Chemistry - notes.engineeringonline.ncsu.edu Review 2019/FE Review Chemistry.pdf4 metals Fundamentals of Engineering Exam Review metalloids non-metals A. Periodic Table (Periodic Table

Chemistry

Fundamentals of Engineering Exam Review

Fundamentals of Engineering

Exam Review

Other Disciplines FE SpecificationsChemistry: 7–11 FE exam problems

A. Periodic Table (e.g., nomenclature, metals and non-metals, atomic

structure of matter)- 8Q’s

B. Oxidation and Reduction – 4 Q’s

C. Acids and Bases – 5 Q’s

D. Equations (e.g., stoichiometry, equilibrium) – 7 Q’s

E. Gas Laws (e.g., Boyle’s and Charles’ Laws, molar volume) – 4 Q’s


A. PERIODIC TABLE:

NOMENCLATURE, METALS AND NON-METALS

The Periodic Table is divided into three regions:

metals, non-metals, and metalloids


3

Proper nomenclature requires identifying if a compound is

composed of: metals & non-metals (ionic)

non-metals & non-metals (molecular)

Materials composed of all the same type of metal or non-metal are

simply named by their element symbol.

A. Periodic Table

4

metals


non-metalsmetalloids

A. Periodic Table

(Periodic Table from chemistry section of FE Reference Handbook)

5

Ionic compounds are composed of metals and non-metals


If the metal is from periodic table column I, column II, or aluminium,

name of the compound with the metal name and the non-metal name

with a substituted -IDE ending

A. Periodic Table

CaCl2 is calcium chlorIDE (chlorine chloride)

Na2O is sodium oxIDE (oxygen oxide)

Al2S3 is aluminium sulfIDE (sulfur sulfide)

6

Ionic compounds are composed of metals and non-metals


If the metal is NOT FROM periodic table column 1, column 2, or

aluminium, name of the compound with the metal name (roman

mumber representing charge) and the non-metal name with a

substituted -IDE ending

A. Periodic Table

FeCl2 is iron (II) chloride (each Cl is -1 so Fe is +2)

TiO2 is titanium (IV) oxide (each O is -2 so Ti is +4)

Tl2S3 is thallium (III) sulfide (each S is -2 so Tl is +3)

7

Ionic compounds are also composed of metals or

ammonium and polyatomic ions, show below.


A. Periodic Table

Cations

NH41+ ammonium ion H3O

1+ hydronium ion

Anions

C2H3O21- acetate ion OH1- hydroxide ion

CO32- carbonate ion NO3

1- nitrate ion

ClO41- perchlorate ion NO2

1- nitrite ion

ClO31- chlorate ion MnO4

1- permanganate ion

ClO21- chlorite ion O2

2- peroxide ion

ClO1- hypochlorite ion PO43- phosphate ion

CrO42- chromate ion SO4

2- sulfate ion

Cr2O72- dichromate ion SO3

2- sulfite ion

CN1- cyanide ion

8


Polyatomics are used in the ionic name without change.

A. Periodic Table

NaOH is sodium hydroxide, and Fe(OH)3 is iron (III) hydroxide

CaSO4 is calcium sulfate, and CuSO4 is copper(II) sulfate

NH4Cl is ammonium chloride, and (NH4)3PO4 is ammonium phosphate

9

Molecular compounds are composed of non-metals


The non-metal on the left of the period table is named as-is and the

non-metal or the right of period table is name with a substituted -IDE

ending.

Greek prefixes are used to indicate how many of each atom type is

present.

A. Periodic Table

# atoms Prefix # atoms prefix

1 mono 6 hexa

2 di 7 hepta

3 tri 8 octa

4 tetra 9 nona

5 penta 10 deca

10

Molecular compounds are composed of non-metals


Greek prefixes are used for all atoms in a molecular compound,

except when there is only one of the first atom type

A. Periodic Table

N2O3 is dinitrogen trioxide

N2O is dinitrogen monoxide

NO2 is nitrogen dioxide

11

Chem A1: Identify the name of Mg(OH)2


A. Periodic Table

A) Magnesium (II) hydroxide

B) Magnesium dihydroxide

C) Magnesium hydroxide

D) Magnesium epoxide

12

Chem A2: Identify the name of MnCO3


A. Periodic Table

A) Manganese carbide

B) Manganese (II) carbonate

C) Manganese (I) carbonate

D) Manganese carbonate

13

Chem A3: Which material is platinum (IV) bromide?


A. Periodic Table

A) Pt4Br

B) PtBr2

C) PtBr3

D) PtBr4

14

Chem A4: Identify the name of SO3


A. Periodic Table

A) Monosulfur trioxide

B) Sulfur (VI) oxide

C) Sulfur trioxide

D) Sulfur trioxygen

15

Chem A5: Which material is dichlorine heptoxide?


A. Periodic Table

A) Cl2O7

B) Cl3O6

C) ClO7

D) Cl4O7

A. PERIODIC TABLE: ATOMIC STRUCTURE OF MATTER


16A. Periodic Table

Proton (p+)

Electron (e-)

Mass (amu)

0.0005 -1

Particle Charge

Neutron (n0) 1.0087

1.0078 +1

0

Major Subatomic Particles

# of protons defines the

element: known as

atomic number

# of protons and

neutrons gives the

atomic mass

# of protons minus the

# of electrons gives the

charge



17A. Periodic Table

For example, iron (Fe) has an atomic mass as 56 for the

most common isotope, and +3 as the most common charge.

How many protons, neutrons and electrons are in 56Fe3+?

Fe26

55.847

The periodic table provides the

number of protons: 26

The atomic mass - #protons gives

the neutrons: 56 – 26 = 30



18A. Periodic Table

For example, iron (Fe) has an atomic mass as 56 for the

most common isotope, and +3 as the most common charge.

How many protons, neutrons and electrons are in 56Fe3+?

Fe26

55.847

Solve the equation to give #electrons:

charge = #protons - #electrons

+ 3 = 26 - # electrons

#electrons = 23

19

Chem A6: The element chlorine has two major isotopes.

The atomic nuclei of these isotopes contain how many

protons?


A. Periodic Table

A) 35.453

B) 17

C) 6

D) 27

Cl17

35.453

20

Chem A7: How many protons, neutrons and electrons are

in 17O2-?


A. Periodic Table

A) 8 protons, 7.999 neutrons, 10 electrons

B) 17 protons, 9 neutrons, 10 electrons

C) 8 protons, 9 neutrons, 10 electrons

D) 8 protons, 9 neutrons, 6 electrons

O8

15.999

21

Chem A8: What is the correct notation for an ion that has

19 protons, 21 neutrons and 18 electrons?


A. Periodic Table

A) 40K1+

B) 39K1+

C) 40K1-

D) 19F1-

K19

39.098F8

19.998

B. OXIDATION AND REDUCTION


22B. Oxidation and Reduction

Knowledge of oxidation states is required to answer questions concerning

oxidation and reduction. Below are some reminders.

1.The oxidation state of an element is 0. Elements are represented as a

single atomic symbol (Cr) or molecule composed of one element type (P4)

2.Fluorine (F) in a compound has a -1 oxidation state. So NaF has F-1.

(F2 is an element and the F is oxidation state 0.)

3. For compounds,

metals in column I have a +1 oxidation state (NaCl Na1+)

metals in column I have a +1 oxidation state (CaS Ca2+)

aluminium has a +3 oxidation state (Al2O3 Al3+)




4. Hydrogen is unusual. It is +1 when bonded to non-metals (NH3 H+1)

and -1 when bonded to metals (NaH H-1)

5. Oxygen(O) in a compound has a -2 oxidation state. So K2O has O-2.

(O2 is an element and the O is oxidation state 0.)

6. Group VII atoms in compounds have a -1 charge. So CrCl3 Cl-1

The oxidation state of other elements in a compound can be determined

algebraically.

In V2O3, O has -2 oxidation state. The entire compound has 0 charge.

Therefor 2 x (V ox state) + (3 x -2) = 0. V must be +3 oxidation state.

In NH41+, H has +1 oxidation state. The entire compound has +1 charge.

Therefor 1 x (N ox state) + (4 x +1) = +1. N must be -3 oxidation state.

24


B. Oxidation and Reduction


Relating oxidation states to the period table will allow easier memorization

-1

-1

-1

-1

+2

+2

+2

+2

+2

+3

+1

+1

+1

+1

+1

-2

-2

-2

-3

-3

-4

+2 or +3

maybe +1, +4, +5, + 6 +7

+1 or +3

+1 or +3 +2 or +4




Oxidation occurs when an atom loses electrons and

becomes more positive in oxidation state.

In a Redox reaction, both reduction and oxidation occur.

Fe + Cu2+ Fe2+ + Cu

Reduction occurs occurs when an atom gains electrons and

becomes more negative in oxidation state.

Oxidation: Fe changes from oxidation state 0 in reactant to +2 in product.

(oxidation state becomes more positive)

Reduction: Cu changes from oxidation state +2 in reactant to 0 in product.

(oxidation state becomes more negative)

26

Chem B1: Which chemical change below represents the

oxidation of nitrogen?


A) N2 NH3

B) NO N2O2


C) N2O NO3-

D) NF3 NO2-

27

Chem B2: Which chemical change below represents the

reduction of carbon?


A) CH4 CO2

B) CO2 CO


C) C2Cl4 CO

D) C C2H2Cl2

28

Chem B3: Which chemical change below is not a

reduction-oxidation reaction?


A) 2K + I2 2 KI

B) SnO2 + C Sn + CO2


C) 4 Fe + 3 O2 2Fe2O3

D) Ca2+ + 2 Cl1- CaCl2

29

Chem B4: Which chemical change below is not a

reduction-oxidation reaction?


A) CH4 + 2O2 CO2 + 2 H2O

B) Ag2S + 2Cl- 2AgCl + S2-


C) Pb + PbO2 + 4H+ → 2Pb2+ + 2 H2O

D) Mg + H2O H2 + MgO

C. ACIDS AND BASESAcids are compounds that when added to water donate a H+

ion to form H3O+.

HCl + H2O Cl- + H3O+


30

Typically the number of H atoms in the front of a chemical

formula are the acidic hydrogens and provide the valence

change.

C. Acids and Bases

HC2H3O2 has 1 acidic H (valence change 1)

H2S has 2 acidic H’s (valence change 2)

H3PO4 has 3 acidic H’s (valence change 3)

C. ACIDS AND BASES

Bases are compounds that when added to water remove a

H+ from water to form OH-.

NH3 + H2O NH4+ + OH-


31

Most bases are negatively charged and exist as ionic compounds

in the solid form. Bases can accept multiple H+ ions

C. Acids and Bases

NaOH (OH- in water) accepts one H+ (valence change 1)

Na2CO3 (CO32- in water) accepts two H+ (valence change 2)

Li3PO4 (PO43- in water) accepts three H+ (valence change 3)

C. ACIDS AND BASES

Acids and Bases can react to form water and an ionic salt in

a neutralization reaction.

HCl + NaOH NaCl + H2O


32C. Acids and Bases

The stoichiometry of the reaction can vary depending on the

materials: H2S + 2 NaOH Na2S + 2H2O

The reaction above requires twice as many moles of NaOH as

H2S, because H2S has two H+ to donate (valence change 2), and

OH- can only accept one H+ (valence change 1)

C. ACIDS AND BASES

Concentration expressed in units of molarity or normality:



For example, 18.23 g of HCl in 1.00 liter (L) of water would

be 0.500 Molarity (M) and 0.500 Normality (N).

Molar mass of HCl is (1.0079 + 35.453 =36.4609 g HCl/mol)

18.25 g HCl x (1 mol HCl / 36.4609) / 1.00 L = 0.500 M

As HCl can only donate 1 H+, it is also 0.500 N

Molarity of Solutions – The number of gram moles of a

substance dissolved in a liter of solution.

Normality of Solutions – The product of the molarity of a

solution and the number of valence changes taking place in

a reaction

(formula from chemistry section of FE Reference Handbook)



Chem C1: What is the Molarity of 10.00 grams of HF dissolved in 2.00

liters of water? The atomic weights of H and F are 1 and 19, respectively.



A) 0.125 M B) 0.250 M C) 0.500 M D) 1.00 M



Chem C2: What is the Normality of 6.80 grams of H2S dissolved in 1.00 liter

of water? The atomic weights of H and S are 1 and 32, respectively.

A) 1.00 N B) 0.400 N C) 0.200 N D) 0.100 N





a reaction



Chem C3: The atomic weights of sodium, oxygen and hydrogen are 23, 16

and 1, respectively. To neutralize 8 g of NaOH dissolved in 1 L of water

requires 1 L of…

A) 0.0500 normal HF solution

B) 0.100 normal HF solution

C) 0.200 normal HF solution

D) 0.800 normal HF solution





a reaction



Chem C4: The atomic weights of sodium, oxygen and hydrogen are 23, 16

and 1, respectively. To neutralize 4 g of NaOH dissolved in 1 L of water

requires 1 L of…

A) 0.0500 normal H2S solution

B) 0.100 normal H2S solution

C) 0.200 normal H2S solution

D) 0.800 normal H2S solution





a reaction



Chem C5: The atomic weights of sodium, carbon and oxygen are 23, 12

and 16, respectively. To neutralize 21.2 g of Na2CO3 dissolved in 1 L of

water requires 1 L of…

A) 0.100 normal HF solution

B) 0.200 normal HF solution

C) 0.400 normal HF solution

D) 0.800 normal HF solution





a reaction

D. EQUATIONS: STOICHIOMETRY


39D. Equations

Step 1: Add one coefficient at a time, starting with most complex compound

Step 2: Add coefficients to the other side of the reaction, based upon your

coefficient in step 1.

Step 3: Go back and forth until all materials have a coefficient

Step 4: Check when finished to ensure atoms in = atoms out

charge in = charge out

1 1 112

2 Cl

4 H

2 O

2 Cl

4 H

2 O

__Cl2 + __H2O __HCl + __OCl1- + __H3O1+


40

Chem D1: Match the numbers with the correct blanks in the equation to

balance the reaction. Some numbers may be used more than once, others

not at all.

__P4O10 + __H2O __H3PO4

1 2 4 6 8 12

D. Equations


41



not at all.

__Ca2+ + __PO43- __Ca3(PO4)2

1 2 3 4 5D. Equations


42



not at all.

1 2 3 4 6 7 12 14

__Cr2O72- + ___Fe + ___H1+

___ Cr3+ + ___Fe2+ + ___H2O

D. Equations

D. EQUATIONS: STOICHIOMETRY


43D. Equations

Molecular or Atomic weight is sum

of particles relative to 12 grams of 12C. This can be determined by

adding up the component masses

using the periodic table.

Formula is the Key – check subscripts for moles of each

HNO2

1 H 1 N 2 O

Ca(NO3)2

1x2=2 N

3x2=6 O

1 Ca

1.0079 + 14.007 + (2x15.999)

47.0129 grams40.078 + (2x14.007) + (6x15.999)

164.085 grams

(definition from chemistry section of FE Reference Handbook)


44

Chem D4: The molecular (or atomic) weight (g/g-mole) of sodium

phosphate (Na3PO4) is most nearly…

D. Equations

A) 96 B) 34 C) 70 D) 164

O8

15.999

P15

30.974

Na11

22.990


45

Chem D5: The molecular (or atomic) weight (g/g-mole) of aluminium sulfate

(Al2(SO4)3) is most nearly…

D. Equations

A) 342 B) 278 C) 426 D) 198

O8

15.999

S16

32.066

Al13

26.981

D. EQUATIONS: EQUILIBRIUM


46

In general, the equilibrium constant, Keq, is equal to

concentration of products ([C] and [D]) divided by reactants

([A] and [B]), raised to their coefficients (a, b, c, d).

D. Equations

(formula from chemistry section of FE Reference Handbook)


47

Chem D6: Consider the following equation:

D. Equations

K = [C]3 [D]2

[A] [B]2

A) 3C + 2D ↔ A + 2B

B) C3 + D2 ↔ A + B2

C) A + B2 ↔ C3 + D2

D) A + 2B ↔ 3C + 2D

The equation above is the formulation of the chemical equilibrium

constant equation for which of the following reactions?


48

Chem D7: Consider the following chemical reaction:

D. Equations

K = [C] [4D]4

[2A] 2 [3B]3

2A + 3B ↔ C + 4D

What is the formula for the equilibrium constant for this reaction?

A)

K = [A]2 [B]3

[C] [D]4

C)

K = [C] [D]4

[A] 2 [B]3

B)

K = [2A] [3B]

[C] [4D]

A)

E. GAS LAWS: BOYLE’S AND CHARLES LAW


49E. Gas Laws

P1n1 P2n2 P = pressure

n = specific volume

T = absolute temperature in Kelvin

In situations where temperature is constant, Boyle’s Law can be used: P1n1 = P2n2

T1 T2

=

In situations where pressure is constant, Charles’ Law can be used: n1 n2

T1 T2

=

(formula from thermodynamics section of FE Reference Handbook)

E. GAS LAWS: MOLAR VOLUME


50E. Gas Laws(From thermodynamics section of FE Reference Handbook)

Reminder: Kelvin = Celsius + 273. 15

(From chemistry section of FE Reference Handbook)


51

Chem E1: The volume (L) of 1 mol of H2O at 348 K and 1.00 atm

pressure is most nearly:

E. Gas Laws

A) 17.6 B) 28.5 C) 22.4 D) 18.7


52

Chem E2: The volume (L) of 1 mol of CH4 at 273.15 K and 2.65 atm

pressure is most nearly:

E. Gas Laws

A) 12.5 B) 22.4 C) 59.4 D) 8.45


53

Chem E3: The pressure of 25 kg of carbon monoxide (CO) at

50°C in a 50-m3 tank is most nearly..

E. Gas Laws

A) 13.3 kPa B) 48.0 kPa C) 4.39 kPa D) 7.42 kPa


54

Chem E4: The volume needed for 700 kg of methane (CH4) at

40°C and 1.10 atm pressure within a reactor is most nearly…..

E. Gas Laws

A) 6.46 x 105 L B) 1.03 x 108 L C) 1.02 x 106 L D) 1.02 x 103 L


55

Best of luck on your FE Exam. I hope you have found this

review module helpful.

E. Gas Laws

Dr. Lori M. Petrovich

NC State University

Department of Chemistry

Teaching Associate Professor and

Director of General Chemistry Labs

Documents

Chemistry - notes.engineeringonline.ncsu.edu Review 2019/FE Review Chemistry.pdf4 metals Fundamentals of Engineering Exam Review metalloids non-metals A. Periodic Table (Periodic Table