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
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
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)
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
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 12C • 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 12C • 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 35CI x % Abundance) + (mass 37CI 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
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 12C x % Abundance) + (Mass 13C 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
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 24Mg x % Abundance) + (Mass 25Mg x % Abundance) + (Mass 26Mg 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
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 particles Relationship between Mole – Mass – Number particles
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
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
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
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
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
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
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
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
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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.
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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
10005.0
25.2
2
05.05.2
2
1
05.0
/5.2
xa
a
a
Answer
Answer
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 0.5625g 0.50g 0.75g
Mass of oxygen atoms in CHO = (Mass CHO – Mass C – Mass O) = 0.5 – (0.1875 + 0.0625) = 0.25g
Empirical formula – C1H4O1 Answer
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%
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