- 1. http://lawrencekok.blogspot.com Prepared by Lawrence Kok
Tutorial on Mole Concept, RAM, RMM Isotopes and Empirical/Molecular
Formula.
2. Mole Concept Mole Unit of measurement to express amount of
particles (atoms, molecules, ions) One mole amt of substance which
contains the same number of particles as in 12g of carbon-12
Correspond to Avogadro constant (NA) - 6.02 x 1023 One mole of iron
element (Fe) contains 6.02 x 1023 Fe particles 6.02 x 1023 Fe atoms
Elements Molecule One mole of CO2 molecule contains 6.02 x 1023 CO2
molecules 6.02 x 1023 C atoms 2 X 6.02 x 1023 O atoms 1 Mole 1 Mole
Ionic compound One mole of ionic compound (NaCI) contains 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 1 Mole One mole of ionic
compound MgCI2 contains 6.02 x 1023 MgCI2 particles 6.02 x 1023
Mg2+ ions 2 X 6.02 x 1023 CI- ions Ionic compound
http://dl.clackamas.edu/ch104/lesson2moles.html
http://forum.nationstates.net/viewtopic.php?f=6&t=231490
http://www.chemistry.wustl.edu/~edudev/LabTutorials/Water/PublicWaterSupply/PublicWaterSupply.html
3. Molar Mass Mass for 1 mole of any substance Symbol M, Unit - g
/mole Molar mass contain 6.02 x 1023 particles 1 mole atom relative
atomic mass in gram ( 1 mole carbon 12g) 1 mole molecule - relative
molecular mass in gram ( 1 mole water 18g) Number moles = mass or
mass RAM/RMM Molar Mass 1 mole of Fe contains 6.02 x 1023 Fe atoms
Element Fe Molecule CO2 1 mole of CO2 contains 6.02 x 1023 CO2
molecules 1 Mole 1 Mole 1 mole of NaCI contains 6.02 x 1023 NaCI
particles 1 Mole 1 Mole 1 mole of MgCI2 contains 6.02 x 1023 MgCI2
particles Molar Mass Ionic Compound NaCI Ionic Compound MgCI2 4.
Molar Mass Mass for 1 mole of any substance Symbol M, Unit - g
/mole Molar mass contain 6.02 x 1023 particles 1 mole atom relative
atomic mass in gram ( 1 mole carbon 12g) 1 mole molecule - relative
molecular mass in gram ( 1 mole water 18g) Number moles = mass or
mass RAM/RMM Molar Mass 1 mole of Fe contains 6.02 x 1023 Fe atoms
Element Fe Molecule CO2 1 mole of CO2 contains 6.02 x 1023 CO2
molecules 1 Mole 1 Mole 1 mole of NaCI contains 6.02 x 1023 NaCI
particles 1 Mole 1 Mole 1 mole of MgCI2 contains 6.02 x 1023 MgCI2
particles Molar Mass Ionic Compound NaCI Ionic Compound MgCI2 5.
Molar Mass Mass for 1 mole of any substance Symbol M, Unit - g
/mole Molar mass contain 6.02 x 1023 particles 1 mole atom relative
atomic mass in gram ( 1 mole carbon 12g) 1 mole molecule - relative
molecular mass in gram ( 1 mole water 18g) Number moles = mass or
mass RAM/RMM Molar Mass 1 mole of Fe contains 6.02 x 1023 Fe atoms
Element Fe Molecule CO2 1 mole of CO2 contains 6.02 x 1023 CO2
molecules 1 Mole 1 Mole 1 mole of NaCI contains 6.02 x 1023 NaCI
particles 1 Mole 1 Mole 1 mole of MgCI2 contains 6.02 x 1023 MgCI2
particles Molar Mass 1 mole of iron element (Fe) weigh RAM = 55.85
Molar Mass = 55.85g Molar Mass 1 mole of CO2 molecule weighs RMM =
12.01 + 16.00 + 16.00 = 44 Molar Mass = 44g Molar Mass Ionic
Compound NaCI 1 mole of NaCI molecule weighs RMM = 22.99 + 35.45 =
55.45 Molar Mass = 55.45g Molar Mass Ionic Compound MgCI2 Molar
Mass 1 mole of MgCI2 molecule weighs RMM = 24..31 + 35.45 + 35.45 =
95.21 Molar Mass = 95.21g TED Video on Mole 6. Relative Atomic Mass
Relative Atomic Mass is used : Impossible to weigh an atom in grams
Compare how heavy one atom is to carbon (standard) One sulphur atom
32x heavier than 1/12 carbon -12 Carbon -12 used as standard No
isotopes are present Mass number = proton + neutron Proton number =
proton Z A Mass number Average atomic mass (atomic mass unit) 7.
Relative Atomic Mass, (Ar) of an element: Number of times one atom
of the element is heavier than one twelfth of the mass of a
carbon-12 Relative atomic mass = Mass of one atom of element 1/12 x
mass of one carbon-12 Relative atomic mass for sulphur = 32 (one
sulphur atom is 32 x heavier than 1/12 of mass of one (C12)
http://www.tutorvista.com/content/science/science-i/atoms-molecules/atom.php
Relative Atomic Mass Relative Atomic Mass is used : Impossible to
weigh an atom in grams Compare how heavy one atom is to carbon
(standard) One sulphur atom 32x heavier than 1/12 carbon -12 Carbon
-12 used as standard 12 6 32 16 No isotopes are present Mass number
= proton + neutron Proton number = proton Z A Mass number Average
atomic mass (atomic mass unit) 8. Relative Atomic Mass, (Ar) of an
element: Number of times one atom of the element is heavier than
one twelfth of the mass of a carbon-12 Relative atomic mass = Mass
of one atom of element 1/12 x mass of one carbon-12 Relative atomic
mass for sulphur = 32 (one sulphur atom is 32 x heavier than 1/12
of mass of one (C12)
http://www.tutorvista.com/content/science/science-i/atoms-molecules/atom.php
Relative Atomic Mass Carbon-12 as standard 1/12 of C12 = 1 unit
Sulphur 32x heavier 1/12 x = 1 unit 32 unit 6 protons + 6 neutrons
16 protons + 16 neutrons Relative Atomic Mass is used : Impossible
to weigh an atom in grams Compare how heavy one atom is to carbon
(standard) One sulphur atom 32x heavier than 1/12 carbon -12 Carbon
-12 used as standard 12 6 32 16 No isotopes are present Mass number
= proton + neutron Proton number = proton Z A Assuming No isotopes
present! Mass number Average atomic mass (atomic mass unit) 9.
Relative Molecular Mass is used : Impossible to weigh an molecules
in grams Compare one molecule to carbon (standard) One H2O is 18 x
heavier than 1/12 carbon -12 Carbon -12 is used as standard
Relative Molecular Mass No isotopes are present Proton number =
proton Mass number = proton + neutron Z A Mass number Average
atomic weight (atomic mass unit) 10. Relative Molecular Mass, (Mr):
Number of times one molecule is heavier than one twelfth of the
mass of a carbon-12 Relative molecular mass = Mass of one molecule
1/12 x mass of one carbon-12 Relative molecular mass for H2O= 18
(one H2O is 18 x heavier than 1/12 of mass of one (C12)
http://www.tutorvista.com/content/science/science-i/atoms-molecules/atom.php
Relative Molecular Mass is used : Impossible to weigh an molecules
in grams Compare one molecule to carbon (standard) One H2O is 18 x
heavier than 1/12 carbon -12 Carbon -12 is used as standard
Relative Molecular Mass No isotopes are present Proton number =
proton Mass number = proton + neutron Z A Mass number Average
atomic weight (atomic mass unit) 11. Relative Molecular Mass, (Mr):
Number of times one molecule is heavier than one twelfth of the
mass of a carbon-12 Relative molecular mass = Mass of one molecule
1/12 x mass of one carbon-12 Relative molecular mass for H2O= 18
(one H2O is 18 x heavier than 1/12 of mass of one (C12)
http://www.tutorvista.com/content/science/science-i/atoms-molecules/atom.php
Relative Molecular Mass is used : Impossible to weigh an molecules
in grams Compare one molecule to carbon (standard) One H2O is 18 x
heavier than 1/12 carbon -12 Carbon -12 is used as standard
Relative Molecular Mass Carbon-12 as standard 1/12 of C12 = 1 unit
H2O 18x heavier 1/12 x = 1 unit 16 unit 2 unit 18 unit + 6 protons
+ 6 neutrons 8 protons + 8 neutrons No isotopes are present Proton
number = proton Mass number = proton + neutron Z A Assuming No
isotopes present! Mass number Average atomic weight (atomic mass
unit) 2 protons 12. Relative Isotopic Mass Isotopes Atoms of same
element with Different number of neutrons Same number of protons
and electrons Due to presence of isotopes, when calculating RAM,
weighted average/mean of all isotopes present is used. Mass number
= proton + neutron Proton number = proton Z = 29 protons A= 29
protons + 35 neutrons = 64 Z Presence of isotopes A Z A 13.
Relative Isotopic Mass Isotopes Atoms of same element with
Different number of neutrons Same number of protons and electrons
Due to presence of isotopes, when calculating RAM, weighted
average/mean of all isotopes present is used. X - No isotopes Mass
number = proton + neutron Proton number = proton Z = 29 protons A=
29 protons + 35 neutrons = 64 Isotopes Y - TWO isotopes CI - TWO
isotopes Z Presence of isotopes A Z A 11 3 17 35 17 37 3 3 10 11
14. Relative Isotopic Mass Isotopes Atoms of same element with
Different number of neutrons Same number of protons and electrons
Due to presence of isotopes, when calculating RAM, weighted
average/mean of all isotopes present is used. X - No isotopes
RAM/Ar X = 11 Mass of 1 atom X Mass of 1/12 of 12 C Mass of 1 atom
X relative to 1/12 mass of 1 atom 12C Relative Abundance 75% 25%
Mass number = proton + neutron Proton number = proton Z = 29
protons A= 29 protons + 35 neutrons = 64 Isotopes Y - TWO isotopes
RAM/Ar Y = 10.5 Average Mass of 1 atom Y Mass of 1/12 of 12 C
Average mass of 1 atom Y relative to 1/12 mass of 1 atom 12C RAM
/Ar, CI = 35.5 Weighted average mass of 2 isotopes present = (mass
35 CI x % Abundance) + (mass 37 CI x % Abundance) = (35 x 75/100) +
(37 x 25/100) = 35.5 CI - TWO isotopes Relative Abundance 50% 50% Z
Presence of isotopes A Z A 11 3 17 35 17 37 3 3 10 11 15. Relative
Atomic Mass Isotopes are present RAM = 12.01 Relative Abundance
98.9% 1.07% 12 13 Why RAM is not a whole number? 16. Relative
Atomic Mass Isotopes are present 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 98.9% 1.07% 12 13 Why RAM is not a whole
number? RAM, C : = (Mass 12 C x % Abundance) + (Mass 13 C x %
Abundance) = (12 x 98.9/100) + (13 x 1.07/100) = 12.01 17.
http://www.tutorvista.com/content/science/science-i/atoms-molecules/atom.php
Relative Atomic Mass Isotopes are present 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 Video on Isotopes RAM = 12.01 Relative Abundance 98.9%
1.07% 12 13 Why RAM is not a whole number? RAM, C : = (Mass 12 C x
% Abundance) + (Mass 13 C x % Abundance) = (12 x 98.9/100) + (13 x
1.07/100) = 12.01 Video on weighted average Weighted average
calculation Video on Isotopes RAM calculation 18. Mg - 3 Isotopes
Pb - 4 Isotopes Relative Atomic Mass 19. Mg - 3 Isotopes 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 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% Convert
relative abundance to % abundance Convert relative abundance to %
abundance Relative Abundance % Abundance Relative Atomic Mass 20.
Mg - 3 Isotopes 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% RAM for Mg : =
(Mass 24 Mg x % Abundance) + (Mass 25 Mg x % Abundance) + (Mass 26
Mg x % Abundance) = (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% RAM for Pb : =
(Mass 204Pb x % Abundance) + (Mass 206Pb x % Abundance) + (Mass
207Pb x % Abundance) + (Mass 208Pb x % Abundance) = (204 x 2/100) +
(206 x 24/100) + (207 x 22/100) + (208 x 52/100) = 207.20 Convert
relative abundance to % abundance Convert relative abundance to %
abundance Relative Abundance % Abundance Relative Atomic Mass 21.
Mass 1 proton or neutron = 1.66x 10-24 g Too small to weigh Molar
Mass Question: Why Molar Mass is used? Why mass for 1 mole of
carbon = (RAM)g 6 protons 6 neutrons Mass for 1 Carbon atom (6
protons + 6 neutrons) 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 1.992 x
10-23 g Too small!!!!!! 22. Mass 1 proton or neutron = 1.66x 10-24
g Too small to weigh Molar Mass Question: Why Molar Mass is used?
Why mass for 1 mole of carbon = (RAM)g 6 protons 6 neutrons Mass
for 1 Carbon atom (6 protons + 6 neutrons) 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
1.992 x 10-23 g Too small!!!!!! Mass for 1 MOLE carbon atoms (6.02
x 1023 carbon atoms) Mass 1 carbon atom = 1.992 x 10-23 g Mass for
1 Mole = 6.02 x 1023 x 1.992 x 10-23 g = 12.00 g 12.00 g RAM in g
23. Mass 1 proton or neutron = 1.66x 10-24 g Too small to weigh
Molar Mass Question: Why Molar Mass is used? Why mass for 1 mole of
carbon = (RAM)g 6 protons 6 neutrons
http://www.aandd.jp/products/weighing/balance/toploader/gx_k.htm
Mass for 1 Carbon atom (6 protons + 6 neutrons) 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 1.992 x 10-23 g Too small!!!!!! Mass for 1 MOLE carbon
atoms (6.02 x 1023 carbon atoms) Mass 1 carbon atom = 1.992 x 10-23
g Mass for 1 Mole = 6.02 x 1023 x 1.992 x 10-23 g = 12.00 g 12.00 g
RAM in g Mole Simulation Relationship between Mole Mass Number
particlesRelationship between Mole Mass Number particles 24.
Calculate the mass for a) 2/3 mole of aluminium atoms b) 0.08 mole
of C6H8O6 molecules c) 0.125 mole Mg(OH)2 Piperazine is used to
kill worms. Molecular formula is C4H6N2. A pill contain 0.005 mole
of piperazine. Determine the mass found in pill. Conversion from
Moles to Mass 25. Calculate the mass for a) 2/3 mole of aluminium
atoms b) 0.08 mole of C6H8O6 molecules c) 0.125 mole Mg(OH)2
Answer: a) 1 mole Al atoms 27g 2/3 mole AI atoms 2/3 x 27 g 18g 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 58g = 7.25g Piperazine is used to kill worms.
Molecular formula is C4H6N2. A pill contain 0.005 mole of
piperazine. Determine the mass found in pill. Answer: a) 1 mole
C4H6N2 4(12) + 6(1) + 2(14) = 82g 0.005 mole C4H6N2 82 x 0.005 =
0.41g Conversion from Moles to Mass Moles Mass Video on Mole
calculation 26. Calculate the moles a) 23.5g of copper(II)nitrate,
Cu(NO3)2 b) 0.97g of caffeine C8H10N4O2 molecules Piperazine is
used to kill worms. Molecular formula is C4H6N2. A pill contain
0.82g of piperazine. Determine the number of moles in pill.
Conversion from Mass to Moles 27. Calculate the moles a) 23.5g of
copper(II)nitrate, Cu(NO3)2 b) 0.97g of caffeine C8H10N4O2
molecules Answer: a) 1 mole copper(II)nitrate, 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
Piperazine is used to kill worms. Molecular formula is C4H6N2. A
pill contain 0.82g of piperazine. Determine the number of moles in
pill. Answer: a) 1 mole C4H6N2 4(12) + 6(1) + 2(14) = 82g 82g
C4H6N2 1 mole 0.82g C4H6N2 1 x 0.82 = 0.01 mol 82 Conversion from
Mass to Moles Mass Moles Video on Mole calculation 28. Calculate
the number of moles in a) 6 x 1021 iron atoms b) 7.5 x 1023 of
water H2O molecules Calculate the number of moles in a) 4.5 x 1023
aluminium oxide, AI2O3 particles b) 7.2 x 1023 magnesium chloride,
MgCI2 particles Conversion from Number of Particles to Moles 29.
Calculate the number of moles in a) 6 x 1021 iron atoms b) 7.5 x
1023 of water H2O molecules Answer: a) 6 x 1023 Fe atoms 1 mole 6 x
1021 Fe atoms 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 Calculate the number of moles in a) 4.5 x 1023 aluminium
oxide, AI2O3 particles b) 7.2 x 1023 magnesium chloride, MgCI2
particles Answer: a) 6 x 1023 AI2O3 particles 1 mole 4.5 x 1023
AI2O3 particles 4.5 x 1023 6 x 1023 = 0.75 mol b) 6 x 1023 MgCI2
particles 1 mole 7.2 x 1023 MgCI2 particles 7.2 x 1023 6 x 1023 =
1.2 mol Conversion from Number of Particles to Moles Particles
Moles Video on Mole calculation 30. Calculate the number of
particles in a) 12.8g of copper atoms, Cu b) 8.5g of ammonia NH3
molecules Calculate the mass in a) 1.2 x 1022 zinc atoms b) 3 x
1023 ethanol, C2H5OH molecules Conversion from Mass to Number of
particles Mass Number particles Conversion from Number of particles
to Mass Moles Mass Moles Number particles 31. Calculate the number
of particles in a) 12.8g of copper atoms, Cu b) 8.5g of ammonia NH3
molecules Answer: a) 64g Cu atoms 1 mole 12.8g Cu atoms 1 x 12.8 =
0.2 mol 64 1 mol Cu 6 x 1021 Cu atoms 0.2 mol Cu 6 x 1021 x 0.2 =
1.2 x 1021 Cu atoms 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 molecules 0.5 mole NH3
0.5 x 6 x 1023 = 3 x 1023 NH3 molecules Calculate the mass in a)
1.2 x 1022 zinc atoms b) 3 x 1023 ethanol, C2H5OH molecules Answer:
a) 6 x 1023 Zn atoms 1 mole 1.2 x 1022 Zn atoms 1.2 x 1022 = 0.02
mol 6 x 1023 1 mol Zn atoms 65g o.o2 mol Zn atoms 0.02 x 65g = 1.3g
b) 6 x 1023 C2H5OH molecules 1 mole 3 x 1023 C2H5OH molecules 3 x
1023 = 0.5mol 6 x 1023 1 mole C2H5OH 46g 0.5 mole C2H5OH 46 x 0.5 =
23g Conversion from Mass to Number of particles Mass Number
particles Conversion from Number of particles to Mass Moles Mass
Moles Number particles 32. Calculate the number of particles in a)
0.75 mole of aluminium atoms b) 1.2 mole of chloride ions, CI- c)
0.07 mole of CO2 molecules Calculate the number of particles in a)
0.75 mole of aluminium oxide, AI2O3 b) 1.2 mole of magnesium
chloride, MgCI2 c) 0.07 mole of CO2 molecules Conversion from Moles
to Number particles 33. Calculate the number of particles in a)
0.75 mole of aluminium atoms b) 1.2 mole of chloride ions, CI- c)
0.07 mole of CO2 molecules Answer: a) 1 mole of AI 6 x 1023 AI
atoms 0.75 mole of AI 0.75 x 6 x 1023 = 4.5 x 1023 atoms b) 1 mole
CI- ions 6 x 1023 CI- ions 1.2 mole CI- ions 1.2 x 6 x 1023 = 7.2 x
1023 ions c) 1 mole of CO2 6 x 1023 CO2 molecules 0.07 mole of CO2
0.07 x 6 x 1023 = 4.2 x 1022 molecules Calculate the number of
particles in a) 0.75 mole of aluminium oxide, AI2O3 b) 1.2 mole of
magnesium chloride, MgCI2 c) 0.07 mole of CO2 molecules Answer: 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 particles 2 mole AI3+
ion and 3 mole O2- ion 4.5 x 1023 AI2O3 particles = 2 x 4.5 x 1023
AI3+ ions = 3 x 4.5 x 1023 O2- ions 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 c) 1
mole of CO2 6 x 1023 CO2 molecules 0.07 mole of CO2 0.42 x 1023 CO2
molecules 1 mole of CO2 particles 1 mole C atoms and 2 mole O atoms
0.42 x 1023 CO2 particles = 0.42 x 1023 C atoms = 2 x 0.42 x 1023 O
atoms Conversion from Moles to Number particles Moles Particles 34.
Chemical formula represent chemical compound show elements present
in compound Empirical formula represent simplest whole number ratio
of atoms of the elements formula obtained by experiment Molecular
formula represent actual number atoms of elements that combine to
form compound Structural formula represent arrangement of atoms in
compound Chemical Compound Ethene 35. 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 atoms of the elements formula obtained by
experiment Molecular formula represent actual number atoms of
elements that combine to form compound Structural formula represent
arrangement of atoms in compound Chemical Compound Ethene C1H1 C2H2
Video tutorial Molecular/empirical formula 36. Empirical Formula
Calculation Step 1: Write mass/ percentage of each element Step 2:
Calculate number of moles of each element (dividing with molar
mass/RAM) Step 3: Divide each by smallest number, obtain simplest
ratio Empirical Formula Calculation Relationship bet Molecular
Formula and Empirical 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 Empirical Formula Calculation
37. Empirical Formula Calculation Step 1: Write mass/ percentage of
each element Step 2: Calculate number of moles of each element
(dividing with molar mass/RAM) Step 3: Divide each by smallest
number, obtain simplest ratio Empirical Formula Calculation
Relationship bet Molecular Formula and Empirical Formula 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. Empirical Formula Calculation Molecular
formula = n x Empirical formula or RMM = n x formula mass of
Empirical formula 38. Empirical Formula Calculation Step 1: Write
mass/ percentage of each element Step 2: Calculate number of moles
of each element (dividing with molar mass/RAM) Step 3: Divide each
by smallest number, obtain simplest ratio Empirical Formula
Calculation Relationship bet Molecular Formula and Empirical
Formula 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. Empirical
Formula Calculation Molecular formula = n x Empirical formula or
RMM = n x formula mass of Empirical formula Video tutorial
Molecular/empirical formula 39. 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. Determine the
empirical formula of boric acid. Empirical Formula Calculation 1
Empirical Formula Calculation Step 1: Write the mass/ percentage of
each element Step 2: Calculate the number of moles of each element
(dividing with molar mass/RAM) Step 3: Divide each by smallest
number, obtain the simplest ratio 2.5 g of X combined with 4 g of Y
to form compound with formula XY2. If the RAM of Y is 80, determine
the relative atomic mass of X. 2 Element X Y Step 1 Mass/g 2.5 4
RAM/RMM a 80 Step 2 Number moles/mol 2.5/a 4/80 = 0.05 Step 3
Simplest ratio 1 2 40. 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. Determine the
empirical formula of boric acid. Empirical formula - H3B1O3.
Empirical Formula Calculation 1 Empirical Formula Calculation Step
1: Write the mass/ percentage of each element Step 2: Calculate the
number of moles of each element (dividing with molar mass/RAM) Step
3: Divide each by smallest number, obtain the simplest ratio 2.5 g
of X combined with 4 g of Y to form compound with formula XY2. If
the RAM of Y is 80, determine the relative atomic mass of X. 2
Element X Y Step 1 Mass/g 2.5 4 RAM/RMM a 80 Step 2 Number
moles/mol 2.5/a 4/80 = 0.05 Step 3 Simplest ratio 1 2 Empirical
formula given as X1Y2. A= 100 100 05.0 25.2 2 05.05.2 2 1 05.0 /5.2
x a a a Answer Answer 41. Gaseous hydrocarbon X contains 85.7% of
carbon by weight. 4.2 g of gas X occupy volume of 3.36 dm3 at stp.
a) Determine the empirical formula of X b) Determine the RMM of X
c) Determine the Molecular formula 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 Challenging Empirical Formula Calculation 3 Compound
X contain carbon, hydrogen and oxygen was analysed. 0.50g of
compound on complete combustion, yielded 0.6875g of carbon dioxide
and 0.5625g of water. Determine the empirical formula. 4 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 4.8/1.6 = 1 1.6/1.6 = 4 4.84/1.6 = 1 42.
Gaseous hydrocarbon X contains 85.7% of carbon by weight. 4.2 g of
gas X occupy volume of 3.36 dm3 at stp. a) Determine the empirical
formula of X b) Determine the RMM of X c) Determine the Molecular
formula 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) Volume of 3.36dm3 at stp Mass, 4.2g Volume of 22.4dm3 at
stp Mass, 4.2g x 22.4/3.36 = 28g RMM of X = 28 c) Assume molecular
formula of X - (CH2)n RMM of X is (12+2)n = 28 n = 2 Molecular
formula of X is C2H4 Challenging Empirical Formula Calculation 3
Answer Compound X contain carbon, hydrogen and oxygen was analysed.
0.50g of compound on complete combustion, yielded 0.6875g of carbon
dioxide and 0.5625g of water. Determine the empirical formula. 4
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 4.8/1.6 = 1 1.6/1.6 = 4 4.84/1.6 = 1
Conservation of mass Mass carbon atoms before = Mass carbon atoms
after Mass hydrogen atom before = Mass oxygen atoms after CHO + O2
CO2 + H2O Moles carbon atoms in CO2 = 0.6875 = 0.0156 mol 44 Mass
carbon = moles x RAM C atoms = 0.015625 x 12 = 0.1875g Moles
hydrogen atoms in H2O = 0.5625 = 0.03125 x 2 mol 18 = 0.0625 mol
Mass hydrogen = moles x RAM H atoms = 0.0625 x 1 = 0.0625g 0.6875g
0.5625g0.50g 0.75g Mass of oxygen atoms in CHO = (Mass CHO Mass C
Mass O) = 0.5 (0.1875 + 0.0625) = 0.25g Empirical formula
C1H4O1Answer 43. RAM /RMM calculation Relative Atomic Mass of
elements X, Y and Z are 12, 16 and 24. a) How much is an atom Z
heavier than an 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? Determine the (RMM) of
each of the following. a) V2O5 b) (NH4)2SO4 c) BaCI2 . 2H2O d)
C31H46O2 Relative formula mass of Y3(PO4)2 is 310 Determine the
relative atomic mass of Y (RAM: O =16, P =31) 1 2 3 Calculate the
percentage by weight of nitrogen in ammonium sulphate (NH4)2SO4 4
44. RAM /RMM calculation Relative Atomic Mass of elements X, Y and
Z are 12, 16 and 24. a) How much is an atom Z heavier than an 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? ANSWER a) Z is heavier than Y by 24/16 = 1.5
times b) Assume n atoms of X has the same mass as the sum of 3
atoms of Y and 2 atoms of Z. 12n = 3(16) + 2(24) 12n = 96 n = 96/12
= 8 atoms. Determine the (RMM) of each of the following. a) V2O5 b)
(NH4)2SO4 c) BaCI2 . 2H2O d) C31H46O2 ANSWER a) RMM for V2O5 = 2
(51) + 5 (16) = 182 b) RMM for (NH4)2SO4 = 2( 14+ 4) + 32 + 4(16) =
132 c) RMM for BaCI2 . 2H2O = 137 + 2(35.5) + 2(2 + 16) = 244 d)
RMM for C31H46O2 = 31(12) + 46(1) + 2(16) = 450 Relative formula
mass of Y3(PO4)2 is 310 Determine the relative atomic mass of Y
(RAM: O =16, P =31) ANSWER Assume RAM for Y = X RMM of Y3(PO4)2 =
310 3Y + 2 [ 31 + 4(16) ] = 310 3Y + 190 = 310 Y = (310 -190)/3 =
40 1 2 3 Calculate the percentage by weight of nitrogen in ammonium
sulphate (NH4)2SO4 ANSWER Total RAM for nitrogen = 14 x 2 = 28 RMM
(NH4)2SO4 = 132 % by weight of N = 28 x 100% 132 = 21.2% 4 45.
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
