http://lawrencekok.blogspot.com

Prepared by

Lawrence Kok

IB Chemistry Mole Concept, RAM, RMM

Isotopes and Empirical/Molecular Formula.

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 MoleOne 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

1 mole Fe contain- 6.02 x 1023 Fe atoms

Element Fe

MoleculeCO2

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

massor

RMM

massor

RAM

massMole

.

Mass for 1 mole of any substanceSymbol – M, Unit - g/mole

Molar mass contain 6.02 x 1023 particles1 mole atom – RAM in gram (1 mole C – 12 g)

1 Mole

1 Mole

1 Mole

Molar Mass

Molar Mass

Molar Mass

Relative Atomic Mass

C-12 as std1/12 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)

Relative Atomic Mass

C-12 as std1/12 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 H1

1

1

1

2 protons

2 unit

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 12C x % Ab) + (Mass 13C 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?

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 24Mg x % Ab) + (Mass 25Mg x % Ab) + (Mass 26Mg 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

Mass Spectrometer

Presence of isotopes and its abundance

Relative atomic mass of an element

Relative Molecular massof a molecule

Structure of organic compound

Distinguish betstructural isomers

structuralformula

Organic structure determination

24 25 26MgMg

CO2Mg Mg Mg Mg24

24 25 26

CH3CH2CH2OH

OHΙ

CH3CHCH3

CH3

|CH3C-CH3

|CH3

72

57

4227

1545432745

3129

% Ab % Ab % Ab

% Ab % Ab% Ab

Video on Mole calculation

12.00 gRAM in g

1.992 x 10-23 gToo small!!

Molar Mass

6 protons6 neutrons

Mass 1 Carbon atom(6 proton + 6 neutron)

1 proton/neutron = 1.66 x 10-24 g12 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 gToo small to weigh

Why Molar Mass is used?Why mass for 1 mole of carbon = (RAM)g

C12

C12

Video on Mole calculation

Chemical formulaRepresent 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 elementsFormula 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

Empirical Formula CalculationStep 1:- Write mass/% of each elementStep 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)

MolecularFormula (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 = 60Molecular 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 formulaor

RMM = n x formula mass of Empirical formula

Relationship bet MF and EF Empirical Formula Calculation

Relative formula mass of Y3(PO4)2 = 310Determine 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 = 28RMM (NH4)2SO4 = 132% by weight of N = 28 x 100%

132= 21.2%

Assume RAM for Y = XRMM Y3(PO4)2 = 3103Y + 2 [ 31 + 4(16) ] = 3103Y + 190 = 310Y = (310 -190)/3

= 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.Find EF of boric acid.

Empirical formula - H3B1O3.

Empirical Formula Calculation

Empirical Formula CalculationStep 1: - Write mass/ % of each elementStep 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

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 Numbermoles/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 gVol of 22.4dm3 at stp – Mass 1 mol (RMM)

3.36 dm3 – 4.2 g22.4 dm3 - (4.2 x 22.4)/3.36 = 28

c) Assume molecular formula of X - (CH2)n

RMM of X is (12+2)n = 28n = 2Molecular 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 0.015620.01562

= 1

0.06250.01562

= 4

0.015620.01562

= 1

Conservation of massMass C atom before = Mass C atom afterMass H atom before = Mass C atom after

Mol C atom in CO2

= 0.6875 = 0.0156 mol44

Mass C = mol x RAM Catom = 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 Hatom = 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

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

0.06090.0122

= 5

0.07300.0122

= 6

0.01220.0122

= 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 → Mass0.3 mol → 98.5 g

1 mol → 98.5/0.3RMM = 328 gmol-1

Assume MF - (C5H6O1)n = 328RMM = ( (5 x 12.01) +(6 x 1.01) + ( 1 x 16.01) )n = 32882.11 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 massMass C atom before = Mass C atom afterMass H atom before = Mass C atom after

Mol C atom in CO2

= 2.68 = 0.0609 mol44

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

P = 101 kNm-2

= 101 x 103 Nm-2

Calculate RMM of gas Mass empty flask = 25.385 gMass flask fill gas = 26.017 g

Mass flask fill water = 231.985 gTemp = 32C, P = 101 kPa

Find molar mass gas by direct weighing, T-23C , P- 97.7 kPaMass empty flask = 183.257 gMass flask + gas = 187.942 gMass flask + water = 987.560 gMass gas = (187.942 – 183.257) = 4.685 gVol gas = Vol water = Mass water = (987.560 – 183.257) = 804.303 cm3

RMM determination

PV = nRTPV = mass x R x T

MM = 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 RTPV

= 0.632 x 8.314 x 305101 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 0.003930.000393

= 10

0.007970.000393

= 20

0.0003930.000393

= 1

Conservation of massMass C atom before = Mass C atom afterMass H atom before = Mass C atom after

CHO + O2 CO2 + H2O

Mol C atom in CO2

= 0.1755 = 0.00393 mol44

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

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

massmol

r

00288.040.233

672.0

X2SO4 + BaCI2 ↔ BaSO4 + 2 XCI

174

00288.0

502.0

r

r

r

M

mol

massM

M

massmol

Mr - X2SO4 = 174

2X + 32 + 4(16) = 174X = 39 ( Potassium)

BaSO4

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

SO42-+ 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

massmol

r

00501.040.233

17.1

1 mol BaSO4 – 1 mol SO4

0.00501 mol BaSO4 – 0.00501 mol SO4

SO42-+ Ba2+ ↔ BaSO4

c. Find amt, of iron in 0.982 g, of Fe(NH4)2(SO4)2.xH2O

Fe(NH4)2(SO4)2. xH2O

molFeAmt

SOAmt

FeAmt

0025.02

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

massmol

r

r

14.085.550025.0

gmass

Mmolmass

M

massmol

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

massmol

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

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 Numbermoles/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 belowCrucible + 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 CuOOxide 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 + CuElement CuO O

Step 1 Mass 6.35 1.60

RAM/RMM 63.55 16.00

Step 2 Numbermoles/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

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%51140

07.302%.

%)(

).(% 52

52

P

OPOPM

PmassP

r

%17%202.94

10.392%.

%)(

).(% 2

2

K

OKOKM

KmassK

r

%18%1814

14%.

%)(

).(%

N

NNM

NmassN

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%10007.62

28%.

%100)(

).(%

22

N

NHCOM

NmassN

r

%2.21%10017.132

28%.

%100)(

).(%

424

N

SONHM

NmassN

r

%22.82%10004.17

14%.

%100)(

).(%

3

N

NHM

NmassN

r

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%10071.330

9.126%.

%100)(

).(%

2

I

CsICIM

ImassI

r

gIMass

I

0773.02015.0100

4.38.

%4.38%10071.330

9.126%.

41009.6

9.126

0773.0

mol

mol

M

massmol

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 .2H2Oc) 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

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)= 176g0.08 mole C6H8O6 → 0.08 x 176 g = 14.08g

c) 1 mole Mg(OH)2 → 24 + 2( 16 + 1) = 58g0.125 mole Mg(OH)2 → 0.125 x 58 g = 7.25g

1 mole C4H6N2 → 4(12) + 6(1) + 2(14) = 82 g0.005 mole C4H6N2→ 82 x 0.005 = 0.41g

Conversion Moles ↔ Mass

Find mass for a) 2/3 mole of AI atomb) 0.08 mole of C6H8O6 moleculesc) 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 mole0.97g C8H10N4O2→ 1 x 0.97 = 0.005 mol

194

Find number of moles ina) 6 x 1021 iron atomb) 7.5 x 1023 of H2O molecule

a) 6 x 1023 Fe atom → 1 mole6 x 1021 Fe atom → 6 x 1021

6 x 1023

= 0.01 mol

b) 6 x 1023 H2O molecules → 1 mole7.5 x 1023 H2O molecules → 7.5 x 1023

6 x 1023

= 1.25 mol

Find the moles ina) 4.5 x 1023 AI2O3 particlesb) 7.2 x 1023 MgCI2 particles

a) 6 x 1023 AI2O3 particle→ 1 mole4.5 x 1023 AI2O3 particle→ 4.5 x 1023

6 x 1023 = 0.75 molb) 6 x 1023 MgCI2 particle→ 1 mole

7.2 x 1023 MgCI2 particle→ 7.2 x 1023

6 x 1023 = 1.2 mol

a) 64 g Cu atom → 1 mole12.8 g Cu atom → 1 x 12.8 = 0.2 mol

64

1 mol Cu→ 6 x 1021 Cu atom0.2 mol Cu→ 6 x 1021 x 0.2 = 1.2 x 1021 Cu

b) 17g NH3 molecules → 1 mole8.5g NH3 molecules → 1 x 8.5 = 0.5 mol

171 mole NH3 → 6 x 1023 NH3 molecule0.5 mole NH3 → 0.5 x 6 x 1023

= 3 x 1023 NH3 molecule

a) 6 x 1023 Zn atom→ 1 mole1.2 x 1022 Zn atom→ 1.2 x 1022 = 0.02 mol

6 x 1023

1 mol Zn atom→ 65go.o2 mol Zn atom→ 0.02 x 65g = 1.3g

b) 6 x 1023 C2H5OH molecule→ 1 mole3 x 1023 C2H5OH molecule→ 3 x 1023 =0.5mol

6 x 1023

1 mole C2H5OH → 46g0.5 mole C2H5OH → 46 x 0.5 = 23g

Mass

Conversion Mol ↔ Mass ↔ Particles

Find number of particles ina) 12.8g of Cub) 8.5g of NH3 molecules

Find the mass ina) 1.2 x 1022 zinc atomb) 3 x 1023 C2H5OH molecules

Find number of particles ina) 0.75 mole of AI atomb) 1.2 mole of CI- ionc) 0.07 mole of CO2 molecule

a) 1 mole of AI → 6 x 1023 AI atom0.75 mole of AI → 0.75 x 6 x 1023

= 4.5 x 1023 atom

b) 1 mole CI- ions → 6 x 1023 CI- ion1.2 mole CI- ions → 1.2 x 6 x 1023

= 7.2 x 1023 ion

c) 1 mole of CO2 → 6 x 1023 CO2 molecule0.07 mole of CO2 → 0.07 x 6 x 1023

= 4.2 x 1022 molecule

Find number of particles ina) 0.75 mol of AI2O3

b) 1.2 mol of MgCI2

a) 1 mole of AI2O3 →6 x 1023 AI2O3 particles0.75 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- ion4.5 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 particles1.2 mole MgCI2 → 7.2 x 1023 MgCI2 particles

1 mole MgCI2 particles → 1 mole Mg2+ and 2 mole CI- ions7.2 x 1023 MgCI2 particles = 7.2 x 1023 Mg2+ ions

= 2 x 7.2 x 1023 CI- ions

Acknowledgements

Thanks to source of pictures and video used in this presentation

Thanks to Creative Commons for excellent contribution on licenseshttp://creativecommons.org/licenses/

Prepared by Lawrence Kok

Check out more video tutorials from my site and hope you enjoy this tutorialhttp://lawrencekok.blogspot.com