SOME BASIC CONCEPTS OF CHEMISTRY

System International de units/ SI units

Significant figures
Significant figures are meaningful digits which are known with certainty plus one which is estimated
or uncertain. The uncertainty is indicated by writing the certain digits and the last uncertain digit.
if we write a result as 11.2 mL, we say the 11 is certain and 2 is uncertain and the uncertainty would
be ±1 in the last digit.
Rules for determining the number of significant figures
i) All non-zero digits are significant
ii) Zeros preceding the first non-zero digit are not significant
iii) Zeros between two non-zero digits are significant.
iv) Zeros at the end or right of the number are significant provided they are on the
right side of the decimal point. But, if otherwise, the zeros are not significant
Standard Prefix
 Laws of Chemical Combination
Law of conservation of Mass [ Antoine Lavoisier]

It states that, mass can neither be created nor be destroyed

MassReactant = MassProduct

Law of
definite proportion [Joseph Proust] –
It state that a given compound always contains elements in a certain proportion by mass.

Example: H2 and O2 are present in H2O in a proportion of 1:8 by mass

1 gram Hydrogen + 8 gram Oxygen à 9 gram water + no biproduct
1 gram Hydrogen + 9 gram Oxygen à 9 gram water + 1 gram excess oxygen.

Question
0.7 g of iron reacts directly with 0.4 g of sulphur to form ferrous sulphide. If 2.8 g of iron is dissolved in
dilute HCl and excess of sodium sulphide solution is added, 4.4 g of iron sulphide is precipitated. Prove
the law of constant composition.
Case 1; formation of FeS from Fe and S Fe (s) + S(s) *⎯⎯, FeS (s)
0.7 g Fe + 0.4 g S = 11.4 g FeS
Definite proportion for Fe : S = 0.7 :0.4 = 7 : 4
Case 2 ; formation of FeS from Fe, dil.HCl and sodium sulphide
Fe (s) + 2HCl(aq) + Na" 𝑆 (𝑎𝑞) *⎯⎯, FeS (s) +2𝑁𝑎𝐶𝑙 (𝑎𝑞) + 𝐻" (𝑔)
Mass of Fe = 2.8
Mass of S = 4.4 - 2.8 = 1.6
g Fe + 0.4 g S = 11.4 g FeS
Definite proportion for Fe : S = 2.8 : 1.6 = 7 : 4
Law of Multiple Proportion [ John Dalton ]
It state that, when two elements combine to form more than one compound, the mass of one element that
combines with fixed mass of the other element, are in the ratio of simplest whole numbers.

For example Hydrogen and Oxygen combine to form water H2O (1:8) and hydrogen peroxide H2O2 (1:16)
in which mass of oxygen combines with fixed mass of hydrogen is in a ratio by 1:2
Avogadro’s Law :
It state that, “equal volumes of all gases under the same conditions of pressure and temperature contain
the same number of molecules."

At standard temperature and pressure, 1 mole of the gas will occupy 22.4L and will contain 6.022 x 1023
particles of the matter.
“At STP 6.022 x 1023 particles of any gas will occupy 22.4L”
The Avogadro constant is named after the early nineteenth century Italian scientist Amedeo Avogadro.
Gay Lussac’s Law
It state that , volume of gases taking part in a chemical reaction show simple whole number ratios to one
another when those volumes are measured at the same temperature and pressure.
Ratio of the volume of the gases= ratio of the number of moles
– N"(g) + 3H" (g) → 2 NH# (g)
– 1L N" : 3L H" : 2L NH#
ATOMIC MASS UNIT
An atomic mass unit or amu is one twelfth of the mass of an unbound atom of carbon-12. It is a unit of
mass used to express atomic masses and molecular masses .( Known As: Unified Atomic Mass Unit (u).
1 amu = 1.6726 × 10 − 24 gram
Mole Concepts

The mole (mol) is the amount of a substance that contains as many elementary entities as there are atoms
in exactly 12.00 grams of 12C

I MOLE sodium = 6.O22 x 10 23 atoms of sodium

I MOLE sodium=23 u x 6.O22 x 10 23 gram of sodium
ie, I MOLE sodium =23 x (1.6726 x 10 -24 gram)x 6.O22 x 10 23 =26 gram of sodium
Thus 23 gram is the molar mass of sodium

Molar mass
It is the mass of 1 mole substance or molecular mass /atomic mass in gram
Eg: Molar mass of H2 is 2g/mol
Molar mass of H atom is 1g/mol
Molar mass of Na element is 23g/mol
Molar mass H2O is 18g/mol
S.No Symbol Atomic / molecular mass ( 1mole ( in Avogadro no. Molar mass (g/mol)
. u) g)
1. O 16 u 16 g 6.022 x 1023 atoms 16 g mol -1
2. N2 28 u 28 g 6.022 x 1023 molecules 28 g mol -1
3. HCl 36.5 u 36.5 g 6.022 x 1023 molecules 36.5 g mol -1

Volume
mass Number of of gas at
particles STP
s
Molar mole
mole mole 𝟐𝟐. 𝟒 𝐋
mass 𝟔. 𝟎𝟐𝟐 × 𝟏𝟎𝟐𝟑 s
s s

Some important Formulas

𝐦𝐚𝐬𝐬 𝐠𝐢𝐯𝐞𝐧
𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐦𝐨𝐥𝐞𝐬 =
𝐦𝐨𝐥𝐚𝐫 𝐦𝐚𝐬𝐬
𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐩𝐚𝐫𝐭𝐢𝐜𝐥𝐞𝐬 𝐠𝐢𝐯𝐞𝐧
𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐦𝐨𝐥𝐞𝐬 =
𝟔.𝟎𝟐𝟐×𝟏𝟎𝟐𝟑

𝐯𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐠𝐚𝐬 𝐚𝐭 𝐒𝐓𝐏

𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐦𝐨𝐥𝐞𝐬 =
𝟐𝟐.𝟒 𝐋
1.How many mole of Al is present in 35 g of it ?
Ans: Molar mass of Al= 27 g/mol
35g
n= = 1.3 mol
27g/mol

ii) How many grams of Si is present in 0.2 mol of SiO2 ?

Ans: 1 mol SiO2 = 1 mol Si + 1mol O2

Then 0.2 mole SiO2 containing 0.2 mol Si
Mass of Si = 0.2mol Si x 24 g/mol = 4.8 g
Percentage composition
𝐌𝐚𝐬𝐬 𝐨𝐟𝐞𝐥𝐞𝐦𝐞𝐧𝐭 𝐱 𝟏𝟎𝟎
%𝐨𝐟 𝐚𝐧 𝐞𝐥𝐞𝐦𝐞𝐧𝐭 𝐢𝐧 𝐚 𝐜𝐨𝐦𝐩𝐨𝐮𝐧𝐝 =
𝐦𝐨𝐥𝐞𝐜𝐮𝐥𝐚𝐫 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐭𝐡𝐞 𝐜𝐨𝐦𝐩𝐨𝐮𝐧𝐝

Q: Calculate the percentage composition of H2 element in C6H12O6

2 x 100
% of Hydrogen element in C$ H%" O$ = = 6.6%
180
Empirical Formula and molecular Formula
Empirical formula is the simplest whole number ratio of each element by mass in a compound and
molecular formula is actual whole number ratio of each element by the mass in a compound

Eg : CH is the empirical formula of benzene C$ H$

CH2O is the empirical formula of glucose C$ H%" O$

Relation between Empirical Formula and molecular Formula

Molecular formula = ratio& Z & × emperical formula
+,-./0-12 +133
Where ratio* Z * =
.+4.25/1- 6,2+0-1 +133
Q: Derive the empirical formula and molecular formula of the compound which contain 54.55% Carbon,
9.06% Hydrogen and 36.9%Oxygen. (Molecular mass of thecompound is 88 amu)
Element Mass % No; of moles 𝑁𝑜; 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 Simple whole number
𝑙𝑒𝑎𝑠𝑡 𝑛𝑢𝑚𝑏𝑒𝑟
C 54.55% '(.''
= 4.54 4.54 2
%" = 1.97
2.30
≈2
H 9.06% 9.06 9.06 4
= 9.06 = 3.93
1 2.30
≈4
O 36.9% 36.9 2.30 1
= 2.30 =1
16 2.30
Empirical formula = C" H( O%
Empirical formula mass = ( 2 × 12) + ( 4 × 1) + (1 × 16) = 44 amu
Molecular mass = 88amu
 88 amu
Ratio& Z & = =𝟐
44 amu

Molecular formula = ( C" H( O% ) × 2 = C( H* O"

Stoichiometric Calculations
Calculation based on definite proportion is called stoichiometric calculation
Question:
How much copper can be obtained from 100 g of copper sulphate (CuSO4) ismixed with Zn granules as
per the equation : Zn + CuSO4 à ZnSO4 + Cu
Mass of copper sulphate = 100 g
Molar mass of copper suphate = 63.5 + 32 + (16 × 4) = 159.5 g/mol
100
Moles of copper sulphate = = 0.627
159.5
Moles of Copper = stoichiometric ratio × 0.627
1
[ stoichiometric ratio for copper to copper sulphate = ]
1
1
Moles of Copper = × 0.627 = 0.627
1
Mass of copper = mole × molar mass = 0.627 × 63.5 = 39.81 g

Limiting Reagent
Limiting reagent can be defined as lesser amount reactant that undergo a complete reaction and can limit
the amount of product is called limitingreagent
Q: Dinitrogen and dihydrogen react with each other to produce ammonia accordingto the following
chemical equation: N2(g) + H2(g) → 2NH3(g)
Calculate the mass of ammonia produced if 10 kg dinitrogen reacts with 50 kg of dihydrogen and also
identify the limiting reagent.
𝐌𝐨𝐥𝐞𝐬 𝐨𝐟 𝐚𝐦𝐦𝐨𝐧𝐢𝐚 𝐟𝐫𝐨𝐦 𝟓𝟎 𝐤𝐠 𝐨𝐟 𝐝𝐢𝐡𝐲𝐝𝐫𝐨𝐠𝐞𝐧
50×103
Moles of H! = = 25 × 103
2
2
[ stoichiometric ratio for ammonia to hydrogen = ]
3
2
Moles of ammonia = 3 × 25 × 103 = 16.6 × 10! moles
𝐌𝐨𝐥𝐞𝐬 𝐨𝐟 𝐚𝐦𝐦𝐨𝐧𝐢𝐚 𝐟𝐫𝐨𝐦 𝟏𝟎 𝐤𝐠 𝐨𝐟 𝐝𝐢𝐧𝐢𝐭𝐫𝐨𝐠𝐞𝐧.
<=×<=#
Moles of N" = = 0.35 × 10$
>?
2
[ stoichiometric ratio for ammonia to nitrogen = ]
1
2
Moles of ammonia = 1 × 0.35 × 103 = 0.7 × 10# moles
Moles of NH# from N" is smaller than moles of NH# from H"
Therefore dinitrogen N" is limiting reagent
And mass of ammonia = 0.7 × 10# × 17 = 11.1 × 10# g = 11.1 kg

Stoichiometric Calculations in Solutions

Solution Homogeneous mixture of two or more components. Eg: sugar solution
Solute + solvent = solution
Solute Lesser amount component which loses its physical state. It will be dissolved in
solvent
 Solvent Higher amount component which will not lose its physical state
Physical state of solvent= physical state of solution, It dissolves solutes
Concentration Expressions methods
WA = weight of solvent, WB =weight of solute nA = no; of moles of solvent
Mole fraction in solution: Ratio of number of mole of one component to total number of moles
𝒏𝑨 𝒏𝑩
𝑿𝑨 = 𝒏𝑨-𝒏𝑩 𝑿𝑩 = 𝒏𝑨-𝒏𝑩 Total mole fraction 𝑿𝑨 + 𝑿𝑩 = 𝟏
Mole fraction doesn’t have any unit.
Molarity: number of moles of solute present in 1L solution
𝑾𝑩 × 𝟏𝟎𝟎𝟎 𝒅 × 𝟏𝟎 × 𝒎𝒂𝒔𝒔%
𝑴𝒐𝒍𝒂𝒓𝒊𝒕𝒚, 𝑴 = 𝑴𝒐𝒍𝒂𝒓𝒊𝒕𝒚, 𝑴 =
𝑴𝑩 × 𝑽 𝑴𝑩
Unit = mol/L or molar or M
Molarity inversely proportional to temperature because volume increases with temperature
Molality: number of moles of solute present in 1kg solvent
Molality is independent on temperature since mass is not depends on temperature.
Unit = mol/kg or molal or m
𝑾𝑩 ×𝟏𝟎𝟎𝟎 𝒎%×𝟏𝟎𝟎𝟎
𝑴𝒐𝒍𝒂𝒍𝒊𝒕𝒚, 𝒎 = 𝑴𝒐𝒍𝒂𝒍𝒊𝒕𝒚, 𝒎 = 𝑴
𝑴𝑩 ×𝑾𝑨 𝑩 ×(𝟏𝟎𝟎7𝒎%)

Relation between molarity, molality and density

𝒅𝒆𝒏𝒔𝒊𝒕𝒚
𝑴𝒐𝒍𝒂𝒓𝒊𝒕𝒚, 𝑴 =
𝟏 𝑴𝑩
(𝒎 + 𝟏𝟎𝟎𝟎 )

1. Calculate the mass percent of different elements present in Sodium Sulphate (Na2SO4).
Solution :
Molar mass of Na2SO4 = [(2 × 23.0) + (32.00) + 4 (16.00)] = 142 g
9:;; <= >?:> @A@B@C> DC E<BF<GCH × %II
Mass percent of an element =
9<A:J B:;; <= >?:> E<BF<GCH
($ × %II
∴ Mass percent of sodium (Na): = 32.39%
%("
#" × %II
Mass percent of sulphur(S): = 22.54%
%("
$( × %II
Mass percent of oxygen:(O): = 45.07%
%("
2. Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen
by mass.
Solution :
Element Mass % No; of 𝑁𝑜; 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 Simple whole number
moles 𝑙𝑒𝑎𝑠𝑡 𝑛𝑢𝑚𝑏𝑒𝑟
Fe 69.9% 𝟔𝟗.𝟗
= 1.25 1 x2 =2
''.*' =𝟏
1.25
1.25
O 30.1% 𝟑𝟎.𝟏
= 𝟑𝟎. 𝟏 1.5 x 2 = 3
%$.II = 𝟏. 𝟓
1.25
1.88
Hence Empirical formula is Fe2O3
3. Calculate the amount of carbon dioxide that could be produced when
(i) 1 mole of carbon is burnt in air.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
(iii) 2 moles of carbon are burnt in 16 g of dioxygen.
Solution :
The balanced reaction of combustion of carbon in dioxygens is:
C(s) + O2(g) → CO2 (g)
1mole 1mole (32g) 1mole (44g)
(i) Here the mole ratio between C and CO2 is 1:1
Hence 1 mole C produces 1 mole CO2
(ii) Here, 1 mole C needed 1 mole O2 for the complete combustion.
%$ N
However, number of moles of O2 in 16 gram is #"N/B<A = 0.5 𝑚𝑜𝑙𝑒 which is not sufficient.
Hence O2 is Limiting reagent.
Therefore, 0.5 mole O2 produce 0.5 mole CO2
(iii) Here, 2 mole C needed 2 mole O2 for the complete combustion.
%$ N
However, number of moles of O2 in 16 gram is #"N/B<A = 0.5 𝑚𝑜𝑙𝑒 which is not sufficient.
Hence O2 is Limiting reagent.
Therefore, 0.5 mole O2 produce 0.5 mole CO2
4. Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar
aqueous solution. Molar mass of sodium acetate is 82.0245g mol-1.
+133%&'()* × <===
Solution : 𝑀 =
+,-12 +133%&'()* × @,-0+.%&'()+&,
𝑚𝑎𝑠𝑠#$%&'( × 1000
0.375 =
82.0245 × 500
=.ABC × ?>.=>DC × C==
𝑚𝑎𝑠𝑠;<AG>@ = = 15.380g
<===

5. Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41
g mL-1 and the mass per cent of nitric acid in it being 69%.
Solution :
𝑚𝑎𝑠𝑠% × 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 × 10
𝑀=
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠;<AG>@
$P × %.(%× %I
𝑀= = 15.44. mol/L
$#
6. How much copper can be obtained from 100 g of copper sulphate (CuSO4 )?
Solution : 1 mole of CuSO4 = 1 mole Cu + 1 mole S + 4 mole O
%II %II
100 g CuSO4 = ($#.') - (#".II) - ((%$.II) = = 0.63 𝑚𝑜𝑙𝑒 CuSO4
%'P.'
0.63 𝑚𝑜𝑙𝑒 CuSO4 = 0.63 mole Cu
0.63 mole Cu = 0.63 x 63.5 =40.5 g
7. Determine the molecular formula of an oxide of iron in which the mass per cent of iron and
oxygen are 69.9 and 30.1 respectively.Given that the molar mass of the oxide is 159.7 g mol-1
Solution :
Element Mass % No; of 𝑁𝑜; 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 Simple whole number
moles 𝑙𝑒𝑎𝑠𝑡 𝑛𝑢𝑚𝑏𝑒𝑟
Fe 69.9% 𝟔𝟗.𝟗
= 1.25 1 x2 =2
''.*' =𝟏
1.25
1.25
O 30.1% 𝟑𝟎.𝟏
= 𝟑𝟎. 𝟏 1.5 x 2 = 3
%$.II = 𝟏. 𝟓
1.25
1.88
Hence Empirical formula is Fe2O3
Empirical formula Mass = ( 2 x 55.85 ) + (3 x 16) =159.7
Molecular mass given is 159.7
B<A@EGA:J B:;;
Then Molecular formula = @BFDJDE:A =<JBGA: B:;; × 𝑒𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎
%'P.Q
Molecular formula = %'P.Q × Fe" O# = Fe" O#
8. Calculate the atomic mass (average) of chlorine using the following data :
% Natural Abundance Molar Mass
35Cl 75.77 34.9689
37Cl 24.23 36.9659
Solution :
(=> ? @>.@@) B (=@ ? CD.C=)
atomic mass (average) = = 35.5
EFF
9. In three moles of ethane (C2H6), calculate the following :
(i) Number of moles of carbon atoms.
(ii) Number of moles of hydrogen atoms.
(iii) Number of molecules of ethane.
Solution :
1 mole of C2H6 = 2 mole C + 6 mole H
(i) 6 mole Carbon atoms
(ii) 18 moles of Hydrogen atoms
(iii) 3×6.02 × 1023= 18.06 ×1023 molecules.
10. What is the concentration of sugar (C12H22O11) in mol L-1 if its 20 g are dissolved in enough
water to make a final volume up to 2L?
Solution :
𝑚𝑎𝑠𝑠;<AG>@ × 1000
𝑀=
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠;<AG>@ × 𝑣𝑜𝑙𝑢𝑚𝑒;<AG>D<C
20 × 1000
M= = 0.0293 𝑚𝑜𝑙/𝐿
342 × 2000
11. If the density of methanol is 0.793 kg L-1 , what is its volume needed for making 2.5 L of its 0.25
M solution?
Solution :
Molar mass of methanol (CH3OH) = (1×12) + (4×1) + (1×16)
= 32 g mol-1 = 0.032 kg mol-1
𝑚𝑎𝑠𝑠;<AG>@ (𝑘𝑔)
𝑀=
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠;<AG>@ (𝑘𝑔) × 𝑣𝑜𝑙𝑢𝑚𝑒;<AG>D<C
I.QP#
𝑀 = I.I#" × %= 24.78 mol/L
Applying, 𝑀% 𝑉% (𝐺𝑖𝑣𝑒𝑛 𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛) = 𝑀" 𝑉" (𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑡𝑜 𝑏𝑒 𝑝𝑟𝑒𝑝𝑎𝑟𝑒𝑑)
24.78 × 𝑉1 = 0.25 × 2.5 𝐿
𝑉1 = 0.02522 𝐿 = 25.22 𝑚𝐿
12. What do you mean by significant figures ?
Solution : Significant figures are meaningful digits which are known with certainty including the last
digit whose value is uncertain.
For example,
In 11.2546 g, there are 6 significant figures but here 11.254 is certain and 6 is uncertain and the
uncertainty would be ±1 in the last digit. Hence last uncertain digit is also included in Significant figures.
13. Express the following in the scientific notation:
(i) 0.0048 (ii) 234,000 (iii) 8008 (iv) 500.0 (v) 6.0012
-3
Solution : (i) 0.0048 = 4.8× 10
(ii) 234, 000 = 2.34× 105
(iii) 8008 = 8.008× 103
(iv) 500.0 = 5.000× 102
(v) 6.0012 = 6.0012× 100
14. How many significant figures are present in the following?
(i) 0.0025 (ii) 208 (iii) 5005 (iv) 126,000 (v) 500.0 (vi) 2.0034
Solution :
(i) 2 (ii) 3 (iii) 4 (iv) 3 (v) 4 (vi) 5
15. The following data are obtained when dinitrogen and dioxygen react together to form different
compounds:
Mass of dinitrogen Mass of dioxygen
(i)14 g 16 g
(ii)14 g 32 g
(iii)28 g 32 g
(iv)28 g 80 g
(a) Which law of chemical combination is obeyed by the above experimental data?Give its
statement.
(b) Fill in the blanks in the following conversions:
(i) 1 km = ...................... mm = ...................... pm
(ii) 1 mg = ...................... kg = ...................... ng
(iii) 1 mL = ...................... L = ...................... dm3
Solution :
(a) law of multiple proportions.
It state that, two elements combines to form two or more compounds in which one element in a fixed
mass and another element in a variable ratio by their mass
(b) i) 1 km = 106 mm = 1015 pm
(ii) 1 mg = 10-6 kg = 10-6 ng
(iii) 1 mL = 10-3 L = 103 dm3
16. In a reaction
A + B2 → AB2
Identify the limiting reagent, if any, in the following reaction mixtures.
(i) 300 atoms of A + 200 molecules of B
(ii) 2 mol A + 3 mol B
(iii) 100 atoms of A + 100 molecules of B
(iv) 5 mol A + 2.5 mol B
(v) 2.5 mol A + 5 mol B
Solution : (i) According to the reaction, 1 atom of A reacts with 1 molecule of B.
∴200 molecules of B will react with 200 atoms of A, thereby leaving 100 atoms of A unreacted. Hence,
B is the limiting reagent.
(ii) According to the reaction, 1 mol of A reacts with 1 mol of B.
∴ 2 mol of A will react with only 2 mol of B leaving 1 mol of B. Hence, A is the limiting reagent.
(iii) 1 atom of A combines with 1 molecule of B.
∴ All 100 atoms of A will combine with all 100 molecules of B. Hence, the mixture is stoichiometric and
ther is no limiting reagent.
(iv) 1 mol of atom A combines with 1 mol of molecule B.
∴ 2.5 mol of B will combine with only 2.5 mol of A. and 2.5 mol of A will be left unreacted. Hence, B is
the limiting reagent.
(v) 1 mol of atom A combines with 1 mol of molecule B.
∴ 2.5 mol of A will combine with only 2.5 mol of B and the remaining 2.5 mol of B will be left. Hence, A
is the limiting reagent.
17. Dinitrogen and dihydrogen react with each other to produce ammonia according to the
following chemical equation:
N2(g) + 3H2(g) → 2NH3(g)
(i) Calculate the mass of ammonia produced if 2.00×103g dinitrogen reacts with 1.00×103g of
dihydrogen.
(ii) Will any of the two reactants remain unreacted?
(iii) If yes, which one and what would be its mass?
Solution :
Moles of ammonia from 1.00×103g of dihydrogen
%.II×%I%
Moles of H2 = = 0.5 × 10#
"
"
Moles of ammonia = # × 0.5 × 10# = 0.33 x 103 moles
Moles of ammonia from 2.00×103g of dinitrogen
".II×%I%
Moles of N2 = = 0.071 × 10#
"*
"
Moles of ammonia = % × 0.071 × 10# = 0.142 x 103 moles
Moles of NH3 from N2 < moles of NH3 from H2
Hence dinitrogen is limiting reagent
Therefore mass of ammonia = 0.142 x 103x 17 = 2.414 x103g
(ii)yes, H2 is the excess reagent and remains unreacted.
(iii) H2 is the excess reagent
Mass of dihydrogen left unreacted = total mass – reacted mass
#
=1000 – (% × 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑁" 𝑟𝑒𝑎𝑐𝑡𝑒𝑑 × 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐻" )
#
=1000 – (% × 0.071 × 10# × 2 ) = 571.4 g
18. How are 0.50mol Na2CO3 and 0.50M Na2 CO3 different?
Solution : Molar mass of Na2CO3 = (2×23) +12.00+(3×16) = 106 g mol-1
∴0.50 mol Na2CO3 means 0.50 ×106g = 53g
0.50 M Na2CO3 means 0.50 mol of Na2CO3 i.e. 53g of Na2CO3 are present in 1litre of the solution.
19. If ten volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of
water vapour would be produced?
Solution : 2H2(g) + O2(g) → 2H2O(g)
2 volume of H2 gives 2 volume water
Then 10 volume of H2 gives 10 volume of water
20. Which one of the following will have largest number of atoms?
(i) 1 g Au (s) (ii) 1 g Na (s)
(iii) 1 g Li (s) (iv) 1 g of Cl2 (g)
Solution :
%
(i) 1 g Au = %PQ × 6.022 × 10"# 𝑎𝑡𝑜𝑚𝑠
%
(ii) 1 g Na = "# × 6.022 × 10"# 𝑎𝑡𝑜𝑚𝑠
 %
(iii) 1 g Li = × 6.022 × 10"# 𝑎𝑡𝑜𝑚𝑠
#
%
(iv) 1 g Cl2 = × 6.022 × 10"# 𝑎𝑡𝑜𝑚𝑠
Q%
Thus, 1 g of Li has the largest number of atoms.
21. What will be the mass of one 12-C atom in g ?
Solution :
1 mol of 12-C atoms = 6.022 ×1023 atoms = 12g
%"
∴ Mass of 1 atom 12C = $.I""×%I&% 𝑔 = 1.9927× 10-23 g
22. Calculate the number of atoms in each of the following
(i) 52 moles of Ar (ii) 52 u of He (iii) 52 g of He.
Solution : (i) 1 mol of Ar = 6.022 × 1023atoms
∴ 52 mol of Ar = 52 × 6.022×1023atoms = 3.131 × 1025atoms
(ii) 1 atom of He = 4 u of He
4 u of He = 1 Atom of He
∴ 52 u of He = 1/4 × 52 = 13 atoms
(iii) 1 mol of He = 4 g = 6.022 × 1023atoms
∴ 52 g of He = (6.022 × 1023/4) × 52 atoms = 7.8286 × 1024atoms
23. A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen
gives 3.38 g carbon dioxide , 0.690 g of water and no other products. A volume of 10.0 L (measured
at STP) of this welding gas is found to weigh 11.6 g. Calculate
(i) empirical formula, (ii) molar mass of the gas, and. (iii) molecular formula.
%"
Solution : Amount of carbon in 3.38 g of CO2 = (( × 3.38 𝑔 = 0.9218 g
"
Amount of hydrogen in 0.690 g H2O = %* × 0.690𝑔 = 0.0767 g
The compound contains only C and H, therefore total mass of the compound
= 0.9218 + 0.0767 = 0.9985 g
I.P"%*
% of C in the compound = I.PP*' × 100 = 92.32%
I.IQ$Q
% of H in the compound = × 100 = 7.68%
I.PP*'
(i) Calculation of empirical formula,
Element Mass % No; of 𝑁𝑜; 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 Simple whole number
moles 𝑙𝑒𝑎𝑠𝑡 𝑛𝑢𝑚𝑏𝑒𝑟
C 92.32% P".#"
= 7.68 7.68 1
%" =1
7.68
H 7.68% Q.$*
= 7.68 7.68 1
% =1
7.68
Hence empirical formula = C1H1
𝟏𝟎.𝟎
(ii) 𝐦𝐨𝐥𝐞𝐬 𝐚𝐭 𝐒𝐓𝐏 = 𝟐𝟐.𝟒 = 𝟎. 𝟒𝟒𝟔𝟒
𝐦𝐚𝐬𝐬 𝟏𝟏.𝟔
Molar mass = = 𝟎.𝟒𝟒𝟔𝟒 = 𝟐𝟔
𝐦𝐨𝐥𝐞𝐬
(iii) Empirical formula Mass = ( 1 x 12 ) + (1 x 1) =13
Molecular mass calculated is 23
B<A@EGA:J B:;;
Then Molecular formula = @BFDJDE:A =<JBGA: B:;; × 𝑒𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎
"$
Molecular formula = %# × CH = C" 𝐻"
24. Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction,
CaCO3 (s) + 2HCl (aq) → CaCl2(aq) + CO2(g) + H2O(l)
What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?
 GHIJKLMN × OHIPQR(QS) F.@>×C>
Solution : 𝑀𝑜𝑙𝑒𝑠 𝑜𝑓 𝐻𝐶𝑙 = = = 0.01875 Moles
EFFF EFFF
%
Moles of CaCO3 = × 0.001875= 0.0093
"
Mass of CaCO3 = 0.0093 x 100 (molarmass of CaCO3 ) =0.93 g
25. Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous
hydrochloric acid according to the reaction
4HCl (aq) + MnO2(s) → 2H2O(l) + MnCl2(aq) + Cl2(g)
How many grams of HCl react with 5.0 g of manganese dioxide?
KLMM C C
Solution : 𝑀𝑜𝑙𝑒𝑠 𝑜𝑓 𝑀𝑛𝑂" =
KNOLP KLMM
= (CCRA>) = ?B = 0.057Moles
D
Moles of HCl = × 0.057= 0.228 moles
<
Mass of HCl = 0.228 x 36.5(molarmass of HCl) =8.322 g
26. Calculate the molality of 3 molar aqueous solution of NaCl if its density is 1.25g/mL.
density
𝑀=
1 molar mass-./012
m+ 1000
1.25
3=
1 (35.5 + 23)
+
𝑚 1000
% '*.'
3 × (B + %III)= 1.25
E >T.> E.C>
+ =
Q EFFF =
EFFFB>T.>Q E.C>
=
EFFF Q =
1.25 × 1000𝑚
1000 + 58.5𝑚 =
3
3000 + 175.5𝑚 = 1250𝑚
3000 = (1250𝑚 − 175.5𝑚) = 1074.5𝑚
#III
m= %IQ(.' = 2.79 𝑚𝑜𝑙/𝑘𝑔
27. A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine. Its molar mass is
98.96 g. What are its empirical and molecular formulas?
Element Mass % No; of moles 𝑁𝑜; 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 Simple whole number
𝑙𝑒𝑎𝑠𝑡 𝑛𝑢𝑚𝑏𝑒𝑟
C 24.27% "(."Q 2.02 1
= 2.02 =1
%" 2.01
H 4.07% (.IQ 4.07 2
= 4.07 = 2.01
% 2.01
Cl 71.65% 71.65 2.02 1
= 2.01 = 1.00
35.5 2.01
Hence Empirical formula is CH2Cl
Empirical formula Mass = ( 1 x 12 ) + (2 x 1) + (1x 35.5) =49.5
Molecular mass given is 98.96
B<A@EGA:J B:;;
Then Molecular formula = @BFDJDE:A =<JBGA: B:;; × 𝑒𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎
P*.P$
Molecular formula = (P.'
× CH" Cl. = C2H4Cl2
28. Calculate the amount of water (g) produced by the combustion of 16 g of methane.
CH4 + 2O2 à CO2 + 2H2O
Moles of water from 16g CH4
%$
Moles of CH4 = =1
%$
 "
Moles of H2O = × 1 = 2 moles
%
Mass of water = 2 x 18 =36 g
29. Calculate the molarity of NaOH in the solution prepared by dissolving its 4 g in enough water to
form 250 mL of the solution.
𝑚𝑎𝑠𝑠;<AG>@ × 1000
𝑀=
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠;<AG>@ × 𝑣𝑜𝑙𝑢𝑚𝑒;<AG>D<C
D × <=== D × <===
𝑀= 𝑀=
(>AR<TR<) ×>C= D= ×>C=
M= 0.4 =mol/
30. Fill in the blanks in the following conversions:
(i) 1 km = ...................... mm = ...................... pm
(ii) 1 mg = ...................... kg = ...................... ng
(iii) 1 mL = ...................... L = ...................... dm3
Solutions:
(i) 𝟏𝐤𝐦 = 𝟏𝟎𝐱 × 𝐦𝐦
1 × 10# m = 10U × 107# m
10U = 10# × 10# = 10$
𝟏𝟎𝟔 𝐦𝐦 = 𝟏𝟎𝐱 × 𝐩𝐦
10$ × 107# m = 10U × 107%" m
10U = 10$ × 107# × 10%" = 10%'
1 km = 𝟏𝟎𝟔 mm = 𝟏𝟎𝟏𝟓 pm

(ii) 𝟏𝐦𝐠 = 𝟏𝟎𝐱 × 𝐤𝐠

1 × 107# g = 10U × 10# g
10U = 107# × 107# = 107$
𝟏𝟎7𝟔 𝐤𝐠 = 𝟏𝟎𝐱 × 𝐧𝐠
107$ × 10# g = 10U × 107P g
10U = 107$ × 10# × 10P = 10$
1 mg = 𝟏𝟎7𝟔 kg = 𝟏𝟎𝟔 ng

(iii) 1 mL = 10U ×L
1 × 107# L = 10U × L
10U = 107#
𝟏𝐋 = 𝟏𝐝𝐦𝟑
1 mL = 10-3 L = 10-3 dm3
MCQ AND ASSERTION REASON QUESTIONS
1. What will be the molarity of a solution, which contains 5.85 g of NaCl(s) per 500 mL?
(i) 4 mol L–1 (ii) 20 mol L–1
(iii) 0.2 mol L–1 (iv) 2 mol L–1
2. If 500 mL of a 5M solution is diluted to 1500 mL, what will be the molarity of the solution obtained?
(i) 1.5 M (ii) 1.66 M
(iii) 0.017 M (iv) 1.59 M
The number of atoms present in one mole of an element is equal to Avogadro number. Which of the
following element contains the greatest number of atoms?
(i) 4g He (ii) 46g Na
(iii) 0.40g Ca (iv) 12g He
3. If the concentration of glucose (C6H12O6) in blood is 0.9 g L–1, what will be the molarity of glucose in
blood?
(i) 5 M (ii) 50 M
(iii) 0.005 M (iv) 0.5 M
4. What will be the molality of the solution containing 18.25 g of HCl gas in 500 g of water?
(i) 0.1 m (ii) 1 M
(iii) 0.5 m (iv) 1 m
5. One mole of any substance contains 6.022 × 1023 atoms/molecules. Number of molecules of H2SO4
present in 100 mL of 0.02M H2SO4 solution is ______.
(i) 12.044 × 1020 molecules (ii) 6.022 × 1023 molecules
(iii) 1 × 1023 molecules (iv) 12.044 × 1023 molecules
6.What is the mass percent of carbon in carbon dioxide?
(i) 0.034% (ii) 27.27%
(iii) 3.4% (iv) 28.7%
7. The empirical formula and molecular mass of a compound are CH2O and 180 g respectively. What will
be the molecular formula of the compound?
(i) C9H18O9 (ii) CH2O
(iii) C6H12O6 (iv) C2H4O2
–1
8. If the density of a solution is 3.12 g mL , the mass of 1.5 mL solution in significant figures is _______.
(i) 4.7g (ii) 4680 × 10–3g
(iii) 4.680g (iv) 46.80g
9. Which of the following statements about a compound is incorrect?
(i) A compound cannot be separated into its constituent elements by physical methods of separation.
(ii) A molecule of a compound has atoms of different elements.
(iii) The ratio of atoms of different elements in a compound is fixed.
(iv) A compound retains the physical properties of its constituent elements.
10. Which of the following statements is correct about the reaction given below:
4Fe(s) + 3O2(g) → 2 Fe2O3 (g)
(i) Total mass of iron and oxygen in reactants = total mass of iron and oxygen in product therefore it follows
law of conservation of mass.
(ii) Amount of Fe2O3 produced will decrease if the amount of any one of the reactants (iron or oxygen) is
taken in excess
(iii) Total mass of reactants = total mass of product; therefore, law of multiplenproportions is followed.
(iv) Amount of Fe2O3 can be increased by taking any one of the reactants(iron or oxygen) in excess.
11. Which of the following reactions is not correct according to the law of conservation of mass.
(i) 2Mg(s) + O2(g) → 2MgO(s) (ii) C3H8(g) + O2(g) → CO2(g) + H2O(g)
(iii) P4(s) + 5O2(g) → P4O10(s) (iv) CH4(g) + 2O2(g) → CO2(g) + 2H2O (g)
12. Which of the following statements indicates that law of multiple proportion is being followed.
(i) Sample of carbon dioxide taken from any source will always have carbon and oxygen in the ratio 1:2.
(ii) Carbon forms two oxides namely CO2 and CO, where masses of oxygen which combine with fixed
mass of carbon are in the simple ratio 2:1.
(iii) When magnesium burns in oxygen, the amount of magnesium taken for the reaction is equal to the
amount of magnesium in magnesium oxide formed.
(iv) At constant temperature and pressure 200 mL of hydrogen will combine with 100 mL oxygen to
produce 200 mL of water vapour.
13. One mole of oxygen gas at STP is equal to _______.
(i) 6.022 × 1023 molecules of oxygen (ii) 6.022 × 1023 atoms of oxygen
(iii) 16 g of oxygen (iv) 32 g of oxygen
(a) (i) and (iv) (b) (ii) and (iii)
(c) (i) , (iii) and (iv) (d) (i) only
14. Sulphuric acid reacts with sodium hydroxide as follows : H2SO4 + 2NaOH → Na2SO4 + 2H2O
When 1L of 0.1M sulphuric acid solution is allowed to react with 1L of 0.1M sodium hydroxide solution,
the amount of sodium sulphate formed and its molarity in the solution obtained is
(i) 0.1 mol L–1 (ii) 7.10 g
–1
(iii) 0.025 mol L (iv) 3.55 g
(a) (i) and (iv) (b) (ii) and (iii)
(c) (i) , (iii) and (iv) (d) (i) only
15. Which of the following pairs have the same number of atoms?
(i) 16 g of O2(g) and 4 g of H2(g) (ii) 16 g of O2 and 44 g of CO2
(iii) 28 g of N2 and 32 g of O2 (iv) 12 g of C(s) and 23 g of Na(s)
(a) (i) and (iv) (b) (ii) and (iii)
(c) (iii) and (iv) (d) (iv) only
16. Which of the following solutions have the same concentration?
(i) 0.5 mol of KCl in 200 mL of solution (ii) 20 g of NaOH in 200 mL of solution
(iii) (iv) 40 g of NaOH in 100 mL of solution (iv) 20 g of KOH in 200 mL of solution
(a) (i) and (ii) (b) (ii) and (iii)
(c) (i) , (iii) and (iv) (d) (i) only
17. 16 g of oxygen has same number of molecules as in
(i) 16 g of CO (ii) 28 g of N2
(iii) 14 g of N2 (iv) 1.0 g of H2
(a) (i) and (iv) (b) (ii) and (iii)
(c) (iii) and (iv) (d) (i) only
18. Which of the following terms are unitless?
(i) Molality (ii) Molarity
(iii) Mole fraction (iv) Mass percent
(a) (i) and (iv) (b) (ii) and (iii)
(c) (iii) and (iv) (d) (i) only
19. One of the statements of Dalton’s atomic theory is given below:
“Compounds are formed when atoms of different elements combine in a fixed ratio”
Which of the following laws is not related to this statement?
(i) Law of conservation of mass (ii) Law of definite proportions
(iii) Law of multiple proportion (iv) Avogadro law
(a) (i) and (iv) (b) (ii) and (iii)
(c) (i) , (iii) and (iv) (d) (i) only
20. Assertion (A) : The empirical mass of ethene is half of its molecular mass.
Reason (R) : The empirical formula represents the simplest whole number ratio of various atoms present
in a compound.
21. Assertion (A) : One atomic mass unit is defined as one twelfth of the mass of one carbon-12 atom.
Reason (R) : Carbon-12 isotope is the most abundunt isotope of carbon and has been chosen as standard.
22. Assertion (A) : Significant figures for 0.200 is 3 where as for 200 it is 1.
Reason (R) : Zero at the end or right of a number are significant provided
24. Assertion (A) : Combustion of 16 g of methane gives 18 g of water.
Reason (R) : In the combustion of methane, water is one of the products