Fundamentals of Engineering Exam Review
Fundamentals of Engineering Exam Review
Chemistry Fundamentals of Engineering Exam Review Other Disciplines FE Specifications Chemistry: 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 Fundamentals of Engineering Exam Review
A. PERIODIC TABLE: NOMENCLATURE, METALS AND NON-METALS
The Periodic Table is divided into three regions: metals, non-metals, and metalloids
Materials composed of all the same type of metal or non-metal are simply named by their element symbol.
Proper nomenclature requires identifying if a compound is composed of: metals & non-metals (ionic) non-metals & non-metals (molecular)
A. Periodic Table 3 Fundamentals of Engineering Exam Review
metals metalloids non-metals
(Periodic Table from chemistry section of FE Reference Handbook)
A. Periodic Table 4 Fundamentals of Engineering Exam Review
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
CaCl2 is calcium chlorIDE (chlorine chloride)
Na2O is sodium oxIDE (oxygen oxide)
Al2S3 is aluminium sulfIDE (sulfur sulfide)
A. Periodic Table 5 Fundamentals of Engineering Exam Review
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
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)
A. Periodic Table 6 Fundamentals of Engineering Exam Review Ionic compounds are also composed of metals or ammonium and polyatomic ions, show below. Cations 1+ 1+ NH4 ammonium ion H3O hydronium ion Anions 1- 1- C2H3O2 acetate ion OH hydroxide ion 2- 1- CO3 carbonate ion NO3 nitrate ion 1- 1- ClO4 perchlorate ion NO2 nitrite ion 1- 1- ClO3 chlorate ion MnO4 permanganate ion 1- 2- ClO2 chlorite ion O2 peroxide ion 1- 3- ClO hypochlorite ion PO4 phosphate ion 2- 2- CrO4 chromate ion SO4 sulfate ion 2- 2- Cr2O7 dichromate ion SO3 sulfite ion 1- A. Periodic Table CN cyanide ion 7 Fundamentals of Engineering Exam Review
Polyatomics are used in the ionic name without change.
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
A. Periodic Table 8 Fundamentals of Engineering Exam Review
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. # atoms Prefix # atoms prefix 1 mono 6 hexa 2 di 7 hepta 3 tri 8 octa 4 tetra 9 nona 5 penta 10 deca A. Periodic Table 9 Fundamentals of Engineering Exam Review
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
N2O3 is dinitrogen trioxide
N2O is dinitrogen monoxide
NO2 is nitrogen dioxide
A. Periodic Table 10 Fundamentals of Engineering Exam Review
Chem A1: Identify the name of Mg(OH)2
A) Magnesium (II) hydroxide
B) Magnesium dihydroxide
C) Magnesium hydroxide
D) Magnesium epoxide
A. Periodic Table 11 Fundamentals of Engineering Exam Review
Chem A2: Identify the name of MnCO3
A) Manganese carbide
B) Manganese (II) carbonate
C) Manganese (I) carbonate
D) Manganese carbonate
A. Periodic Table 12 Fundamentals of Engineering Exam Review
Chem A3: Which material is platinum (IV) bromide?
A) Pt4Br
B) PtBr2
C) PtBr3
D) PtBr4
A. Periodic Table 13 Fundamentals of Engineering Exam Review
Chem A4: Identify the name of SO3
A) Monosulfur trioxide
B) Sulfur (VI) oxide
D) Sulfur trioxygen
A. Periodic Table 14 Fundamentals of Engineering Exam Review
Chem A5: Which material is dichlorine heptoxide?
A) Cl2O7
B) Cl3O6
C) ClO7
D) Cl4O7
A. Periodic Table 15 Fundamentals of Engineering Exam Review A. PERIODIC TABLE: ATOMIC STRUCTURE OF MATTER # of protons defines the element: known as Major Subatomic Particles atomic number Particle Mass (amu) Charge # of protons and neutrons gives the Electron (e-) 0.0005 -1 atomic mass Proton (p+) 1.0078 +1 # of protons minus the Neutron (n0) 1.0087 0 # of electrons gives the charge
A. Periodic Table 16 Fundamentals of Engineering Exam Review A. PERIODIC TABLE: ATOMIC STRUCTURE OF MATTER 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+?
The periodic table provides the 26 number of protons: 26
The atomic mass - #protons gives Fe the neutrons: 56 – 26 = 30 55.847
A. Periodic Table 17 Fundamentals of Engineering Exam Review A. PERIODIC TABLE: ATOMIC STRUCTURE OF MATTER 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+?
Solve the equation to give #electrons: 26 charge = #protons - #electrons + 3 = 26 - # electrons Fe #electrons = 23 55.847
A. Periodic Table 18 Fundamentals of Engineering Exam Review
Chem A6: The element chlorine has two major isotopes. The atomic nuclei of these isotopes contain how many protons?
A) 35.453
B) 17 17
C) 6 Cl 35.453 D) 27
A. Periodic Table 19 Fundamentals of Engineering Exam Review
Chem A7: How many protons, neutrons and electrons are 17 2- in O ? 8 A) 8 protons, 7.999 neutrons, 10 electrons O 15.999 B) 17 protons, 9 neutrons, 10 electrons
C) 8 protons, 9 neutrons, 10 electrons
D) 8 protons, 9 neutrons, 6 electrons
A. Periodic Table 20 Fundamentals of Engineering Exam Review
Chem A8: What is the correct notation for an ion that has 19 protons, 21 neutrons and 18 electrons? 8 19 A) 40K1+ F K 19.998 39.098 B) 39K1+
C) 40K1-
D) 19F1-
A. Periodic Table 21 Fundamentals of Engineering Exam Review B. 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+) 3+ aluminium has a +3 oxidation state (Al2O3 Al )
B. Oxidation and Reduction 22 Fundamentals of Engineering Exam Review B. OXIDATION AND REDUCTION +1 4. Hydrogen is unusual. It is +1 when bonded to non-metals (NH3 H ) and -1 when bonded to metals (NaH H-1) -2 5. Oxygen(O) in a compound has a -2 oxidation state. So K2O has O . (O2 is an element and the O is oxidation state 0.) -1 6. Group VII atoms in compounds have a -1 charge. So CrCl3 Cl 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.
1+ In NH4 , 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.
B. Oxidation and Reduction 23 Fundamentals of Engineering Exam Review B. Oxidation and Reduction Relating oxidation states to the period table will allow easier memorization
+1+2 +1 or +3 -4 -3 -2 -1 +1+2 +3 -3 -2 -1 +1+2 +2 or +3 -2 -1 +1+2 maybe +1, +4, +5, + 6 +7 -1 +1+2
+1 or +3 +2 or +4
B. Oxidation and Reduction 24 Fundamentals of Engineering Exam Review B. OXIDATION AND REDUCTION Oxidation occurs when an atom loses electrons and becomes more positive in oxidation state. Reduction occurs occurs when an atom gains electrons and becomes more negative in oxidation state. In a Redox reaction, both reduction and oxidation occur. Fe + Cu2+ Fe2+ + Cu 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)
B. Oxidation and Reduction 25 Fundamentals of Engineering Exam Review
Chem B1: Which chemical change below represents the oxidation of nitrogen?
- A) N2 NH3 C) N2O NO3
- B) NO N2O2 D) NF3 NO2
B. Oxidation and Reduction 26 Fundamentals of Engineering Exam Review
Chem B2: Which chemical change below represents the reduction of carbon?
A) CH4 CO2 C) C2Cl4 CO
B) CO2 CO D) C C2H2Cl2
B. Oxidation and Reduction 27 Fundamentals of Engineering Exam Review
Chem B3: Which chemical change below is not a reduction-oxidation reaction?
A) 2K + I2 2 KI C) 4 Fe + 3 O2 2Fe2O3
2+ 1- B) SnO2 + C Sn + CO2 D) Ca + 2 Cl CaCl2
B. Oxidation and Reduction 28 Fundamentals of Engineering Exam Review
Chem B4: Which chemical change below is not a reduction-oxidation reaction? 2+ A) CH4 + 2O2 CO2 + 2 H2O C) Pb + PbO2 + 4H+ → 2Pb + 2 H2O
- 2- B) Ag2S + 2Cl 2AgCl + S D) Mg + H2O H2 + MgO
B. Oxidation and Reduction 29 Fundamentals of Engineering Exam Review C. ACIDS AND BASES Acids are compounds that when added to water donate a H+ + ion to form H3O . - + HCl + H2O Cl + H3O Typically the number of H atoms in the front of a chemical formula are the acidic hydrogens and provide the valence change.
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 30 Fundamentals of Engineering Exam Review C. ACIDS AND BASES Bases are compounds that when added to water remove a H+ from water to form OH-. + - NH3 + H2O NH4 + OH Most bases are negatively charged and exist as ionic compounds in the solid form. Bases can accept multiple H+ ions
NaOH (OH- in water) accepts one H+ (valence change 1)
2- + Na2CO3 (CO3 in water) accepts two H (valence change 2) 3- + Li3PO4 (PO4 in water) accepts three H (valence change 3)
C. Acids and Bases 31 Fundamentals of Engineering Exam Review C. ACIDS AND BASES Acids and Bases can react to form water and an ionic salt in a neutralization reaction.
HCl + NaOH NaCl + H2O
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 32 Fundamentals of Engineering Exam Review C. ACIDS AND BASES Concentration expressed in units of molarity or normality: 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) 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 + C. Acids and Bases As HCl can only donate 1 H , it is also 0.500 N 33 Fundamentals of Engineering Exam Review
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
Molarity of Solutions – The number of gram moles of a substance dissolved in a liter of solution.
C. Acids and Bases 34 Fundamentals of Engineering Exam Review
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
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
C. Acids and Bases 35 Fundamentals of Engineering Exam Review 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… Molarity of Solutions – The number of gram moles of a A) 0.0500 normal HF solution substance dissolved in a liter of solution. B) 0.100 normal HF solution Normality of Solutions – The product of the molarity of a C) 0.200 normal HF solution solution and the number of valence changes taking place in a reaction D) 0.800 normal HF solution
C. Acids and Bases 36 Fundamentals of Engineering Exam Review 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… Molarity of Solutions – The number of gram moles of a A) 0.0500 normal H2S solution substance dissolved in a liter of solution. Normality of Solutions – The product of the molarity of a B) 0.100 normal H2S solution C) 0.200 normal H S solution solution and the number of valence changes taking place in 2 a reaction D) 0.800 normal H2S solution
C. Acids and Bases 37 Fundamentals of Engineering Exam Review 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… Molarity of Solutions – The number of gram moles of a A) 0.100 normal HF solution substance dissolved in a liter of solution. B) 0.200 normal HF solution Normality of Solutions – The product of the molarity of a C) 0.400 normal HF solution solution and the number of valence changes taking place in a reaction D) 0.800 normal HF solution
C. Acids and Bases 38 Fundamentals of Engineering Exam Review D. EQUATIONS: STOICHIOMETRY 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+ __Cl1 2 + __H2 2O __1 HCl + __OCl1 + __H1 3O
2 Cl 2 Cl 4 H 4 H 2 O 2 O
D. Equations 39 Fundamentals of Engineering Exam Review 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 40 Fundamentals of Engineering Exam Review Chem D2: 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. 2+ 3- __Ca + __PO4 __Ca3(PO4)2
1 2 3 4 5 D. Equations 41 Fundamentals of Engineering Exam Review Chem D3: 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. 2- 1+ 3+ 2+ __Cr2O7 + ___Fe + ___H ___ Cr + ___Fe + ___H2O
1 2 3 4 6 7 12 14
D. Equations 42 Fundamentals of Engineering Exam Review D. EQUATIONS: STOICHIOMETRY 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. (definition from chemistry section of FE Reference Handbook) Formula is the Key – check subscripts for moles of each 3x2=6 O HNO2 Ca(NO3)2 1 H 1 N 2 O 1 Ca 1x2=2 N 1.0079 + 14.007 + (2x15.999) 40.078 + (2x14.007) + (6x15.999) 47.0129 grams 164.085 grams D. Equations 43 Fundamentals of Engineering Exam Review
Chem D4: The molecular (or atomic) weight (g/g-mole) of sodium
phosphate (Na3PO4) is most nearly… A) 96 B) 34 C) 70 D) 164
8 11 15 O Na P 15.999 22.990 30.974
D. Equations 44 Fundamentals of Engineering Exam Review
Chem D5: The molecular (or atomic) weight (g/g-mole) of aluminium sulfate
(Al2(SO4)3) is most nearly… A) 342 B) 278 C) 426 D) 198
8 13 16 O Al S 15.999 26.981 32.066
D. Equations 45 Fundamentals of Engineering Exam Review D. EQUATIONS: EQUILIBRIUM
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).
(formula from chemistry section of FE Reference Handbook)
D. Equations 46 Fundamentals of Engineering Exam Review Chem D6: Consider the following equation: [C]3 [D]2 K = [A] [B]2
The equation above is the formulation of the chemical equilibrium constant equation for which of the following reactions? A) 3C + 2D ↔ A + 2B
B) C3 + D2 ↔ A + B2
C) A + B2 ↔ C3 + D2 D) A + 2B ↔ 3C + 2D
D. Equations 47 Fundamentals of Engineering Exam Review
Chem D7: Consider the following chemical reaction:
2A + 3B ↔ C + 4D
What is the formula for the equilibrium constant for this reaction?
A) [C] [4D]4 B) [C] [D]4 K = K = [2A] 2 [3B]3 [A] 2 [B]3
C) [A]2 [B]3 A) [2A] [3B] K = K = [C] [D]4 [C] [4D]
D. Equations 48 Fundamentals of Engineering Exam Review E. GAS LAWS: BOYLE’S AND CHARLES LAW P = pressure P1n1 P2n2 = n = specific volume T1 T2 T = absolute temperature in Kelvin In situations where temperature is constant, Boyle’s Law can
be used: P1n1 = P2n2 In situations where pressure is constant, Charles’ Law can be used: n1 n2 = T1 T2
E. Gas Laws (formula from thermodynamics section of FE Reference Handbook) 49 Fundamentals of Engineering Exam Review E. GAS LAWS: MOLAR VOLUME
(From chemistry section of FE Reference Handbook)
Reminder: Kelvin = Celsius + 273. 15
(From thermodynamics section of FE Reference Handbook) E. Gas Laws 50 Fundamentals of Engineering Exam Review
Chem E1: The volume (L) of 1 mol of H2O at 348 K and 1.00 atm pressure is most nearly: A) 17.6 B) 28.5 C) 22.4 D) 18.7
E. Gas Laws 51 Fundamentals of Engineering Exam Review
Chem E2: The volume (L) of 1 mol of CH4 at 273.15 K and 2.65 atm pressure is most nearly: A) 12.5 B) 22.4 C) 59.4 D) 8.45
E. Gas Laws 52 Fundamentals of Engineering Exam Review
Chem E3: The pressure of 25 kg of carbon monoxide (CO) at 50°C in a 50-m3 tank is most nearly..
A) 13.3 kPa B) 48.0 kPa C) 4.39 kPa D) 7.42 kPa
E. Gas Laws 53 Fundamentals of Engineering Exam Review
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…..
A) 6.46 x 105 L B) 1.03 x 108 L C) 1.02 x 106 L D) 1.02 x 103 L
E. Gas Laws 54 Fundamentals of Engineering Exam Review
Best of luck on your FE Exam. I hope you have found this review module helpful.
Dr. Lori M. Petrovich NC State University Department of Chemistry Teaching Associate Professor and Director of General Chemistry Labs
E. Gas Laws 55