2.7 Density and Specific Gravity
2.7.4 Brix Scale
This scale is used in sugar industry and 1 degree Brix is equal to 1% sugar in solution.
Degree Brix = 400 G – 400
WORKED EXAMPLES
2.1 Convert 5000 ppm into weight %.
6
5000 100 10
× = 0.5%
2.2 The strength of H3PO4 was found to be 35% P2O5. Find the weight % of the acid.
The acid can be split into
3 4 2 5 2
(3 18) (2 98) 142
2H PO P O + 3H O
×
× →
196 units of the acid contains 142 units of the pentaoxide.
The weight % of pentaoxide is (142/196) which is 72.5% for pure acid.
When the strength of pentoxide is 35%, the weight % of acid is 35 100
72.5
⎛ ⎞
⎜ × ⎟
⎝ ⎠ = 48.3%
2.3 What is the volume of 25 kg of chlorine at standard condition?
25 kg Cl2 = 25
2×35.46 kmoles of Cl2 Volume = 25 22.414
2 35.46
×× = 7.9 m3
2.4 How many grams of liquid propane will be formed by the liquefaction of 500 litres. of the gas at NTP? Molecular weight of propane (C3H8) is 44
1 g mole of any gas occupies 22.414 litres at NTP
500 litres of propane at NTP = 500
22.414 = 22.31 g moles 22.31 g moles of propane weighs = 22.31 ´ 44 = 981.52 g.
2.5 Find the volume of (a) 100 kg of hydrogen and (b) 100 lb of hydrogen at standard conditions?
(a) 100 kg of H2 = 50 kmoles of hydrogen
volume occupied by 50 kmoles of hydrogen º 50 ´ 22.414 = 1120.7 m3 (b) 100 lb of H2 º 50 lb moles of hydrogen
Volume occupied by 50 lb moles of hydrogen º 50 ´ 359 = 17950 ft3 2.6 A solution of naphthalene in benzene contains 25 mole %
Naphtha-lene. Express the composition in weight %.
Basis: 100 g moles of solution
Component Molecular Weight, Actual Composition in weight g mole weight, g weight percent Naphthalene C10H8 128 25 25´ 128 = 3200 3200 ´ 100/9050 = 35.35 Benzene C6H6 78 75 75´ 78 = 5850 5850 ´ 100/9050 = 64.65
Total 9050 100.00
2.7 What is the weight of one litre of methane CH4 at standard conditions?
22.414 litres of any gas at NTP is equivalent to 1 g mole of that gas
\ 1 litre of methane º 1 ´ 1
22.414 = 0.0446 g mole
\ Weight of one litre methane = 0.0446 ´ 16 = 0.714 g
2.8 A compound whose molecular weight is 103 analyses C : 81.5, H : 4.9 and N : 13.6 by weight. What is the formula?
Basis: 100 g of substance
Element Atomic Weight, g Weight, Rounding of Weight of
weight g atom atoms each element
Carbon 12 81.5 81.5/12 = 6.8 7 84
Hydrogen 1 4.9 4.9/1 = 4.9 5 5
Nitrogen 14 13.6 13.6/14 = 0.9 1 14
Total 103
Hence the formula obtained after rounding is correct.
So the molecular formula is C7H5N
2.9 An analysis of a glass sample yields the following data. Find the mole %.
Na2O : 7.8%, MgO : 7.0%, ZnO : 9.7%, Al2O3: 2.0%, B2O3: 8.5%
and rest SiO2.
Basis: 100 g of glass sample
Component Weight, g Molecular weight g mole mole %
Na2O 7.8 62.0 0.1258 7.665
MgO 7.0 40.3 0.1737 10.583
ZnO 9.7 81.4 0.1192 7.262
Al2O3 2.0 102.0 0.0196 1.194
B2O3 8.5 69.6 0.1221 7.439
SiO2 65.0 60.1 1.0815 65.857
Total 100.0 — 1.6419 100.0
2.10 A gaseous mixture analyzing CH4: 10%, C2H6: 30% and rest H2 at 15 °C and 1.5 atm is flowing through an equipment at the rate of 2.5 m3/min. Find (a) the average molecular weight of the gas mixture, (b) weight % and (c) the mass flow rate.
Basis: 100 g moles of the gaseous mixture.
Component Weight, Molecular Weight, Weight %
g mole weight g
CH4 10 16 160 13.56
C2H6 30 30 900 76.27
H2 60 2 120 10.17
Total 100 — 1180 100
The average molecular weight = 1180
100 = 11.8
Volumetric flow rate at standard conditions = 2.5¥ 1.5 1
⎛ ⎞
⎜ ⎟
⎝ ⎠ ¥ (273/288) = 3.555 m3/min
Moles of the gas = 3.555
22.414 = 0.156 kmole
Mass flow rate = moles ¥ average molecular weight = 0.156 ¥ 11.8 = 1.84 kg/min 2.11 In an evaporator a dilute solution of 4% NaOH is concentrated to 25%
NaOH. Calculate the evaporation of water per kg of feed.
Basis: 1 kg of feed.
NaOH present is 0.04 kg, which appears as 25% in the thick liquor formed
Weight of thick liquor formed = 0.04
0.25 = 0.16 kg Weight of water evaporated = (1 – 0.16) = 0.84 kg Water evaporated per kg of feed = 0.84 kg
2.12 The average molecular weight of a flue gas sample is calculated by two different engineers. One engineer used the correct molecular weight of N2 as 28, while the other used an incorrect value of 14.
They got the average molecular weight as 30.08 and the incorrect one as 18.74. Calculate the % volume of N2 in the flue gases. If the remaining gases are CO2 and O2 calculate their composition also.
Basis: 100 g moles of flue gas
Component g mole I Engineer II Engineer
N2 x 28x 14x
CO2 y 44y 44y
O2 z 32z 32z
Total 100 3008 1874
x + y + z = 100 (i)
28x + 44y + 32z = 3008 (ii)
14x + 44y + 32z = 1874 (iii)
Solving Eqs. (i), (ii) and (iii), we get x = Moles of nitrogen = 81%
y = Moles of carbon dioxide = 11%
z = Moles of oxygen = 8%
2.13 An aqueous solution contains 40% of Na2CO3 by weight. Express the composition in mole percent.
Basis: 100 g of solution
Component grams Molecular g mole Composition in
weight mole %
Na2CO3 40 106 40/106 = 0.377 0.377´ 100/3.71 = 10.16 Water 60 18 60/18 = 3.333 3.333´ 100/3.71 = 89.84
Total 3.710 100.00
2.14 What is the weight of iron and water required for the production of
Method 1 (Based on absolute mass)
167.52 kg of Fe is required for producing 8 kg of H2
\ For producing 100 kg of H2 (by stoichiometry) Iron (Fe) required = 100 167.52
8
The total weight of reactants is
2094 kg Fe and 900 kg H2O = 2994 kg Method 2 (Based on moles)
100 kg of H2 = 50 kmoles
4 kmoles H2 comes from = 3 katoms of Fe (by stoichiometry) 50 kmoles of H2 comes from = 50 Total weight of reactants = 2994 kg
Total weight of products = 2994 kg
2.15 How much super phosphate fertilizer can be made from one ton of calcium phosphate 93.5% pure?
Atomic weights are: Ca : 40, P : 31, O : 16, S : 32 Ca3(PO4)2 + 2H2SO4 Æ CaH4(PO4)2 + 2CaSO4
310 (2 ¥ 98) 234 (2 ¥ 136)
506 506
One ton of raw calcium phosphate contains 0.935 tons of pure calcium phosphate
\ Weight of super phosphate formed is = 234 ¥ 0.935
310 = 0.70577 tonne 2.16 SO2 is produced by the reaction between copper and sulphuric acid.
How much Cu must be used to get 10 kg of SO2?
2 4 4 2 2
64 63.54
Cu + 2H SO →CuSO + SO + 2H O
64 kg of sulphur dioxide is obtained from 63.54 kg of copper.
10 kg of sulphur dioxide will be obtained from 9.93 kg of copper.
2.17 How much potassium chlorate must be taken to produce the same amount of oxygen that will be produced by 2.3 g of mercuric oxide?
2KClO3 Æ 2KCl + 3O2
(2¥ 122.46) (2¥ 74.46) (6¥ 16)
244.92 148.92 96
2HgO Æ 2Hg + O2
(2¥ 216.6) (2¥ 200.6) (2¥ 16)
433.2 g HgO gives 32 g of oxygen
2.3 g of HgO will give (32 ¥ 2.3/433.2) = 0.1698 g of O2
0.1698 g O2 is obtained from (244.92 ¥ 0.1698)/96 = 0.4332 g of KClO3. 2.18 Ammonium phosphomolybdate is made up of the radicals NH3, H2O, P2O5 and MoO3. What is % composition of the molecule with respect to these radicals?
The formula of ammonium phosphomolybdate is (NH4)3PO4 ◊ 12MoO3 ◊ 3H2O First let us form the final product from the radicals:
3NH3 + 4.5H2O + 12MoO3 + ½P2O5Æ (NH4)3PO4 12MoO3· 3H2O
(3¥ 17 = 51) (4.5 ¥ 18 = 81) (12 ¥ 144 = 1728) (½ ¥ 142 = 71) 1931
% of NH3 = 51 ¥ 100
1931 = 2.64
% of H2O = 81 ¥ 100
1931 = 4.19
% of MoO3 = 1728 ´ 100
1931 = 89.49
% of P2O5 = 71 ´ 100
1931 = 3.68
Total = 100.00
2.19 How many grams of salt are required to make 2500 g of salt cake?
How much Glauber’s salt can be obtained from this?
The molecular formula of Glauber’s salt is Na2SO4× 10H2O (142 + 180 = 322)
2 4 2 4
(2 58.46 =116.92) 98 142 (2 36.46)
2NaCl H SO Na SO + 2HCl
× + → ×
Thus, 142 g of Na2SO4 is obtained from 116.92 g NaCl.
2500 g of salt cake is obtained from 116.92 ´ 2500/142 = 2058.45 g NaCl
Hence, salt needed is 2058.45 g
Glauber’s salt (Na2SO4× 10H2O) obtained is 2500 ´ 322/142 = 5669 g 2.20 (a) How many grams of K2Cr2O7 are equivalent to 5 g KMnO4?
(b) How many grams of KMnO4 are equivalent to 5 g K2Cr2O7? 2KMnO4 + 8H2SO4 + 10FeSO4 ® 5Fe2(SO4)3 + K2SO4 + 2MnSO4
+ 8H2O
K2Cr2O7 + 7H2SO4 + 6FeSO4 ® 3Fe2(SO4)3 + K2SO4 + (Cr2SO4)3 + 7H2O
2KMnO4 gives 5Fe2(SO4)3
(2´ 158 = 316) (5 ´ 400 = 2000)
K2Cr2O7 gives 3Fe2(SO4)3
(294) (3´ 400 = 1200)
\ 294 g K2Cr2O7 is equivalent to 1200 316 2000
t = 189.6 g KMnO4
\ 3 g K2Cr2O7 º 3 189.6 294
t = 1.935 g KMnO4
Similarly, 5 g KMnO4º 5 294 189.6
t = 7.75 g K2Cr2O7
Alternatively, 5 3 1.935
t = 7.75 g K2Cr2O7
2.21 If 45 g of iron react with H2SO4, how many litres of hydrogen are liberated at standard condition?
There are two possible reactions in this case:
(a) Case I
2 4 4 2
(55.85) (2)
Fe +H SO →FeSO + H (i)
The weight of hydrogen formed by reaction (i) is = 45 ¥ 2
55.85 = 1.611 g, i.e. 1.611
2 = 0.806 g mole
0.806 g mole ∫ 0.806 ¥ 22.414 = 18.06 litres (b) Case II
2 4 2 4 3 2
(111.7) (6)
2Fe +3H SO →Fe (SO ) + 3H (ii)
The moles of hydrogen formed by reaction (ii) is 45 ¥ 6
111.7 = 2.418 g 2.418 g H2 = 1.209 g moles of hydrogen ∫ 1.209 ¥ 22.414 ∫ 27.1 litres
2.22 A natural gas has the following composition by volume CH4: 83.5%, C2H6: 12.5%, and N2: 4%. Calculate the following:
(a) composition in mole % (b) composition in weight % (c) average molecular weight (AVMWT) (d) density at standard condition (kg/m3) Basis: 100 kmoles of gas mixture
Component Molecular mole % Weight, kg Weight % weight
CH4 16 83.5 83.5¥ 16 = 1336 1336 ¥ 100/1823 = 73.29 C2H6 30 12.5 12.5¥ 30 = 375 375¥ 100/1823 = 20.57
N2 28 4.0 4.0¥ 28 = 112 112¥ 100/1823 = 6.14
Total 1823 100.0
(c) Average molecular weight = 1823
100 = 18.23
Volume at standard condition = 00 ¥ 22.414 = 2241.4 m3 (d) Density of gas at standard condition = 1823
2241.4 = 0.813 kg/m3
2.23 Convert 54.75 g/litre of HCl into molarity.
Molarity = g moles/litre of solution = 54.75
36.45 = 1.5
2.24 A solution of NaCl in water contains 230 g of NaCl per litre at 20 °C.
The density of the solution at this temperature is 1.148 g/cc. Find the composition in (a) weight % (b) volume % of water (c) mole % (d) atomic % (e) molality and (f) g NaCl/g water.
Basis: (a) 1 litre of solution has a weight of 1148 g
Component Molecular Weight, g Weight, % g mole mole % weight
NaCl 58.5 230 20.03 230/58.5 = 3.93 3.93/54.93 = 7.15 Water 18 918 79.97 918/18 = 51.00 51/54.93 = 92.85
Total 1148 100.00 54.93 100.00
(b) Volume % of water: 918 g is present in 1 litre of solution, i.e.
918 cc water is present in 1000 cc. of solution (density of water is 1 g/cc)
Volume % = 91.8%
(d)
Element g atoms Atomic %
Na 3.93 2.443
Cl 3.93 2.443
H 102.00 63.409
O 51.00 31.705
Total 160.86 100.000
(All are based on the molecular formula)
(e) Molality = g moles of solute in 1 kg of solvent (3.93 ´ 1000/918) or, 3.93 g moles of NaCl is present in 918 g of water
(i.e.) 4.28 g moles/1000 g of solvent Molality = 4.28
Molarity = Moles of solute per litre of the solution = 3.93 (f) g NaCl/g water = 230
918 = 0.252
2.25 A benzene solution of anthracene contains 10% by weight of the solute. Find the composition in terms of (a) molality (b) mole fraction.
Basis: 100 g of solution
Component Molecular Weight, g Weight, mole
weight g mole fraction
Anthracene 178 10 (10/178) 0.046
0.0562
Benzene 78 90 (90/78) 0.954
1.1538
Total 1.2100 1.00
Molality = g moles of anthracene in 1000 g benzene = 0.0562
90 ¥ 1000 = 0.624
2.26 Calculate the weight of NaCl that should be placed in a 1 litre volumetric flask to prepare a solution of 1.8 molality. Density of this solution is 1.06 g/cc
Molality = g moles of NaCl/1000 g of water = 1.8
or, 1.8 g moles NaCl = (1.8 ¥ 58.46) = 105.228 g
Component Weight, g Weight %
NaCl 105.228 9.52
H2O 1000.000 90.48
1105.228 100.00
Density of this solution = 1.06 g/cc
Volume of this solution, i.e. mass/density = 1105.228
1.06 = 1042.67 cc or 1042.67 cc of this solution contains 105.228 g of NaCl
\ 1000 cc of this solution will have = 1000 ¥ 105.228 1042.67 = 100.92 g of NaCl.
NaCl needed = 100.92 g
2.27 For the operation of a refrigeration plant it is desired to prepare a solution of 20% by weight of NaCl solution.
(a) Find the weight of salt that should be added to one gallon of water at 30°C?
(b) What is the volume of this solution?
Basis: 100 lb of solution
It will have 20 lb NaCl and 80 lb water
80 lb water = 1.28 ft3 (since the density of water is 62.47 lb/ft3) We know that 1 ft3 = 7.48 gallons
Therefore, 1.28 ft3 = 9.57 gallons.
(a) Weight of salt per gallon of water = 20 9.57
¦ µ
§ ¶
¨ · = 2.09 lb.
(b) Specific gravity of NaCl solution at 30 °C = 1.14
\ Density of solution = 1.14 ´ 62.4 = 71.14 lb/ft3 Weight of 1 gallon of water = 62.47
7.48 = 8.35 lb.
Total weight of solution = weight of water + weight of salt = 8.35 + 2.09 = 10.44 lb.
Hence, volume of the above solution = 10.44
71.14 = 0.147 ft3 = 1.1 gallons.
2.28 (a) A solution has 100° Tw gravity. What is its specific gravity and
°Be’?
(b) An oil has a specific gravity of 0.79. Find °API and °Be’
(a) 100 = 200 (G – 1) \ G = 1.5
2.29 An aqueous solution contains 15% ethyl alcohol by volume. Express the composition in weight % and mole %. Density of ethyl alcohol and water are 790 kg/m3 and 1000 kg/m3 respectively.
Basis: 1 m3 of solution.
Compound Molecular Volume, Density, Weight, Number Weight mole
weight m3 kg/m3 kg of moles % %
Ethanol 46 0.15 790 118.5 2.576 12.235 5.173
Water 18 0.85 1000 850 47.222 87.765 94.827
Total 1.00 968.5 49.798 100 100
2.30 The quality of urea is expressed in terms of nitrogen content. If the nitrogen content in the sample is only 40%, estimate the purity of sample in terms of urea content.
The molecular weight of urea (NH2CONH2) is 60 and that of N2 is 28.
Basis: 100 kg of sample
60 kg of urea has 28 kg of N2
100 kg of urea will have = 28 × 100
60 = 46.67 kg of N2 (Theoretically)
The given sample has 40% N2 Hence, the % purity is = 40 × 100
46.67 = 85.71%
2.31 If the nitrogen content in ammonium nitrate sample is 28%, estimate the purity of ammonium nitrate.
Molecular weight of ammonium nitrate, NH4NO3 = 80
% Nitrogen in pure ammonium nitrate = 28 × 100
80 = 35%
The % of nitrogen in the sample is 28
Hence, the purity of ammonium nitrate is 28 35
¦ µ
§ ¶
¨ · × 100 = 80%
2.32 Nitrobenzene is produced by reacting nitrating mixture with benzene.
The nitrating mixture contains 31.5% HNO3, 60% H2SO4 and 8.5%
H2O. A charge contains 663 kg of benzene and 1700 kg of nitrating mixture which sent into the reactor. If the reaction is 95%, then calculate the amount of nitrobenzene and spent acid produced.
The reaction is
C6H6 + HNO3 ® C6H5NO2 + H2O Feed, C6H6 : 663/78 = 8.5 kmole
HNO3 : 31.5% of 1700 kg = 535.5 kg = 8.5 kmoles H2SO4 : 60 % of 1700 kg = 1020 kg = 10.408 kmoles H2O : 8.5 % of 1700 kg = 144.5 kg = 8.028 kmoles
Reaction is 95% complete
Hence, HNO3 unreacted : 0.05 × 8.5 = 0.425 kmole C6H6 unreacted : 0.05 × 8.5 = 0.425 kmole H2SO4 unreacted : 10.408 kmoles
H2O unreacted : 8.028 kmoles
H2O formed : 8.5 × 0.95 = 8.075 kmoles Nitrobenzene formed : 8.5 × 0.95 = 8.075 kmoles
Component Weight, Molecular Weight, Weight,
kmole weight kg %
HNO3 0.425 63 26.775 1.133
H2SO4 10.408 98 1020.000 43.165
H2O (8.075 + 8.028) = 16.103 18 289.850 12.266
Nitrobenzene 8.075 123 993.225 42.032
C6H6 0.425 78 33.150 1.403
Total 2363.00 100%
Nitrobenzene produced = 993.225 kg
Spent acid = 26.775 + 1020.000 + 289.85 = 1336.63 kg
2.33 A sample of caustic soda flake contains 74.6% Na2O by weight.
Estimate the purity of flakes.
Reaction is as follows:
2NaOH ® Na2O + H2O
Amount of Na2O in pure flakes = 62 × 100/80 = 77.5%
% Purity = 0.746/0.775 × 100 = 96.26%
2.34 Two kg of CaCO3 and MgCO3 was heated to a constant weight of 1.1 kg. Calculate the % amount of CaCO3 and MgCO3 in reacting mixture.
Reaction is as follows:
CaCO3 ® CaO + CO2
(100) (56) (44)
MgCO3® MgO + CO2
(84) (40) (44)
Let, x be the amount of CaCO3.
Therefore, (2 – x) be the weight of MgCO3 100 kg of CaCO3 gives 56 kg of CaO
Therefore, x kg of CaCO3 gives 56 100
x = 0.56x kg of CaO.
Similarly, 84 kg of MgCO3 gives 40 kg of MgO Therefore, (2 – x) kg of MgCO3 gives 40
84
© ¸ ª ¹
« º × (2 – x) kg of MgO The weight of product left behind is 1.1 kg, i.e. weight of MgO + CaO left behind
0.56x + (0.4672)(2 – x) = 1.1 0.0838x = 1.1 – 0.96524 Therefore, x = 1.761 kg
Component Weight, kg Weight, %
CaCO3 1.761 88.05
MgCO3 0.239 11.95
Total 2.000 100.00
2.35 The composition of NPK fertilizer is expressed in terms of N2, P2O5 and K2O each of about 15 weight %. Anhydrous ammonia, 100%
phosphoric acid and 100% KCl are mixed to get 1 ton of fertilizer.
Estimate the amount of filler in the NPK fertilizer.
Basis: 1000 kg of fertilizer Reactions are:
2NH3 ® N2 + 3H2
(34) (28) (6)
2H3PO4 ® P2O5 + 3H2O
(196) (142) (54)
2KCl + H2O ® K2O + 2HCl
(149) (18) (94) (73)
N2, K2O and P2O5 are each equivalent to 15 weight % = 150 kg each Ammonia reacted = 34 × 150
28 = 182.14 kg H3PO4 needed = 196 × 150
142 = 207.04 kg KCl needed = 149 × 150
94 = 237.77 kg
The amount of inert material/filler = 1000 – 626.95 = 373.05 kg 2.36 A solution whose specific gravity is 1 contains 35% A by weight and
the rest is B. If the specific gravity of A is 0.7, find the specific gravity of B.
Basis: 1000 kg of solution Weight of A: 350 kg Weight of B: 650 kg
Volume of solution = 1 m3 (since density is 1000 kg/m3 due to specific gravity being unity)
Mass/volume = density Assuming ideal behaviour
B
350 650 700 S = 1
rB = 1300 kg/m3
Therefore, specific gravity of B = 1.3
2.37 An aqueous solution contains 47% of A on volume basis. If the den-sity of A is 1250 kg/m3, express the composition of A in weight %.
Basis: 1 m3 of solution
Volume of A in solution = 1 × 0.47 = 0.47 m3 Weight of A = 0.47 × 1250 = 587.5 kg Volume of water = (1 – 0.47) = 0.53 m3
Therefore, the weight of water = 0.53 m3 × 1000 = 530 kg Hence, weight % of A = 587.5
530 587.5 = 52.57%
2.38 An aqueous solution contains 43 g of K2CO3 in 100 g of water. The density of solution is 1.3 g/cc. Find the composition in molarity and molality.
Basis: 100 g of solvent Weight of K2CO3 = 43 g Weight of solution = 143 g Density of solution = 1.3 g/cc Volume of solution = 143
1.3 = 110 cc
Moles of solute = Weight/Molecular weight = 43
138 = 312 g moles Molarity = g mole/volume of solution in lit = 0.312
0.11 = 2.833M Molality = g mole/kg of solvent = 0.312
0.1 = 3.12 g moles/kg solvent.
2.39 A gaseous mixture contains ethylene: 30.6%, benzene: 24.5%, O2: 1.3%, ethane: 25%, N2: 3.1% and methane: 15.5% in volume basis.
Estimate the composition in mole %, weight %, average molecular weight and density.
Basis: 100 kmoles of mixture
Compound mole % Molecular weight Weight, kg Weight %
C2H4 30.6 28 856.8 22.00
C6H6 24.5 78 1911.0 49.07
O2 1.3 32 41.6 1.07
CH4 15.5 16 248.0 6.37
C2H6 25 30 750.0 19.26
N2 3.1 28 86.8 2.23
Total 3894.2 100.00
Density = Weight/Volume = 3894.2
100 × 22.414 = 1.737 kg/m3 = 1.737 g/l.
2.40 A compound has a composition of 9.76% Mg, 13.01% S, 26.01%
O2 and 57.22% H2O by weight. Find the molecular formula of this compound.
Basis: 100 g of compound
Compound Weight, Atomic weight Number Converting to g or Molecular of moles whole numbers
weight dividing by 0.41
Mg 9.76 24 0.410 1
S 13.01 32 0.410 1
O 26.01 16 1.615 3.94
H2O 57.22 18 2.8738 6.92
Therefore, molecular formula of the compound = MgSO4.7H2O 2.41 A substance on analysis gave 1.978 g of Ag, 0.293 g of S and 0.587 g
of O2. Find the molecular formula of the compound.
Compound Weight, Atomic weight Number Converting to g or Molecular of moles whole numbers
weight dividing by
9.156 × 10–3
Ag 1.978 108 0.0183 2
S 0.293 32 9.156 × 10–3 1
O2 0.587 16 0.0367 4.04
Molecular formula = Ag2SO4
2.42 Two engineers are estimating the average molecular weight of gas containing oxygen and another gas. One uses the molecular weight as 32 and finds the average molecular weight as 39.8 and the other uses the atomic weight of oxygen as 16 and finds the average molecular weight as 33.4. Estimate the composition of the gas mixture.
By using the atomic weight of oxygen as 16, the value is 33.4 and by using the molecular weight of oxygen, the value is 39.8.
Let x be the mole fraction of oxygen in the mixture and the molecular weight of the other gas be M
39.8 = (x) × (32) + (1 – x) × (M) 33.4 = (x) × (16) + (1 – x) × (M) Solving, we get x = 0.4
i.e the fraction of oxygen in the mixture is 0.4
2.43 A mixture of methane and ethane has an average molecular weight of 21.6. Find the composition.
Let the mole fraction of methane be X
21.6 = (Molecular weight of CH4)(X) + (Molecular weight of C2H6) (1 – X)
21.6 = 16 × X + 30 × (1 – X) Solving, we get X = 0.6
2.44 A mixture of FeO and Fe3O4 was heated in air and is found to gain 5% in mass. Find the composition of initial mixture.
Reactions involved are:
2FeO + 0.5 O2 ® Fe2O3 2Fe3O4 + 0.5 O2 ® 3 Fe2O3 Basis: 100 kg of feed mixture
Let X be the weight of FeO in the mixture From 144 kg of FeO, Fe2O3 formed is 160 kg
Therefore, from X kg of FeO, Fe2O3 formed is 160 × 144
X
From 232 kg of Fe3O4, Fe2O3 formed is 480 kg
Therefore, from (100 – X) of Fe3O4, Fe2O3 formed is (100 – X) × (480) 464 Since 5% gain in mass is observed, the weight of final product is 105 kg, i.e.
160 (480)
100 105
144 464
X X
t t
Solving, X, the weight of FeO = 20.25 kg Fe3O4 = (100 – 83.45) = 79.75 kg
2.45 A sample of lime stone has 54.5% CaO. Find the weight % of lime stone.
Basis: 100 kg of lime stone
100 kg of CaCO3 will have 56% CaO
If the CaO is 54.5%, then % of CaCO3 in the sample is 54.5 × 100/50 = 97.32%
2.46 Express the composition of magnesite in mole %.
Compound Weight %
MgCO3 81
SiO2 14
H2O 5
Compound Weight % Molecular weight moles mole %
MgCO3 81 84 0.964 65.34
SiO2 14 60 0.233 15.82
H2O 5 18 0.278 18.83
2.47 The concentration of H3PO4 is expressed in terms of P2O5 content. If 35% P2O5 is reported, find the composition of H3PO4 by weight.
P2O5 + 3H2O ® 2H3PO4
(142) (54) (98)
i.e. 142 kg of P2O5 º 196 kg of H3PO4 Therefore, 35 kg of P2O5 º 35 × 196
142 = 48.3 H3PO4 i.e. H3PO4 is 48.3%
2.48 Ten kg of PbS and 3 kg of oxygen react to yield 6 kg of Pb and 1 kg of PbO2 according to the reaction shown below:
PbS + O2 ® Pb + SO2 (1)
PbS + 2O2 ® PbO2 + SO2 (2)
Estimate (i) unreacted PbS, (ii) % excess oxygen supplied, (iii) total SO2 formed, and (iv) the % conversion of PbS to Pb.
PbS + O2 ® Pb + SO2 (1)
(239.2) (32) (207.2) (64)
PbS + 2O2® PbO2 + SO2 (2)
(239.2) (64) (239.2) (64)
207.2 kg of Pb comes from 239.2 kg of PbS [from stoichiometry Eq. (1)]
Therefore, 6 kg of Pb comes from 239.2 × 6
207.2 = 6.927 kg of PbS 239.2 kg of PbO2 comes from 239.2 kg of PbS
Therefore, 1 kg of PbO2 comes from 1 kg of PbS Therefore, total PbS reacted [from Eqs. (1) and (2)]
= 6.927 + 1 = 7.927 kg Unreacted PbS = 10 – 7.927 = 2.073 kg O2 required for this process
From Reaction 1:
32 kg of oxygen is needed to produce 207.2 kg of Pb Therefore, to produce 6 kg of PbO2, oxygen needed = 6 × 32
207.2 = 0.927 kg From Reaction 2:
239.2 kg of PbO2 requires 64 kg of oxygen
Therefore, to produce 1 kg of PbO2,oxygen required is 268 kg Therefore, total oxygen used = 0.927 + 0.268 = 1.195 kg Percentage excess O2 supplied = (3 1.195)
1.195
© ¸
ª ¹
« º × 100 = 151%
Amount of SO2 formed
If 207.2 kg of Pb is formed, SO2 formed is 64 kg If 6 kg of Pb is formed, SO2 formed is 64 × 6
207.2 = 1.853 kg If 239.2 kg of PbO2 is formed, SO2 formed is 64 kg
If 1 kg of PbO2 is formed, SO2 formed is 64
239.2 = 0.268 kg Total SO2 formed = 1.853 + 0.268 = 2.121 kg
% conversion of PbS fed to Pb = Mass of PbS converted to Pb/Total mass of PbS
= 6.927 × 100
10 = 69.27%
2.49 The composition of a liquid mixture containing A, B and C is peculiarly given as 11 kg of A, 0.5 kmole of B and 10 wt of % C.
The molecular weights of A, B and C are 40, 50 and 60 respectively and their densities are 0.75 g/cc, 0.8 g/cc and 0.9 g/cc respectively.
Express the composition in weight %, mole %. Also give its average molecular weight and density assuming ideal behaviours.
Let the weight of mixture be W kg Weight of A = 11 kg
Weight of C (10%) = 0.1W kg
Weight of B = W – 11 – 0.1W = 0.5 kmole = 0.5 × 50 = 25 kg
i.e. weight of B = W – 11 – 0.1W = 25 kg 0.9W = 36 kg
W = 40 kg
i.e. total weight of mixture is 40 kg.
Component Weight, Weight, Molecular moles, mole Density, Volume,
kg % weight kmole % kg/m3 m3
A 11 27.5 40 0.275 32.66 750 0.0147
B 25 62.5 50 0.500 59.38 800 0.0313
C 4 10.0 60 0.067 7.96 900 0.0044
Total 40 100.00 0.842 100.00 0.0504
Average density = Mass/Volume = 40
0.0504 = 793.65 kg/m3 Average molecular weight = Weight/Total moles = 40
0.842 = 47.5 (Check: Average molecular weight
= 40 × 0.3266 + 50 × 0.5938 + 60 × 0.0796 = 47.5)
EXERCISES
2.1 How many g moles are equivalent to 1.0 kg of hydrogen?
2.2 How many kilograms of charcoal is required to reduce 3 kg of arsenic trioxide?
As2O3 + 3C Æ 3CO + 2As
2.3 Oxygen is prepared according to the following equation:
2KClO3 Æ 2KCl + 3O2.What is the yield of oxygen when 9.12 g of potassium chlorate is decomposed? How many grams of potassium chlorate must be decomposed to get 5 g of oxygen?
2.4 An aqueous solution of sodium chloride contains 28 g of NaCl per 100 cc of solution at 293 K. Express the composition in (a) percentage NaCl by weight (b) mole fraction of NaCl and (c) molality. Density of solution is 1.17 g/cc.
2.5 An aqueous solution has 20% sodium carbonate by weight. Express the composition by mole ratio and mole percent.
2.6 A solution of caustic soda in water contains 20% NaOH by weight.
The density of the solution is 1196 kg/m3. Find the molarity, normality and molality of the solution.
2.7 A saturated solution of salicylic acid in methanol contains 64 kg salicylic acid per 100 kg methanol at 298 K. Find the composition by weight % and volume %.
2.8 A solution of sodium chloride in water contains 270 g per litre at 323 K. The density of this solution is 1.16 g/cc. Estimate the composition by weight %, volume %, mole %, atomic %, molality and kg of salt per kg of water.
2.9 A mixture of gases has the following composition by weight at 298 K and 740 mm Hg.
Chlorine: 60%, Bromine: 25% and Nitrogen: 15%.
Express the composition by mole % and determine the average molecular weight.
2.10 Wine making involves a series of very complex reactions most of which are performed by microorganisms. The initial concentration of sugar determines the final alcohol content and sweetness of the wine.
The general convention is to adjust the specific gravity of the starting stock to achieve a desired quality of wine. The starting solution has a specific gravity of 1.075 and contains 12.7 weight % of sugar. If all the sugar is assumed to be C12H22O11, determine
(a) kg sugar/kg H2O (b) kg solution/m3 solution (c) g sugar/litre solution
2.11 The synthesis of ammonia proceeds according to the following reaction
2.11 The synthesis of ammonia proceeds according to the following reaction