Mole Concept And Stoichiometry Exe-5(C) Concise Class-10 ICSE Chemistry Selina Solutions Chapter-5. We Provide Step by Step Answer of Exe-5(C) Questions of Exercise-5 for ICSE Class-10. The given Solutions is according to the Latest editions. Visit official Website CISCE for detail information about ICSE Board Class-10.

Mole Concept And Stoichiometry Exe-5(C) Concise Class-10 ICSE Chemistry Selina Solutions Chapter-5

Board	ICSE
Book / Publication	Concise / Selina
Subject	Chemistry
Class	10th
Writer	Dr SP Singh
Chapter-5	Mole Concept And Stoichiometry
Topics	Exe-5(C)
Edition	2025-2026

Exe-5(C) Questions on Mole Concept And Stoichiometry

Page-91

Que-1: Give the empirical formula of:

(a) C₆H₆
(b) C₆H₁₂O₆
(c) C₂H₂
(d) CH₃COOH

Ans:
(a) CH
(b) CH₂O
(c) CH
(d) CH₂O

Que-2: Find the percentage of water of crystallisation in CuSO₄.5H₂O. (At. Mass Cu = 64, H = 1, O = 16, S = 32) 

Ans: Relative molecular mass of CuSO4, 5H2O
= 64 + 32 + 4 x 16 + 5(2 +16)
= 160 + 90
= 250
250 g of CuSO4, 5H2O contain 90g of water of crystallsation.
= (90/250) x 100
= 36%

Que-3: Calculate the percentage of phosphorus in

(a) Calcium hydrogen phosphate Ca(H₂PO₄)₂
(b) Calcium phosphate Ca₃(PO₄)₂

Ans:
(a) Molecular mass of Ca(H₂PO₄)₂ = 234
So, % of P = 2 31 100/234 = 26.5%
(b) Molecular mass of Ca₃(PO₄)₂ = 310
% of P = 2 31 100/310 = 20%

Que-4: Calculate the percent composition of Potassium chlorate KClO₃.

Ans: Molecular mass of KClO₃ = 122.5 g
And , % of K = 39 /122.5 = 31.8%
So % of Cl = 35.5/122.5 = 28.98%
Hence % of O = 3 16/122.5 = 39.18%

Que-5: Find the empirical formula of the compounds with the following percentage composition:

Pb = 62.5%, N = 8.5%, O = 29.0%

Ans: Element % At. mass Atomic ratio Simple ratio
Pb 62.5/207 = 0.3019
N 8.5/14 =0.6071
O 29.0/16  = 1.81
So, Pb(NO₃)₂ is the empirical formula.

Que-6: Calculate the mass of iron in 10 kg of iron ore which contains 80% of pure ferric oxide.

Ans:  ₂O₃ , Fe = 56 and O = 16
Molecular mass of Fe₂O₃ = 2 56 + 3 16 = 160 g
Iron present in 80% of Fe₂O₃ = (112/160) x 80 = 56g
So, mass of iron in 100 g of ore = 56 g
mass of Fe in 10000 g of ore = 56 10000/100
= 5.6 kg

Que-7: If the empirical formula of two compounds is CH and their Vapour densities are 13 to 39 respectively, find their molecular formula. (2015)

Ans:
For acetylene , molecular mass = 2 V.D = 2 13 = 26 g
The empirical mass = 12(C) + 1(H) = 13 g
n = molecular formula mass/ empirical formula mass= 26/13 = 2
Molecular formula of acetylene= 2 Empirical formula =C₂H₂
Similarly, for benzene molecular mass= 2 V.D = 2 39 = 78
n = 78/13=6
So, the molecular formula = C₆H₆

Que-8: Find the empirical formula of a compound containing 17.7% hydrogen and 82.3% nitrogen. (2018)

Ans:
Element % At. mass Atomic ratio Simple ratio
H 17.7 1  (17.7/1) = 17.7 (17.7/5.87) = 3
N 82.3 14 (82.3/14) = 5.87 (5.87/5.87) = 1
So, the empirical formula = NH₃

Que-9: On analysis, a substance was found to contain

C = 54.54%, H = 9.09%, O = 36.36%
The vapour density of the substance is 44,calculate;
(a) its empirical formula, and
(b) its molecular formula

Ans: Element % at. mass atomic ratio simple ratio
C 54.54 12 (54.54/12) = 4.55   2
H 9.09 1 (9.09/1) =9.09   4
O 36.36 16  (36.36/16) = 2.27  1
(a) So, its empirical formula = C₂H₄O
(b) empirical formula mass = 44
Since, vapour density = 44
So, molecular mass = 2 V.D = 88
Or n = 2
so, molecular formula = (C₂H₄O)₂ = C₄H₈O₂

Que-10: An organic compound ,whose vapour density is 45, has the following percentage composition

H=2.22%, O = 71.19% and remaining carbon. Calculate ,
(a) its empirical formula, and
(b) its molecular formula

Ans: Element % at. mass atomic ratio simple ratio
C 26.59 12 1
H 2.22 1 1
O 71.19 16 2
(a) its empirical formula = CHO₂
(b) empirical formula mass = 45
Vapour density = 45
So, molecular mass = V.D 2 = 90
so, molecular formula = C₂H₂O₄

Que-11: An organic compound contains H = 4.07%, Cl = 71.65% chlorine and remaining carbon. Its molar mass = 98.96. Find,

(a) Empirical formula, and
(b) Molecular formula

Ans: Element % at. mass atomic ratio simple ratio
Cl 71.6535.5 (71.65/35.5) = 2.01   1
H 4.071 ( 4.07/1) = 4.07   2
C 24.2812 (24.28/12) = 2.02   1
(a) its empirical formula = CH₂Cl
(b) empirical formula mass = 49.5
Since, molecular mass = 98.96
so, molecular formula = (CH₂Cl)₂ = C₂H₄Cl₂

Que-12: A hydrocarbon contains 4.8g of carbon per gram of hydrogen. Calculate

(a) the g atom of each
(b) find the empirical formula
(c) Find molecular formula, if its vapour density is 29.

Ans:
(a) The g atom of carbon = 4.8/12 = 0.4 and g atom of hydrogen = 1/1=1
(b) Element Given mass At. mass Gram atom Ratio
C 4.8 12 0.4 1 2
H 1 1 1 2.5 5
So, the empirical formula = C₂H₅
(c) Empirical formula mass = 29
Molecular mass = V.D 2 = 29 2 = 58
So, molecular formula = C₄H₁₀

Que-13: 0.2 g atom of silicon Combine with 21.3 g of chlorine. Find the empirical formula of the compound formed.

Ans:
Since, g atom of Si = given mass/mol. Mass
so, given mass = 0.2 28 = 5.6 g
Element mass At. mass Gram atom Ratio
Si5.6280.21
Cl 21.335.5 (21.3/35.5) = 0.6  3
Empirical formula = SiCl₃

Que-14: A gaseous hydrocarbon contains 82.76% of carbon. Given that its vapour density is 29, find its molecular formula. (2016)

Ans:
% of carbon = 82.76%
% of hydrogen = 100 – 82.76 = 17.24%

Element	% Weight	Atomic Weight	Relative No. of Moles	Simplest Ratio
C	82.76	12	82.76/12 = 6.89	6.89/6.89 = 1 x 2 = 2
H	17.24	1	17.24/1 = 17.24	17.24/6.89 = 2.5 x 2 = 5

Empirical formula = C₂H₅
Empirical formula weight = 2 x 12 + 1 x 5 = 24 + 5 = 29
Vapour Density = 29
Relative molecular mass = 29 x 2 = 58
N = Relative molecular mass/ Empirical Weight = 58/29 = 2
Molecular formula = n x empirical formula
= 2 x C₂H₅
= C₄H₁₀

Que-15: In a compound of magnesium (Mg = 24) and nitrogen (N = 14), 18 g of magnesium combines with 7g of nitrogen. Deduce the simplest formula by answering the following questions.

(a) How many gram- atoms of magnesium are equal to 18g?
(b) How many gram- atoms of nitrogen are equal to 7g of nitrogen?
(c) Calculate simple ratio of gram- atoms of magnesium to gram-atoms of nitrogen and hence the simplest formula of the compound formed.

Ans:
(a) G atoms of magnesium = 18/24 = 0.75 or g- atom of Mg
(b) G atoms of nitrogen = 7/14 = 0.5 or 1/2 g- atoms of N
(c) Ratio of gram-atoms of N and Mg = 1:1.5 or 2:3
So, the formula is Mg₃ N₂

Que-16: Barium chloride crystals contain 14.8% water of crystallization. Find the number of molecules of water of crystallization per molecule.

Ans: Barium chloride = BaCl₂.x H₂O
Ba + 2Cl + x[H₂ + O]
=137+ 235.5 + x [2+16]
=[208 + 18x] contains water = 14.8% water in BaCl₂.x H₂O
=[208 + 18 x] 14.8/100 = 18x
=[104 + 9x] 2148=18000x
=[104+9x] 37=250x
=3848 + 333x =2250x
1917x =3848
x = 2 molecules of water

Que-17: Urea is very important nitrogenous fertilizer. Its formula is CON₂H₄.Calculate the percentage of nitrogen in urea. (C=12,O=16 ,N=14 and H=1).

Ans: Molar mass of urea; CON₂H₄ = 60 g
So, % of Nitrogen = 28 100/60 = 46.66%

Que-18: Determine the formula of the organic compound if its molecule contains 12 atoms of carbon. The percentage compositions of hydrogen and oxygen are 6.48 and 51.42 respectively.

Ans: Element % At. mass Atomic ratio Simple ratio
C 42.1 12 3.5 1
H 6.48 1 6.48 2
O 51.42 16 3.2 1
The empirical formula is CH₂O
Since the compound has 12 atoms of carbon, so the formula is
C₁₂ H₂₄ O_12.

Que-19:

(a) A compound with empirical formula AB₂, has the vapour density equal to its empirical formula weight. Find its molecular formula.
(b) A compound with empirical formula AB has vapour density 3 times its empirical formula weight. Find the molecular formula.
(c)10.47 g of a compound contained 6.25 g of metal A and rest non-metal B. Calculate the empirical formula of the compound [At. wt of A = 207, B = 35.5]

Ans:
(a) Now since the empirical formula is equal to vapour density and we know that vapour density is half of the molecular mass i.e. we have n=2 so, the molecular formula is A₂B₄.
(b) Since molecular mass is 2 times the vapour density, so Mol. Mass = 2 V.D
Empirical formula weight = V.D/3
So, n = molecular mass/ Empirical formula weight = 6
Hence, the molecular formula is A₆B₆
(c) Given:
Wt. of the compound: 10.47g
Wt. of metal A: 6.25g
Wt. of non-metal B: 10.47 – 6.25 = 4.22g

Element	mass	At. Wt.	Relative no. of atoms	Simplest ratio
A	6.25g	207	6.25/207=0.03	0.03/0.03=1
B	4.22g	35.5	4.26/35.5=0.12	0.12/0.03=4

 

Hence, the empirical formula is AB₄

Que-20: A hydride of nitrogen contains 87.5% per cent by mass of nitrogen. Determine the empirical formula of this compound.

Ans:
Atomic ratio of N = 87.5/14 =6.25
Atomic ratio of H= 12.5/1 = 12.5
This gives us the simplest ratio as 1:2
So, the molecular formula is NH₂

Que-21: A compound has O=61.32%, S= 11.15%, H=4.88% and Zn=22.65%.The relative molecular mass of the compound is 287 amu. Find the molecular formula of the compound, assuming that all the hydrogen is present as water of crystallization.

Ans:
Element % at. mass atomic ratio simple ratio
Zn 22.65 65 0.348 1
H 4.88 1 4.88 14
S 11.15 32 0.348 1
O 61.32 16 3.83 11
Empirical formula of the given compound =ZnSH₁₄O₁₁
Empirical formula mass = 65.37+32+141+11+16=287.37
Molecular mass = 287
n = Molecular mass/Empirical formula mass = 287/287=1
Molecular formula = ZnSO₁₁H₁₄
=ZnSO₄.7H₂O

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