Section 1 Atoms, Molecules & Stoichiometry
Atomic Structure
particle
atomic symbol
relative mass
charge
position
electron (e − )
0 e
−1
1/1840
-1 unit
(-1.6021 x 10 −19 C )
outside the nucleus,
moving very fast in
‘shells’ (or more
precisely orbitals, see
Section 2) that
occupy
approximately the
whole size of the
atom
paroton (p − )
1 or 1
p
H
neutron (n)
1
+1 unit
(1.6021 x 10 −19 C )
1
1
1
in the nucleus
(of radius ≈ 10 −15 m)
1
0
n
0
Atomic Mass Unit, Relative Isotopic Mass, RAM, RMM
y
y
y
y
Carbon-12 scale: the relative mass of a 12 C atom is defined to be 1200000
6
12
A C atom has a mass of 12 atomic mass units (a.m.u.)
6
1 atomic mass unit (a.m.u.) = 1/12 (mass of a 12 C atom)
6
The relative isotopic mass (or simply the isotopic mass) of an isotope is the
mass of that isotope in the 12 C = 1200000 scale.
6
y
The relative atomic mass (or simply atomic mass) of an element is the
weighted average of all the isotopic masses of all the natural occurring
isotopes.
E.g. relative abundances of isotopes of oxygen:
isotope
relative isotopic mass
relative abundance
16
O
15.995
99.76%
17
O
16.999
0.04%
18
O
17.999
0.20%
RAM = 15.995(0.9976) + 16.999(0.0004) + 17.999(0.002) = 16.00
z
relative formula mass of a compound (any compound)
= sum of RAM of all elements in + formula unit of that compound
y
relative molecular mass of a molecule (molecular substances only)
= sum of RAM of all atoms present in the molecule
Mass Spectrometer – Principle
y
Mass Spectrometer for
y
determining the mass-to-charge ratio (m/e) of a particle
(cation of an element, e.g. C1 1 ) or molecular ion, e.g.
+
+
2
2
CO , C1 ) – if the charge is +1, the mass of the isotope or
molecule can be determined from the mass/charge ratio
y
y
the analysis of the sample (useful in organic chemistry)
The sample is vaporized by (1) heating in the vaporization chamber (A)
AND (2) under reduced pressure.
y
The sample is bombarded by fast-moving e _ to form positively charged ions
X + (ions of an element or molecular ions) in the ionization chamber (B)
(usually +1 ions are formed).
y
The pressure inside the mass spectrometer is very low to prevent ions from
colliding with air molecules.
z
The +ve ions are accelerated by the electric field (C)
y
The ions of different masses are deflected by the magnetic field (D) which is
adjusted so that ions of a particular mass-to- charge ratio (m/e) are focused
onto the ion detector (E)
y
The ion detector responds to the mass-to-charge ratio (NOT mass!) of the
ionized species
y
From (1) the strength of the magnetic field, (2) the accelerating voltage &
(3) the radius of the path, the (mass/charge) ratio of the particle can be
calculated.
y
If the charge of an electron (-1 unit) is known, the mass of the particle can
be found.
y
Mass spectrum: y-axis D peak height (or relative intensity) is directly
proportional to the % abundance of the ionized species (e.g.
x-axis D m/e ratio
y
When the sample is not monoatomic, many peaks may arise:
14
+
N + or CO ),
2
Given: Mass spectrum of chlorine gas
(a) Explain why there are 5 peaks in the mass spectrum.
(b) Relative atomic mass of chorine.
(c) Ratio of relative abundance of C to D to E.
D Solution:
(a) Since there are 2 isotopes of C1, 3 types of chlorine molecular ions exist:
35
C1+ , 35 C1− 37 C1+ , 37 C − 37 C1+
Since C1 2 molecule can dissociate during the electron
bombardment in the ionization chamber, atomic ions 35C1+
& 37 C1+ will also be detected. Total no. of peaks = 3+2 = 5
D Indicate the +ve charge on the ionized species!
35
C1 -
(b) RAM of C1 = (35)(3/4) + (37)(1/4) = 35.5 (a.m.u.)
(c) Abundance of C ∝ 3/4 x 3/4 = 9/16
Abundance of D ∝ 2 x 1/4 x 3/4 = 6/16
Abundance of E ∝ 1/4 x 1/4 = 1/16
Ratio of relative abundance of C : D : E = 9 : 6 : 1
D Given: mass spectrum of a carbon- containing compound:
Ion
C+
13 +
C
Mass / a.m.u.
12.000
13.003
12
Relative intensity
100.00
1.12
D Find: relative atomic mass of C
12000 x100 + 13.003x1.12
D Solution: RAM =
= 12.011 (NO unit!)
(100 + 1.12)
D Given: Elements X & Y form a compound with molecular
formula XY 3 with the following mass spectrum:
D Find: (a) How is the molecular ion XY 3 + formed?
(b) Relative atomic masses of X & Y = ?
D Solution:
(a) In the ionization chamber of the mass spectrometer, XY 3
molecules were bombarded by fast-moving electrons to give
XY 3 + :
_
e
2c
XY 3 (g) +
t XY 3 + (g) +
fast
slow
−
(b) The peak at m/e ratio = 31 is due to X + ions.
The peak at m/e ratio = 19 is due to Y + ions.
∴ RAM of x = 31, RAM of Y = 19
Mass Spectrum – Composition of a Mixture
z
Application of Mass Spectroscopy in the analysis of the composition of a
mixture: If a gas mixture containing 10% He, 30% Ne, 50% Ar & 10% Xe
by volume is analyzed by mass spectroscopy, the following peaks of the
molecular ion can be found (assuming each noble gas consists of 1 stable
isotope only):
m/e radio
ion
Relative intensity
D
D
D
4
He +
10
20
Ne +
30
40
Ar +
50
131
Xe +
10
Given: volume composition of air N 2 78%, O 2 21% &
CO 2 1% (RAM: C= 12.0, N = 14.0, O = 16.0)
Find: mass spectrum of air (molecular ions)
Solution: volume composition = mole composition
Mole, P-V-T relationships of gases, Ideal Gas Law (Equation)
z
z
z
z
z
z
Boyle’s law: P-V relationship at constant T (absolute temp.!) D
PV = constant for a fixed amount of gas
Charles’ law: V-T relationship at constant P D V/T is constant
for a fixed amount of gas
P-T relationship at constant V D P/T = constant for a fixed
amount of gas
Avogadro’s law: equal volumes of gases at same conditions D
equal no. of moles
PV
Ideal gas equation:
= nR or PV = nRT
T
Relationship between density & molar mass (NOTE: the
UNITS of mass must be consistent!)- The ideal gas equation
can be rewritten for the calculation of the gas density ( ρ , in g m −3 [or kg
m −3 ]) For a gas of mass in (in g [or kg]) & a molar mass of M (in g
mol −1 [or kg mol −1 ]), a volume of V (in m 3 ) at a pressure P (N m −2 , Pa):
PV = nRT & ρ = m/V
m
ρ
) RT ∴ P = (
) RT
M
M
PM
ρRT
DM=
OR ρ =
P
RT
D PV = (
D
Beware of the units:
R = 8.31 J K −1 mol −1 ;
For pressure: 1 Pa = 1 N m −2 [1 atm. = 101 kPa]
For mass: 1 kg = 1000g
For volume: 1 m 3 = 1000 dm 3 ; 1 dm 3 = 1000 cm 3
For density: 1 g dm −3 = 1 kg m −3 = 10 −3 g cm −3
Dalton’s law of partial pressures
z
z
In a gas mixture, the partial pressure of each component gas is the pressure that
the gas, would exert if it were the only gas occupying the total volume at that
temperature.
Dalton’s law can be derived from the ideal gas equation, i.e.
RT
RT
P total = (n total )
= (n 1 + n 2 + n 3 +…)
V
V
RT
RT
RT
+ n2
+ n3
+ …..
= n1
V
V
V
P total = P 1 + P 2 + P 3 + … [P 1 = partial pressure of gas i]
z
The mole fraction (X i ) of a component in a mixture is defined as the no. of
moles of that component over the total no. of moles of gases, i.e.
Xi =
z
ni
ntotal
PiV
P V
& n total = total , if the total pressure of the gas mixture is known,
RT
RT
the partial pressure of component gas i is given by:
Since n i =
Xi =
ni
P
= i
ntotal
Ptotal
∴ Pi -
ni
P total - X i P total
ntotal
D Given: Gaseous compound X contains mass % N (21.6),
O (49.2), F (29.2), at 298 K & 1.01 x 10 5 N m −3 , density of X= 2.65 g dm −3
[RAM: N = 14.0: O = 16.0; F = 19.0]
D Find: (a) empirical formula & (b) molecular formula of x
D Solution:
21.6 49.2 29.2
:
:
= 1.543 : 3.075 : 1.537
(a) mole ratio N: O: F =
14
16
19
= 1 : 2 : 1 D empirical formula NO 2 F
(c) For an ideal gas, PV = nRT, M r =
ρ
P
RT
2.65 x10 3
molar mass of X =
x 8.31 x 298 = 64.97 g mol −1
5
1.01x10
Let molecular formula of X be (NO 2 F) n
(14.0+16.0x2+19.0)n = 64.97 ∴ n = 1, molecular formula: NO 2 F
D Given: aluminium chloride gas, T = 433 K, m – 0.569g,
V = 96.0 cm 3 , P = 80.0 kPa [RAM: Al=27.0, Cl = 35.5]
D Find: [Assuming ideal gas law is valid]
(a) apparent molar mass of aluminium chloride
(b) apparent molecular formula of aluminium chloride
(c) structure of aluminium chloride in the vapour
D Solution:
(a) PV = nRT
m
0.569
M=
RT =
x 8.31 x 433 = 266.6 g mol −1
3
−6
PV
80 x10 x96 x10
(b) empirical formula mass of AICI 3 = 27.0 + 3(35.5) = 133.5
molecular formula = (AICI 3 )n; n = 266.6/ 133.5 ≠ 2
i.e. molecular formula AI 2 CI 6 ∴ exist as dinner
(c)
D Given: At 500K & 70.0 kPa, 6.0 g of ethanoic acid
vapour occupies a volume of 4.90 dm 3 [RAM: H=1.0,
C – 12.0: O = 16.0]
D Find: [Assuming ideal gas law is valid]
(a) apparent relative molecular mass of ethanoic acid
(b) reason for the departure from expected RMM of CH 3 COOH
D Given: 0.5 mole of H 2 and 0.5 mole of O 2 are introduced into a flask of constant
Volume 10 dm 3 at 600‫ﹾ‬C, after an electrical sparking, some water vapour is
formed.
D Find:
(a) composition of the mixture in terms of mole fractions
(b) partial pressure of each gas in mixture
D Given: In a flask of constant volume, the pressure of a
mixture of gases X & Y at 200‫ﹾ‬C is 4.5 atm. At -40‫ﹾ‬C,
X is condensed & the pressure becomes 1.5 atm.
D Find: mole fraction of X in the mixture
D Given: Gas containers A and B each contains an ideal
gas at low pressure & 25‫ﹾ‬C. Volume of container A = 2
x volume of container B, but the no. of moles of ideal
gas contained in A is only half of that in B
D Given: hydrocarbon X, density of 1.15 g dm −3 at 95.3 kPa
pressure & 298 K (1 kPa = 1 x 10 3 N m −2 )
D Find: molar mass & molecular formula of X
D Give: At 100 kPa, a certain temperature & the presence of a catalyst, a certain
volume of C 2 H 6 (g) is cracked into C 2 H 4 (g) & H 2 (g). The volume of the gas
increases by 30% after cracking.
D Find: (Under the same conditions)
(a) Mole fractions of C 2 H 6 (g), C 2 H 4 (g) & H 2 (g) respectively
(b) Partial pressures of C 2 H 6 (g), C 2 H 4 (g) & H 2 (g) respectively
Stoichiometry, Limiting Reagents & Yield
z
Stoichiometry is the numerical relationship between the amounts of reactants &
products in a chemical reaction.
z
The exact amounts of reactants as given by the balanced equation are called the
stoichiometric amounts.
z
The yield is the amount of product formed.
z
The theoretical yield is the amount of product formed as predicted from the
stoichiometric of the reactants if the reaction goes completely.
z
The actual yield is the actual amount of product obtained.
z
percentage yield =
z
If the % yield is 100%, the reaction is said to be quantitative,
or produces a ‘quantitative yield’,
z
If the % yield is less than 100%, the reaction is said to be non-stoichiometric. A
low yield may be caused by
z
the presence of reversible reaction(s),
z
impure reactants,
z
side reactions &/or
z
the loss of products during purification.
z
If the reactants are not in stoichiometric amounts, the one that is NOT in excess is
called the limiting reactant (or reagent), the one that is in excess is called the
excess reactant (or reagent).
z
Note that the theoretical yield is calculated from the amount of the limiting
reactant.
x 100%
D Given: 15.00 g 4-nitrobenzoic, acid (O 2 N-COOH)
reacted with excess PCl 5 to give 13.60 g of 4-nitrobenzoyl
chloride (O 2 N-COCl) [RAM: H=1.0, C= 12.0, N- 14.0, O-16.0, Cl= 35.5]
D Find: % yield of the reaction
D Given 20.0 g of (CH 3 ) 3 COH reacted with 35.0 g of HCl to give 18.0 g of
(CH 3 ) 3 CCl
D Find: (a) the limiting reactant
(b) percentage yield of (CH 3 ) 3 CCl
Calculation involving Equations
D Given: 5200 g of a hydrated sodium carbonate (Na 2 CO 3 xH 2 O) were dissolved
in water. Excess CaCl 2 (aq) is added to the resulting solution to give 4.20 g of a
precipitate. [RAM: H=1.0, C=12.0, O=16.0 Na=23.0, Ca=40.1]
D Find:
(a) formula of the hydrated sodium carbonate
(b) give the assumption
D Given: An air bag in car contains 130 g of sodium azide N a N 3 When a car
crash occurs, the crash sensor in car sends an electric signal to an ignitor to heat
up the sodium azide. Within 0.03 sec, the sodium azide decomposes to give
sodium metal & nitrogen gas, which inflates the car’s air bags. [RAM: N=14.0;
Na=23.0]
D Find:
(a) maximum volume of the N 2 formed at 298K & 101 kPa
(b) hazard of sodium vapour formed
Formula, Chemical Equations, Calculation & Titration
z
z
A primary standard is a substance that can be used to prepare a standard solution
without standardization (i.e. by dissolving a known mass of that substance in a
solvent to give a known volume of the solution, the morality of that solution
can be calculated)
z
Usually solid
z
Must be very pure / with constant composition, does not
absorb moisture nor lose water easily, stable in air, soluble in water &
with high relative formula mass
Examples of primary standards:
z
(COOH) 2 2H 2 O(s) [acid]
z
constant boiling hydrochloric acid [acid] (NOT HNO 3 )
z
aminosulphonic acid (sulphamic acid) NH 2 SO 3 H [acid]
z
anhydrous Na 2 CO 3 (s) [base]
z
K 2 Cr 2 O 7 (s) [OA]
z
KIO 3 (s), KBrO 3 (s) [OA]
z
Na 2 C 2 O 4 (s) [RA]
z
Fe(NH 4 ) 2 (SO 4 ) 2 6H 2 O ammonium iron(II) sulphate(VI)-6-water [RA]
z
Fe(s) wire [RA]
The following are NOT primary standards:
z
NaOH(s): absorbs moisture in air
z
l 2 (s), Br 2 (1): volatile, reacts with RA (SO 2 /H 2 S) in air
cannot be prepared in highly pure form solid
z
AgNO 3 (s): decomposes by light
z
Na 2 S 2 O 3 (s): decomposes by acids, bases, OA, microorganism
z
Standardization involving redox reagents: ﹡primary standards
−
2−
5C 2 O (aq) * + 2MnO (aq) + 16H + (aq) t 2Mn 2+ (aq) +8H 2 O(1)+10CO 2 (g)
4
4
2−
2−
3C 2 O (aq) * + Cr 2 O
(aq) * + 14H + (aq) t 2Cr 3+ (aq) + 7H 2 O(1) + 6CO 2 (g)
7
4
−
5Fe 2+ (aq) * + MnO (aq) + 8H + (aq) t 5Fe 3+ (aq) + Mn 2+ (aq) + 4H 2 O(1)
4
−
5H 2 O 2 (aq) + 2MnO (aq) + 6H + (aq) t 2Mn 2+ (aq) + 5O 2 (g) + 8H 2 O(1)
4
2+
z
Fe (aq) is easily oxidized by air in alkaline / neutral medium,
must be used in acidic medium [dilute H 2 SO 4 (aq) must be used]
z
KMnO 4 is acidified with dilute H 2 SO 4 (aq), NOT HNO 3 [OA], nor HCl(aq)
[ Cl will be oxidized to give C1 2 ], K 2 Cr 2 O 7 can also be acidified will dilute
2−
HCl(aq) [Cl − cannot be oxidized by Cr 2 O
]
7
z
z
Preparation of standard I 2 (aq)
I 2 (aq) + I − (aq) t I 3 (aq)
−
IO (aq) + 5I − (aq) + 6H + (aq) t 3I 2 (aq) + 3H 2 O(1)
3
z
Since iodine is volatile, we cannot prepare standard solution of iodine directly
by accurately weighing a certain amount of I 2 (s) & dissolving the I 2 (s) in
water.
z
z
z
z
The standard I 2 solution is prepared (& usually used immediately) by the
addition of a known amount of potassium iodate(V) KIO 3 into excess I − in an
acidic medium (must be dilute H 2 SO 4 , NOT conc. H 2 SO 4 , NOT HNO 3 ,
NOT HCI)
I 2 (aq) is unstable in alkaline medium [disproportionates to I = (aq) &
−
IO (aq)], I 2 must be used in acidic medium.
3
Standard Br 2 (aq) can be similarly prepared from KBrO 3 & excess Br = (aq)
in acidic medium (dilute H 2 SO 4 )
Titration of iodine (in flask) against thiosulphate (in burette)
z
Standard I 2 (in conical flask) can be used to standardize Na 2 S 2 O 3 (in
burette) using starch indicator (in flask).
2−
2−
(aq) t S 4 O
(aq) + 2I − (aq)
I 2 (aq) + 2S 2 O
3
6
Starch + I 2 (aq) <=> blue ‘complex’
z
Since starch irreversibly combines with iodine at a high concentration of
I 2 (aq) (so that the I 2 will not be released from starch at the equivalence
point), starch solution should be added at the later stage of the titration
(when the solution has just turned from brown to pale yellow).
After the addition of starch, the mixture turns deep blue.
z
The end point is indicated by the complete disappearance of the blue
colour of the starch-iodine complex
Back Titration
z
Back titration: ammonium salt – The amount of NH 4 NO 3 (or other
ammonium salt) in a sample is determined by the addition of an excess
known volume of standard alkali, e.g. 50.0 cm 3 of 1.00M KOH. The
reaction mixture is then warmed to drive off the NH 3 formed. After the
reaction has completed (no alkaline gas evolved), the excess KOH can be
determined by titration with standard acid, e.g. 1.00 M HCI.
NH 4 NO 3 (aq) + KOH(aq) t KNO 3 (aq) + NH 3 (g) + H 2 O(1)
KOH(aq) + HCI(aq) t KCI(aq) + H 2 O(1)
no of moles of KOH = no. of moles of NH 4 NO 3 + no. of moles of HCI