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

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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)

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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)

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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

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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

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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

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

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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