DAWSON COLLEGE DEPARTMENT OF CHEMISTRY & CHEMICAL TECHNOLOGY
FINAL EXAMINATION CHEMISTRY 202NYB05 May 21, 2010 9:30 12:30
Print your Name: __________________________________________
Student Number: ____________________________
INSTRUCTORS: Please circle the name of your instructor:
J. Ali I. Dionne M. Haniff
D. Baril M. Di Stefano S. Holden
O. Behar N. Duxin / YS. Uh S. Mutic
INSTRUCTIONS:
This exam set consists of 16 questions. Please ensure that your copy of this examination is complete.
Answer all questions in the space provided.
1. Calculators may not be shared. Programmable calculators are not permitted.
2. No books or extra paper are permitted.
3. In order to obtain full credit, you must show the method used to solve all problems involving calculations and express your answers to the correct number of significant figures.
4. Your attention is drawn to the College policy on cheating.
5. A Periodic Table is provided. (last page).
6. If a mathematical equation is used to solve a problem, the equation should be clearly written.
7. Write your answer in the appropriate space when required.
USEFUL DATA: Avogadros Number NA = 6.022 x 1023 mol1 Gas Constant = 0.08206 LatmK1mol1 R = 8.314 LkPaK1mol1 = 8.314 JK1mol1 1 atm = 101.3 kPa = 760 mmHg = 760 torr 1 J = 1 kgm2s2 101.3 J = 1 Latm
MARK DISTRIBUTION
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2. / 5
3. / 3
4. / 6
5. / 5
6. / 8
7. / 6
8. / 6
9. / 9
10. / 7
11. / 6
12. / 7
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14. / 6
15. / 6
16. / 5
TOTAL /100
 2 
Question 1
Ethanol is the common alcohol with molecular formula C2H5OH. An alcoholwater solution is prepared by dissolving 10.00 cm3 of ethanol, with density dethanol = 0.789 g/cm
3, in a sufficient volume of water to produce 100.00 cm3 of solution. Density of solution is dsoln = 0.932 g/cm
3. For a given solution calculate the following for ethanol: a. the mass percent
Ans. Mass%: _____________
b. the molarity
Ans. molarity: _____________
c. the molality
Ans. molality: _____________
d. the mole fraction.
Ans. Mole fraction: _____________
(2 marks)
(2 marks)
(2 marks)
(2 marks)
For 100.00 cm3 of solution: 100.00 cm3 x 0.932
g
cm3 = 93.2 g solution
In this solution 10.00 cm3 ethanol are present: 10.00 cm3 x 0.789
g
cm3 = 7.89 g ethanol
Mass percent =
7.89 g93.2 g
x 100% = 8.46% ethanol
For 1.00 L of solution: 1.00 L x 0.932
gmL
x 1000 mLL
= 932 g solution
In this solution 8.46% is ethanol then: 932 g solution x
7.80 g ethanol100. g solution
= 78.8g ethanol
Molarity = mol soluteL solution
=
78.8 g x 1 mol ethanol46.07 g ethanol1.00 L
= 1.71 M ethanol
Mole of ethanol for 1.00L solution = 78.8 g x
1 mol46.07 g
= 1.71 mol
Number of kilogram of solvent in one liter of solution: 932 g solution  78.8 g ethanol = 853 g solvent
Molality =
mol solutekg solvent
=
1.71 mol
853 g x 1 kg
1000 g solvent
= 2.00 molal ethanol
Mole of ethanol for 1.00 L solution = 78.8 g x
1 mol46.07 g
= 1.71 mol
Mole of solvent for 1.00 L solution = 853 g solvent
1 mol18.015 g
= 47.3 mol
Mole fraction =
mol solutetotal mol solution
=
1.71 mol1.71 mol + 47.3 mol
= 3.49x102 ethanol
 3 
Question 2
Toluene, C7H8 is a component of gasoline (octane, C8H18). It is present in gasoline as an octane booster at concentrations between 3 to 5% by mass (25% in racing cars gasoline). Consider a solution of octane with 20.% by mass of toluene at 20C a. Calculate the total vapor pressure of this solution
Data: Poctane = 10.5 mm Hg at 20C, Tb = 126C Ptoluene 22 mm Hg at 20C, Tb = 111C
ans. total vapor pressure:_____________
b. Calculate the mole ratio of toluene to octane in the vapor phase above the solution
ans. mole ratio: toluene/octane:_____________
c. If the actual vapor pressure measured is 15.2 mm Hg, will the boiling point of this solution be higher
or lower than the one expected from Raoults law? Explain.
(3 marks)
(1 mark)
(1 mark)
for 100 g of solution you have : (keep one extra sig.fig. for calculations)
100. g solution x 20 % toluene = 20. g toluene 100. g solution  20. g toluene = 80. g octane
ntoluene = 20 g x
1 mol toluene92.14 g
= 0.217 mol noctane = 80 g x
1 mol octane114.23 g
= 0.700 mol
toluene =
0.217 mol0.217 mol + 0.700 mol
= 0.237 octane = 1  0.237 = 0.763 (only 2 components)
Total vapor pressure = Ptolueneo x toluene + Poctane
o x octane =
= (22 mm Hg x 0.237) + (10.5 mm Hg x 0.763) = 13 mm Hg
Molar ratio =
ntoluene Vap.noctane Vap.
Since PV = nRT then P n therefore:
ntoluene Vap.noctane Vap.
=
PtoluenePoctane
=
22 mm Hg x 0.23710.5 mm Hg x 0.763
=
5.21 mm Hg 8.01 mm Hg
= 0.65
The actual vapor pressure is higher than the ideal one (calculated) therefore, the boiling point will be
lower than the one expected. The intermolecular forces between toluene  octane are weaker than the
one of the pure compounds.
 4 
Question 3
A 0.461 g sample of cumene, a nonvolatile nonionic compound, is dissolved in 10.0 g cyclohexane (C6H12), producing a solution that freezes at 1.25C. Cyclohexane has a normal freezing point of 6.50C and a freezing point depression constant of 20.2C/m. What is the molar mass of cumene?
Ans. Mol. mass cumene: _____________
(3 marks)
Colligative properties involving the freezing point depression of a non ionic solute:
Tf = Kf x m where Tf = 6.50C  (1.25C) = 7.25C
The concentration of cumene can be found:
Tf Kf
= m =
7.75C
20.2C.m1 = 0.384 m
The number of mole of cumene is:
molality =
mole solutekg solvent
Mole solute = molality x kg solvent = 0.384 m x 10.0 g x
1 kg1000 g
= 3.84x103 mol
Finally, the molar mass of cumene =
mass cumenemole cumene
=
0.461 g0.00384 mol
= 120.
gmol
(or 1.20x102
gmol
)
 5 
Question 4
Hydrofluoric acid, (HF) is a weak acid that can be used in the fluoridation of water. An aqueous solution of 0.100 M HF has an osmotic pressure of 2.64 atm at 25C. a. Calculate the vant Hoff factor for HF at this concentration
Ans. vant Hoff factor:_____________
b. Does it differ from the maximum vant Hoff factor expected for a monoprotic acid? If so, explain why. c. What is the percent ionization of HF at this concentration?
Ans. % ionization:_____________
(2 marks)
(2 mark)
(2 marks)
Colligative properties involving the osmotic pressure of an ionic compound in solution: = i M R T
i =
M R T
=
2.64 atm
(0.100 molL
)(0.0821L.atmK.mol
)(25 + 273)K = 1.08
Yes it differs from the expected van't Hoff factor of 2 for a monoprotic acid.
Therefore. HF is a week acid since its dissociation is partial because of ion pairing at this concentration
corresponding to an incomplete dissociation of the acid (weak acid).
HF H+ + F
Cx x x
i =
mole of particle in solutionmole solute added
=
2x + (c  x)mole solute added
=
x + CC
Since I = 1.08 and C = 0.100 M 1.08 =
x + 0.1000.100
then x = 0.008
% ionization =
xC
x 100% =
0.008 M0.100 M
x 100% = 8%
 6 
Question 5
Iodide ion is oxidized in acidic solution to triiodide ion I3 by hydrogen peroxide.
H2O2(aq) + 3 I(aq) + 2 H+(aq) I3
(aq) + 2H2O(l)
A series of four experiments was run at different concentrations, and the initial rates of I3
formation were
determined (see table).
Initial concentration
(molL1)
Initial concentration (molL1)
Initial concentration (molL1)
Initial rate
(molL1s1)
H2O2 I H+
Exp 1 0.010 0.010 0.00050 1.15x106 Exp 2 0.020 0.010 0.00050 2.30x106 Exp 3 0.010 0.020 0.00050 2.30x106 Exp 4 0.010 0.010 0.00100 1.15x106
a. From the table above, obtain the reaction orders with respect to each of the following species: H2O2, I
, H+.
Ans. Reaction order: H2O2: ______ I
: ______ H+: ______ b. Find the rate constant with its units.
Ans. rate constant: _____________________
(3 marks)
(2 marks)
The reaction rate is given by: rate = k [H2O2] [I] [H+]
For :
Exp.2Exp.1
=
2.30x106
1.15x106 =
k(0.020) (0.010)(0.00050)
k(0.010)(0.010)(0.00050) 2 = 2 then = 1
For :
Exp.3Exp.1
=
2.30x106
1.15x106 =
k(0.010) (0.020)(0.00050)
k(0.010)(0.010)(0.00050) 2 = 2 then = 1
For :
Exp.4Exp.1
=
1.15x106
1.15x106 =
k(0.010) (0.010)(0.00100)
k(0.010)(0.010)(0.00050) 1 = 2 then = 0
The reaction orders are: H2O2: ___1___ I_: ___1___ H+: ___0___
The reaction rate is given by: rate = k [H2O2] [I]
Using Exp.4 : 1.16x106 M.s1 = k (0.010 M) (0.010 M)
k =
1.16x106 M.s1
(1.0x102 M)2 = 1.2x102 M1s1
 7 
Question 6
The reaction below was monitored as a function of time at a temperature of 400 K:
2NOC (g) 2NO(g) + C2(g) A plot of 1/[NOC] against time yielded a straight line with slope of 6.7x104 Lmol1s1. a. Write the rate law for the reaction. b. What is the halflife for the reaction if the initial concentration of NOC is 0.20 M?
Ans. halflife: _____________________ c. If the initial concentration of NOC is 0.35 M, what is the concentration of NOC after 5.0 min?
Ans. [NOC] after 5.0 min: _____________________
d. If the initial concentration of NOC is 0.35 M, How long will it take for the concentration to drop to 20%
of its original value?
Ans. time after 20% drop: _____________________
(2 marks)
(2 marks)
(2 marks)
(2 marks)
rate = 6.7x104 Lmol1s1 [NOCl]2 (second order)
t 1
2=
1
k[A]o
t1
2=
1
6.7x104L.mol1s1 [0.20 mol.L1] = 7.5x103 s.
1[NOC_]
= k t +
1[NOC_]o
1[NOC_]
= ( 6.7x104L.mol1s1 ) x 5.0 min x
60 s1 min
+
10.35 M
= 3.06 M1 or [NOC] = 0.33 M
1[NOC_]
= k t +
1[NOC_]o
t =
1k
x (
1[NOC_]

1[NOC_]o
) =
1
6.7x104L.mol1s1 x (
1
0.35 M x 20/100 
10.35 M
) = 1.7x104 s.
 8 
Question 7
a. Consider the potential energy profiles for three different chemical reactions.
Indicate which reaction is the slowest one. Explain your choice
b. Consider the potential energy profiles for a chemical reaction.
Mechanism 1
2A + B C C D
Mechanism 2
2A C
B + C D Mechanism 3
A C A + B + C 2D
Circle the proposed mechanism that is consistent with the reaction profile shown and explain your choice.
c. Beside concentration and pressure, give two parameters you can change that could affect the reaction
rate of a chemical reaction: i. ii.
(2 marks)
2A + B D
(2 marks)
(2 marks)
Reaction 1 is the slowest one since its activation energy is the highest .
The curve is consistent with mechanism 2 since the slow step is the first one (highest Ea).
The rate law will be rate = k [A]2
Temperature Catalyst or enzyme
 9 
Question 8
At elevated temperature (997C) limestone dissociates according to the equation
CaCO3(s) CaO(s) + CO2(g) H = +42.5 kJ a. If 50.0 g CaCO3 (100.1 g/mol) is placed in an evacuated 4.00 L container and heated up to 997C,
how many grams of CaCO3 will decompose if the pressure at equilibrium is 392 kPa? b. If the volume of the container is expanded to 10.0 L at 997C, what will be the CO2 pressure at
equilibrium? c. Calculate Kc for this reaction at 997C d. Predict the effect of each of the following changes will have on the equilibrium position.
equilibrium position shift
change to the left no change to the right
i. CO2 is added
ii. CaCO3 is added
iii. Pressure is increased (adding N2 gas, volume unchanged)
iv. The temperature is increased
(2 marks)
(1 mark)
(1 mark)
(2 marks)
Mole of CO2 produced = PV = nRT
n = RTPV =
273)K(977 xmol K.
L.kPa 8.31
L 4.00 x kPa 392
+= 0.149 mole
Since it is a 1:1 ratio between CO2 and CaCO3 then:
Mass of CaCO3 consumed = 0.149 mol CO2 x 2
3CO 1
CaCO 1 x 100.1 g.mol1 = 14.9 g
K = [CO2]. The concentration of CO2 remains the same whatever the volume For a gas, concentration = pressure. Therefore, pressure does not change = 392 kPa or 3.87 atm.
K = [CO2] since the other reactants are all solids.
In part A, the number of mole of CO2 for a container of 4.00 L was 0.149 mole.
K =
0.149 mol4.00 L
= 0.0373 also K = RTKp =
RTP = where P is also obtained from PV = nRT
 10 
Question 9
Consider the following set of data:
Formula Ka (at 25C)
[Al(H2O)6]3+ 1.4105 HNO2 4.0104
HF 7.2104
a. What is the strongest acid in the table? _______HF____________ b. With the help of the table, arrange the following in order of most basic to least basic: H2O, NO2
, [Al(H2O)5OH]2+
Most basic ____[Al(H2O)5OH]
2+ ___ > ___ NO2___ > ____ H2O ____ Least basic
c. What is the value of Kb for F
at 25C ?
Ans. Kb : ___________
d. Write the chemical reaction represented by the Kb for F
in water and place the species involved in
the appropriate place __H2O_____ + __F
 __ __ HF__ + ___OH__ Acid Base Conjugate Conjugate acid base
e. At 40C, Kw = 2.9x10
14. What is the neutral pH of water at this temperature?
Ans. ___________
(1 mark)
(2 marks)
(2 marks)
(2 marks)
(2 marks)
Kb = 414
7.2x1010 = 1.4x1011
H2O OH + H+ then Kw = [H
+][OH]
at neutral pH: [H+] = [OH] then kw = [H+] = 2.9x10
14 = 1.7x107
pH = log(1.7x107) = 6.77
 11 
Question 10
a. A solution of the basic oxide CaO is prepared by adding water to 0.28 g CaO to make 0.50 L of solution.
i. Write the equations for the reactions that occur when CaO is dissolved in water ii. Assuming that ionpairing is nonexistent, what is the expected pH of this solution?
ans. pH:____________
b. For which of the following salts will the solubility depend on pH? pH sensitive pH independent
i. KCO4 ii. Pb(OH)2 iii. AgF iv. Ba(NO3)2
c. For each of the following salts dissolved in water, predict whether the aqueous solution will be acidic,
neutral or basic. acid neutral basic
i. RbOH ii. NaIO iii. NH4OH iv. LiCO3
(1 mark)
(2 marks)
(2 marks)
(2 marks)
1) CaO(s) + H2O(l) Ca(OH)2(aq) 2) Ca(OH)2(aq) Ca
2+(aq) + 2OH(aq) or
1) CaO(s) Ca2+(aq) + O2 (aq) 2) O2(aq) + H2O(l) 2OH
(aq)
Mole of OH in solution: 0.28 g CaO x
1 mol56.08 g
x
Ca(OH)2CaO
x
2 OH
Ca(OH)2 = 0.010 mol OH
Concentration of OH in solution:
0.010 mol OH
0.50 L = 0.020 M
Finally, the pH will be: pOH = log(0.020 M) = 1.70 pH = 14  pOH = 12.30
 12 
Question 11
a. Consider 0.500 L of a buffer that consists of 1.50 M KCO (Ka HCO = 3.5x108) and 0.50 M HCO. What will be the pH of this buffer after the addition of 250 mL of 1.0 M HNO3?
ans. pH:____________
b. Which of the following mixtures would result in a buffer solution when 100 mL of each of the two solutions are mixed together?
buffer not a buffer
i. 0.1 M KOH and 0.2 M NH3
ii. 0.2 M HC and 0.2 M NH3
iii. 0.2 M HNO3 and 0.4 M NaNO3
iv. 0.1 M HNO3 and 0.2 M NaF
(4 marks)
(2 marks)
The system is the following: HClO(aq) H+(aq) + ClO(aq) (note H+ = H3O
+)
Mol of HClO = 0.500 L x 0.50
molL
= 0.25 mol
Mol of ClO = 0.500 L x 1.50
molL
= 0.750 mol
Mol of H+ added = 0.250 L x 1.0
molL
= 0.25 mol
First we have a reaction between a strong acid and a week base: H+ + ClO HClO Initial 0.25 0.75 0.25 Reaction 0.25 0.25 +0.25 After reaction 0 0.50 0.50 After, the system reach equilibrium: HClO H
+ + ClO I 0.50 0 0.50 C x +x +x E 0.50x +x 0.50+x Since we have a buffer, x
 13 
Question 12
Consider the following titration curve of trimethylamine (C3H9N) a weak base with 0.100 M HC at 23C. Initial solution: 50.0 mL of C3H9N, 4.00x10
2 M a. Draw on the graph the shape of the titration curve if this base had a smaller Kb value.
b. Which letter (A to F) on the graph corresponds to each of the following?
letter
The equivalence point D
The point of halfneutralization B
The point corresponding to the pKa of C3H9NH+ B
c. When 15.0 mL of 0.100 M HC is added, the pH of the solution is 9.255. Calculate Kb of
trimethylamine.
ans.
Kb:_______
_____
Let's call trimethylamine TMA an its conjugated acid TMAH+ Mole of TMA = 4.00x102 M x 0.0500 L = 2.00x103 mol Mole of acid added = 1.00x101 M x 0.0150 L = 1.50x103 mol The HCl reacts to completion (strong acid) with TMA according to:
TMA + H+ TMAH+
Initial 2.00x103 1.50x103 0 Reaction 1.50x103 1.50x103 +1.50x103 After reaction 0.50x103 0 1.50x103
Then, the system will reach equilibrium from those initial conditions: TMA + H2O TMAH
+ + OH
0.50x103 x mole 1.50x103 + x mole x mole In a buffer regime, x is small compared to [base] or [conjugated acid] and can be neglected. In this case ONLY the ratio of mole of TMAH+ / TMA is the same as the one of concentrations. Therefore:
Kb =
[TMAH+][OH ][TMA]
=
(1.50x103)(10(149.255) )
0.50x103 = 5.4x105
smaller Kb value
 14 
(1 mark)
(3 marks)
(3 marks)
 15 
Question 13
a. A saturated aqueous solution of Mg(OH)2 has a pH of 10.08, what is the Ksp of Mg(OH)2?
ans. Ksp:____________
b. The Ksp of cobalt(III) hydroxide is 2.5x10
43. Calculate the solubility of Co(OH)3 in water in mol/L
ans. solubility (mol/L):____________
c. Does a precipitate form when 25 mL of 0.10 M lithium nitrate LiNO3, is mixed with 35 mL of 0.75 M
sodium carbonate Na2CO3? (Ksp Li2CO3 = 8.15x104) Show your work.
ans: yes no
(2 marks)
(2 marks)
(3 marks)
Mg(OH)2(s) Mg2+(aq) + 2 OH(aq) Ksp = [Mg
2+][OH]2 I Solid 0 0 C x +x +2x E solid x +2x Ksp = (x)(2x)
2 = 4x3 We can use the pH to know [OH] that is equal to 2x pOH = 14  10.08 = 3.92, [OH] = 103.92 = 1.2x104 M. since 2x = 1.2x104 M. then x = 6.0x105. finally: Ksp = 4 (6.0x10
5)3
Ksp = 8.6x1013
According to the Ksp value, the concentration of OH generated by the salt will be small compared to the one of natural ionization of water [OH] = 107 M therefore, common ion effect.
Co(OH)3(s) Co
3+(aq) + 3OH(aq) Ksp = [Co3+][OH]3
I Solid 0 107 C x +x +3x E solid x 3x+107 if x
 16 
Question 14
a. A system is made of a cylinder of gas with a piston. When 4.0 kJ of heat is transferred from the surroundings to the system, the gas in the piston expands from 12 L to 27 L and performs work on the surroundings. If the system gains 201 J of internal energy from this process, against what constant external pressure, in atmospheres, is the piston working?
Ans. pressure (atm): ____________
b. Bromine is a liquid at room temperature. Calculate the freezing point of bromine if its heat of fusion is + 5.79 kJmol1 and its entropy of fusion is 21.8 JK1mol1.
Ans. Tf bromine: ____________
(3 marks)
(3 marks)
H = E + PV (first law: conservation of the energy) H  E = PV
H  E = [+4000 J  (+ 201 J) ] x
1 L.atm101.3 J
= 37.50 L.atm
37.5 L.atm = PV and V = Vfinal  Vinitial
P =
37.5 L.atm27L  12L
= 2.5 atm
G = H  TS (second law: spontaneity of a system) The freezing point occurs when all the heat added to a system is converted to entropy. At this point, the system is highly reversible (equilibrium condition, therefore G = 0)
The equation becomes: 0 = H  TS and
HS
= T
T =
HS
=
5.79x103 J.mol
21.8 J.K1 . mol = 266K or 7C
 17 
Question 15
a. Circle the substance in each of the following pairs that would have the greater entropy. i. H2O (, 1 mol, 75C, 1 atm) or H2O (g, 1 mol, 75C, 1 atm) ii. Fe (s, 50.0 g, 5C, 1 atm) or Fe (s, 0.80 mol, 5C, 1 atm) iii. Br2 (, 1 mol, 8C, 1 atm) or Br2 (s, 1 mol, 8C, 1 atm) iv. SO2 (g, 0.312 mol, 32.5C, 0.110 atm) or SO2 (g, 0.284 mol, 22.3C, 15 atm)
b. Methyl isothiocyanate, CH3NCS, is a highly irritating pesticide. It can be prepared by reacting
carbon disulfide with methylamine. Given the thermodynamic data at 25C below, calculate the standard molar entropy of methyl isothiocyanate.
CS2 (g) + CH3NH2 (g) CH3NCS (g) + H2S (g)
G (kJmol1) 67.15 32.09 144.35 33.56
H (kJmol1) 117.36 22.98 130.96 20.63
S (Jmol1K1) 237.73 243.30 ? 205.69
Ans: __________________:
(2 marks)
(4 marks)
G = (33.56) +(144.35)  [(67.15) + (32.09)] = 11.55 kJ H = (20.63) +(130.96)  [(117.36) + (22.98)] = 15.95 kJ
S =
H  GT
=
(15950  11550)J(25 + 273)K
= 14.8 J/K
finally: S = Sproduct  Sreactant 14.8 J/K = (205.69 + x) J/K (237.73 + 243.30)J/K x = SCH3NSC = 290.1 J/K =
 18 
Question 16
In the laboratory experiment 4, you want to determine the activation energy of the following reaction:
2I(aq) + S2O82(aq) I2 (aq) + 2S2O4
2
Where the reaction rate is: Rate =  [I]/2t and the rate law for this reaction is: Rate = k [I][S2O82]
By recording the reaction rate of several experiments at different temperatures, the following graph based on the linear form of the Arrhenius equation is obtained.
Arrhenius plot for the determination of the activation energy for the reaction of iodide with peroxydisulfate
From this graph, calculate the activation energy (with units) for this reaction.
Ans. Ea : _________________
(5 marks)
We will be using the linear form of the Arrhenius equation: Ln k = 
EaR
1T
+ constant
The graph of Ln k vs.
1T
gives a slope =
EaR
Slope =
Y2 Y1X2 X1
=
(4.80) (5.60)
(3.2x103  3.35x103)K1 =
0.80
1.5x104K1 = 5.3x103 K
Since: slope =
EaR
then Ea = (8.31
JK.mol
) x 5.3x103 K = 44x103
Jmol
or 44
kJmol
.
 19 
1A Periodic Table of the Elements 8A 1 2
1 H He 1.008 2A 3A 4A 5A 6A 7A 4.003
3 4 5 6 7 8 9 10 2 Li Be B C N O F Ne 6.941 9.012 10.81 12.01 14.01 16.00 19.00 20.18
11 12 13 14 15 16 17 18 3 Na Mg Al Si P S Cl Ar 22.99 24.31 3B 4B 5B 6B 7B 8B 9B 10B 1B 2B 26.98 28.09 30.97 32.07 35.45 39.95
19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 4 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 39.10 40.08 44.96 47.87 50.94 52.00 54.94 55.85 58.93 58.69 63.55 65.39 69.72 72.61 74.92 78.96 79.90 83.80
37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 85.47 87.62 88.91 91.22 92.91 95.94 98.00 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3
55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 6 Cs Ba La* Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn 132.9 137.3 138.9 178.5 181.0 183.8 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 209.0 210.0 222.0
87 88 89 104 105 106 107 108 109 110 111 112 7 Fr Ra Ac Rf Db Sg Bh Hs Mt Uun Uuu Uub = metalloid 223.0 226.0 227.0 261.0 262.0 263.0 262.0 265.0 266.0 269.0 272.0 277.0
58 59 60 61 62 63 64 65 66 67 68 69 70 71 *Lanthanides Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu
140 141 144 145 150 152 157 159 163 165 167 169 173 175
90 91 92 93 94 95 96 97 98 99 100 101 102 103 Actinides Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr
232 231 238 237.1 244 243 247 247 251 252 257 258 259 260