IB Chemistry Mole Concept and Empirical Formula

1. http://lawrencekok.blogspot.com
Prepared by
Lawrence Kok
IB Chemistry Mole Concept, RAM, RMM
Isotopes and Empirical/Molecular Formula.

2. Mole
One mole of iron element (Fe) contain
- 6.02 x 1023 Fe particles
- 6.02 x 1023 Fe atoms
Elements
Molecule
One mole of CO2 molecule contain
- 6.02 x 1023 CO2 molecules
- 6.02 x 1023 C atoms
- 2 x 6.02 x 1023 O atoms
Ionic compound
One mole of ionic sub(NaCI) contain
- 6.02 x 1023 NaCI particles
- 6.02 x 1023 Na+ ions
- 6.02 x 1023 CI- ions
- 2 x 6.02 x 1023 CI- and Na+ ions
1 Mole
One mole of ionic sub MgCI2 contain
- 6.02 x 1023 MgCI2 particles
- 6.02 x 1023 Mg2+ ion
- 2 x 6.02 x 1023 CI- ion
Measurement to express amt of particles (atoms, molecules, ions)
One mole – amt substance contain same number of particles in 12g of C-12
Correspond to Avogadro constant (NA) - 6.02 x 10 23

3. 1 mole Fe contain
- 6.02 x 1023 Fe atoms
Element
Fe
Molecule
CO2
1 mole CO2 contain
- 6.02 x 1023 CO2 molecule
1 Mole
1 mole NaCI contain
- 6.02 x 1023 NaCI particles
1 mole MgCI2 contain
- 6.02 x 1023 MgCI2 particles
Molar Mass
1 mole (Fe) weigh
- RAM = 55.85
- Molar Mass = 55.85 g
Molar Mass
1 mole CO2 weigh
- RMM = 12.01 + 16.00 + 16.00 = 44
- Molar Mass = 44g
Ionic
Compound
NaCI
1 mole NaCI molecule weigh
- RMM = 22.99 + 35.45 = 55.45
- Molar Mass = 55.45 g
Ionic
Compound
MgCI2
1 mole MgCI2 weigh
- RMM = 24..31 + 35.45 + 35.45 = 95.21
- Molar Mass = 95.21 g
TED Video on MolemassMolar
mass
or
RMM
mass
or
RAM
mass
Mole
.

Mass for 1 mole of any substance
Symbol – M, Unit - g/mole
Molar mass contain 6.02 x 1023 particles
1 mole atom – RAM in gram (1 mole C – 12 g)
1 Mole
1 Mole
1 Mole
Molar Mass
Molar Mass
Molar Mass

4. Relative Atomic Mass
C-12 as std
1/12 of C12 = 1 unit
Sulphur – 32x heavier
1/12 x = 1 unit
32 unit
6 proton + 6 neutron
16 proton + 16 neutron
12
6
32
16
Mass number =
proton + neutron
Proton number =
proton
No isotopes are present
Relative Atomic Mass is used :
↓
- Impossible to weigh an atom in g
- Compare how heavy one atom is to carbon (std)
- One Sulfur atom 32x heavier than 1/12 C-12
- C- 12 used as std
Relative Atomic Mass, (Ar) of element:
- Number of times one atom of element is heavier than 1/12 of C-12
- Relative atomic mass = Mass of one atom
1/12 x mass C-12
- RAM, S = 32 (one sulfur atom is 32 x heavier than 1/12 of mass (C- 12)
C
S
Mass number ≠ Average atomic mass (atomic mass unit)

5. Relative Atomic Mass
C-12 as std
1/12 of C12 = 1 unit
H2O – 18 x heavier
1/12 x = 1 unit
16 unit
6 proton + 6 neutron
8 protons + 8 neutrons
12
6
16
8
Mass number =
proton + neutron
Proton number =
proton
No isotopes are present
Relative Atomic Mass is used :
↓
- Impossible to weigh an atom in g
- Compare how heavy one atom is to carbon (std)
- One Sulfur atom 32x heavier than 1/12 C-12
- C- 12 used as std
Relative Atomic Mass, (Ar) of element:
- Number of times one atom of element is heavier than 1/12 of C-12
- Relative atomic mass = Mass of one atom
1/12 x mass C-12
- RAM, S = 32 (one sulfur atom is 32 x heavier than 1/12 of mass (C- 12)
C
O
Mass number ≠ Average atomic mass (atomic mass unit)
H H
1
1
1
1
2 protons
2 unit

6. Click here spectroscopy database (NIST)
Weighted average calculationRAM calculationVideo on IsotopesVideo on weighted average
Relative Atomic Mass
Weighted average mass- due to presence of isotopes
Relative Isotopic Mass, (Ar) of an element:
•Relative isotopic mass = Average mass of one atom of element
1/12 x mass of one carbon-12
• Relative isotopic mass, carbon = 12.01
RAM = 12.01 Relative Abundance
13
Relative Isotopic Mass:
= (Mass 12
C x % Ab) + (Mass 13
C x % Ab)
= (12 x 98.9/100) + (13 x 1.07/100) = 12.01
Video on Isotopes
12
Isotopes are present
CCC12.01
98.9% 1.07%
Click here spectroscopy database (Ohio State)
Why RAM not a whole number?

7. 24 Mg – (100/127.2) x 100% - 78.6%
25 Mg – (12.8/127.2) x 100% - 10.0%
26 Mg – (14.4/127.2) x 100% - 11.3%
Relative Isotopic Mass
= (Mass 24
Mg x % Ab) + (Mass 25
Mg x % Ab) + (Mass 26
Mg x % Ab)
= (24 x 78.6/100) + (25 x 10.0/100) + (26 x 11.3/100) = 24.30
Relative Abundance % Abundance
Pb - 4 Isotopes
204Pb – (0.2/10) x 100% - 2%
206Pb – (2.4/10) x 100% - 24%
207Pb – (2.2/10) x 100% - 22%
208Pb – (5.2/10) x 100% - 52%
Relative Isotopic Mass
= (Mass 204Pb x % Ab) + (Mass 206Pb x % Ab) + (Mass 207Pb x % Ab) + (Mass 208Pb x % Ab)
= (204 x 2/100) + (206 x 24/100) + (207 x 22/100) + (208 x 52/100) = 207.20
Convert relative abundance to % Ab
Convert relative abundance to % Ab
Relative Abundance % Abundance
Relative Isotopic Mass
24 25 2624 25 26 MgMg
Mg - 3 Isotopes

8. Mass Spectrometer
Presence of isotopes
and its abundance
Relative atomic mass
of an element
Relative Molecular mass
of a molecule
Structure of organic
compound
Distinguish bet
structural isomers
structural
formula
Organic structure
determination
24 25 26Mg
Mg
CO2Mg Mg Mg Mg
24
24 25 26
CH3CH2CH2OH
OH
Ι
CH3CHCH3
CH3
|
CH3C-CH3
|
CH3
72
57
42
27
15
45432745
3129
% Ab % Ab % Ab
% Ab % Ab % Ab

9. Video on Mole calculation
12.00 g
RAM in g
1.992 x 10-23
g
Too small!!
Molar Mass
6 protons
6 neutrons
Mass 1 Carbon atom
(6 proton + 6 neutron)
1 proton/neutron = 1.66 x 10-24
g
12 proton/neutron = 12 x 1.66 x 10-24
g
= 1.992 x 10-23
g
Mass 1 MOLE carbon atom
(6.02 x 1023 carbon atom) Mass 1 C atom = 1.992 x 10-23
g
Mass 1 Mole C = 6.02 x 1023 x 1.992 x 10-23
g
= 12.00 g
Mole Simulation
Relationship bet Mole – Mass – Number particles
Mass 1 proton/neutron = 1.66x 10-24
g
Too small to weigh
Why Molar Mass is used?
Why mass for 1 mole of carbon = (RAM)g
C12
C
12
Video on Mole calculation

10. Chemical formula
Represent chemical compound
Show elements present in compound
Name
compound
Chemical
Formula
Name of each element
Sulphuric acid H2SO4 2 Hydrogen, 1 Sulphur, 4 Oxygen
Ammonia NH3 1 Nitrogen, 3 Hydrogen
Hydrogen
Chloride
HCI 1 Hydrogen, 1 Chlorine
Nitric Acid HNO3 1 Hydrogen, 1 Nitrogen, 3 Oxygen
Empirical formula
Represent simplest whole number
Ratio of atom of the elements
Formula obtained by expt
Molecular formula
Actual number atom of elements
Structural formula
Arrangement of atoms in compound
Chemical Compound
Ethene
C2H2
C1H1
Video tutorial
Molecular/empirical formula

11. Empirical Formula Calculation
Step 1:- Write mass/% of each element
Step 2: - Cal number of moles of each element
(divide with molar mass/RAM)
Step 3: - Divide each by smallest number, obtain simplest ratio
Empirical Formula Calculation
Empirical formula RMM Molecular formula
Compound Empirical
Formula (RMM)
Molecular
Formula (RMM)
Ethene C1H2- 14 C2H4 - 28
Phosphorus (V) oxide P2O5- 142 P4O10 – 284
Hydrogen Peroxide H1O1- 17 H2O2 – 34
Ethanoic acid C1H2O1 – 30 C2H4O2- 60
(C1H2O1) n = 60
(30) n = 60
n = 2
(C1H2O1)2 = 60
Molecular Formula = C2H4O2
Element M combines with O to form oxide, MO.
Find the empirical formula for MO.
Element M O
Step 1 Mass/g 2.4 1.6
RAM/RMM 48 16
Step 2 Number
moles/mol
2.4/48
= 0.05
1.6/16
= 0.1
Step 3 Simplest
ratio
0.05/0.05
= 1
0.1/0.05
= 2
Empirical formula - M1O2.
Molecular formula = n x Empirical formula
or
RMM = n x formula mass of Empirical formula
Relationship bet MF and EF Empirical Formula Calculation
Relative formula mass of Y3(PO4)2 = 310
Determine relative atomic mass of Y (RAM: O =16, P =31)
Find % by weight of nitrogen in (NH4)2SO4
Total RAM for nitrogen = 14 x 2 = 28
RMM (NH4)2SO4 = 132
% by weight of N = 28 x 100%
132
= 21.2%
Assume RAM for Y = X
RMM Y3(PO4)2 = 310
3Y + 2 [ 31 + 4(16) ] = 310
3Y + 190 = 310
Y = (310 -190)/3
= 40

12. Element H B O
Step 1 Percentage/% 4.8% 17.7% 77.5%
RAM/RMM 1 11 16
Step 2 Number
moles/mol
4.8/1
= 4.8
17.7/11
= 1.6
77.5/16
= 4.84
Step 3 Simplest ratio 4.8/1.6
= 3
1.6/1.6
= 1
4.84/1.6
= 3
Boric acid used to preserve food contains 4.8% hydrogen, 17.7% boron and rest is oxygen.
Find EF of boric acid.
Empirical formula - H3B1O3.
Empirical Formula Calculation
Empirical Formula Calculation
Step 1: - Write mass/ % of each element
Step 2: - Find number of moles of each element (divide with RAM)
Step 3: - Obtain the simplest ratio
2.5 g of X combined with 4 g of Y to form compound formula XY2.
RAM of Y is 80, Find RAM of X.
Element X Y
Step 1 Mass/g 2.5 4
RAM/RMM RAM 80
Step 2 Number
moles/mol
2.5/RAM
= ?
4/80
= 0.05
Step 3 Simplest ratio 1 2 Empirical formula given as X1Y2.
RAM = 100
100
2
05.05.2
2
1
05.0
/5.2



RAM
RAM
RAM

13. CHO + O2 CO2 + H2O
X contain 85.7% of carbon by weight. 4.2 g of gas X occupy vol of 3.36 dm3 at stp.
Find EF, RMM and MF of X
Element C H
Step 1 Percentage/% 85.7 14.3
RAM/RMM 12 1
Step 2 Number
moles/mol
85.7/12
= 7.14
14.3/1
= 14.3
Step 3 Simplest ratio 7.14/7.14
= 1
14.3/7.14
= 2
a) Empirical Formula = C1H2
b) Vol of 3.36 dm3 at stp – Mass, 4.2 g
Vol of 22.4dm3 at stp – Mass 1 mol (RMM)
3.36 dm3 – 4.2 g
22.4 dm3 - (4.2 x 22.4)/3.36 = 28
c) Assume molecular formula of X - (CH2)n
RMM of X is (12+2)n = 28
n = 2
Molecular formula X = C2H4
X contain carbon, hydrogen and oxygen. 0.50 g of compound on combustion, yield
0.6875 g of carbon dioxide and 0.5625 g of water. Find EF.
Element C H O
Step 1 Mass/g 0.1875 0.0625 0.25
RAM/RMM 12 1 16
Step 2 Number
moles/mol
0.1875/1 2
= 0.01562
0.0625/1
= 0.0625
0.25/16
= 0.01562
Step 3 Simplest ratio 0.01562
0.01562
= 1
0.0625
0.01562
= 4
0.01562
0.01562
= 1
Conservation of mass
Mass C atom before = Mass C atom after
Mass H atom before = Mass C atom after
Mol C atom in CO2
= 0.6875 = 0.0156 mol
44
Mass C = mol x RAM C
atom = 0.015625 x 12
= 0.1875 g
Mol H atom in H2O
= 0.5625 = 0.03125 x 2 = 0.0625 mol
18
Mass H = mol x RAM H
atom = 0.0625 x 1
= 0.0625 g
0.6875g 0.5625g0.50g 0.75g
Mass of O = (Mass CHO – Mass C – Mass H)
= 0.5 – 0.1875 - 0.0625 = 0.25 g
Empirical formula – C1H4O1
Empirical Formula Calculation

14. CHO + O2 CO2 + H2O
Element C H O
Step 1 Mass/g 0.731 0.0730 0.195
RAM 12.01 1.01 16.01
Step 2 Number
moles/mol
0.731/12.01
= 0.0609
0.0730/1.01
= 0.0730
0.195/16.01
= 0.0122
Step 3 Simplest
ratio
0.0609
0.0122
= 5
0.0730
0.0122
= 6
0.0122
0.0122
= 1
2.68 g 0.657 g1.00g
Empirical formula – C5H6O1
X contain elements carbon, hydrogen and oxygen.
Find EF of X, if 1 g X form 2.68 g of carbon dioxide and 0.657 g water when combust with O2.
Find MF of X, if 0.3 mol has mass of 98.5 g.
Empirical Formula = C5H6O1
Mole → Mass
0.3 mol → 98.5 g
1 mol → 98.5/0.3
RMM = 328 gmol-1
Assume MF - (C5H6O1)n = 328
RMM = ( (5 x 12.01) +(6 x 1.01) + ( 1 x 16.01) )n = 328
82.11 x n = 328 = 4
MF = (C5H6O1)4 C20H24O4
Find MF, given 0.3 mol X has mass of 98.5 g.
Empirical Formula Calculation
Conservation of mass
Mass C atom before = Mass C atom after
Mass H atom before = Mass C atom after
Mol C atom in CO2
= 2.68 = 0.0609 mol
44
Mass C = mol x RAM C
= 0.0609 x 12
= 0.731 g
Mol H atom in H2O
= 0.657 x 2 = 0.0729 mol
18
Mass H = mol x RAM H
= 0.0729 x 1
= 0.0736 g
Mass O = (Mass CHO – Mass C – Mass H)
= 1.0 – 0.731 - 0.0736 = 0.195 g

15. P = 101 kNm-2
= 101 x 103 Nm-2
Calculate RMM of gas
Mass empty flask = 25.385 g
Mass flask fill gas = 26.017 g
Mass flask fill water = 231.985 g
Temp = 32C, P = 101 kPa
Find molar mass gas by direct weighing, T-23C , P- 97.7 kPa
Mass empty flask = 183.257 g
Mass flask + gas = 187.942 g
Mass flask + water = 987.560 g
Mass gas = (187.942 – 183.257) = 4.685 g
Vol gas = Vol water = Mass water = (987.560 – 183.257) = 804.303 cm3
RMM determination
PV = nRT
PV = mass x R x T
M
M = mass x R x T
PV
= 4.685 x 8.314 x 296
97700 x 804.303 x 10-6
= 146.7
Vol gas = 804.303 cm3
= 804.303 x 10-6 m3
P = 97.7 kPa
= 97700 Pa
Density water = 1g/cm3
M = m x RT
PV
= 0.632 x 8.314 x 305
101 x 103 x 206 x 10-6
= 76.8
m gas = (26.017 – 25.385)
= 0.632 g
vol gas = (231.985 – 25.385)
= 206 x 10-6 m3
X contain C, H and O. 0.06234 g of X combusted,
0.1755 g of CO2 and 0.07187 g of H2O produced.
Find EF of X
Element C H O
Step 1 Mass/g 0.0479 0.00805 0.006384
RAM/RMM 12 1 16
Step 2 Number
moles/mol
0.0479/1 2
= 0.00393
0.00805/1
= 0.00797
0.006384/16
= 0.000393
Step 3 Simplest ratio 0.00393
0.000393
= 10
0.00797
0.000393
= 20
0.000393
0.000393
= 1
Conservation of mass
Mass C atom before = Mass C atom after
Mass H atom before = Mass C atom after
CHO + O2 CO2 + H2O
Mol C atom in CO2
= 0.1755 = 0.00393 mol
44
Mass C = mol x RAM C
= 0.00393 x 12
= 0.0479 g
Mol H atom in H2O
= 0.07187 = 0.0039 x 2 = 0.00797 mol
18
Mass H = mol x RAM H
= 0.00797 x 1.01
= 0.00805 g
Mass of O = (Mass CHO – Mass C – Mass H)
= 0.06234 – 0.0479 - 0.00805 = 0.006384 g
0.06234 g 0.1755 g 0.07187 g
Empirical formula – C10H20O1

16. 0.502 g of alkali metal sulfate dissolve in water and excess barium chloride sol, BaCl2(aq)
added to precipitate all sulfate ion as barium sulfate, BaSO4(s).
Precipitate is filtered, dried and weigh 0.672 g.
a) Find amt, mol BaSO4 formed
Mr (BaSO4)= 233.40
b) Find amt, mol of alkali metal sulfate present.
1 mol BaSO4 – 1 mol X2SO4
0.00288 mol BaSO4 – 0.00288 mol X2SO4
c) Find molar mass of alkali metal sulfate
(d) Deduce the identity of alkali metal,
X2SO4 + BaCI2 ↔ BaSO4 + 2 XCI
X2SO4 BaCl2
precipitate filter weighed
Precipitate = 0.672 g
X2SO4 + BaCI2 ↔ BaSO4 + 2 XCI
molmol
M
mass
mol
r
00288.0
40.233
672.0


X2SO4 + BaCI2 ↔ BaSO4 + 2 XCI
174
00288.0
502.0



r
r
r
M
mol
mass
M
M
mass
mol
Mr - X2SO4 = 174
2X + 32 + 4(16) = 174
X = 39 ( Potassium)
BaSO4

17. x in Fe(NH4)2(SO4)2. xH2O be found from the amt, of sulfate form in compound.
0.982 g sample dissolved in water, excess BaCl2 was added.
Precipitate of BaSO4 was separated, dried and weigh 1.17 g.
b. Find amt, of sulfate in 0.982 g, of Fe(NH4)2(SO4)2.xH2O
SO4
2-+ Ba2+ ↔ BaSO4
BaCl2
Fe(NH4)2(SO4)2. xH2O BaCl2
Fe(NH4)2(SO4)2. xH2O
Precipitate = 1.17 g
a) Find amt, mol BaSO4 in 1.17g ppt.
Mr (BaSO4)= 233.40
molmol
M
mass
mol
r
00501.0
40.233
17.1


1 mol BaSO4 – 1 mol SO4
0.00501 mol BaSO4 – 0.00501 mol SO4
SO4
2-+ Ba2+ ↔ BaSO4
c. Find amt, of iron in 0.982 g, of Fe(NH4)2(SO4)2.xH2O
Fe(NH4)2(SO4)2. xH2O
mol
FeAmt
SOAmt
FeAmt
0025.0
2
1
00501.0
).(
2
1
).(
).(
4


d. Find mass of Fe in 0.982g of Fe(NH4)2(SO4)2.xH2O
e. Find mass NH4 in 0.982g of Fe(NH4)2(SO4)2.xH2O
gmass
Mmolmass
M
mass
mol
r
r
14.085.550025.0 


gmass
Mmolmass
M
mass
mol
r
r
0904.005.1800501.0 

 RMM NH4 = 18.05
RMM Fe = 55.85
f. Find mass SO4 in 0.982g of Fe(NH4)2(SO4)2.xH2O
gmass
Mmolmass
M
mass
mol
r
r
481.006.9600501.0 


g. Find mass H2O in 0.982g of Fe(NH4)2(SO4)2.xH2O
Mass H2O = Mass sample – Mass (Fe + NH4 + SO4)
= 0.982 – 0.711 = 0.271 g
BaSO4
RMM SO4 = 96.06

18. Mg ribbon heated in crucible. Find EF for MgO
Mass/g
Mass crucible + lid 19.777
Mass crucible + lid + Mg 19.820
Mass crucible + lid + white solid 19.849
Element Mg O
Step 1 Mass 0.043 0.029
RAM/RMM 24.31 16.00
Step 2 Number
moles/mol
0.043/24.31
= 0.00177
0.029/16
=0.00181
Step 3 Simplest ratio 0.00177/0.00177
= 1.00
0.00181/0.00177
= 1.02
Data shown in table below
Crucible + lid
Mg
Mass Mg = 19.820 – 19.777
= 0.043
Mass O = 19.849 – 19.820
= 0.029
Empirical formula – Mg1O1.02
Find EF for CuO
Oxide of copper reduced by hydrogen shown below.
Mass/g
Mass crucible 13.80
Mass crucible + CuO bef heat 21.75
Mass crucible + CuO after
heat/cool
20.15
CuO + H2 ↔ H2O + Cu
Element CuO O
Step 1 Mass 6.35 1.60
RAM/RMM 63.55 16.00
Step 2 Number
moles/mol
6.35/63.55
= 0.10
1.60/16.00
=0.10
Step 3 Simplest ratio 0.10/0.10
= 1
0.10/0.10
= 1.
Empirical formula – Cu1O1
Crucible + CuO
H2
Mass CuO = 21.75 – 13.80
= 6.35
Mass O = 21.75 – 20.15
= 1.60

19. Fertilizer contain N-P-K rating of 18-51-20.
Find % by mass of N, P and K.
18% - N 51% - P2O5 20% - K2O
%22%51
140
07.302
%.
%
)(
).(
% 52
52




P
OP
OPM
Pmass
P
r
%17%20
2.94
10.392
%.
%
)(
).(
% 2
2




K
OK
OKM
Kmass
K
r
%18%18
14
14
%.
%
)(
).(
%


N
N
NM
Nmass
N
r
Fertilizer contain nitrogen used as below.
Find % by mass of N for NH3, CO(NH2)2, (NH4)2SO4
NH3 CO(NH2)2 (NH4)2SO4
%14.45%100
07.62
28
%.
%100
)(
).(
%
22


N
NHCOM
Nmass
N
r
%2.21%100
17.132
28
%.
%100
)(
).(
%
424


N
SONHM
Nmass
N
r
%22.82%100
04.17
14
%.
%100
)(
).(
%
3


N
NHM
Nmass
N
r

20. CsICl2 , is used in cancer treatment.
0.2015 g of CsICl2 was used for expt.
Find % iodine by mass in CsICI2
Find mass of iodine atoms in CsICI2
Mass/g
Mass CsICI2 0.2015
%4.38%100
71.330
9.126
%.
%100
)(
).(
%
2


I
CsICIM
Imass
I
r
gIMass
I
0773.02015.0
100
4.38
.
%4.38%100
71.330
9.126
%.


4
1009.6
9.126
0773.0




mol
mol
M
mass
mol
r
Relative Atomic Mass of X, Y , Z are 12, 16 and 24.
a) How much is an atom Z heavier than atom Y?
b) How many atoms of X will have same mass as
the sum of 3 atoms of Y and 2 atoms of Z?
Find amt of iodine in CsICI2
a) Z is heavier than Y by 24/16 = 1.5 times
b) Assume n atom of X has same mass as
the sum of 3 atom of Y + 2 atom of Z.
12n = 3(16) + 2(24)
12n = 96
n = 96/12 = 8 atoms.
Determine (RMM) of each of following.
a) (NH4)2SO4
b) BaCI2 .2H2O
c) C31H46O2
a) RMM for (NH4)2SO4
= 2( 14+ 4) + 32 + 4(16)= 132
b) RMM for BaCI2 . 2H2O
= 137 + 2(35.5) + 2(2 + 16) = 244
c) RMM for C31H46O2
= 31(12) + 46(1) + 2(16)= 450

21. a) 1 mole Al atoms → 27g
2/3 mole AI atoms → 2/3 x 27 = 18 g
b) 1 mole,C6H8O6 → 6(12)+8(1)+6(16)= 176g
0.08 mole C6H8O6 → 0.08 x 176 g = 14.08g
c) 1 mole Mg(OH)2 → 24 + 2( 16 + 1) = 58g
0.125 mole Mg(OH)2 → 0.125 x 58 g = 7.25g
1 mole C4H6N2 → 4(12) + 6(1) + 2(14) = 82 g
0.005 mole C4H6N2 → 82 x 0.005 = 0.41g
Conversion Moles ↔ Mass
Find mass for
a) 2/3 mole of AI atom
b) 0.08 mole of C6H8O6 molecules
c) 0.125 mole Mg(OH)2
Piperazine MF - C4H6N2. Find mass of
0.005 mole of piperazine in pill
Find the moles
a) 23.5 g of Cu(NO3)2
b) 0.97g of caffeine C8H10N4O2 molecules
a) 1 mole Cu(NO3)2 → 64 + 2 [14 + 3(16)]
= 188g
188g Cu(NO3)2 → 1 mole
23.5g Cu(NO3)2 → 1 x 23.5 = 0.125 mol
188
b) 1 mole, C8H10N4O2 → 8(12)+10+4(14)+2(16)
= 194g
194g C8H10N4O2 → 1 mole
0.97g C8H10N4O2 → 1 x 0.97 = 0.005 mol
194
Find number of moles in
a) 6 x 1021 iron atom
b) 7.5 x 1023 of H2O molecule
a) 6 x 1023 Fe atom → 1 mole
6 x 1021 Fe atom → 6 x 1021
6 x 1023
= 0.01 mol
b) 6 x 1023 H2O molecules → 1 mole
7.5 x 1023 H2O molecules → 7.5 x 1023
6 x 1023
= 1.25 mol
Find the moles in
a) 4.5 x 1023 AI2O3 particles
b) 7.2 x 1023 MgCI2 particles
a) 6 x 1023 AI2O3 particle→ 1 mole
4.5 x 1023 AI2O3 particle→ 4.5 x 1023
6 x 1023 = 0.75 mol
b) 6 x 1023 MgCI2 particle→ 1 mole
7.2 x 1023 MgCI2 particle→ 7.2 x 1023
6 x 1023 = 1.2 mol

22. a) 64 g Cu atom → 1 mole
12.8 g Cu atom → 1 x 12.8 = 0.2 mol
64
1 mol Cu→ 6 x 1021 Cu atom
0.2 mol Cu→ 6 x 1021 x 0.2 = 1.2 x 1021 Cu
b) 17g NH3 molecules → 1 mole
8.5g NH3 molecules → 1 x 8.5 = 0.5 mol
17
1 mole NH3 → 6 x 1023 NH3 molecule
0.5 mole NH3 → 0.5 x 6 x 1023
= 3 x 1023 NH3 molecule
a) 6 x 1023 Zn atom→ 1 mole
1.2 x 1022 Zn atom→ 1.2 x 1022 = 0.02 mol
6 x 1023
1 mol Zn atom→ 65g
o.o2 mol Zn atom→ 0.02 x 65g = 1.3g
b) 6 x 1023 C2H5OH molecule→ 1 mole
3 x 1023 C2H5OH molecule→ 3 x 1023 =0.5mol
6 x 1023
1 mole C2H5OH → 46g
0.5 mole C2H5OH → 46 x 0.5 = 23g
Mass
Conversion Mol ↔ Mass ↔ Particles
Find number of particles in
a) 12.8g of Cu
b) 8.5g of NH3 molecules
Find the mass in
a) 1.2 x 1022 zinc atom
b) 3 x 1023 C2H5OH molecules
Find number of particles in
a) 0.75 mole of AI atom
b) 1.2 mole of CI- ion
c) 0.07 mole of CO2 molecule
a) 1 mole of AI → 6 x 1023 AI atom
0.75 mole of AI → 0.75 x 6 x 1023
= 4.5 x 1023 atom
b) 1 mole CI- ions → 6 x 1023 CI- ion
1.2 mole CI- ions → 1.2 x 6 x 1023
= 7.2 x 1023 ion
c) 1 mole of CO2 → 6 x 1023 CO2 molecule
0.07 mole of CO2 → 0.07 x 6 x 1023
= 4.2 x 1022 molecule
Find number of particles in
a) 0.75 mol of AI2O3
b) 1.2 mol of MgCI2
a) 1 mole of AI2O3 →6 x 1023 AI2O3 particles
0.75 mole of AI2O3 → 0.75 x 6 x 1023
= 4.5 x 1023 particles
1 mole AI2O3 particle → 2 mole AI3+ ion and 3 mole O2- ion
4.5 x 1023 AI2O3 particle = 2 x 4.5 x 1023 AI3+ ion
= 3 x 4.5 x 1023 O2- ion
b) 1 mole MgCI2 → 6 x 1023 MgCI2 particles
1.2 mole MgCI2 → 7.2 x 1023 MgCI2 particles
1 mole MgCI2 particles → 1 mole Mg2+ and 2 mole CI- ions
7.2 x 1023 MgCI2 particles = 7.2 x 1023 Mg2+ ions
= 2 x 7.2 x 1023 CI- ions

23. Acknowledgements
Thanks to source of pictures and video used in this presentation
Thanks to Creative Commons for excellent contribution on licenses
http://creativecommons.org/licenses/
Prepared by Lawrence Kok
Check out more video tutorials from my site and hope you enjoy this tutorial
http://lawrencekok.blogspot.com