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1Pb(NO3)2(aq) + 2KI(aq) → 1PbI2(s) + 2KNO3(aq) Lead + Potassium → Lead + Potassium Nitrate iodide iodide nitrate
Chemical Eqn
Reactant – Left Product – Right Conservation Mass
Total Mass reactant = Total Mass product
Mole Ratio – Coefficient of reactant/product
1Pb(NO3)2(aq) + 2KI(aq) → 1PbI2(s) + 2KNO3(aq)
1 : 2 → 1 : 2
Word equation Chemical formula
Calcium + hydrochloric → Calcium + carbon + water carbonate acid chloride dioxide
1CaCO3(s) + 2HCI(aq) → 1CaCI2(aq) + 1CO2(g) + 1H2O(l)
Product – Right Reactant – Left Conservation Mass Total Mass reactant = Total Mass product
1 : 2 → 1 : 1 : 1 Mole Ratio - Coefficient of reactant/product
Physical state/ symbol (s) – solid (I) - liq (g) – gas (aq) – aqueous ∆ - heat ppt – precipitate/solid ↔ - reversible
Physical state/ symbol (s) – solid (I) - liq (g) – gas (aq) – aqueous ∆ - heat ppt – precipitate/solid ↔ - reversible
Mass reactants (Pb(NO3)2 + KI) = 15.82
Mass products (PbI3 + KNO3) = 15.82
After Before
Chemical rxn Matter is neither created nor destroyed Undergoes physical/chemical change.
LAW of conservation of mass.
Mole proportion/ratio (reactant) → (product) 1 : 2 → 1 : 2
Concept Map
represented by
1Pb(NO3)2(s) + 2NaI(aq) → 1PbI2(s) + 2NaNO3 (aq)
Limiting reactant Use up first
Limit products form Rxn stop if all used up
Excess reactant left over
remains behind
Percentage Yield mass of Actual Yield x 100% mass of Theoretical Yield
Theoretical yield Max amt prod form if rxn complete
Stoichiometry ratio Assume all limiting reagent used up
Actual yield Amt of prod formed experimentally
Less than theoretical yield due to experimental error
Rxn Stoichiometry Quantitative relationship bet quantities react/ prod
Find quantities/amt (mass, mole, vol) Predict how much react and amt prod form
Chemical rxn react in definite ratio
Chemical Change
Chemical Equation
Balanced Chemical equation
Molecular Eqn
Complete Ionic Eqn
Net Ionic Eqn
1Pb(NO3)2(s) + 2NaI(aq) → 1PbI2(s) + 2NaNO3 (aq)
1Pb2+(aq) + 2NO3
-(aq) + 2Na+
(aq) + 2I-(aq) → 1PbI2(s) + 2Na+
(aq) + 2NO3-(aq)
1Pb2+(aq) + 2CI-
(aq) → 1PbCI2(s)
Limiting and Excess
Which is limiting and excess ?
How many hot dog with 6 bun and 3 hot dog?
Stoichiometric ratio 1 mol (bun) : 1 mol (hot dog) → 1 mol
+ 5 5 5
+
No Excess No limiting
Excess - Bun Limiting - Hot dog are used up
Both hot dog and bun used up
How many burger with 12 bun and 6 patties?
+ +
Stoichiometric ratio 2 mol (bun) : 1 mol (burger) → 1 mol
No Excess No limiting
Limiting reactant Use up first, limit the prod form
Rxn stop if all used up
Excess reactant Left over, remain behind
1Zn (s) + 2HCI (aq) → 1ZnCI2(aq) + 1H2(g)
Mole ratio
1 : 2 → 1: 1
Moles reactant given, which is limiting and excess ?
Which is limiting and excess ?
1st method
2nd method
0.30 mol Zn + 0.52 mol HCl add
1 mol Zn → 2 mol HCI 0.30 mol Zn → 2 x 0.30 mol HCI = 0.60 mol HCI needed = 0.52 mol HCI added (HCI – limiting)
0.52 mol HCI
0.30 mol Zn
Reactant that produce least amt product → will be limiting
Assume Zn limiting 1 mol Zn → 1 mol H2 gas 0.3 mol Zn → 1 x 0.3 = 0.3 mol H2
Assume HCI limiting 2 mol HCI → 1 mol H2 gas 0.52 mol HCI → 1 x 0.52 = 0.26 mol H2
2
Simulation on limiting/excess
1Pb(NO3)2(s) + 2NaI(aq) → 1PbI2(s) + 2NaNO3 (aq)
Mole ratio
1 : 2 → 1: 2
Simulation on limiting/excess
10 g Pb(NO3)2 + 10 g NaI added
0.0302 mol Pb(NO3)2 + 0.0667 mol NaI
Which is limiting and excess ?
1 mol Pb(NO3)2 → 2 mol NaI 0.0302 mol Pb(NO3)2 → 2 x 0.0302 mol NaI = 0.0604 mol NaI needed = 0.0667 mol NaI add (NaI excess)
Mass = 10.0 RMM 149.9 = 0.0667 mol
Mass = 10.0 RMM 331.2 = 0.0302 mol
1st method
2nd method
Reactant that produce least amt product → will be limiting
Assume Pb(NO3)2 limiting 1 mol Pb(NO3)2→ 1 mol PbI2 0.0302 mol Pb(NO3)2→ 1x0.0302 mol PbI2
= 0.0302 mol PbI2
Assume NaI limiting 2 mol NaI → 1 mol PbI2 0.0667 mol NaI → 1 x 0.0667 = 0.0334 mol PbI2
2
3rd method
Mole ratio method
1 : 2
57.052.0
3.0
).(
).(
5.02
1
).(
).(
HCIMole
ZnMole
HCIMole
ZnMolefrom eqn
given mass
Ratio higher ↓
Zn excess/HCI limit
1 : 2 → 1 : 1
Limiting and excess ?
Which is limiting and excess ?
1st method
2nd method
1 mol Mg → 2 mol HCI 0.0256 mol Mg → 0.0512 mol HCI = 0.0512 mol HCI needed = 0.0341 mol HCI added (HCI – limiting)
27.3 ml, 1.25M HCI 0.623 g Mg
Reactant that produce least amt product → will be limiting
Assume Mg limiting 1 mol Mg → 1 mol H2 0.0256 mol Mg → 0.0256 mol H2
Assume HCI limiting 2 mol HCI → 1 mol H2 0.0341 mol HCI → 0.01705 mol H2
Simulation on limiting/excess
Simulation on limiting/excess
0.02 mol NaOH + 0.025 mol H2SO4
Which is limiting and excess ?
2 mol NaOH → 1 mol H2SO4
0.02 mol NaOH → 0.01 mol H2SO4
= 0.01 mol H2SO4 needed = 0.025 mol H2SO4 add (H2SO4 excess)
1st method
2nd method
Reactant that produce least amt product → will be limiting
Assume NaOH limiting 2 mol NaOH → 1 mol H2O 0.02 mol NaOH → 0.01 mol H2O
Assume H2SO4 limiting 1 mol H2SO4 → 1 mol H2O 0.025 mol H2SO4 → = 0.025 mol H2O
Mg(s) + 2HCI(aq) → MgCI2(aq) + H2 (g)
0.623 g Mg + 27.3 ml, 1.25M HCI add
Mass = 0.623 RMM 24 = 0.0256 mol
Mol = M x V 1000 = 1.25 x 0.0273 = 0.0341 mol
0.0256 mol Mg + 0.0341 mol HCI
2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + H2O(l)
1 : 2 → 1 : 1 2 : 1 → 1 : 1
100 ml, 0.2M, NaOH + 50 ml, 0.5M H2SO4 add
Mol = M x V 1000 = 0.2 x 0.1 = 0.02 mol
Mol = M x V 1000 = 0.5 x 0.05 = 0.025 mol
3rd method
Mole ratio method
1 : 2
75.00341.0
0256.0
).(
).(
5.02
1
).(
).(
HCIMole
MgMole
HCIMole
MgMolefrom eqn
given mass
Ratio higher ↓
Mg excess/HCI limit
Limiting and excess ?
Which is limiting and excess ?
1st method
2nd method
2 mol CO → 1 mol O2
2 mol CO → 1 mol O2
= 1 mol O2 needed = 0.5 mol O2 added (O2 – limiting)
Reactant that produce least amt product → will be limiting
Assume CO limiting 2 mol CO → 2 mol CO2 2 mol CO → 2 mol CO2
Assume O2 limiting 1 mol O2 → 2 mol CO2 0.5 mol O2 → 1 mol CO2
Simulation on limiting/excess
Simulation on limiting/excess
0.02 mol NaOH + 0.025 mol H2SO4
Which is limiting and excess ?
2 mol NaOH → 1 mol H2SO4
0.02 mol NaOH → 0.01 mol H2SO4
= 0.01 mol H2SO4 needed = 0.025 mol H2SO4 add (H2SO4 excess)
1st method
2nd method
Reactant that produce least amt product → will be limiting
Assume NaOH limiting 2 mol NaOH → 1 mol H2O 0.02 mol NaOH → 0.01 mol H2O
Assume H2SO4 limiting 1 mol H2SO4 → 1 mol H2O 0.025 mol H2SO4 → = 0.025 mol H2O
Mol = Volume Molar vol = 45.42 = 2 mol 22.4
Mol = Volume Molar vol = 11.36 = 0.5 mol 22.4
2 mol CO + 0.5 mol O2
2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + H2O(l)
2 : 1 → 2 2 : 1 → 1 : 1
100ml, 0.2M, NaOH + 50 ml, 0.5M H2SO4 add
Mol = M x V 1000 = 0.2 x 0.1 = 0.02 mol
Mol = M x V 1000 = 0.5 x 0.05 = 0.025 mol
2CO(g) + 1O2(g) → 2CO2 (g)
45.42 L CO + 11.36 L O2 add
3rd method
Mole ratio method
2 : 1
45.0
2
).(
).(
21
2
).(
).(
2
2
OMole
COMole
OMole
COMolefrom eqn
given mass
Ratio higher ↓
CO excess/O2 limit
Which is limiting and excess ?
1st method
2nd method
0.30 mol Zn + 0.52 mol HCl add
1 mol Zn → 2 mol HCI 0.30 mol Zn → 2 x 0.30 mol HCI = 0.60 mol HCI needed = 0.52 mol HCI added (HCI – limiting)
0.52 mol HCI
0.30 mol Zn
Reactant that produce least amt product → will be limiting
Assume Zn limiting 1 mol Zn → 1 mol H2 gas 0.3 mol Zn → 1 x 0.3 = 0.3 mol H2
Assume HCI limiting 2 mol HCI → 1 mol H2 gas 0.52 mol HCI → 1 x 0.52 = 0.26 mol H2
2
Simulation on limiting/excess
Simulation on limiting/excess
3rd method
Mole ratio method
1 : 2
57.052.0
3.0
).(
).(
5.02
1
).(
).(
HCIMole
ZnMole
HCIMole
ZnMolefrom eqn
given mass
Ratio higher ↓
Zn excess/HCI limit
Theoretical, Actual and % Yield
1Zn (s) + 2HCI (aq) → 1ZnCI2(aq) + 1H2(g)
1 : 2 → 1 : 1
Mole ratio
1 : 2 → 1: 1
Find theoretical yield, cm3 for H2 gas Find % yield if expt yield is 5800 cm3
%2.98%1005902
5800%
%100.
.exp%.
yield
yieldltheoretica
yieldtyield
HCI limiting ↓
Mole ratio 2 mol HCI : 1 mol H2
1Zn + 2HCI → 1ZnCI2 + 1H2
2 mol HCI → 1 mol H2
0.52 mol HCI→ 0.26 mol H2
Theoretical yield 1 mol H2 – 22700 cm3
0.26 mol H2 – 5902 cm3
Expt yield = 5800 cm3
Click here tutorial on austute
Which is limiting and excess ?
1st method
2nd method
1 mol Mg → 2 mol HCI 0.0256 mol Mg → 0.0512 mol HCI = 0.0512 mol HCI needed = 0.0341 mol HCI added (HCI – limiting)
27.3 ml, 1.25M HCI 0.623 g Mg
Reactant that produce least amt product → will be limiting
Assume Mg limiting 1 mol Mg → 1 mol H2 0.0256 mol Mg → 0.0256 mol H2
Assume HCI limiting 2 mol HCI → 1 mol H2 0.0341 mol HCI → 0.01705 mol H2
Simulation on limiting/excess
Simulation on limiting/excess
Mg(s) + 2HCI(aq) → MgCI2(aq) + H2 (g)
0.623 g Mg + 27.3 ml, 1.25M HCI add
Mass = 0.623 RMM 24 = 0.0256 mol
Mol = M x V 1000 = 1.25 x 0.0273 = 0.0341 mol
0.0256 mol Mg + 0.0341 mol HCI
1 : 2 → 1 : 1
3rd method
Mole ratio method
1 : 2
75.00341.0
0256.0
).(
).(
5.02
1
).(
).(
HCIMole
MgMole
HCIMole
MgMolefrom eqn
given mass
Ratio higher ↓
Mg excess/HCI limit
Theoretical, Actual and % Yield Find theoretical yield, cm3 for H2 gas Find % yield if expt yield is 300 cm3
Mg + 2HCI → MgCI2 + H2
HCI limiting ↓
Mole ratio 2 mol HCI : 1 mol H2
2 mol HCI → 1 mol H2
0.0341 mol HCI→ 0.01705 mol H2
Theoretical yield 1 mol H2 – 22700 cm3
0.01705 mol H2 – 387 cm3
Expt yield = 5800 cm3
%5.77%100387
300%
%100.
.exp%.
yield
yieldltheoretica
yieldtyield
Click here tutorial on chemwiki
Which is limiting and excess ?
1st method
2nd method
2 mol CO → 1 mol O2
2 mol CO → 1 mol O2
= 1 mol O2 needed = 0.5 mol O2 added (O2 – limiting)
Reactant that produce least amt product → will be limiting
Assume CO limiting 2 mol CO → 2 mol CO2 2 mol CO → 2 mol CO2
Assume O2 limiting 1 mol O2 → 2 mol CO2 0.5 mol O2 → 1 mol CO2
Simulation on limiting/excess
Simulation on limiting/excess
Mol = Volume Molar vol = 45.42 = 2 mol 22.4
Mol = Volume Molar vol = 11.36 = 0.5 mol 22.4
2 mol CO + 0.5 mol O2
2 : 1 → 2
2CO(g) + 1O2(g) → 2CO2 (g)
45.42 L CO + 11.36 L O2 add
3rd method
Mole ratio method
2 : 1
45.0
2
).(
).(
21
2
).(
).(
2
2
OMole
COMole
OMole
COMolefrom eqn
given mass
Ratio higher ↓
CO excess/O2 limit
Theoretical, Actual and % Yield Find theoretical yield, g for CO2 gas
Find % yield if expt yield is 30 g
2CO + 1O2 → 2CO2
O2 limiting ↓
Mole ratio 1 mol O2 : 2 mol CO2
1 mol O2 → 2 mol CO2
0.5 mol O2 → 1 mol CO2
Theoretical yield 1 mol CO2 – 44 g
%1.68%10044
30%
%100.
.exp%.
yield
yieldltheoretica
yieldtyield
Expt yield = 30 g
Click here tutorial on chemtamu
1st method
2nd method
1 mol CH3COOH → 1 mol C5H11OH 0.0596 CH3COOH → 0.0596 C5H11OH = 0.0596 C5H11OH need = 0.0539 C5H11OH add (C5H11OH – limiting)
0.0539 mol C5H11OH
0.0596 mol CH3COOH
Simulation on limiting/excess
Simulation on limiting/excess
Mole ratio method
1 : 1
1.10539.0
0596.0
).(
).(
1
1
).(
).(
115
3
115
3
OHHCMole
COOHCHMole
OHHCMole
COOHCHMolefrom eqn
given mass
Ratio higher ↓
CH3COOH excess/C5H11OH limit
Theoretical, Actual and % Yield
CH3COOH + C5H11OH → Ester + H2O
1 : 1 → 1 : 1
Mole ratio
1 : 1 → 1: 1
Find mass of ester forms if it has 45% yield
gyieldt
yieldltheoretica
yieldtyield
15.3%1007
.exp%45
%100.
.exp%.
C5H11OH limiting ↓
Mole ratio 1 mol C5H11OH : 1 mol Ester
1 mol C5H11OH → 1 mol Ester
0.0539 mol C5H11OH→ 0.0539 Ester
Theoretical yield 1 mol Ester - 130 g
0.0539 mol Ester – 7 g
Expt yield = ????
Click here tutorial on austute
Which is limiting and excess ?
CH3COOH + C5H11OH → Ester + H2O
1st method
2nd method
1 mol CH3COOH → 1 mol C5H11OH 0.0596 CH3COOH → 0.0596 C5H11OH = 0.0596 C5H11OH need = 0.0539 C5H11OH add (C5H11OH – limiting)
0.0539 mol C5H11OH
0.0596 mol CH3COOH
Simulation on limiting/excess
Simulation on limiting/excess
Mole ratio method
1 : 1
1.10539.0
0596.0
).(
).(
1
1
).(
).(
115
3
115
3
OHHCMole
COOHCHMole
OHHCMole
COOHCHMolefrom eqn
given mass
Ratio higher ↓
CH3COOH excess/C5H11OH limit
Theoretical, Actual and % Yield
CH3COOH + C5H11OH → Ester + H2O
1 : 1 → 1 : 1
Mole ratio
1 : 1 → 1: 1
Find mole of C5H11OH used if 0.888 mol Ester is needed with a 65% yield
molx
yieldltheoretica
yieldtyield
37.1%100888.0
%65
%100.
.exp%.
Mole ratio 1 mol C5H11OH : 1 mol Ester
1 mol C5H11OH → 1 mol Ester
0.888 mol C5H11OH → 0.888 Ester
Expt yield = 0.888 mol
Click here tutorial on austute
Which is limiting and excess ?
CH3COOH + C5H11OH → Ester + H2O
Aspirin, widely used drugs, prepared below Student reacted salicylic acid with excess ethanoic anhydride. Impure solid aspirin was obtained by filtering. Pure aspirin was obtained by recrystallization.
Find amt, mol, of salicylic acid, C6H4(OH)COOH, used.
Find theoretical yield, in g, of aspirin, C6H4(OCOCH3)COOH.
Find % yield of pure aspirin.
Find % uncertainty in mass of aspirin.
Salicylic acid ethanoic anhydride aspirin
Mass salicylic acid 3.15 ± 0.02g
Mass pure aspirin 2.50 ± 0.02g
%8.60%10011.4
50.2%
%100.
.exp%.
yield
yieldltheoretica
yieldtyield
molMole
M
gMassMole
r
0228.013.138
15.3
).(
Mole ratio 1 mol salicylic acid : 1 mol aspirin
0.0228 mol salicylic acid : 0.0228 mol aspirin
gMass
MMoleMass
M
gMassMole
r
r
11.47.1800228.0
).(
%80.0int%.
%10050.2
02.0int%.
yuncerta
yuncerta
limiting
100 g zinc react with 100 g of iodine producing zinc iodide.
Zn + I2 → ZnI2
Find mass of ZnI2 produced
Find, amt of Zn and I2, and determine which reactant is in excess
molZnMole
M
gMassZnMole
r
53.137.65
100).(
).().(
molIMole
M
gMassIMole
r
394.08.253
100).(
).().(
2
2
Mole ratio 1 mol I2 : 1 mol Zn
0.394 mol I2 : 0.394 mol Zn
I2 – limiting Zn - excess
Mole – Zn Mole I2
Mole ratio 1 mol I2 : 1 mol ZnI2
0.394 mol I2 : 0.394 mol ZnI2
gZnIMass
MMoleZnIMass
M
gMassZnIMole
r
r
8.125319394.0).(
).(
).().(
2
2
2
Copper metal produced by copper(I) oxide and copper(I) sulfide shown below Mixture of 10 kg of copper(I) oxide and 5 kg of copper(I) sulfide was heated.
2Cu2O + Cu2S → 6Cu + SO2
Find limiting reagent.
Mole – Cu2O Mole Cu2S
molOCuMole
M
gMassOCuMole
r
9.69143
10000).(
).().(
2
2
molSCuMole
M
gMassSCuMole
r
4.31159
5000).(
).().(
2
2
Find maximum mass Cu produced
Mole ratio 2 mol Cu2O : 1 mol Cu2S
69.9 mol Cu2O: 35 mol Cu2S
Cu2S – limiting Cu2O - excess
Mole ratio 1 mol Cu2S : 6 mol Cu
31.4 mol Cu2S : 188 mol Cu
gCuMass
MMoleCuMass
M
gMassCuMole
r
r
1190055.63188).(
).(
).().(
Student determine Mr of solid monoprotic acid, HA, by titrating with a known mass of acid. Data shown below.
Find mass of acid and determine its absolute and % uncertainty
Known mass of acid, HA, was dissolved in water to form a 100ml sol in volumetric flask. 25 ml sample of sol reacted with 12.1 ml of 0.1M NaOH. Find molar mass of acid.
Mass bottle 1.737 ± 0.001 g
Mass bottle + acid, HA 2.412 ± 0.001 g
Mass acid = ( 2.412 – 1.737) ± 0.002 = (0.675 ±0.002) g % uncertainty = (0.002) x 100% = 0.3% 0.675
NaOH + HA → NaA + H2O
NaOH + HA → NaA + H2O M = 0.1M M = ? V = 12.1 ml V = 25 ml
MM
M
VM
VM
a
a
aa
bb
0484.0
1
1
25
1.121.0
1
1
? HA 100 ml water added
25 ml transfer
NaOH M = 0.1M V = 12.1 ml
HA
M = ?
Amt acid in 1000 ml = 4.84 x 10-2 mol Amt acid in 100 ml = 4.84 x 10 -3 mol
1391084.4
675.0
).(
3
r
r
M
M
gMassMole
Cations/Metals/+ve ions
Gp 1 Gp 2 Gp 3 Transition metals ions ( variable oxidation states)
Oxidation state
+1
Oxidation
state
+2
Oxidation state
+3
Sc +3
Ti +2 +3
V +2 +3
Cr +2 +3 +6
Mn +2 +3 +6 +7
Fe +2 +3
Co +2 +3
Ni +2
Cu +1 +2
Zn +2
Li 1+ Be2+ Sc 3+ Ti 2+
Ti 3+
V 2+
V 3+
Cr 2+
C r3+
Cr6+
Mn 2+
Mn 3+
Mn 6+
Mn 7+
Fe 2+
Fe 3+
Co2+
Co 3+
Ni2+ Cu1+
Cu2+
Zn2+
Na 1+ Mg2+ Al 3+
K 1+ Ca2+
Anion/Non metal
Gp 5 Gp 6 Gp 7
Oxidation state
Oxidation state
Oxidation state
-3 -2 -1
N3- O2- F-1
P3- S2- CI-1
Br-1
I-1
Ionic Compound
Li2O MgCI2 Al2O3 FeO Iron(II) oxide
NiO Nickel(II) oxide
CuO Copper(II) oxide
Li3N Mg3N2 AlN Fe3N2
Iron(II) nitride Ni3N2
Nickel(II) nitride Cu3N2
Copper(II) nitride
Oxidation state/Charge ion → Li1+ O2-
Formula compound Li2 O1
Video on polyatomic ions
Writing Chemical Formula
Step 1 : Write Oxidation state/charge
Step 2 : Balance it, (electrically neutral) by cross multiply – as subscript
Metal/Cations/+ve ion Non Metal/
Anion/-ve ion
Polyatomic ions
Group of non-metals bonded together
Oxidation state
Oxidation state
Oxidation state
-1/+1 -2 -3
(OH)-1
Hydroxide (SO4)2-
Sulphate (PO4)3-
Phosphate
(CN)-1
Cyanide (SO3)
2-
Sulphite
(SCN)-1
Thiocyanate (CO3)
2-
Carbonate
(NO3)-1
Nitrate (S2O3)
2-
Thiosulphate
(NO2)-1
Nitrite (Cr2O7)
2-
Dichromate
(NH4)+1
Ammonium
Li2(CO3) Mg(CO3) Al2(CO3)3 Fe(CO3) Ni(CO3) Cu(CO3)
Li(OH) Mg(OH)2 Al(OH)3 Fe(OH)2 Ni(OH)2 Cu(OH)2
Li2(SO4) Mg(SO4) Al2(SO4)3 FeSO4 Ni(SO4) Cu(SO4) Video on polyatomic ions
Ionic Compound
Cations/Metals/+ve ions
Gp1 Gp 2 Gp3 Transition metals ions ( variable oxidation states)
Oxidation state
+1
Oxidation
state
+2
Oxidation state
+3
Sc +3
Ti +2 +3
V +2 +3
Cr +2 +3 +6
Mn +2 +3 +6 +7
Fe +2 +3
Co +2 +3
Ni +2
Cu +1 +2
Li 1+ Be2+ Sc 3+ Ti 2+
Ti 3+
V 2+
V 3+
Cr 2+
C r3+
Cr6+
Mn 2+
Mn 3+
Mn 6+
Mn 7+
Fe 2+
Fe 3+
Co2+
Co3+
Ni2+ Cu1+
Cu2+
Na 1+ Mg2+ Al 3+
K 1+ Ca2+
Oxidation state/Charge ion → Li1+ (CO3)2-
Formula compound Li2 (CO3)1
Step 1 : Write Oxidation state/charge ion
Step 2 : Balance it, (electrically neutral) by cross multiply – as subscript
Writing Chemical Formula
Metal/Cations/+ve ion Polyatomic ions
Acids Alkali Metal Hydroxide
Metal oxides Salts Gas
HCI Hydrochloric acid
KOH Potassium hydroxide
CuO Copper(II) oxide
CaCO3
Calcium carbonate CO
Carbon monoxide
HNO3
Nitric acid NaOH
Sodium hydroxide MgO
Magnesium oxide Na2CO3
Sodium carbonate CO2
Carbon dioxide
H2SO3
Sulphurous acid Ca(OH)2
Calcium Hydroxide ZnO
Zinc oxide NaHCO3
Sodium bicarbonate SO2
Sulphur dioxide
HCOOH Methanoic acid
NH3
Ammonia Na2O
Sodium oxide KNO3
Potassium nitrate SO3
Sulphur trioxide
CH3COOH Ethanoic acid
Mg(OH)2
Magnesium hydroxide Al2O3
Aluminium oxide Pb(NO3)2
Lead (II) Nitrate NO2
Nitrogen dioxide
H3PO4
Phosphoric acid Cu(OH)2
Copper (II) hydroxide Fe2O3
Iron(III) oxide NaNO3
Sodium nitrate CH4
Methane
H2CO3
Carbonic acid Al(OH)3
Aluminium hydroxide K2S
Potassium sulphide PbI2
Lead (II) nitrate H2S
Hydrogen sulphide
HNO2
Nitrous acid Fe(OH)2
Iron (II) hydroxide PbS
Lead(II) sulphide AgCI
Silver chloride O2
Oxygen
HF Hydrofluoric acid
Fe(OH)3
Iron (III) hydroxide ZnS
Zinc sulphide MgSO4
Magnesium sulphate N2
Nitrogen
HCIO Hypochlorous acid
Zn(OH)2
Zinc hydroxide AI2S3
Aluminium sulphide Na2S2O3
Sodium thiosulphate CI2
Chlorine
Chemical Formula for common chemicals
Naming chemical compound Writing chemical formula Writing chemical formula
VIDEO TUTORIALS
Chemical Eqn
1Pb(NO3)2(aq) + 2NaCI(aq) → 1PbCI2(s) + 2NaNO3(aq)
1Pb2+(aq) + 2NO3
-(aq) + 2Na+
(aq) + 2CI-(aq) → 1PbCI2(s) + 2Na+
(aq) + 2NO3-(aq)
unchanged
1Pb2+(aq) + 2CI-
(aq) → 1PbCI2(s)
1Pb2+(aq) + 2NO3
-(aq) + 2Na+
(aq) + 2CI-(aq) → 1PbCI2(s )+ 2Na+
(aq) + 2NO3-(aq)
Break aq → ions
Cancel out Cancel out
Spectators ions- don’t participate in rxn Cancel out ions from both sides of eqn
Only ions involved in rxn
Net ionic eqn Complete ionic eqn Molecular eqn
Break down electrolytes, (aq) → ion Leave sol, liq, gas unchanged
Molecular eqn
Complete ionic eqn
Net ionic eqn
Na2CO3(aq) + 2HNO3(aq) → 2NaNO3(aq) + H2O(l) + CO2(g)
2Na+(aq) + CO3
2-(aq) + 2H+
(aq) + 2NO3-(aq) → 2Na+
(aq) + 2NO3-(aq) + H2O(l) + CO2 (g)
2Na+(aq) + CO3
2-(aq) + 2H+
(aq) + 2NO3-(aq) → 2Na+
(aq) + 2NO3-(aq) + H2O(l) + CO2 (g)
CO32-
(aq) + 2H+(aq) →H2O(l) + CO2(g)
Break aq → ions
Net ionic eqn
Complete ionic eqn
Molecular eqn
Cancel out
Chemical Eqn
Cancel out
Net ionic eqn Complete ionic eqn Molecular eqn
Molecular eqn
Complete ionic eqn
Net ionic eqn
Net ionic eqn
Complete ionic eqn
Molecular eqn
2 Na3PO4(aq) + 3CaCI2(aq) → 6NaCI(aq) + Ca3(PO4)2(s)
6Na+(aq) + 2PO4
3-(aq) + 3Ca2
(aq) + 6CI-(aq) → 6Na+
(aq) + 6CI-(aq) + Ca3(PO4)2(s)
6Na+(aq) + 2PO4
3-(aq) + 3Ca2
(aq) + 6CI-(aq) → 6Na+
(aq) + 6CI-(aq) + Ca3(PO4)2(s)
2PO43-
(aq) + 3Ca2+(aq) → Ca3(PO4)2(s)
Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)
Zn(s) + Cu2+(aq) + SO4
2-(aq) → Zn2+
(aq) + SO42-
(aq) + Cu(s)
Zn(s) + Cu2+(aq) + SO4
2- (aq) → Zn2+
(aq) + SO42-
(aq) + Cu(s)
Zn (s) + Cu2+(aq) → Zn2+
(aq) + Cu(s)
Cancel out
SO42-
(aq) + Ba2+(aq) → BaSO4 (s)
2K+(aq) + SO4
2-(aq) + Ba2+
(aq) + 2Cl-(aq) → BaSO4(s) + 2K+
(aq) + 2Cl-(aq)
2K+(aq) + SO4
2-(aq) + Ba2+
(aq) + 2Cl-(aq) →BaSO4(s) + 2K+
(aq) + 2Cl-(aq)
Mg(s) + 2H+(aq) +2Cl-
(aq) → Mg2+(aq) + 2Cl-
(aq) + H2(g)
Mg(s) + 2H+(aq) + 2Cl-
(aq) → Mg2+(aq) + 2Cl-
(aq) +H2(g)
Cancel out
Chemical Eqn
Cancel out
Net ionic eqn Complete ionic eqn Molecular eqn
Molecular eqn
Complete ionic eqn
Net ionic eqn
Net ionic eqn
Complete ionic eqn
Molecular eqn
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
Mg(s) + 2H+(aq) → Mg2+
(aq) + H2(g)
K2SO4(aq) + BaCl2 (aq) → BaSO4(s) + 2KCl(aq)
3CO32-
(aq) + 2Al3+(aq) → Al2(CO3)3(s)
6NH4+
(aq) + 3CO32-
(aq) + 2Al3+(aq) + 6NO3
-(aq) → 6NH4
+(aq) + 6NO3
-(aq) + Al2(CO3)3(s)
6NH4+
(aq) + 3CO32-
(aq) + 2Al3+(aq) + 6NO3
-(aq) → 6NH4
+(aq) + 6NO3
-(aq) + Al2(CO3)3(s)
OH-(aq) + H+
(aq) → H2O(l)
2Na+(aq) + 2OH-
(aq) + 2H+(aq) + SO4
2-(aq) → 2Na+
(aq) + SO42-
(aq) + 2H2O(l)
2Na+(aq) + 2OH-
(aq) + 2H+(aq) + SO4
2-(aq) → 2Na+
(aq) + SO42-
(aq) + 2H2O(l)
Cancel out
Chemical Eqn
Cancel out
Net ionic eqn Complete ionic eqn Molecular eqn
Molecular eqn
Complete ionic eqn
Net ionic eqn
Net ionic eqn
Complete ionic eqn
Molecular eqn
2NaOH(aq) + H2SO4 (aq) → Na2SO4(aq) + 2H2O(l)
3(NH4)2CO3(aq) + 2Al(NO3)3(aq) → 6NH4NO3(aq) + Al2(CO3)3(s)
Ca2+(aq) + CO3
2-(aq) → CaCO3 (s)
Ca2+(aq)+ 2Cl- (aq) + 2Na+
(aq)+ CO32- (aq)→ 2Na+
(aq) + 2Cl-(aq) + CaCO3 (s)
Ca2+(aq)+ 2Cl-
(aq) + 2Na+(aq)+ CO3
2-(aq) → 2Na+
(aq) + 2Cl-(aq) + CaCO3 (s)
OH-(aq) + H+
(aq) → H2O(l)
2Na+(aq) + 2OH-
(aq) + 2H+(aq) + SO4
2-(aq) → 2Na+
(aq) + SO42-
(aq) + 2H2O(l)
2Na+(aq) + 2OH-
(aq) + 2H+(aq) + SO4
2-(aq) → 2Na+
(aq) + SO42-
(aq) + 2H2O(l)
Cancel out
Chemical Eqn
Cancel out
Net ionic eqn Complete ionic eqn Molecular eqn
Molecular eqn
Complete ionic eqn
Net ionic eqn
Net ionic eqn
Complete ionic eqn
Molecular eqn
2NaOH(aq) + H2SO4 (aq) → Na2SO4(aq) + 2H2O(l)
CaCl2 (aq) + Na2CO3(aq) → 2 NaCl(aq) + CaCO3(s)
4 g H2 508 g I2 512 g HI 150 g NO2 300 g N2O4
Simulation conservation mass/balancing eqn
Click to view animation
+
2 moles H2 2 moles I2 4 moles HI
150 g NO2
3 moles NO2 3 moles NO2 3 moles N2O4 +
+
3NO2 + 3NO2 → 3N2O4 2H2 + 2I2 → 4HI
= +
Video on stoichiometry
=
Concept Map
Chemical Rxn Chemical change Chemical eqn Balancing chemical eqn
Molecular Eqn 1Pb(NO3)2(s) + 2KI(aq) → 1PbI2(s) + 2KNO3 (aq)
1Pb2+(aq) + 2NO3
-(aq) + 2K+
(aq) + 2I-(aq) → 1PbI2(s) + 2K+
(aq) + 2NO3-(aq)
1Pb2+(aq) + 2CI-
(aq) → 1PbCI2(s)
Coefficient Mole ratio
(reactant) → (product) 1 : 2 → 1 : 2
Complete ionic eqn
Net ionic eqn
Reaction Stoichiometry Quantitative relationship bet quantities react/ prod
Find quantities/amt (mass, mole, vol) Predicts how much react and amt prod form
Chemical rxn react in definite ratio
Exercise
Write balanced eqn for following rxn 1. Rxn of sulphur dioxide and oxygen to form sulphur trioxide 2SO2(g) + O2(g) → 2SO3(g)
2. Neutralization bet potassium hydroxide and sulphuric acid to form potassium sulphate and water 2KOH(aq) + H2SO4(aq) → K2SO4(aq) + 2H2O(I)
3. Combustion of ethane (C2H6) to form carbon dioxide and water 2C2H6(g) + 7O2(g ) → 4CO2(g) + 6H2O(I)
4. Displacement rxn bet zinc metal and copper(II) sulphate sol to form copper metal and zinc sulphate Zn(s) + CuSO4(aq) → Cu(s) + ZnSO4(aq)
5. Decomposition of zinc carbonate to form zinc oxide and carbon dioxide when heated ZnCO3(s) → ZnO(s) + CO2(g)
6. Ammonia react with oxygen to form nitrogen monoxide and water. 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(I)
7. Manganese(IV) oxide react with hydrochloric acid to form manganese(II) chloride sol, chlorine and water MnO2(s) + 4HCI(aq) → MnCI2(aq) + CI2(g) + 2H2O(I)
8.Neutralization bet aq ammonia with hydrochloric acid to form ammonium chloride and water. NH4OH(aq) + HCI(aq) → NH4CI(aq) + H2O(I)
