THE MOLE MARKING SCHEME
1. 1989 Q3
Mass of oxygen = 0.318 – 0.254 = 0.064
M O
0.254 0.064
63.5 16
Formula = MO
2. 1989 Q29
2KOH(aq) + H2SO4(aq) 2H2O(l) + K2SO4(aq)
Moles of KOH = 20 x 1
1000
= 0.02
Moles of H2SO4 = ½ (Moles of KOH)
= ½ (0.02)
= 0.01 moles
5cm3 contain 0.01 moles
1000cm3 will contain 1000 x 0.01
5
= 2moles
3. 1990 Q5
Moles of NaOH on 200cm3 = 80 x 200 = 4 x 10-3
40 100
4 -3
Moles of acid = ½ x /10
= 2 x10
RFM of acid = 0.81 x 90
= 2 x10-3
4. 1990 Q9
Na O
590 410
23 16
Empirical formula = NaO
Empirical formula mass =23 + 16
=39
5. 1990 Q12
RAM of H = 10 x 18.7/100 + 11 x 81.3/100
= 1.87 + 8.94
= 10.8
6. 1990 Q20
Zn2+ + 2HCl (aq) Zn2+ (aq) + 2HCl-(aq) + H2(g)
7. 1992 Q7
Moles of CuCl2 = 2.50 x 10-3 = 0.25
No. Of chloride ions = 0.25 x 6.0 x 1023 x 2 = 3.0 x 1023
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8. 1992 Q29
(a) (i) C H O
64.9 13.5 21.6
=5.4 =13.5 =1.35
12 1 16
5. 4 13.5 1.3 5
Ration =4 =1 0 =1
1.35 1.35 1 .35
(ii) Empirical formula = C4H10O1
(12 x 4) + ( 1 x 10) + (16 x 1) x n =74
74n = 74
n= 1
Molecular formula = C4H10O or C4H10O1
(iii) I P= Alcohols/alkanols R=ester
II S = Hydrogen gas
III CH3CH2CH2CH2OH or isomers
(iv) 2CH3CH2CH2CH2OH + 2Na CH3CH2CH2CH2ONa + H2
or
2C4H9OH(l) + 2Na(s) 2Cl4H9ONa(l) + H2 (g)
(v) M= Less soluble reasons :- Longer carbon chain/any, solubility of organic
compounds decreases any molecular mass increases
(b) 2CH3-C-OH + Na2CO3 2CH3—C—ONa+ CO2H2O
9. 1993 Q7
% of H = 100 - 79.9=20.1
C H
20.1
79.9
1
66.6 20.1
or
6.66 6.66
1 3
10. 1993 Q9
4
Conc of NaOH =
10
= 0.1M
Moles of NaOH = 0.1 x 22.2 x 10
= 2.22 x 10
1 −3
Moles of dibasic acid = (2.22 x 10 )
2
= 1.11 x 10-3
2
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0.1
RF mass of acid = −3
1.11 x 10
= 90.90/90.1
11. 1994 Q3
Mass of water 34.8-15.9=18.9g
18.9 18 x
= x = 18.9 x 106 = 2003.4
15.9 1 06
18 x 15.9 286.2
NaCO3= (2 x 23) + 12 + 42 = 106
12. 1994 Q16
2 Moles of H2 (g) = 2 Moles of H2O
= 100cm3 of H2 (g) = 50cm3 O2 (g) 100cm3 H2O
Oxygen is excess by 50cm3
13. 1995 Q3 P1
1 mole CaCO3 2 moles of HCL
Therefore 0.1(1/2) mole CaCO3 0.2 Mole (½)
CaCO3 = 40 + 12 + 48 = 100g (½)
Therefore 15g CaCO3 = 15 = 0.15Moles
100g
Excess moles 0.15 – 0.05 (½)
Excess mass= (0.05) x 100 (½) = 5g (3 marks)
14. 1995 Q14 P1
a) (C3H6O)n = 116
(3 x 12 + 6 + 16)n =116 (1/2 ) Molecular formulae = 2(CHO)
58n = 116 (1/2) = C3H12O2 (1/2)
1
n = 116 = 2 ( /2) (2 marks)
58
a) Percentage of Carbon = 12 x 6 x 1000(1/2 ) = 62.07 (1/2 ) Range (62.05 – 62)
116
OR
3 x 12 x 100 (1/2) = 62.07 (1/2) (mark consequently)
58
15. 1996 Q24 P1
A gas with a smell of rotten eggs is formed H2S gas is formed / A greenish solution is formed?
Effervescence / A gas is produced / Black solid dissolves. (1 mark)
16. 1997 Q19 P1
(a)-Ca(OH)2(aq) + Ca(HCO3)2(aq) →2CaCO3 (s) + H2O(l)
(b) Moles = Volume x Morality
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1000
Moles of CO2+ = 90 x 0.01
1000
= 0.009 moles
(c) - It forms scum initially then produces lather
- All the Ca2+ had not been precipitated.
- Water was still hard
17. 1997 Q28 P1
10
No. of moles of hydrogen H2 = /2 = 5 Moles
No. of moles of Nitrogen dioxide NO2 = 46
Relative molecular mass of NO2 = 46
1 Mole of No2 = 5 x 46
5 Moles = 30g
18. 1998 Q6 P1
ALT 1
CxHy + O2 x CO2 + y/2 H3O
y
XCO2 /2 H2O
3:52 1:44
r:3.52 = 0.08 1.44 = 0.08
44 44
= 0.08 =1 0.08 = 1
0.08 0.08
X = 1 y/2 = 1
=E.F = CH2 y = 2
E.F.M = 14
N= 56 =4
14
M.F. (CH2)4 = C4H8
Mass of C = 12 x 3.52 = 0.96
44
Mass of H = 2 x 1.44 = 0.16g
18
Moles of C = 0.96 = 0.08
12
Moles of H = 0.16 = 0.16
1
Ratio 0.08 : 0.16
0.08 : 0.08
1 2
EF : CH2
N : 4
MF = (CH2)4 = C4 H8
19. 1999 Q7 P1
4
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C = 2: 3 Valency C2S3 or C2S5
Accept B2S3 OR B2S3 if the candidate identifies C as Boron (B)
20. 1999 Q10 P1
Moles of nitric acid =50× 2 = 0.1
1000
Moles of KOH IN 50CM3 =0.1
moles OF KOH IN 50cm3 = 0.1 1/2
moles of KOH IN 100cm3 = 0.1 1/2
=0.2× 56
=11.2g
21. 1999 Q4b P2
Fe(s) + Cu2- (aq)=Fe 2+ (s) + Cu (s)
Moles of Fe (s) = 3.36 = 0.06 (1/2)
56(I)
Moles of Cu = 0.06
Mass of CU = 0.06 63.5 (1) = 3.81 (g)
(Mark consequentially from equation given (1/2) for units)
21. 2000 Q11 P1
Alt 1 Alt. 2
Molarity of NaOH = 4 = 0.1M H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) +H2O(l)
Moles of NaOH = 20 x 0.1 Molarity of NaOH = 4 = 0.1M
1000 40
= 0.002 MaVa = 1 ma x8 = 1
mbVb 2 0.1 x20 2
moles of H2SO4 = 0.001
8cm3 = 0.001 Ma = 0.1x20
1000cm3=? 8x2
= 0.1235M = 0.125M
22. 2001 Q10 P1
R.m.m of H2O = 2 +16 = 18 16 + 18 = 100%
R.m.m of Na2CO3 = 46 + 12 + 48 = 106 18n = 14.5
Moles H2O = 14.5 = 0.805 106 + 18n 100
18
Moles of Na2CO3 = 85.5 = 0.866 18n x(-100) = 14.5 (106 + 18n)
100
Mole ration Na2CO3:H2O 1800n = 1537 + 261n
1: 1 1539n = 1537
1537 = 09987
1539
23. 2003 Q6 P1
Na2SO3(s) + 2 HCl (aq) 2NaCl (aq) + SO2 (g) + H2O(l)
Moles of So2 = 160 / 2400 Mass of NaSO3
=0.04 0.04 x 126
Moles ratio 1:1 =5.04 gm
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Moles of NaSo3 = 0.04
24. 2004 Q14 P1
a) Mg2+(aq) + CO32 (aq) MgCO3(s) (1) penalize ½ more for missing
state symbols
b) RFM of Mg CO3 = 24 + 12 + 48 =84 (½)
= 24 + 12 + 16 x 3 (½)
2+
Moles of Mg = 8.4 = 0.1
8 (½)
xx0.5 = 0.1 (½)
1000
X = 1000x0.1
0.5
25. 2005 Q9 P1
GCO3(s) + 2HCI (aq) →GCl2(aq) + CO2(g) + H2O(l)
1 mol 2 mol
Moles of acid used = 20 x 1 = 0.02 moles
1000
Of the carbonate = ½ of acid = 0.01 moles
0.01 moles = 1 g
1 mole = 1 x 1 = 100g
0.01
Molar mass of GCO3 = G + 16 x 3
100 = G + 60
G = 40
R.A.M of G = 40
26. 2005 Q2b P2
b) i) 2NaOH(aq) + H2SO4 (aq) → Na2SO4(aq) + 2H2O(l)
2 Mol 1 Mol
ii) No. of moles of H2SO4 used = 40 x 0.5 moles
1000
= 0.02 moles
No. of moles of NaOH = 0.02 x 2
= 0.04 moles
0.5 x 2 mole = 1.0 moles will react with 1 litre of the solution of the acid
100 cm3 = 0.04 moles of NaOH
1000 cm2 = 0.04 x 1000 = 0.4 moles
100
Molar mass of NaOH = 23 +16 +1
= 40
1 mole = 40
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0.4 moles = 0.4 x 40
= 16g
Mass of the unreacted = 17.6 – 16
= 1.6g
27. 2006 Q8 (P1), 7c-g p2
8. Mass of water 94.5 – 51.3 = 43.2
R.M.M. of Ba(OH)2 = 171
R.M.M of H2O = 18
51.3 43.2 = 8
171 18
0.3 = 1 2.4 8
0.3 0.3
7.c)
d) i) 360 cm3 (Correct value read from graph) (1mark)
ii) 40 cm3 ( Correct value read from graph) (1mark)
e) i) Moles of lead = 2.07
2.07
1 mole of lead = 40
0.01
= 4000cm (2marks)
ii) 480 = 48000cm3 (2marks)
0.01
f) i) Moles of nitric acid = 4000
That react with 1 mole of lead 1000
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=4 (1mark)
ii) Moles of nitrogen dioxide = 48000 =2 (1mark)
24000
g) Pb(s) + 4HNO3(aq) pb(NO3)2(aq) + 2H2O(l) + 2NO2(g)
28. 2007 Q20 P1
(a)
Element C H O
% 64.9 21.6 13.5
12 16 1
Moles 5.41 1.35 13.5
Ration 4 1 10
[E.F.= C4H9OH]
(b)
H H H H
│ │ │ │
H ─ C ─ C ─ C ─ C ─ O─ H
│ │ │ │
H H H H
29. 2007 Q22 P1
Al2(SO4)3 → 3SO4-2 + 2Al3+
Moles a2 Al2 (SO4)3 = 6.84 = 0.02
342
2 -2
Moles a SO4 = 0.02 x 3 = 0.06
30. 2008 Q2 P1
Mass of hydrated salt = (33.111 – 30.296)= 2.815g
Mass of anhydrous salt = 32.781 – 30.296) = 2.485g
E.F = CaSo4 33. 111g
32 781g = 0.330
Mass of water = (2.815 – 2.485) = 0.330g
Accept any correct method
CaSO4 x H20
Mass 2.485 0.320
Moles 2. 485 = 0.0183 0.330/18 = 0.0183
Ration 0.0183/0.0183 = 0.0183/0.0183
Or; CaSo4. XH2O → CaSo4 + XH2O
2.815g = 2.485g
CaSo4 x H2O 136
Y= 2.815 x 136 = 154
2.485
CaSo4 x H2O= 154
136 + 18x = 154
18x = 154 – 136 = 18
X= 18/18 = 1
31. 2008 Q5 P1
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RFM of Ca3(PO4)2 Ca=40 x 3 = 120
P = 31 x 2 = 62
O= 16 x 8 = 128
310
H3PO4 H=1 x 3 = 3
P = 31 x 1 = 31 1 mole Co3(PO4)2 gives moles of H3PO4
O = 16 x 4 = 64/98 310g Co3 (PO4) 2 gives 2.98 g
155 x 100g Co3 (PO4) gives 2.98 x 155 x 100
310
= 98000g
= 98kg
32. 2008 Q27 P1
(a) Mg (OH)2(aq) + 2 HCL (aq) → Mg Cl2 (aq) + H2O(l)
Mole ration (1:2)
No of moles of acid = 0.1 x 23 = 0.0023
1000
No of moles of Mg (OH)2 = ½ x 0/1 x 23
1000 = 0.00115
Mass of Mg (OH) w in antacid = 0.00115 x 58 = 0.067g
33. 2009 Q13 P1
% of Mg (OH)2 in anti- acid
Mg (OH) 2 = 0.67 x 100 = 13.34%
0.50
Moles of oxygen = 0.83 = 0.026 (½) / 0.0259375
Moles of NaNO3 = 2 x 0.026 / 0.051875
0.05 (½ ) / 0.051875
R. M .M NaCO3 = 85 (½)
Mass of NaNO3 = converted 0.052 x 85 / 4.4094 (½ )
4.41
4.41
8.53
51. 693%5
Or 183
51.7% (3 marks)
34. 2011 Q26 P1
(i) Metal M (½ mark)
19.352 – 10.240 = 0.112g
(ii) Oxygen (½ mark)
10.400 – 10.352 = 0.048g
Element M O
Mole ratio 0.112 0.048
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56 16
0.0020 0.0030
Simple ratio 2 3
E.F M2O3
35. 2012 Q3 P2
(a) (i) concentrated sulphuric (vi) acid (1mark)
if sulphuric acid ½ / sulphuric (vi) acid penalise ½ mark
(ii) potassium nitrate √ or KNO3 ½
(iii) is condense the nitric acid fumes into liquid or to cool nitric acid vapour. Reject
to cool the nitric acid (1mark)
(b) (i) Nitric acid attacks rubber/ corrodes√ rej reacts with rubber
(ii) The reaction produces nitrogen √ ½ (ii) oxide which readily gets oxidised
by air to form nitrogen oxide that as brown.
Identifying that brown gas is NO2 √ ½
(c) (i) -Rej Electrolysis of brine / steam alkanes
- From natural gas, crude oil, watergas, biogas.
(ii) equation √1 HNO3 (g) + NH(g) NH4NO3(l) or (aq)√ 1
Moles of NH4NO3 17kg 80kg 17 x 4800
Moles of NH3 ? 4800 80 = 1020
Use of ratio 6 x 17 x 104 = 1020kg √ ½
1000
Answers √ ½
(iii) - Manufacture of explosives - etching of metals (2marks)
- manufacture of dyes - clearing of gas wave
- Manufacture of royal water
- Manufacture of drugs
- As an oxidising agent
36. 2012 Q8 P1
Moles of NaOH =
= 0.0036
H2SO4 (aq) + 2NaOH(aq) NaSO4(aq) + H2O(l)
Or mole ratio 1:2
Or
= 0.0018 (1.8 x 10-3)
0.0018 x 100
10 0.018 moles in a 100cm3
10
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RFM of H2SO4 = 98
Mass of acid = 98 x 0.018
= 1.764g
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