio), it is possible to calculate the moles

of another substance present in the equation. Use this method;

1. Write a correct and balanced equation.

2. Find the number of moles present by using a moles relationship for one substance.

3. Use the stoichiometric coefficients in the equation to find the reacting ratio of the moles. Use

this relationship to find the number of moles of the unknown substance.

4. Re-apply a moles relationship for the unknown substance.

Task 04e

The combustion of methane (CH4) can be summarized by the equation below.

CH4 + 2O2 CO2 + 2H2O

1. Calculate the mass of O2 required to produce 2.23 g of carbon dioxide.

2. Calculate the mass of water produced when 34.0 g of CH4 is burned.


8/9/2016 7:40 PM Page 10 of 14

Volumetric analysis and moles

Chemical reactions are often carried out between substances that are in solution (dissolved in a

solvent, usually water). The concentration of a solution can be measured in terms of the number

of grams of the solute (solid) that has been dissolved in a particular volume of the solution (where

water is usually the solvent), or more usually, in terms of the number of moles of the solute in a

particular volume of the solution. Typical units are g dm-3

or g/dm3 or mol dm

-3 or mol/dm

3 or

mol/L or mol L-1

.

The method of expressing the concentration of a solution in mol L-1

is the most common (and

most useful) and is referred to as Molarity (M). So, for example, a solution that has a

concentration of 0.250 mol L-1

can be referred to as 0.250 M solution, or a 0.250 “molar” solution.

When concentration is measured in mol L-1

, or M, and volume in L, then, for solutions;

Moles (concentration) (volume)

Task 04f

1. What mass of solute (solid sodium carbonate) must be used in order to prepare 275

mL of 1.20 mol L-1

sodium carbonate solution?

2. A sample of copper (II) sulfate pentahydrate with a mass of 8.512 g is dissolved in

500.0 mL of water. A 25.00 mL portion completely reacts with 20.00 mL of a 0.1702 mol

L-1

solution of iodide ions. In what molar ratio do Cu2+

and iodide ions react?

Questions #3-5 require balanced equations before they can be solved.

3. Carbonates, in the form of antacid tablets, can be used to neutralize stomach acid. 25.0

mL of 0.100 mol L-1

sodium carbonate solution completely reacts with 35.3 mL of HCl

in such a simulated neutralization. What is the concentration (molarity) of the acid?

4. Some metals will react vigorously with acids to produce hydrogen gas. What mass of

zinc metal, will react completely with 75.0 mL of 0.200 mol L-1

sulfuric acid?

5. Hydroxides can be used to neutralize acids. What volume of 1.00 mol L-1

NaOH, would

be required to completely neutralize 25.0 mL of 2.00 mol L-1

HCI?


8/9/2016 7:40 PM Page 11 of 14

Dilution

Often, solutions are prepared by adding water to (diluting) more concentrated ones. For example,

if 4.0 L of a 2.0 M solution was required, it could be made by diluting some 10. M solution.

Calculations involving dilution problems involve three steps.

1. Calculate the number of moles present in the final, diluted solution, by applying moles =

(concentration) (volume).

2. Calculate the volume of the starting, more concentrated solution that supplies this

number of moles by applying moles = (concentration) (volume).

3. The volume of water that must be added to the concentrated solution is simply the

difference between the volume of the final, diluted solution and the volume of the

concentrated solution.

Worked Example

Calculate the volume of water that must be added to prepare 2.0 L of 3.0 M KOH from a stock

solution that has a concentration of 8.0 mol L-1

.

1. Final solution must contain (2.0 L)(3 mol/L) mols = 6.0 mols of KOH.

2. Since moles = (concentration) (volume), the volume (in L) of the stock (concentrated)

solution that contains 6.0 mols of KOH = 6.0 mol

8.0 mol/L = 0.75 L.

3. So, by taking 0.75 L of the stock solution and adding 1.25 L of water to make the solution

up to 2.00 L, the final, diluted solution, will have a concentration (molarity) = 6.0 mol

2.0 L =

3.0 mol L-1

or 3.0 M.

Task 04g

1. Calculate the volume of 3.25 M nitric acid that must be diluted with water to produce

500. mL of 1.25 M nitric acid.

2. Calculate the volume of 2.60 M KOH that must be diluted with water to produce 250.

mL of 2.00 M KOH.

3. What volume of water must be added to 4.0 M HCl in order to produce 2.0 L of 0.5 M

HCl?


8/9/2016 7:40 PM Page 12 of 14

Combustion Analysis

Compounds that contain carbon and hydrogen only, when burned completely in oxygen, will yield

only carbon dioxide and water. Analysis of the mass of CO2 and H2O produced can be used to

determine the empirical formula of the substance in question. This method assumes that all the

carbon in CO2 originated from the carbon in the original compound, and all the hydrogen in the

water originated from the hydrogen in the original compound. The method is as follows.

1. Calculate the moles of CO2 produced. Since there is one carbon atom in one molecule of

CO2 this is also the number of moles of C atoms present in the original compound.

2. Calculate the moles of H2O produced. Since there are two hydrogen atoms in one

molecule of H2O multiply this number by two to calculate the number of moles of H atoms

present in the original compound.

3. Calculate the mass of C and H present in the combusted sample by multiplying the moles

of each by their RAM’s.

4. If there is another element present (typically O) in the combusted substance then

calculate its mass by subtracting the mass of C and H from the total mass of the

combusted sample. Turn this mass into moles by dividing by the appropriate RAM.

Find the smallest number of moles calculated and divide all the numbers of moles by that

number. (This gives the molar ratio). N.B. Avoid rounding up or down too much at this

stage and be lenient with significant figures.

5. The results from #4 should be in a convenient ratio and give the empirical formula. N.B. It

may be that the ratio includes a fraction such as .500, .250 or .333 etc. If so then multiply

all numbers by 2, 4 or 3 as appropriate to remove the fraction.

6. If necessary use the Molar Mass to turn the empirical formula into a molecular formula.

Task 04h

When 4-ketopentenoic acid is analyzed by combustion, it is found that a 0.3000 g sample

produces 0.579 g of CO2 and 0.142 g of H2O. The acid contains only carbon, hydrogen, and

oxygen. What is the empirical formula of the acid?


8/9/2016 7:40 PM Page 13 of 14

Analysis of hydrates

Hydrates are formula units with water associated with them. The water molecules are

incorporated into the solid structure. For example, CuSO4.5H2O, copper(II) sulfate pentahydrate.

Strong heating can evaporate the water. When water is removed the salts are called anhydrous

(without water).

Task 04i

1. A sample of the hydrated salt CoCl2.xH2O, with a mass of 11.73 g is weighed, heated to

drive off the water of crystallization, cooled and reweighed until constant mass (6.410

g) is achieved. Calculate the value of x.

2. What is meant by the term, “constant mass” in question #1?

Limiting Reactant

When all the reactants in a chemical reaction are completely consumed, i.e. they are all

converted to products, then the reactants are said to be in stoichiometric proportions. On other

occasions it may be necessary to ensure that only one particular reactant is completely used up.

This is achieved by using an excess of all the other reactants. The reactant that is completely

consumed is called the limiting reagent and it, is what determines the quantities of products that

form.

Task 04j

The two non-metals, sulfur and chlorine, react according to the equation below.

S(s) + 3Cl2(g) SCl6(l)

If 202 g of Sulfur are allowed to react with 303 g of CI2 in the reaction above, which is the

limiting reactant, how much product will be produced and what mass of the excess

reactant will be left over?


8/9/2016 7:40 PM Page 14 of 14

Percentage yield

In all chemical syntheses the yield of the product will be less than 100%. The % yield is given as

actual yield of product% Yield = X 100

theoretical yield of product

The yield is usually less than 100% since the reactants are often not pure, some of the product is

lost during purification, the reaction may be reversible and/or side reactions may give by-

products.

Task 04k

Aluminum will react with oxygen gas according to the equation below;

4Al + 3O2 2Al2O3

In one such reaction, 23.4 g of Al are allowed to burn in excess oxygen. 39.3 g of

aluminum oxide are formed. What is the percentage yield?

© Adrian Dingle’s Chemistry Pages and ChemEducator LLC 2013 – All rights reserved

1

AP WORKSHEET 00a: Significant Figures 1. Determine the number of significant figures in each of the following. (6) (a) 0.7680 (b) 1230.00 (c) 1000.01 (d) 120.0 (e) 1.09 x 10

4

(f) 0.0080060 2. Use a calculator to find the results of the following and then round the answer to the correct

number of significant figures. (6)

(a) 34.66 + 333.0 (b) 1.23 + 9.66 (c) 445 - 1.22 (d) 18.2 + 1.998 (e) 10.2 – 1.34 (f) 100 - 23 3. State the significant figure rule that is associated with “captive zeros”. (1) 4. State the significant figure rule that is associated with “leading zeros”. (1) 5. State the significant figure rule that is associated with “trailing zeros”. (1) 6. State the significant figure rule that is associated with addition and subtraction operations. (1) 7. State the significant figure rule that is associated with multiplication and division operations.

(1)


2

8. Use a calculator to find the results of the following calculations and then round the answer to

the correct number of significant figures. (6)

(a) 12 x 11.45 (b) (1.23 x 10

3) x (6.4 x 10

2)

(c) 5.233 x 6.324 (d) 34 / 22 (e) (1.8 x 10

5) / 14

(f) 100.23 / 5.22 9. Round each of the following to three significant figures. (6) (a) 167.789 (b) 0.0000456922 (c) 23.00567 (d) 3.4569 (e) 7903.0005 (f) 11.044 10. How many significant figures in each of the following? (6) (a) 654.001 nm (b) 6.02 x 10

23 particles

(c) 1.0079 g (d) 13 neutrons (e) 11.22201 mg (f) 0.004504 g


1

AP WORKSHEET 00b: Unit Conversions This worksheet utilizes the conversions given at this web site http://www.onlineconversion.com 1. Perform the following conversions. In each case show the full, dimensional analysis. Source

any conversion factors from the web site above. An example is given below. (6)

Question: 3.00 cm to mm.

Answer: 3.00 cm

10 mm

1 cm

= 30 mm

(a) 120 J to MJ (b) 3 m to cm (c) 400 miles to km (d) 25 hectares to acres (e) 34 inches to ft (f) 289 s to hrs 2. Perform the following conversions. In each case you do NOT need to show the full,

dimensional analysis. Source any conversion factors from the web site above. (6)

(a) 120000 J to kJ (b) 13 kg to lbs (c) 83.2 K to

oC

(d) 48 mins to ms (e) 34

oF to

oC

(f) 13.2 kg to lbs


2

3. Perform the following sequences of conversions. In each case show the full, dimensional

analysis. Source any conversion factors from the web site above. An example is given below.

(6)

Question: 3.00 cm to m VIA mm.

Answer: 3.00 cm

10 mm

1 cm

0.001 m

1 mm

= 0.03 m

(a) 679 nm to cm VIA m (b) 23 miles to m VIA km (c) 567 feet to m VIA yd (d) 12 L to UK gal VIA mL (e) 8 MJ to J VIA kJ (f) 418 s to hrs VIA min


1

AP WORKSHEET 00c: Atomic Structure & Ions Use the periodic table here; http://www.adriandingleschemistrypages.com/apptable.pdf to help you answer this worksheet. 1. What is the charge on a sodium atom? (1) 2. What is the charge on a sodium nucleus? (1) 3. What is the atomic number of potassium? (1) 4. How many protons are there in the nucleus of a potassium atom? (1) 5. How many electrons in the potassium nucleus? (1) 6. What is the most likely charge (the most common charge) on an ion of sulfur? (1) 7. If a chloride ion and a strontium ion were to form a compound, what would its formula be? (1) 8. What do all the ions of the transition metals have in common with one another? (1)


1

AP WORKSHEET 00d: Elements & Symbols Although you will always have access to a periodic table in tests and exams, the periodic table will NOT have element names on it. In that light it will be extremely helpful to you if you can begin to recognize as many of the element names and symbols as possible.

1. Assign the 50 elements below to one of the four lists based upon their symbols. (50) List A: Elements that have symbols that are only the first letter of the element's name List B: Elements that have symbols that are the first two letters of the element's name List C: Elements that have symbols that are the first letter and another letter of the element’s name List D: Elements that have symbols that are rooted in another language or other source

ELEMENT NAME ELEMENT SYMBOL LIST A, B, C or D?

aluminum

antimony

argon

arsenic

barium

beryllium

bismuth

boron

bromine

cadmium

calcium

carbon

cerium

cesium

chlorine

chromium

cobalt

copper

fluorine

gallium

gold

helium

hydrogen

iodine

iron


2

ELEMENT NAME ELEMENT SYMBOL LIST A, B, C or D?

lead

lithium

magnesium

manganese

mercury

neon

nickel

oxygen

phosphorus

platinum

potassium

radium

selenium

silicon

silver

sodium

strontium

sulfur

tin

titanium

tungsten

uranium

vanadium

xenon

zinc 2. Using the periodic table, find one more (i.e. not those listed in question #1.) element for each

list. (4) A. _______________

B. _______________ C. _______________ D. _______________


1

AP WORKSHEET 00e: Inorganic Nomenclature I 1. The following compounds are all binary compounds. Give the name of each one. (6)

(a) SrO (b) K2O (c) Na2S (d) Cs3P (e) AlCl3 (f) Mg3N2

2. Some of the following name and formula combinations are incorrect. Identify the correct combinations. For the others, suggest corrected combinations. (13)

(a) barium hydroxide, BaOH2 (b) sodium oxide, SoO2 (c) barium chloride, BCl3 (d) strontium oxide SrO2 (e) boron trifluoride, BoFl6 (f) vanadium (III) chloride, VCl3 (g) magnesium oxide, MgO4

3. Write the name of the following compounds. Use Roman numerals in the names. (7) (a) FeI3 (b) MnCl2 (c) HgO

(d) Cu2S (e) CuS (f) SnI4 (g) MnBr2


2

4. Write the name of each of the following. To help get the correct name, use the periodic table to determine which elements are metals, which are non-metals and which compounds should include Roman numerals in their names. (16) (a) N2Br5 (b) P2S5 (c) Ge2O3 (d) N2O5 (e) SiO2 (f) AlH3 (g) FeO (h) CuCl2 (i) OCl2 (j) XeF6 (k) RaCl2 (l) SeCl2 (m) PCl5 (n) Na3P (o) CuF (p) V2O5


1

AP WORKSHEET 00f: Inorganic Nomenclature II Add either a name or a formula to complete each table. (100)

1. Potassium dichromate

2. Lithium sulfide

3. Potassium bromide

4. Cesium iodide

5. Calcium phosphide

6. Sodium fluoride

7. Strontium oxide

8. Beryllium sulfide

9. Magnesium bromide

10. Lithium oxide

11. Strontium chloride

12. Barium bromide

13. Magnesium sulfide

14. Magnesium iodide

15. Hydrogen fluoride (Hydrogen monofluoride)

16. Barium phosphide

17. Sodium hydrogen phosphate

18. Potassium chloride

19. Lithium nitride

20. Calcium sulfide

21. Rubidium oxide

22. Strontium nitride

23. Cesium phosphide

24. Magnesium carbonate

25. Beryllium sulfate


2

26. Dinitrogen Tetraoxide

27. Carbon dioxide

28. Mercury(I) chloride

29. Hydroiodic acid

30. Iodic acid

31. Perbromic acid

32. Hypobromous acid

33. Phosphorus pentachloride

34. Iodine monochloride

35. Antimony(III) fluoride

36. Bromine monofluoride

37. Bromine dioxide

38. Dinitrogen pentoxide

39. Carbon monosulfide

40. Tellurium dioxide

41. Phosphorus tribromide

42. Carbon tetraiodide

43. Vanadium(V) chromate

44. Zinc carbonate

45. Silver hydroxide

46. Vanadium(III) chromate

47. Mercury(II) iodide

48. Uranium(V) nitrate

49. Nickel (III) nitride

50. Sulfuric acid


3

51. ScCl3

52. HCl

53. PtO2

54. Sb(ClO3)5

55. GeS2

56. ZnO

57. VSO4

58. CuCl2

59. TiO2

60. NiN

61. Ni3(PO4)2

62. CoF3

63. Au2O3

64. Zn3P2

65. Cr(NO3)6

66. NaIO2

67. NaIO3

68. NaI

69. H2SO3

70. H2CO3

71. AlN

72. AlH3

73. Li3AsO4

74. NaCN

75. Na2O2


4

76. Li3PO3

77. KHCO3

78. HF

79. AuI2

80. KMnO4

81. Na2Cr2O7

82. Ag2CrO4

83. AgCl

84. NaCH3COO

85. RaF2

86. KSCN

87. FeS

88. Fe2(SO3)3

89. FeSO4

90. MgS

91. Na2S2O3

92. RbCl

93. Cu(OH)2

94. Mg3N2

95. Cu3N

96. LiH

97. K2O

98. K2O2

99. Li3N

100. DsCl3


1

AP WORKSHEET 00g: Inorganic Nomenclature III (Acids)

1. Write the formula of each of the following acids. (14)

(a) Nitric acid

(b) Chloric acid

(c) Hydrochloric acid

(d) Sulfurous acid

(e) Chlorous acid

(f) Hydrobromic acid

(g) Phosphoric acid

(h) Nitrous acid

(i) Perchloric acid

(j) Hydrofluoric acid

(k) Perbromic acid

(l) Sulfuric acid

(m) Bromic acid

(n) Hypoiodous acid


2

2. Name the following acids. (14)

(a) HClO3

(b) H3PO4

(c) HI

(d) H2SO3

(e) HNO3

(f) HF

(g) HC2H3O2

(h) HBr

(i) H3PO3

(j) HClO

(k) H2CO3

(l) H2SO4

(m) HBrO2

(n) HNO2


1

AP WORKSHEET 00h: Inorganic Nomenclature IV Formula Name

1 HClO2

2 NaBr

3 K2CO3

4 Cu(HCO3)2

5 Al2(SO3)3

6 SrH2

7 BI3

8 N2O4

9 SeF6

10 P2O5

11 FeCl2

12 OF

13 Cu2SO4

14 ClF5

15 MgSO4

16 MnO

17 FeO

18 Mg(OH)2

19 CrO3

20 Rb2O

© Adrian Dingle’s Chemistry Pages 2004, 2005, 2006, 2007, 2008, 2009, 2010, 2011, 2012. All rights reserved. These materials may NOT be copied or redistributed in any way, except for individual class instruction.

Revised August 2011

Macintosh HD:Users:adriand:Dropbox:ADCP:hwsheet04s.doc Page 1 of 8

HONORS WORKSHEET 4s: Stoichiometry Summary

• TYPE 1: Those involving Avogadro’s number (the mole concept). Question 1 A sample of Ge is found to contain 9.7 x 1023 atoms of Ge. How many moles of Ge atoms are in the sample? (1) Question 2 How many W atoms are found in 0.43 moles of pure W? (1)


Revised August 2011


• TYPE 2: Those involving the relationship between mass, moles and molar mass (RAM, RFM, RMM).

Question 3 What is the mass in grams of 0.531 moles of Sn? (1) Question 4 How many moles of Ca are in 2.03 g of Ca? (1) Question 5 5.00 moles of a binary, group II oxide are found to have a mass of 521 g. Identify the group II metal. (2)


Revised August 2011


• TYPE 3: Those combining types #1 & #2.

Question 6 How many Ta atoms are found in a 1.231 g sample of Ta? (2) Question 7 What is the mass of 8.11 x 1023 atoms of Sulfur? (2) Question 8 What mass of Cu atoms have the same number of atoms as there are in a 4.21g sample of Si? (2)


Revised August 2011


• TYPE 4: % by Mass Composition.

Question 9 Calculate the percent by mass composition of dimethylether, CH3OCH3. (2)

Question 10 What is the percent by mass composition of aluminum sulfate? (2) Question 11 A compound that contains a complex ion has the formula Al4[Fe(CN)6]3. What is the percent by mass composition of this compound? (2) • TYPE 5: Empirical formula.

Question 12 A compound containing silver and chlorine contains 75.3% Ag. What is the empirical formula of the compound? (2)

Question 13 In a vigorous chemical reaction, 1.403 g of sodium metal is completely reacted with 1.159 g of fluorine gas. What is the empirical formula of the compound formed? (3)


Revised August 2011


• TYPE 6: Molecular formulae from empirical formulae.

Question 14 What is the molecular formula of hydrocarbon that has an empirical formula of CH and a molecular mass of 78 gmol-1? (1) Question 15 A compound contains 48.65% carbon, 8.108% hydrogen, and the remainder oxygen. The molecular mass of this compound is approximately 74.00 g/mol. What is the empirical formula? What is the molecular formula? (3) • TYPE 7: Combustion analysis.

Question 16 The combustion of 4.000 g of a compound that contains only C, H, N and Br yields 3.826 g of CO2 and 2.087 g of H2O. Another sample of the compound with a mass of 3.111 g is found to contain 1.803 grams of Br. What is the empirical formula of the compound? (6)


Revised August 2011


• TYPE 8: % Yield.

Question 17 Propane will combust according to the reaction below. If 11.1 g of Propane produces 23.3 g of CO2 when burned in excess oxygen, what is the % yield? (3)

C3H8 + 5O2 è 3CO2 + 4H2O

• TYPE 9: Limiting reactant.

Question 18 Consider the reaction between Iron and anhydrous Copper (II) sulfate that produces Iron (II) sulfate and Copper metal.

(a) Write an equation for the reaction. (2) (b) If 120. g of Fe are reacted with 200. g of Copper (II) sulfate, identify the limiting

reagent. Which reagent is in excess? (2)

(c) Calculate the mass of Copper formed. (2)

(d) How much of the excess reagent is left over at the end of the reaction? (2)


Revised August 2011


• Type 10: Analysis of hydrated salts.

Question 19 Barium Chloride is found as a hydrated salt, BaCl2.xH2O. A student carefully heats 2.50 g of the salt to a constant mass of 2.13 g. Find x. (4) • TYPE 11: Moles and reacting ratios (including solutions).

Question 20 Calcium hydrogen carbonate, Ca(HCO3)2, reacts with HCl according to the equation below.

Ca(HCO3)2 + 2HCl è CaCl2 + 2CO2 + 2H2O

(a) What volume of 0.235 M HCl solution must be present to totally react with 0.140 moles of the calcium compound? (2)

(b) How many moles of water are produced when 0.491 g of the calcium compound

combines with excess HCl? (2)


Revised August 2011


• TYPE 12: Dilution.

Question 21 Calculate the volume of 0.120 M sulfuric acid that must be diluted with water to produce 3.00 L of 0.018 M sulfuric acid. (2)


1

AP WORKSHEET 00s: Preamble Summary 1. Classify the following as either chemical or physical changes. (3)

(a) Ice melting (b) Gasoline burning (c) Evaporation of perfume from an open bottle 2. Mercury is a liquid metal that has a density of 13.58 g/mL. Calculate the volume of mercury

that must be poured out in order to obtain 0.5000 g of Mercury. (2)

3. Classify the following as either quantitative or qualitative observations. (4)

(a) My eyes are brown (b) My neck size is 17 inches

(c) My average grade last year was 79%

(d) Physics is a difficult subject 4. Give an example of a natural law (other than the law of conservation of mass). (1)

5. Convert these numbers to scientific notation. (2)

(a) 35800000000000

(b) 0.00000000821


2

6. Round the following numbers to four figures. (6)

(a) 2.16347 x 105

(b) 4.000574 x 106

(c) 3.682417

(d) 7.2518

(e) 375.6523

(f) 21.860051 7. Perform the following conversions. (5)

(a) 0.75 kg to milligrams

(b) 1500 millimeters to km

(c) 2390 g to kg

(d) 0.52 km to meters

(e) 65 kg to g 8. Complete the following table of temperatures, performing the appropriate conversions. (18)

Kelvin Fahrenheit Celsius

200.

23.0

0.000

180.

45.0

500.

350.

97.0

30.0


3

9. An experiment is performed in which the molar mass of a gas is found to be 48.45 g mol-1

.

The published (actual) value is 52.98 g mol-1

. Calculate the percentage error. (2)

10. Distinguish carefully between precision and accuracy. (2) 11. In the table below, match the scientist with the experiment. (2)

Scientist Experiment

Crookes Oil Drop

Millikan Cathode Ray

Rutherford Gold Foil

12. Consider the following pairs; does either pair represent a pair of isotopes? Explain. (4) (a)

11Na23 and

11Na24

(b)

11Na24 and

12Mg24

13. Determine the number of protons, electrons and neutrons in each of the following

isotopes. (3)

(a) 79

Au171 (b)

79Au182

(c) 35

Br-

79


4

14. In the following question give the missing formula or name. (10)

Formula Name

CaS

Pb3N2

AlP

HBrO4

(NH4)2CO3

Calcium ethanoate

Phosphorous pentachloride

Strontium bromite

Potassium hydrogen carbonate

Chloric acid

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