Some Basic Concept of Chemistry

( by : Jitender Sir)
Chemistry is the science of molecules and their transformations. It is the science not so much of the one
hundred elements but of the infinite variety of molecules that may be built from them ...
Roald Hoffmann

Base Physical Name of Definition

Quantity SI Unit

Length Metre The metre is the length of the path travelled by light in vacuum during a time
1
interval of of a second
299792458
Mass kilogram The kilogram is the unit of mass; it is equal to the mass of the international
prototype of the kilogram.
Time Second The second is the duration of 9 192 631 770 periods of the radiation
corresponding to the transition between the two hyperfine levels of the
ground state of the caesium-133 atom.
Electric Current ampere The ampere is that constant current which, if maintained in two straight
parallel conductors of infinite length, of negligible circular cross-section, and
placed 1 metre apart in vacuum, would produce between these conductors a
force equal to 2 × 10–7 newton per metre of length.
Thermodynamic kelvin 1
The kelvin, unit of thermodynamic temperature, is the fraction of
temperature 273.16
the thermodynamic temperature of the triple point of water.
Amount of mole 1. The mole is the amount of substance of a system which contains as many
Substance elementary entities as there are atoms in 0.012 kilogram of carbon-12; its
symbol is “mol.”
2. When the mole is used, the elementary entities must be specified and may
be atoms, molecules, ions, electrons, other particles, or specified groups of
such particles.
Luminous candela The candela is the luminous intensity, in a given direction, of a source that
intensity emits monochromatic radiation of frequency 540 × 1012 hertz and that has a
1
radiant intensity in that direction of watt per steradian.
683

Q)What is the composition of 20 carat gold ?

Ans)Pure gold is 24 carat . means 20 carat gold contains 20 parts gold and 4 parts Cu .
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Precision: refers to the closeness of various measurements for the same quantity.
Accuracy: is the agreement of a particular value to the true value of the result.
True Value Students- A Student-B Student-C

2.00g 1.95g 1.94g 2.01g

2.00g 1.93g 2.05g 1.99g
Precise not accurate Neither precise nor Both Precise and accurate
Accurate

Q) Explain Matter?
Ans) Anything that occupies space, possesses mass and can be felt by anyone or more of our senses.
Matter has been classified into two types: Physical and Chemical

b) Chemical classification of matter: (based upon the composition of matter)

Pure substances: A single substance which cannot be separated into other kinds of matter by any physical
process.
Pure substances have been further classified as elements and compounds.
(A)Elements: The simplest form of a pure substance which can neither be broken into nor built from simpler
substances by ordinary physical and chemical methods.
Types of elements: i) Metals ii) nonmetal iii) semimetal or metalloid
(B)Compounds: A pure substance containing two or more elements combined together in a fixed proportion by
weight and can be decomposed into these elements by suitable chemical methods . Types:
i)Inorganic compound: compounds which are obtained from non-living sources such as rocks and minerals
eg : common salt , marble etc.
ii)Organic compound: compounds which are present in plants and animals. All the organic compound have
been found to contain carbon as their essential constituent eg: Protein, oils and fats etc.
(C) Mixture: The combination of two or more elements or compounds which are not chemically combined and
may also by present in any proportion. The constituents of a mixture do not lose their identity and properties.
Types:
Homogeneous Mixture: if it has uniform composition throughout and there are no visible boundaries of
separation between the constituents Moreover, the constituents cannot be seen even by a microscope.
Heterogeneous Mixture : if it does not have uniform composition throughout and there are no visible
boundaries of separation between the constituents .Moreover , the constituents can be seen even with naked eyes

Laws of chemical combination:

(a) Law of conservation of mass
(b) Law of Multiple proportions
(c) Law of constant composition /definite proportions
(d) Law of Gay Lussac`s Law of combining volume

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i)Law of conservation of mass: In all physical changes and chemical reactions, the total mass of the products is
the same as the total mass of the reactants.
ii)Law of constant composition: A pure chemical compound always consists of the same elements combined
in a fixed proportion by weight.

Q1. The mass of copper oxide obtained by heating 2.16g of metallic copper with nitric acid and
subsequent ignition was found to be 2.7g .In another experiment, 1.15g of copper oxide on reduction
yielded 0.92g of copper . Show that the data illustrates the law of constant composition.

Problem 1: One mole of V2O5 contains 5 moles of oxygen atoms, as it is clear from its formula. What is the %
oxygen by weight in V2O5? (Atomic weights are; V = 50.9 g/mol, O = 16 g/mol) (Ans: 44%)
Problem 2: What is the experimental percent of oxygen in CO2 if 42.0 g of carbon reacted completely with 112.0
g of oxygen? (Ans: 72.7%)
Problem 3: What is the theoretical percent of aluminum in aluminum oxide? (Ans: 52.9%).
Problem 5 : When 0.0976 g of magnesium was heated in air, 0.1618 g of magnesium oxide (MgO) was
produced.
(a) What is the percent of Mg in MgO?(Ans: 60.3%)
(b) Using only law of definite proportions, what mass of oxygen was needed to combine with the magnesium?
(Ans: 0.0642%)

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iii)Law of multiple proportion: If two elements combine to form two or more compounds, the weights of
one of the elements which combine with a fixed weight of the other in these compounds ,bear a simple
whole nuber ratio by weight
Oxide of No. of parts by No. of parts 14 parts of No. of parts by weight of oxygen
nitrogen weight of by weight of nitrogen as which combine with 14 parts by
nitrogen oxygen fixed weight weight of nitrogen

N2O 28 16 14 8

NO 14 16 14 16

N2O3 28 48 14 24

N2O4 28 64 14 32

N2O5 28 80 14 40

Thus, the ratio between the weights of oxygen in all the above listed oxides which combined with 14 parts
by weight of nitrogen is 1: 2 : 3 : 4 : 5

Q: Two different compounds are formed by the elements carbon and oxygen. The first compound
contains 42.9% by mass carbon and 57.1% by mass oxygen. The second compound contains 27.3% by
mass carbon and 72.7% by mass oxygen. Show that the data are consistent with the law of multiple
proportions.
Ans: the masses of oxygen in the two compounds that combine with a fixed mass of carbon should be in a
whole number ratio. In 100 grams of the first compound (100 is chosen to make calculations easier), there
are 57.1 grams oxygen and 42.9 grams carbon. The mass of oxygen (O) per gram of carbon (C) is:

57.1 g O / 42.9 g C = 1.33 g O per g C

In the 100 grams of the second compound, there are 72.7 grams of oxygen (O) and 27.3 grams of carbon (C).
The mass of oxygen per gram of carbon is:

72.7 g O / 27.3 g C = 2.66 g O per g C

Dividing the mass O per g C of the second (larger value) compound:

2.66 / 1.33 = 2

This means that the masses of oxygen that combine with carbon are in a 2:1 ratio. The whole-number ratio is
consistent with the law of multiple proportions.

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Numerical : Copper gives two oxides . On heating 1g of each in hydrogen, we get .888g and 0.798g of the
metal respectively . Show that these results agree with the Law of Multiple proportions.
Solution:
Let us fix 1g of copper as the fixed weight in the two oxides .
In the first oxide : weight of copper = 0.888g
Weight of oxygen = 1 – 0.888 = .112g
Now , 0.888g of copper combined with oxygen = 0.112g
Then 1g of copper combined with oxygen = 0.112/0.888 = 0.126g
In second oxide : : weight of copper = 0.798g
Weight of oxygen = 1 – 0.798 = 0.202g
Now , 0.798g of copper combined with oxygen = 0.202g
Then 1g of copper combined with oxygen = 0.202/0.798 = 0.253g
Ratio by weight of oxygen which combine with 1g of copper in the two oxide is 0.126 : 0.253 = 1 : 2 , As
the ratio is simple whole number in nature , the Law of Multiple proportions is proved .

Q:Three common gaseous compounds of nitrogen and oxygen of different elementary composition are known,
(A) laughing gas containing 63.65% nitrogen, (B) a colorless gas containing 46.68% nitrogen, and (C) a
brown, toxic gas containing 30.45% nitrogen. Show how these data illustrate the law of multiple proportions.
Solution:
Compounds → (A) (B) (C)
Amount of Nitrogen (in gm) 63.65 46.68 30.45
Amount of Oxygen (in gm) 36.35 53.32 69.55
Amount of Nitrogen 1.7510 0.8755 0.4378
Ratio of
Amount of Oxygen

The relative amounts are not affected if all three amounts are set up in the form of a ratio and then divided by
the smallest of the relative amounts.
1.7510 0.8755 0.4378
= = = 4 : 2: 1
0.4378 0.4378 0.4378

Q1. In an experiment, 2.4g of iron oxide on reduction with hydrogen yield 1.68g of iron. In another
experiment 2.9g of iron oxide gives 2.03g of iron on reduction with hydrogen. Show that the above data
illustrate the law of constant proportions.
Q2.Two compounds each containing only tin and oxygen had the following composition :
Mass% of tin Mass% of Oxygen
Compound A 78.77 21.23
Compound B 88.12 11.88
Show how data illustrate the law of multiple proportions ?

iv)Gay lussac`s law of combining volumes : Under similar conditions of temperature and pressure ,
whenever gases combine, they do so in volumes which bear simple whole number ratio with each other and
also with the gaseous products .

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Dalton`s Atomic theory :
1)Matter is made up of extremely small particles called atoms.
2)Atoms of the same element are identical in all respect i.e., in shape, size, mass and also in properties .
3)Atoms of the different elements differ in all respects and have also different properties.
4)Atom is the smallest portion of matter which can take part in chemical combination.
5)Atoms of the same or different elements combine to form compound atoms ( later these were called
molecule )
6)When atoms combine with one another to form compound atoms or molecules, they do so in simple whole
number ratios .
7) Atoms can neither be created nor be destroyed in a physical change or chemical reaction.

Important Definition
Atomic mass unit (amu):
An atomic mass unit (amu) is one twelfth ( 1/12 ) of the mass of one C-12 isotopes .
Today, ‘amu’ has been replaced by ‘u’ which is known as unified mass.
Mass of 1 atom of an element
Relative atomic mass of an element =
1
 Mass of 1 atom of Carbon
12
One Mole:
One mole is the amount of a substance that contains as many particles or entities as there are atoms in
exactly 12 g of the 12C isotope.
Molar Mass:
The mass of one mole of a substance in grams is called its molar mass.
Gram atomic mass:
Thus gram-atomic mass can be defined as the absolute mass in gram of 6.022 x 1023 atoms of any element.
Molecular mass: It is a sum of the atomic masses of all the element present in a molecule.
Formula unit mass: It is a sum of the atomic masses of all element in one formula unit of an ionic
compound.

% abundance  Atomic Mass

Average relative mass of an element =
100

Q1.9: Calculate the atomic mass (average) of chlorine using the following data :
% natural Abundance Molar Mas
35
Cℓ 75.77 34.9689
37
Cℓ 24.23 36.9659

  75.77 
Ans: Average atomic mass of Chlorine = 
  100 
 ( 34.9689u )   24.23 
+ 
 100 
 ( 36.9659) 



= 26.4959 + 8.9568 = 35.4527u
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 Q:Use tdata given in the following table to calculate the molar mass of naturally occuring argon isotopes:
Isotope Isotopic molar mass Abundance
36Ar 35.96755 g mol–1 0.337%
38Ar 37.96272 g mol–1 0.063%
40Ar 39.9624 g mol–1 99.600%

Avogadro`s hypothesis : Under similar conditions of temperature and pressure(STP condition, means,
273K , 1 ba pressure) equal volumes of all gases contain equal number of molecules .
1 mol of nitrogen gas( N2) contain= 22.4L at STP = 6.022 x 1023 molecules of N2
1 mol of nitrogen gas( O2) contain = 22.4L at STP = 6.022 x 1023 molecules of O2

Some tips to solve the numerical of mole concept:

One mole = 6.022 x 1023 particles (atom , molecule ,electron ,proton , neutron etc depend on given value )
If atomic mass of any element is given in gm then it is the mass of one mole of atom of that element
Given mass Given Volume Given no. of atoms
No of moles = = =
Molar mass 22.4 L at STP 6.022  1023
(atomic mass in gm is also known as gram atomic mass)
Gram atomic mass = mass of 6.022 x 1023 atom
For eg: 12gm Carbon = 6.022 x 1023 atoms of carbon
14gm Nitrogen = 6.022 x 1023 atoms of nitrogen
16gm Oxygen = 6.022 x 1023 atoms of Oxygen
If molecular mass of any molecule is given in gm again it means mass of one mole of that molecule
Or it is also know as gram molecular mass
Gases always study in form of volumes and molecules
*Diatomic molecule = O2 , N2 , Cl2 etc (comprises of two atoms )
Monoatomic gases = He , Ne , Ar , Kr , Xe an Rn (noble gases)
Volume of 1mole of any gas at STP (273K and 1atm ) = 22.4L
Volume of 1mole of any gas at SATP (289.15K and 1bar ) = 24.789L

Standard ambient temperature and pressure (SATP):

Unit of volume = 1m3 = 103L =103dm3 =106ml =106 cm3 =106cc
Unit of pressure = 1atm = 760mm =760torr = 76cm = 1.013bar = 1.01325 x 105 Pa 0r Nm-2
Unit of temperature = 0oC = 273K
P1V1 P2 V2
= (P1 , V1 and T1 are at STP condition whereas P2 , V2 and T2 are at given condition)
T1 T2
Atomic number = no. of proton = no. of electron (but not in case of ions for electrons )

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Molecular mass of some important molecules :
S.No. compound Mol. mass S.No. compound Mol. mass

1 H2SO4 98gm 7. MgCO3 84gm

2 HCl 36.5gm 8. Na2CO3 106gm
3 HNO3 63gm 9. NaHCO3 84gm
4 NaOH 40gm 10. NH2CONH2 (urea ) 60gm
5 CaCl2 111gm 11. C6H12O6 (glucose ) 180gm
6 CaCO3 100gm 12. C12H22O11 (Sucrose or cane sugar ) 342gm

Numericals to solve
1. Calculate the mass in grams of the following :
(i) 6.022 x 1023 atoms of oxygen (ii) 1.0 x 1023 molecules of H2S
(iii) 6.022 x 1023 molecules of oxygen (iv) 1.5 moles of H2SO4 (Ans: 16g , 5.645g , 32g and 147g )
2. How many atoms of oxygen and hydrogen are present in 0.15mole of water?
(Ans :1.81 x 1023 atoms of H , 9.03 x 1022 atoms of O )
3. Calculate the no. of moles of phosphorus in 92.9g of phosphorus assuming that molecular formula of
phosphorus is P4 . Also calculate the no of atoms and molecules of phosphorus in the sample.
(Ans:0.75mol ,4.52 x 1023molecules ,1.8 x 1024atoms)
4. Which of the following weighs most ?
(i) 50g of Iron (ii) 5g atom of N (iii) 0.1 g atom of silver (iv) 1023 atoms of C
(Ans : 5 g atom of N weigh most )
5. Calculate the total no. of electron present in 1.6g of methane? (Ans: 1 mole electron)
6. What is the number of molecules of CO2 which contain 8g of O2 ? (Ans : 1.5 x 1023molecules)
7. Which of the following has largest number of atoms ?
(i)1g of Au (ii) 1g of Na (iii) 1g of Li (iv) 1g of CI2
8. A cylinder of compressed gas contains nitrogen and oxygen in the molar ratio of 3:1 . If the cylinder
contains 2.5 x 104g of oxygen, what is the total mass of the gas in the mixture ? (Ans : 9.0625 x 104 g )
9. In a bank, there are as many coins as the number of molecules in 1.6 µg of CH4. How many coins are
there in the bank ? (ans : 6.022 x 1016 )
10. What weight of calcium contains the same number of atoms as are presents in 3.2 g of sulphur ?
(ans : 4g)
11. From 200mg of CO2, 1021 molecules are removed. How many moles of CO2 are left ?
(ans:2.89 x 10-3 mol)
12. The density of water at room temperature is 1.0g/ml . How many molecules are there in a drop of water if
its volume is 0.05ml . (ans :1.68 x 10-2 molecule )

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13. Calculate the weight of carbon monoxide having the same number of oxygen atoms as are present in
22g of carbon dioxide . (ans:28gm)
14. The red colour of blood is due to haemoglobin. It contains 0.335 percent of iron, Four atoms of iron are
present in one molecules of haemoglobin . What is the molecular mass of haemoglobin? (ans :66675u)
15. The cost of table salt (NaCI) and table sugar ( C12H22O11) are Rs 2 per kg and Rs. 6 per Kg respectively
Calculate their cost per mole. (ans: 11.7 paise NaCI , 2.0 Rupees table sugar)
16. Chlorophyll, the green coloring matter of plants responsible for photosynthesis, contains 2.68% of
magnesium by weight. Calculate the number of magnesium atoms in 2.0g of chlorophyll.
(Ans: 1.345 x 1021)
17. Calculate the following:
a) What is the mass percent of carbon in CO2?
b) Find out the number of atoms in 28 g of N2 and 32 g of O2.
c) How many moles are present in 88 g of CO2.
18. Aspirin has the formula C9H8O4 How many moles of aspirin are in a tablet weighing 500mg?
(Ans: 4.8184 x 1024 atom)
19. Which sample contains the greatest number of atoms? Two moles of helium atoms or half mole of
methane molecules CH4.
20. The atomic weights of two elements (A and B) are 20 and 40 respectively. If x g of A contains y atoms,
how many atoms are present in 2x g of B? (Ans: no of atoms of B = y)
21. Calculate the number of molecules present in 350 cm3 of NH3 gas at 273K and 2 atmosphere pressure.
(Ans : 1.882 x 1022 molecules )
22. Naturally occurring neon consists of three isotopes, 20Ne , 21Ne , 22Ne; with masses 19.92, 20.99, and
21.99 amu and their % abundance in nature are 90.51, 0.27 and 9.22 respectively . Calculate the average
atomic mass of neon. ( Ans : 20.10amu )
23. An atom of some element Y weighs 6.644×10−23 g. Calculate the number of gram-atoms in 40 kg of it.
(Ans: 1000)
24. The atomic masses of two elements (P and Q) are 20 and 40 respectively. X g of P contains Y atoms, how
many atoms are present in 2X g of Q? (Ans: No of Q atoms = Y)

Limiting Reagents
If in any given chemical reaction, after  time, one of the reactant will be completely consumed while the
other would be left in excess. Thus, the reactant which is completely consumed when a reaction goes to
completion and which decides the yield of the product is called limiting reagent.

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Q: How much zinc should be added to 0.01 mol AgNO3 solution to displace all the silver in the solution?
Solution: The involved balanced reaction would be
Zn + 2AgNO3 ⎯→ Zn(NO3)2 + 2Ag
Moles of AgNO3 in the solution = 0.01
Moles of Zn to be added to solution = 0.005
(since AgNO3 and Zn are reacting in the molar ratio of 2 : 1)
 Mass of Zn to be added to solution = 0.005  65.4 = 0.327 g

Numerical based on Limiting reagent

1. Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction:
2Al + 3Cl2 →2AlCl3
How many grams of aluminum chloride could be produced from 34.0 g of aluminum and 39.0 g of
chlorine gas?
2. The reaction of 4.25 g of Cl2 with 2.20 g of P4 produces 4.28 g of PCl5. Find out the limiting reagent?
3. Take the reaction: NH3 + O2 NO + H2O. In an experiment, 3.25 g of NH3 can react
with 3.50 g of O2.
a. Which reactant is the limiting reagent?
b. How many grams of NO are formed?
c. How much of the excess reactant remains after the reaction?
4. If 4.95 g of ethylene (C2H4) are combusted with 3.25 g of oxygen.
a. What is the limiting reagent?
b. How many grams of CO2 are formed?
5. A reaction container holds 5.77 g of P4 and 5.77 g of O2.
The following reaction occurs: P4 + O2 P4O6. If enough oxygen is available, then the P4O6 reacts
further: P4O6 + O2 P4O10.
a. What is the limiting reagent for the formation of P4O10?
b. What mass of P4O10 is produced?
c. What mass of excess reactant is left in the reaction container?
( In this problem, always remember what has to happen in order for P4O10 to be formed. All of the
P4O6 and any left over oxygen from the first reaction react to form the final product. This means that if
oxygen is the limiting reagent in the first reaction then no P4O10 will be formed.)
Ozone (O3) reacts with nitric oxide (NO) discharged from jet planes to form oxygen gas and nitrogen
dioxide. 0.740 g of ozone reacts with 0.670 g of nitric oxide. Determine the identity and quantity of the
reactant supplied in excess.
Reaction takes place: NO + O3 → NO2 + O2

Problem based on mass - mass relation

1. How much CO2 is evolved when 50gm of CaCO3 is heated?
2. How much of CaCl2 we will get when 25gm of CaCO3 is reacting with 18.25gm of HCl?
3. How much of CO2 and H2O we get when 6.5gm of Acetylene(Ethyne) is reacting with excess of oxygen.
4. What mass of copper oxide will be obtained by heating 12.35g of copper carbonates ? (ans : 7.95g)
5. Calculate the weight of lime (CaO) obtained by heating 200kg of 95% pure lime stone (CaCO3) .
(ans:106.4 kg)
6. Zinc and hydrochloric acid react according to the reaction:
Zn + 2HCℓ → ZnCℓ2 + H2
If 0.30 mol Zn are added to hydrochloric acid containing 0.52 mol of HCℓ, how many moles of H2 are
produced? (Ans : H2 formed = 0.26 mol )
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7. 1 gram of pyrolusite (MnO2) was boiled with excess of concentrated HCI and the issuing gas was passed
through a solution of potassium iodide when 1.27g of iodine were liberated. What is the percentage of
pure MnO2 in the sample?
MnO2 + 4HCI + 2KI → MnCI2 + 2KCI + 2H2O + I2 (ans:43.5%)
8. What weight of AgCI will be precipitate when a solution containing 4.77 g of NaCI is added to a
solution of 5.77 g of AgNO3? (Ans: 4.87 g of AgCI )
9. 1.84g of a mixture of CaCO3 and MgCO3 is strongly heated till no further loss in mass takes place. The
residue is found to weigh 0.96g. Calculate the percentage of each component in the mixture.
(ans:54.35%CaCO3 and 45.65% MgCO3)
10. An impure sample of sodium chloride which weighed 1.2gram gave on treatment with excess of silver
nitrate solution 2.4g of silver chloride as the precipitate. Calculate the percentage purity of the sample.
(81.5%pure)

Q) Define solution?
Ans) Solutions are homogeneous mixtures of two or more than two components. By homogenous mixture we
mean that its composition and properties are uniform throughout the mixture. Generally, the component that
is present in the largest quantity is known as solvent and in less quantity is known as solute.

Types of solution Solute Solvent Common example os solution

Gaseous solution Gas Gas Mixture of oxygen and nitrogen gases ,air
Liquid Gas Chloroform mixed with nitrogen gas , fog
Solid Gas Camphor in nitrogen gas
Liquid solution Gas Liquid Oxygen dissolved in water ,soft drink
Liquid Liquid Ethanol dissolved in water
Solid Liquid Glucose dissolved in water
Solid solution Gas Solid Solution of hydrogen in palladium ,ZnO
Liquid Solid Amalgam of mercury with sodium, hydrated salt
solid Solid Copper dissolved in gold , alloy

1 = solvent , 2 = Solute W2 = Given mass of solute ; Vml = Volume of solution in ml

W1 = Given mass of solvent ; n2 = number of moles of soute
M2 =molar mass of solute ; M1 = molar mass of solvent

Eg: 20% H2SO4 by mass(W/W) (density of solution is 2gm/cm3) means :

20gm H2SO4 dissolve in 100gm solution and only 80gm solvent is present
*20gmH2SO4 + 80gm solvent ( water) = 100 gm solution

Mass 100
Density of solution = Volume of solution = = 50cm3
Volume 2

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ii) Mass by volume percentage (w/v):
Another unit which is commonly used in medicine and pharmacy is mass by volume percentage. It is the
mass of solute dissolved in 100 mL of the solution.
Eg: 20% H2SO4 (w/v) density of solution is 2 gm/cm3 :
20gm H2SO4 as solute dissolve in 100cm3 of the solution
Mass of solution = 100 x 2 = 200 gm
so Mass of solvent = 180gm ( because : mass = volume x density )

mass of A
Mass % of component A = ×100
mass of A + mass of B

Molarity(M) : number of moles of the solute dissolve in 1000ml or 1litre of the solution.

WB × 1000ml Given mass of solute× 1000ml

Molarity (M) = =
M B × Vml Molar mass of solute × Volume of solution in ml

For dilution:
%
%by
bymass
massx xdensity of solution
density x 10 x 10
Molarity = M1V1 = M2V2
Molar Mass
Molar of solute
mass of solute

Molality(m) : number of moles of the solute dissolve in 1000gm or 1Kg of the solvent

WB × 1000gm Given mass of solute × 1000gm

Molality (m) = =
M B × WA Molar mass of solute × mass of solvent in gm

*Molality = no. of moles of solute when no of moles of water is Taken as 55.56.

2MH2SO4 Solution with a density of 1.2gm/ cm3 . What is its means : its means?

2moles of H2SO4 present in = 1000ml solution

2 x 98 = 196gm H2SO4 present in = 1000ml solution
Mass of solution = 1000 x 1.2 = 1200gm
Mass of solvent = 1200 – 196 = 1004gm

2mH2SO4 Solution with a density of 1.2gm/ cm3 . What is its means : its means?

2moles of H2SO4 present in = 1000gm solvent

2 x 98 = 196gm H2SO4 present in = 1000gm solvent
Mass of solution i= 1000 + 196 = 1196gm
1196
Volume of solution = = 996.7ml
1.2

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 *Mole fraction(X): ( XA + XB =1 )
nB nA
Mole fraction of solute ( XB ) = ; Mole fraction of solvent ( XA ) =
nA + nB nA + nB

Mole fraction of A component Partial Vapour pressure of A component (PA)

in Vapour phase =
Total Vapour pressure of solution ( PA+PB)

*Parts per million: When a solute is present in trace quantities, it is convenient to express concentration in
parts per million (ppm) and is defined as:
Number of parts of the component  106
Parts per million =
Total number of parts of all componentof the solution
The conc. of pollutants in water or atmosphere is often expressed in terms of μg mL–1 or ppm.

Some important questions based on molarity , molality and mole fraction :

Q1. A 10 cm3 sample of human urine was found to have 5 milligrams of urea on analysis .Calculate the
molarity of the given sample with respect to urea . (0.0083moles/litre)
Q2. The concentration of H2SO4 in a bottle labelled ‘conc sulphuric acid’ is 18M . The solution has a
density of 1.84g/cm3 . What is the mole fraction and weight % of H2SO4 in the solution ? (ans =0.81)
Q3. A sugar syrup of weight 214.2 g contains 34.2 g of cane sugar .Calculate (i)molal concentration.
(ii)mole fraction of sugar in the syrup . (ans = 0.556 , 0.0099 )
Q5. Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass)
aqueous KI is 1.202 g mL-1. (ans= 37 g )
Q6. Concentrated HNO3 used in laboratory work is 68% nitric acid by mass in aqueous solution. What
should be the molarity of a sample of the acid if the density of the solution is 1.504 g mL–1?
(ans: 16.23M)
Q7. A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of
each component in the solution? If the density of solution is 1.2 g mL–1, then what shall be the molarity
of the solution? (ans: 0.617m , 0.67M , mole fraction of glucose = 0.01 )
Q8. An antifreeze solution is prepared from 222.6 g of ethylene glycol (C2H6O2) and 200 g of water.
Calculate the molality of the solution. If the density of the solution is 1.072 g mL–1, then what shall be
the molarity of the solution? ( 17.95m , 9.11M)

Empirical and molecular formula :

The empirical formula of a compound is the chemical formula which expresses the simplest whole number
ratio of the atoms of the various elements presents in one molecule of the compound .
Eg: the empirical formula of benzene is CH that of hydrogen peroxide is HO and that of glucose is CH2O .

Molecular formula : The molecular formula of a compound is the chemical formula which represents the
true formula of its molecule . It expresses the actual .
Molecular formula = n x empirical formula
Molecular Mass = 2 x Vapour density

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Q)An organic compound containing carbon , hydrogen and oxygen gave the following percentage
composition : C = 40.687% ; H = 5.085% ; O= 54.228%
The vapour density of the compound is 59. Calculate the molecular formula of the compound.
Element % of At. mass of Moles of Simplest Simplest whole no. ratio
element element element molar ratio
C 40.687 12 3.390 1 2
H 5.085 1 5.085 1.5 3
O 54.228 16 3.389 1 2
Empirical formula is C2H3O2 (so empirical formula mass = 59 )
Molecular mass = 2 x 59 = 118
n =2 ; so molecular formula is C4H6O4

Q:Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.
% of Iron by mass 69.9
= = = 1.25
Answer: relative moles of iron in iron oxide= Atomic mass of Iron 55.85
% of Oxygen by mass 30.1
= = = 1.88
Answer: relative moles of oxygen in iron oxide= Atomic mass of Oxygen 16
Simplest Molar Ratio of Iron to Oxygen : 1.25 : 1.88 = 1 : 1.5 = 2:3
The empirical formula of the iron oxide is Fe2O3.

Questions :
Q1) 2.38 g of uranium was heated strongly in a current of air . The resulting oxide weighed 2.806g
Determine the empirical formula of the oxide .(At.mass of U = 238 ) (ans : U3O8 )
Q2) A compound contains 4.07 % hydrogen, 24.27 % carbon and 71.65 % chlorine. Its molar mass is
98.96g. What are its empirical and molecular formulas ?( Ans: EF=CH2Cl, MF=C2H4Cl2).
Q3. A compound is found to contain 23.3% magnesium, 30.7% sulfur and 46.0% oxygen. What is the
empirical formula of this compound? ( Ans: MgSO3).
Q4. What is the empirical formula for a compound containing 38.8% carbon, 16.2% hydrogen and 45.1%
nitrogen? ( Ans: H2NCH3)

Home assignments :
1.1) Calculate the molecular mass of the following : (i) H2O (ii) CO2 (iii) CH4
1.2) Calculate the mass per cent of different elements present in sodium sulphate (Na2SO4).
1.4) 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.
1.5) Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous
solution. Molar mass of sodium acetate is 82.0245 g mol–1. (Ans:15.38g)
1.6) Calculate the concentration of HNO3 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%.( Ans: 15.44mol/L)
1.7) How much Copper can be obtained from 100 g of (CuSO4) ? ( ans : 39.81g)
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1.10) 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.
1.11) 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? ( ans: 0.02925 mol/L-1)
1.12) 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? ( Ans: 25.22ml)
1.14) What is the SI unit of mass? How is it defined?
1.16) A sample of drinking water was found to be severely contaminated with chloroform, CHCl3, supposed
to be carcinogenic in nature. The level of contamination was 15 ppm (by mass).
(i) Express this in percent by mass. (ii) Determine the molality of chloroform in the water sample.
1.19) The following data are obtained when dinitrogen and dioxygen react together to form different
compounds : Mass of dinitrogen and 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.
1.20) If the speed of light is 3.0 × 108 m s–1, calculate the distance covered by light in 2.00 ns(nanosecond)
1.21) 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
1.22) Dinitrogen and dihydrogen react with each other to produce ammonia according to the following
chemical equation: N2 (g) + H2 (g) → 2NH3 (g)
(i)Calculate the mass of ammonia produced if 2.00 ×103 g dinitrogen reacts with 1.00 ×103 g of dihydrogen.
(ii) Will any of the two reactants remain unreacted?
(iii) If yes, which one and what would be its mass?
1.23) How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different?
1.24) If ten volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water
vapour would be produced?
1.26) 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)
1.27) Calculate the molality of a solution of ethanol in water in which the mole fraction of ethanol is 0.040.
1.28) What will be the mass of one 12C atom in g ?
1.29) 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.
1.31) Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction,
CaCO3 (s) + 2 HCl (aq) → CaCl2 (aq) + CO2(g) + H2O(l)
What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?
1.32) Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric
acid according to the reaction
4 HCl (aq) + MnO2(s) → 2H2O (l) + MnCl2(aq) + Cl2 (g)
How many grams of HCl react with 5.0 g of manganese dioxide? ( molar mass of Mn = 55)

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