1. introduction 2. dalton’s atomic theory

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1. MOLE CONCEPT AND STOICHIOMETRY

1. Introduction

As we have learned about the units and various conversions, it’s time that we start using them. Before starting with the core calculations, we need to know the laws which have been formulated with the help of the facts known from experimental evidences. So now, get ready to play.

2. Dalton’s Atomic Theory

Flowchart 1.1: Postulates of Dalton theory

1.2 Mole Concept and Stoichiometry

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PLANCESS CONCEPTS

Many unexplained chemical phenomena were quickly explained by Dalton with his theory. Dalton’s theory quickly became the theoretical foundation in chemistry.

Vipul Singh

AIR 1, NSTSE 2009

2.1 Law of Conservation of Mass

A. Lavoisier in 1789 introduced the concept of mass conservation. He stated that “In all physical changes and chemical reactions, the total mass of the products is the same as the total mass of the reactants.” He is considered as “Father of Modern Chemistry”.

Modi�cation to the Law of Conservation of mass

�e total sum of mass and energy of the system remains constant.

PLANCESS CONCEPTS

�e above law will remain constant over time, provided the system is isolated.

A similar statement is that mass cannot be created/destroyed, although it may be rearranged in space, and changed into di�erent types of particles.

Sir Lavoisier was the one who coined the word “oxygen” meaning “acid former” and came up with the combustion process through his experiment.

Vipul Singh

AIR 1, NCO

2.2 Law of Definite Proportions or Constant Composition

Law of Constant Composition was given by Sir Joseph Proust in 1799, �e law states that “A pure chemical compound always consists of the same elements combined together in a �xed (de�nite) proportion by weight.”

Illustration 1: �e mass of copper oxide obtained by heating 2.16 g of metallic copper with nitric acid and subsequent ignition was found to be 2.7 g. In another experiment, 1.15 g of copper oxide on reduction yielded 0.92 g of copper. Show that the data illustrates the law of constant composition.

1.3 Foundation for Chemistry

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Sol:

In the �rst experiment In the second experiment

Weight of copper = 2.16 g

Weight of copper oxide= 2.70 g

Weight of oxygen = 2.70 – 2.16 = 0.54 g

% copper = ×Weightof copper

100Weightof copperoxide

×(2.16g)

100 80%(2.70g)

% oxygen = ×Weightof oxygen


×(0.54g)

100 20%(2.70g)

Weight of copper oxide = 1.15 g

Weight of oxygen = 1.15 – 0.92 = 0.23 g

% copper ×Weightof copper


×(0.92g)

100 80%(1.15g)

% oxygen = ×Weightof oxygen


×(0.23g)

100 20%(1.15g)

Since the ratio by weights of copper and oxygen in the two compounds remains the same, the law of de�nite composition is illustrated.

2.3 Law of Multiple Proportions

In 1803 Sir John Dalton discovered the law of multiple proportions. According to this law, “When one element combines with the other element to form two or more di�erent compounds, the mass of one element, which combines with a constant mass of the other, bears a simple ratio to one another.”

Simple ratio means the ratio between small natural numbers, such as 1:1, 1:2, 1:3.

It is one of the basic law of stoichiometry and it is used by Dalton to establish the atomic theory.

Figure 1.1: Illustration of law of multiple proportion


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Illustration 2: Carbon is found to form two oxides, which contain 42.8% and 27.27% of carbon respectively. Show that these �gures illustrate the law of multiple proportions.

Sol:

For the �rst oxide: For the second oxide:

% of carbon in �rst oxide = 42.8

% of oxygen in �rst oxide = 100 – 42.8 = 57.2

Mass of oxygen in grams that

combines with 42.8 g of carbon = 57.2

∴ Mass of oxygen that combines with 1 g of

carbon = 57.242.8

= 1.34 g

% of carbon in second oxide = 27.27

% of oxygen in second oxide =100 – 27.27= 72.73

Mass of oxygen in grams that

combines with 27.27g of carbon =72.73

∴ Mass of oxygen that combines with 1 g of

carbon = 72.7327.27

= 2.68 g

Ratio between the masses of oxygen that combine with a �xed mass (1 g) of carbon to the two oxides = 1.34:2.68 or 1:2 which is a simple ratio. Hence, this illustrates the law of multiple proportion.

Illustration 3: Carbon and hydrogen combine to form three compounds A, B and C. �e percentage of hydrogen in these compounds are: 25, 14.3 and 7.7 respectively. Show that the data illustrates the Law of Multiple Proportions.

Sol: Let us �x 1g of hydrogen as the �xed weight in the three compounds.

In the �rst compound In the second compound In the third compound

Weight of hydrogen=25.0 g

Weight of carbon

=100 –25 = 75.0g

25.0 g of hydrogen have

combined with carbon=75.0 g

1.0 g of hydrogen has

combined with carbon =75.0

g 3g25.0

Weight of hydrogen = 14.3 g

Weight of carbon

= 100 – 14.3 = 85.7g

14.3g of hydrogen have

combined with carbon

= 85.7 g

1.0g of hydrogen has


= 85.7/14.3 = 6 g

Weight of hydrogen = 7.7g

Weight of carbon

= 100 – 7.7 = 92.3g

7.7g of hydrogen have

combined with carbon = 92.3g

1.0 g of hydrogen has


= 92.3/7.7 =12 g

Ratio of weight of carbon which combine with 1g of hydrogen in the three compounds is 3 : 6 : 12 or 1 : 2 : 4. As the ratio is simple whole number in nature, the law of multiple proportions is proved.


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PLANCESS CONCEPTS

Dalton became the �rst scientist to explain the behavior of atoms in terms of the measurement of weight.

Vipul Singh

AIR 1, NSO

2.4 Law of Reciprocal Proportions

Jeremiasa Richter German scientist in 1972 proposed a law known as “Law of Reciprocal Proportions “ �e Law States that , “�e ratio of the weights of two elements, A and B which combine separately with a �xed weight of the third element C is either the same or some simple multiple of the ratio of the weights in which A and B combine directly with each other.”

He introduced the term “Stoichiometry.”

�e elements C and O combine separately with the third element H to form CH4 and H2O and they combine directly with each other to form CO2.

In CH4, 12 parts by weight of carbon combine with 4 parts by weight of hydrogen. In H2O, 2 parts by weight of hydrogen combine with 16 parts by weight of oxygen. �us, the weight of C and O which combine with �xed weight of hydrogen (say 4 parts by weight) and 12 and 32 i.e., they are in the ratio 12:32 or 3:8. Now in CO2, 12 parts by weight of carbon combine directly with 32 parts by weight of oxygen i.e., they combine directly in the ratio 12:32 or 3: 8 which is the same as the �rst ratio.

Illustration 4: Ammonia contains 82.35% of nitrogen and 17.65% of hydrogen. Water contain 88.90% of oxygen and 11.10% of hydrogen. Nitrogen trioxide contains 63.15% of oxygen and 36.85% of nitrogen. Show that this data illustrates the law of reciprocal proportions.

Sol:

In NH3, 17.65 g of H combine with 82.35 g N

∴ 1 g of H combine with N = 82.35g

17.65 = 4.67 g

In H2O, 11.10 g of H combine with 88.90 g O

∴ 1 g H combine with O = 88.90g

11.10= 8.01 g

∴ Ratio of the weights of N and O which

combine with �xed weight (1g) of H

= 4.67: 8.01 = 1: 1.72

In N2O3, ratio of weights of N and O

which combine with each other =36.85: 63.15

= 1: 1.71

�us, the two ratios are the same. Hence, it illustrates the law of reciprocal proportions.


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Illustration 5: Phosphorus trichloride (PCl3) contains 22.57% of phosphorus, phosphine (PH3) contains 91.18% of phosphorus and Hydrochloric acid (HCl) contains 97.23% of chlorine. Show that the data is according to the Law of Reciprocal Proportions.

Sol: Let us �x 1g of phosphorus (P) as the �xed weight.

In phosphorus trichloride (PCl3) In phosphine (PH3)

Weight of phosphorus = 22.57g

Weight of chlorine = 100 – 22.57 = 77.43g

22.57g of phosphorus have combined

with chlorine = 77.43g

1.0g of phosphorus has combined

with chlorine = 77.43g 3.43g

22.57

Weight of phosphorus = 91.18g

Weight of hydrogen = 100 – 91.18

= 8.82g

91.18g of phosphorus have combined

with hydrogen = 8.82g

1.0g of phosphorus has combined

with hydrogen = 8.82g 0.097g

91.18

�us, the ratios by weights of chlorine and hydrogen combining with a �xed weight of phosphorus in the two compounds is 3.43:0.097 (�is ratio is not the same) ....(i)

In hydrochloric acid gas (HCl), Weight of chlorine = 97.23g

Weight of hydrogen = 100 – 97.23 = 2.77g

�us, chlorine and hydrogen have combined in the ratio 97.23:2.77 ....(ii)

Let us compare the two ratios.

�ese are related to each other as 3.43 97.23: 35.37:35.10

0.097 2.77 or 1:1 (approximately)

2.5 Gay Lussac’s Law of Combining Volumes

Sir Gay Lussac found the relationship existing between the volumes of the gaseous reactants and their products. In 1808, he put forward a generalization known as the Gay Lussac’s Law of combining volumes. According to this Law “When gases react together, they always do so in volumes which bear a simple ratio to one another and to the volumes of the product. If these are also gases, provided all measurements of volumes are done under similar conditions of temperature and pressure.”

It implies that one molecule of hydrogen chloride gas is made up of 1 atom of hydrogen and 1atom of chlorine.

2.6 Avogadro’s Hypothesis

According to this hypothesis “Equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.”


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�is hypothesis explains elegantly all the gaseous reactions and is now widely recognized as a law or a principle known as Avogadro’s Law or Avogadro’s principle.

�e reaction between hydrogen and chlorine can be explained on the basis of Avogadro’s Law as follows:

Applying Avogadro’s hypothesis:

Hydrogen + Oxygen Water Vapour

No. of molecules - 2n n 2n

�us, 1 molecule of water contains 12

molecule of oxygen. But, 1 molecule of water contains 1 atom of oxygen. Hence, 1 molecule of oxygen = 2 atoms of oxygen i.e., atomicity of oxygen = 2

2.6.1 Applications of Avogadro’s hypothesis

(A) In the calculation of atomicity of elementary gases

e.g. 2 volumes of hydrogen combine with 1 volume of oxygen to form two volumes of water vapour.

Word equation for this reaction can be written as

Hydrogen + Oxygen Water Vapour

2 vol. 1 vol. 2 vol.

Example: Combination between hydrogen and chlorine to form hydrogen chloride gas. One volume of hydrogen and one volume of chlorine always combine to form two volumes of hydrochloric acid gas.

Word equation for this reaction can be written as

H2 (g) + Cl2(g) ⟶ 2HCl (g)

1 vol. 1 vol. 2 vol.

�us proved, the ratio between the volume of the reactants and the product in this reaction is simple, i.e., 1: 1: 2. Hence it illustrates the law of combining volumes.

(B) To �nd the relationship between molecular mass and vapour density of a gas

Vapour density (VD) = Densityof gas Massof acertainvolumeof thegasDensityofhydrogen Massof thesamevolumeofhydrogenat

thesametempandpressure

If n molecules are present in the given volume of a gas and hydrogen under similar conditions of temperature and pressure.

VD = Mass of n molecules of the gas Mass of 1 molecule of the gasMass of n molecules of hydrogen Mass of 1 molecule of hydrogen


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= Mass of 1 molecule of the gas Molecular massMass of 1 molecule of hydrogen 2

(Since molecular mass of hydrogen is 2). Hence, Molecular mass = 2 × vapour density

In a single reactant reaction, the calculations are arrived out with only that amount of the reactant which has converted into the product.

In the reactions, where more than one reactant is involved, one has to �rst identify the limiting reactant, i.e., the reactant which is completely consumed. All calculations are to be carried out with the amount of the limiting reactant only. For this, limiting reactant should be identi�ed. �e following example would help us.

A + 2B ⟶ 4C

Initially — 5 moles 12 moles 0 moles

If A is the limiting reactant, moles of C produced = 20

If B is the limiting reactant, moles of C produced = 24

�e reactant producing the least number of moles of the product is the limiting reactant and hence A is the limiting reactant.

�us,

A + 2B ⟶ 4C

Initially — 5 moles 12 moles 0 moles

Finally — 0 moles 2 moles 20 moles

Figure 1.2: Illustration of limiting reagent

PLANCESS CONCEPTS

In a chemical reaction, not all reactants are necessarily consumed. One of the reactants may be in excess and the other may be limited. �e reactant that is completely consumed is called limiting reactant, whereas unreacted reactants are called excess reactants.


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Amounts of substances produced are called yields. �e amounts calculated according to stoichiometry are called theoretical yields whereas the actual amounts are called actual yields. �e actual yields are often expressed in percentage, and they are often called percent yields.

Neeraj Toshniwal

Gold Medalist, INChO

3. Main Drawbacks of Dalton’s Atomic Theory

Flowchart 1.2: Drawbacks of Dalton atomic theory


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PLANCESS CONCEPTS

Many unexplained chemical phenomena were quickly explained by Dalton with his theory. Dalton’s theory quickly became the theoretical foundation in chemistry.

Neeraj Toshniwal

AIR 23 , NSO

4. Modern Atomic Theory

Flowchart 1.3: Postulates of modern atomic theory

(i) Atom is no longer indivisible: It is composed of sub-atomic particles which are electrons, protons and neutrons.

(ii) All atoms of an element may not be similar: �e atoms of the same element may have di�erent atomic masses. For example, the atoms of chlorine element have been found to possess atomic masses of 35 and 37. �ese are called isotopes.

(iii) Atoms of the di�erent element may have same atomic masses:

It is interesting to note that atomic mass of calcium (Ca) and argon (Ar) which represent di�erent elements, is 40. �ese are called isobars.

(iv) Atoms may not always combine in simple whole number ratios: Contrary to Dalton’s atomic theory, in many cases, atoms do not combine in simple whole number ratios. For example, in sucrose (C12H22O11), the element carbon, hydrogen and oxygen are present in the ratio of 12:22:11 and the ratio is not a simple whole number ratio.

(v) Atom is no longer indestructible: In many nuclear reactions, a certain mass of the nucleus is converted into energy in the form of , and rays. �us, atom is no longer indestructible as suggested by Dalton.


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PLANCESS CONCEPTS

Modern atomic theory is, of course, a little more involved than Dalton’s theory but the essence of Dalton’s theory remains valid. Today, we know that atoms can be destroyed via nuclear reactions but not by chemical reactions.

Shivam Agarwal

Gold Medalist, INChO

5. Important Formulae of Mole Concept

(i) 1 mole of atoms = 6.023 × 1023 atoms

Gram Atomic mass = mass of 6.023 × 1023 atoms

(ii) 1 mole of molecules = 6.023 × 1023 molecules

Gram molecular mass = mass of 6.023 × 1023 molecules

(iii) Number of moles of atoms = Mass of element in gramsGram atomic mass of element

(iv) Number of moles =A

No.of atom of element NAvogadro's no.of atoms N

(v) Number of moles in molecules = Mass of substance in gramsGram molecular mass of substance

(vi) Number of moles of molecules = A

No.ofmolecules of element NAvogadro's no.of atoms N


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(vii) 1 gram formula mass = Mass of 1 mole of an ionic compound.

(viii) Molarity (M) = Number of moles of solute

Volume of the solution in litres

= Mass of solute in gram /Gram moleular mass of solute

Volume of solution in litres

(ix) Normality (N) = Number of grams equivalent of solute


(x) N = Weight of solute in gram/equivalent weight of solute


Where N = normality of solution

V = volume of solution in ml.

(xi) If weight of substance is given,

milliequivalent (NV) = ×W 1000E

Where, W = weight of substance in grams, E = equivalent weight of substance �e ratio between the moles of solute or solvent to the total moles of solution is called mole fraction.

Mole fraction of solute = Molesof solvent n w /mMoles of solution n M w /m W /M

Moles fraction of solvent = Molesof solvent N W /MMoles of solution n N w /m W /M

where, n = number of moles of solute , N = number of moles of solvent m = molecular weight of solute, M = molecular weight of solvent , w = weight of solute W = weight of solvent Sum of mole fraction of solute and solvent is always equal to one.

6. Chemical Formulae

Smallest particle of elements or compound which can exist independently is known as molecule. Formula of a substance is a group of symbols of the elements which represents one molecule of the substance.E.g. - Water molecule (H2O), Chlorine molecule (Cl2), Hydrogen molecule (H2) etc.Chemical formula of water molecule is H2O.�is indicates that one water molecule contains two atoms of hydrogen and one atom of oxygen. �us in simple words it is a way of expressing proportions of atoms


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that are present in one single Molecule.

6.1 Types of Formulae

(A) Empirical Formula: Formula which represents the simplest relative whole number ratio of atoms of each element present in one molecule of that substance is known as empirical formula.

(B) Molecular Formula: Formula which represents the actual number of atoms of each element present in one molecule of the substance is called molecular formula.

(C) Structural formula: From the molecular formula of element or compound, we get no information regarding the bonds between the atoms. Formula of element or compound that shows bonds between its atoms is known as structural formula.

PLANCESS CONCEPTS

Many compounds may have the same empirical formula but their molecular and structural formula may be di�erent.

e.g.: CH2O is the empirical formula of formaldehyde (HCHO), acetic acid (CH3COOH) and glucose (C6H12O6).

Anand K

AIR 1, NSO 2011

6.1.1 Determination of Empirical and Molecular Formulae

�e following steps are involved in determining the empirical formula of a compound.

1. �e percentage composition of each element is divided by its atomic mass. It gives atomic ratio of the element present in the compound.

2. �e atomic ratio of each element is divided by the minimum value of atomic ratio so as to get the simplest ratio of the atoms of elements present in the compound.

3. If the simplest ratio is fractional, then values of simplest ratio of each element is multiplied by smallest integer to get the simplest whole number for each of the element.

4. To get the empirical formula, symbols of various elements present are written side by side with their respective whole number ratio as a subscript to the lower right hand corner of the symbol.

Flowchart 1.4: Types of chemical formulae


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5. �e molecular formula of a substance may be determined from the empirical formula if the molecular mass of the substance is known. �e molecular formula is always a simple multiple of empirical formula and the value of simple multiple (n) is obtained by dividing molecular mass with empirical formula mass.

Molecular Mass

nEmpirical Formula Mass

Illustration 6: Calculate the percentage composition of following

(a) Potassium chloride KClO3

(b) Sodium carbonate Na2CO3

Sol: (a) Gram molecular weight of the compound = K + Cl + 3(O) = 39 + 35 + 48 = 122 g

Weight of potassium = 39

% composition of potassium = (39/122) × 100 = 31.96%

% composition of oxygen =(16/122) × 100 = 39.34%

% composition of chlorine = 100 – (31.96 + 39.34) = 28.7%

(b) Gram molecular weight of compound Na2CO3 = 2(Na) + C + 3(O) = 2(23) + 12 + 3(16)

= 46 + 12 + 48 = 106g

Weight of Na in compound = 2 × 23 = 46; % composition of Na in compound =(46/106) × 100 = 43.4%

Weight of C in compound =12; % composition of C in compound =(12/106)x100= 11.32

Weight of O in compound =48; % composition of Oxygen in compound =(48/106)x100 = 45.3%

Illustration 7: Find the empirical formula of a compound whose molecular weight is 90, has the percentage composition of C = 26.59%, H = 2.22%, O = 71.19%. Also calculate its molecular formula.

Sol:

Element Carbon Hydrogen Oxygen

Symbol C H O

Relative atomic mass 12 1 16

% composition 26.59 2.22 71.19


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Element Carbon Hydrogen Oxygen

Relative no. of atoms 26.592.21

122.22

2.221

71.194.44

16

Simple ratio 2.211

2.212.22

12.21

4.442

2.21

Simple whole no. of ratio 1 1 2

(i) �e empirical formula = CHO2

(ii) Empirical formula weight = 12 + 1 + 2 ×16 = 12 + 1 + 32 = 45

Ratio of molecular weight to empirical formula weight = 902

45

Hence, molecular formula = Empirical formula × 2 = (CHO2) ×2 = C2H2O4.

n (whole number positive integer ) Molecular WeightEmpirical weight

.

Illustration 8: How many litres of CO2 is produced from 10 g of decomposition of calcium carbonate at S.T.P?

Sol: From the equation:

CaCO3 + 2HCl → CaCl2 + H2O + CO2

100g 111g

(i) 100g of CaCO3 produces 111g of CaCl2

10g of CaCO3 will produce × 2111

10g 11.1gof CaCl100

11.1 g of CaCl2

(ii) 100g of CaCO3 produces 22.4 l of CO2

10g of CaCO3 will produce CO2×22.4 10

2.24l100

2.24 l

7. Stoichiometry

�e term Stoichiometry was introduced by a German scientist Jeremias Richter.

De�nition: Calculations based on chemical equation are known as stoichiometry. A chemical equation is the symbolic representation of a chemical change.


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It gives following information used in solving problems based on a chemical equation:

1. It indicates the number of moles of the reactants involved and number of moles of the products formed.

2. It gives the relative weight of the reactants and products.

3. It gives the volume of the gaseous reactants and products.

For example, the equation →3 2CaCO CaO CO indicates that

(a) 1 mole of CaCO3 gives 1 mole of CaO and 1 mole of CO2(b) 100g (mol.wt. of CaCO3) of CaCO3 gives 56g of CaO and 44g (or 22.4 litres) of carbon

dioxide.

Problems based on chemical equations are dealt under three headings for easy understanding.

(a) Problems involving mass-mass relationship.

(b) Problems involving mass-volume relationship.

(c) Problems involving volume-volume relationship.

PLANCESS CONCEPTS

�e word stoichiometry is just a fancy way of saying “the method you use to �gure out how much of a chemical you can make, or how much you need, during a reaction.”

For example, if you are doing a reaction and want to make 88.5 grams of the product, you would do a bunch of calculations to �gure out how much of each reagent you would need. �ose calculations are stoichiometry.

Vaibhav Gupta

AIR 2, NSO

Illustration 9: Calculate the percentage by weight of each element in calcium carbonate.

[At. wt. Ca = 40, C = 12, O = 16]

Sol: Mol. wt. (wt. of 1 mole) of CaCO3= 40 + 12 + 48 = 100

From formula, it is obvious that 100g CaCO3 has 40g Ca

100g CaCO3 has ×40

100 40%100

Similarly, % of C in CaCO3 ×12

100 12%100

% of O in CaCO3 ×48

100 48%100


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Illustration 10: 1g of Mg is burnt in a closed vessel which contains 0.5g of O2

(i) Which reactant is left in excess?

(ii) Find the weight of the excess reactant.

(iii) How many millilitres of 0.5N H2SO4 will dissolve the residue in the vessel?

Sol: �e relevant chemical equation is

2Mg + O2 ⟶ 2MgO

2 ×24 2×16 2(24+16)

= 48g = 32g = 80g

(i) From the equation, it is clear that 48g of Mg require = 32g of O2

1g of Mg requires = 32g

48 = 0.66g of O2

But only 0.5g of oxygen is available, hence, whole of Mg will not burn and a part of it will remain as such.

∴ Magnesium will be left in excess.

(ii) Weight of excess reactant

From the equation, 32g of O2 reacts with 48g of Mg

0.5g of O2 reacts with = ×48

0.5g32

= 0.75g of Mg

Wt. of Mg present = 1g

Wt. of Mg left unreacted = 1 – 0.75 = 0.25g

Hence, the wt. of excess Mg = 0.25g

(iii) Calculation of 0.5N H2SO4 required to dissolve residue, which consists of MgO and unreacted Mg.

Mg + H2SO4 ⟶ MgSO4

MgO + H2SO4 ⟶ MgSO4 + H2O

Since, in present case, it is only the magnesium atom that reacts with H2SO4 hence, total amount of Mg, i.e., 1g (whether as Mg or MgO) will react with H2SO4.

Now, we know that 1000ml of 1N H2SO4 (i.e., 1eq. wt.) = 1eq. of Mg

or 1 eq. wt. (12g) of mg = 1000mL of 1N H2SO4

1g of Mg = ×1000

112

mL of N H2SO4 = 83.3 mL of N H2SO4 =83.30.5

mL of 0.5 N H2SO4

= 166.6 mL of 0.5 N H2SO4


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Hence, the amount of 0.5N H2SO4 required to dissolve the residue (Mg and MgO) = 166.6 Weight of solid substance can be compared with the volume of gases with the help of the fact

that 1 mole or 1gm molecule of a gas occupies 22.4 litres or 22400 cc. at NTP. �us in short, the weight-volume relationship can be represented as below

Mg + 2HCl ⟶ MgCl2 + H2

By mole 1 2 1 1

By a.m.u. 24 73 95 2

By gm. wt. 24g 73g 95g 2g

By wt. or vol. 24g 73g 95g 22.4 litres at N.T.P.

It is important to note that if other conditions are not mentioned, the chemical reaction is assumed to take place at NTP i.e., at 0°C and 760 mm (1 atmospheric) pressure.

For solving problems involving mass-volume relationship, proceed according to the following instructions

(a) Write down the relevant balanced chemical equation(s).(b) Write the weights of various solid reactants and products.

In case of gases

Gases are usually expressed in terms of volumes. In case the volume of the gas is measured at room temperature and pressure (or any condition other than NTP), convert it into NTP by applying gas laws. Volume of a gas at any temperature and pressure can be converted into its weight and vice-versa with the help of the relation,

PV = gM

×RT where g is weight of gas, M is mol. wt. of gas, R is gas constant.

Calculate the unknown factor by unitary method.

Illustration 11: Two grams of sulphur is completely burnt in oxygen to form SO2. What is the volume (in L) of oxygen consumed at STP?

Sol: S + O2 ⟶ SO2

32g 22.4 L at NTP 32g of S react with 22.4 L of O2 at NTP

2g of S react with = 22.432

×2 = 1.4 L at NTP

Illustration 12: What volume of hydrogen gas, at 273 K and 1 atm pressure will be consumed in obtaining 21.6 of elemental boron (atomic mass = 10.8) from the reduction of boron trichloride by hydrogen?

Sol: 2BCl3 + 3H2 ⟶ 2B + HCl 3 ×22.4 L at NTP 2 ×10.8g = 21.6g


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From the above balanced equation it is clear that 21.6g of boron is obtained by 67.2 l of hydrogen at NTP.

Such problems can be solved by following steps

(i) Write down the relevant balanced chemical equation.

(ii) Write down the volume of reactants and products below the formula to each reactant and product with the help of the fact that 1gm molecule of every gaseous substance occupies 22.4 litres at NTP.

(iii) In case, volume of the gas is measured under particular (or room) temperature, convert it to NTP by using ideal gas equation.

Take help of Avogadro’s hypothesis “Gases under similar conditions of temperature and pressure contain the same number of molecules”. �us, under similar conditions of temperature and pressure, the volume of reacting gases are proportional to the number of moles of the gases in the balanced equation. For example,

N2 (g) + 3H2 (g) ⟶ 2NH3 (g)

1 mole 3 moles 2 moles

22.4 litres 3 ×22.4 litres 2 ×22.4 litres

1 volume 3 volumes 2 volumes

1x volume 3x volumes 2x volumes

From the above relation, it is obvious that 22.4 litres of N2 will react with 3 ×22.4 litres of H2 to form 2 ×22.4 litres of NH3.

Illustration 13: Calculate the volumetric composition of heavy water, D2O [At. wt. D = 2, O = 16]

Sol: �e relevant balanced chemical equation is

2D2O ⟶ 2D2 + O2

2 moles 2 moles 1 mole

2 vol 2 vol 1 vol

�us, according to equation, 3 vol. of gaseous products of D2O have 2 vol. of D2

100 vol. of gaseous products of D2O have 23

×100 = 66.66 vol. of D2

Similarly, 3 vol. of gaseous products of D2O have 1 vol. of O2

100 vol. of gaseous products of D2O = 13

×100 = 33.33 vol. of O2

�us, the volumetric composition of D2O = Volume % of D2 : volume % of O2 = 66.66% : 33.33% = 2:1


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SUMMARY

Terminologies

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.”

Law of De�nite Proportions or Constant Composition

“A pure chemical compound always consists of the same elements combined together in a �xed (de�nite) proportion by weight.”

Law of Multiple Proportions

“When one element combines with the other element to form two or more di�erent compounds, the mass of one element, which combines with a constant mass of the other, bears a simple ratio to one another.”

Law of Reciprocal Proportions

“�e ratio of the weights of two elements, A and B which combine separately with a �xed weight of the third element C is either the same or some simple multiple of the ratio of the weights in which A and B combine directly with each other.”

Gay Lussac’s Law of Combining Volumes

“When gases react together, they always do so in volumes which bear a simple ratio to one another and to the volumes of the product. If these are also gases, provided all measurements of volumes are done under similar conditions of temperature and pressure.”

Avogadro’s Hypothesis “Equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.”

Molarity (M) Number of moles of soluteVolume of the solution in litres

Normality(N) Weight of solute in gram/equivalent weight of soluteVolume of the solution in litres

Milliequivalent mEq = ×W 1000

E (E is the equivalent weight)

Mole fraction of solute olesof solute n w / mMolesof solution n M w /m W/M


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Moles fraction of solvent olesof solvent N W/ M

Molesof solution n N w /m W/M

Whole number positive integer(n)

MolecularMassn

Empirical Mass

PV = gM

×RT where g is weight of gas, M is mol. wt. of gas, R is gas constant.

PV = nRT n = no. of moles.

To �nd the relationship between molecular mass and vapour density of a gas.

Vapour density (VD) = Density of gas Mass of a certain volume of the gasDensity of hydrogen Mass of the same volume of hydrogen at

the same temp,and pressure

Molecular mass = 2 × vapour density

1 mole of atoms = NA = 6.023 × 1023 atoms.

GAM = mass of NA atoms.

1 mole of molecules = NA molecules.

GMM = mass of NA molecules.

Number of moles in molecules = Mass of substance in gramsGram molecular mass of substance


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SOLVED EXAMPLESExample 1: Find the percentage of water of CRYSTALLISATION in the following

(a) Copper sulphate crystals CuSO4.5H2O (b) Washing soda crystals Na2CO3.10H2O

Sol:

Copper sulphate crystals CuSO4.5H

2O Washing soda crystals Na

2CO

3.10H

2O

Gram molecular weight of compound

= Cu + S + 4(O) + 5 (H2O)

= 63 + 32 + 4(16) + 5(18) = 249 g

Weight of water in the compound

= 5 ×18 = 90g

% of water of crystallisation in CuSO4.5H2O

×

90100 36.14%

249

Gram-molecular weight of compound

Na2CO3.10H2O

= 2(Na) +C +3(O)+10(H2O)

= 2 ×23 + 12+3 ×16 +10 × 18 =286 g

Wt. of water in the compound

=10×18= 180g

% of water of crystallisation in the compound Na2CO3.10H2O

×180

100 62.94%286

Example 2: Calculate the volume of oxygen at S.T.P. required for the complete combustion of 100 l of Carbon monoxide at the same temperature and pressure.

Sol: 2CO (l) + O2 (g) 2CO2 (g)

2 (22.4l) 22.4l

44.8 l of CO requires 22.4 l of O2 for combustion.

100 l of CO will require O2 = 22.4 100 l 50 l44.8

×

Example 3: How many litres of ammonia are present in 3.4 kg of it?

Sol: At S.T.P., Molar mass has volume equal to 22.4

Molar mass of ammonia (NH3) = 17g

17g of NH3 has volume = 22.4 l

3.4 kg or 3400 g will have volume = 22.4 3400 4480 l17

× = 4480 l


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Example 4: Find out the mole fraction of solute in 10% (by weight) urea solution, urea (NH2(CO)NH2).

Sol: Weight of solute (urea) = 10g

Weight of solution = 100 g

Weight of solvent (water) = 100 –10 = 90g

Mole fraction of solute olesof solute w / mMolesof solution w /m W/M

10 / 600.032

10 / 60 90 / 18

Example 5: A compound of carbon, hydrogen and nitrogen contains these elements in the ratio of 9:1:3.5. Calculate the empirical formula. If its molecular mass is 108, what is the molecular formula?

Sol:

Element Element Ratio Atomic Mass Relative Number of atoms Simplest Ratio

Carbon 9 12 90.75

120.75

30.25

Hydrogen 1 1 11

11

40.25

Nitrogen 3.5 14 3.5 0.2514

0.251

0.25

Empirical formula = C3H4N

Empirical formula mass = (3×12) + (4×1) + 14 = 54

MolecularMass 108n 2Empirical Mass 54

�us, molecular formula of the compound = (empirical formula)n = (C3H4N)2 = C6H8N2

Example 6: A compound on analysis, was found to have the following composition:(i) Sodium = 14.31%, (ii) Sulphur = 9.97%, (iii) Oxygen = 69.50%, (iv)Hydrogen = 6.22%. Calculate the molecular formula of the compound assuming that whole of hydrogen in the compound is present as water of crystallisation. Molecular mass of the compound is 322.

Sol:

Element % Atomic Mass Relative Number of atoms

Simplest Ratio

Sodium 14.31 23 0.622 0.6222

0.311 .


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Element % Atomic Mass Relative Number of atoms

Simplest Ratio

Sulphur 9.97 32 0.311 0.3111

0.311

Hydrogen 6.22 1 6.22 6.2220

0.311

Oxygen 69.50 16 4.34 4.3414

0.311

Empirical formula = Na2SH20O14

Empirical formula mass = 23 × 2 + 32 + 20 + 14 × 16 = 322

molecular massn

empirical mass = 1

∴ As all H-atoms are as water of crystallisation.

Na2SO4.10H2O

�is is the molecular formula.

Example 7: How many oxygen can be obtained from 90 kg of water?

Sol: �e concerned chemical equation is

2H2O → 2H2 + O2

2[2+16] 2×16

=36 = 32

�us, 36kg of water gives oxygen = 32kg

90kg of water gives oxygen = ×32

90 80kg36

Example 8: Calculate the weight of 90% pure sulphuric acid that would be required for neutralising 60g of sodium hydroxide.

Sol: �e concerned chemical equation is

2NaOH + H2SO4 ⟶ Na2SO4 + 2H2O

2[23+16+1]= 80 2 + 32 + 64= 98

�us, 80g of NaOH requires H2SO4 = 98g

60g of NaOH requires H2SO4 = ×98

60 73.5g80


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�is is the weight of 100% H2SO4.

Hence, wt. of H2SO4 required ×100

73.5 81.67g90

of H2SO4

Example 9: A metal oxide has the formula Z2O3. It can be reduced by hydrogen to give free metal and water. 0.1596g of the metal oxide requires 6 mg of hydrogen for complete reduction. What is the atomic weight of the metal?

Sol: Writing the concerned equation

Z2O3 + 3H2 ⟶ 2Z + 3H2O

0.1596g 6mg = 0.006 g

�us, 0.006g of H2 reduces = 0.1596g of Z2O3

To calculate weight of 1 mole of Z2O3 we take 3 mole H2 (6 gram of H2)

6g of H2 will reduce = × 2 30.1596

6gof Z O0.006

= 159.6g of Z2O3

Molecular weight of Z2O3 = 159.6, in other words, 2Z + 48 = 159.6

2Z =159.6 – 48 = 111.6; Z = 111.6/2 = 55.8

Example 10: How many moles of impure potassium chlorate of 75% purity are required to produce 24g of oxygen?

Sol: ×

→heat3 2

3[2 16]2moles96g

2KClO 2KCl 3O

�us, 96g of oxygen are produced from KClO3 = 2 moles

24g of oxygen are produced from ×2

24 0.5mole96

Now, ∵ 75 moles of pure KClO3 are present in 100 moles of KClO3

0.5 moles of pure KClO3 = ×100

0.575

= 0.667 moles

Example 11: 5.82g of a silver coin was dissolved in strong nitric acid, and excess of sodium chloride solution was added. �e silver chloride precipitate was dried and weighed 7.2g. Calculate the percentage of silver in the coin. �e reaction is

Ag + 2HNO3 ⟶ AgNO3 + NO2 + H2O; AgNO3 + NaCl ⟶ AgCl + NaNO3


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Sol: �e relevant chemical reactions are as below

Ag + 2HNO3 ⟶ AgNO3 + NO2 + H2O; AgNO3 + NaCl ⟶ AgCl + NaNO3

On adding,

Ag + 2HNO3 + NaCl ⟶ AgCl + NaNO3 + NO2 + H2O

108 108 + 35.5

= 143.5

�us, 143.5g of AgCl is formed by 108g of Ag

7.2g of AgCl is formed by ×108

7.2g143.5

= 5.418g of Ag;

�us, 5.82g of silver coin contains = 5.418g of Ag

100g of silver contains ×5.418 1005.82

= 93.1g of Ag

Percentage of silver in coins = 93.1%

Example 12: Determine the composition of the residual mixture when 30g of magnesium is reacted with 30g of O2.

Sol: 2Mg + O2 ⟶ 2MgO

2 ×24 = 48g 2×16 = 32g 2[24+16] = 80g

�us, 48g of Mg requires = 32g of O2 ;

30g of Mg requires = ×32

3048

20g of O2

Amount of O2 left unreacted = 30 – 20 = 10g

32g of O2 requires 48g of Mg;

30g of O2 requires ×48

3032

= 45g

However, only 30g of Mg is present, therefore whole of Mg would react. 32g of O2 give = 80g of MgO; 20g of O2 give = ×80 20

32=50g

Example 13: A sample of hard water contains 20 mg of Ca2+ ions per litre. How many milliequivalents of Na2CO3 would be required to remove the Ca+2 ions from 1 litre of sample? �e given reaction is

Ca2+ + Na2CO3 ⟶ CaCO3 + 2Na+


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Ca2+ + Na2CO3 ⟶ CaCO3 + 2Na+

40g 106g

�us, 40 g Ca2+ = 106g Na2CO3

Converting these into required unit; 40g Ca2+ = 2 equivalents of Na2CO3

Mol.wt. 106

No.ofEq.Eq.wt. 53

40 × 1000 mg of Ca2+ = 2 eq. of Na2CO3 = 2 × 1000 milliequivalents of Na2CO3

20 mg of Ca2+ = ×2 100040000

×20 meq. of Na2CO3 = 1 milliequivalent of Na2CO3

Hence, No. of meq. required to soften 1 litre of sample of water = 1

Example 14: Determine the volume occupied by 11.2gm of a gas at NTP whose vapour density is 11.2.

Sol: Mol. wt. of the gas = 2 × V.D. = 2 × 11.2 = 22.4

Now we know that,

22.4 litres of every gas at NTP weighs equal to its 1 gm mole (Mol. wt.in gm). �us here

22.4 gm of the gas = 22.4 litres ; 11.2 gm of the gas = 11.2 litres

Example 15: Calculate the weight of bromine in grams absorbed by 0.1 mole of ethylene, atomic weight of bromine is 80.

Sol: �e concerned chemical reaction is-

→

2

22 2 2

CH BrCH CH Br |

CH Br

1 mol 160 g

1 mole of ethylene absorbs 160g of bromine; 0.1 mole of ethylene will absorb 16g of bromine

Example 16: 25.4 g of iodine and 14.2g of chlorine are made to react completely to yield a mixture of ICl and ICl3. Calculate the number of moles of ICl and ICl3 formed.

Sol: �e relevant balanced equation is

I2 + 2Cl2 ⟶ ICl + ICl3

1 mol 2 mol 1 mol 1 mol


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ICl ICl3

According to equation,

254 g of I2 form = 1 mole of ICl

25.4g I2 form = 1254

×25.4 mole of ICl

= 0.1 mole of ICl

According to equation

254g of I2 form = 1 mole of ICl3

25.4g of I2 form = 1254

×25.4 mole of

ICl3= 0.1 moles of ICl3

Note: In this example there is no limiting reagent but we should always check for limiting reagent in chemical reaction.

Example 17: What volume of hydrogen would be obtained by treating 0.65g of pure zinc with su�cient amount of dilute hydrochloric acid (At. wt. of Zn is 65)?

Sol: �e relevant equation is

Zn + 2HCl ⟶ Cl2 + H2

65g 22400 ml at NTP

�us, 65g of Zn evolve = 22400 ml of H2 at NTP

0.65g of Zn evolve ×22400

0.65ml65

= 224 ml at NTP

Example 18: What volume of air containing 21% of oxygen (by volume) is required to completely burn 900gm of sulphur?

Sol: Writing the relevant chemical equation,

S + O2 ⟶ SO2

32g 22.4 litres

�us, 32g of sulphur require 22.4 litres of O2

900g of sulphur requires 22.432

×900 litres = 630 l of pure O2

But, percentage of oxygen in air by volume = 21%

In other words, 21 litres of O2 are present in 100 litres of air

630 litres of O2 is present in 10021

×630 litres = 3000 litres air

Volume of air required = 3000 litres


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Example 19: What volume of oxygen gas is necessary for the complete combustion of 20.0 litres of propane (CH3CH2CH3)?


C3H8 + 5O2 ⟶ 3CO2 + 4H2O

1 mole 5 moles

1 vol. 5 vol.

1 litre 5 litres

�us, 1 litre of propane require 5 litres of O2. 20 litres of propane require 51

×20 litres = 100 litres of O2

Example 20: In Haber’s process, 20 litres of dihydrogen and 30 litres of dinitrogen were taken for reaction which yielded only 50% of the expected to form Ammonia. What will be the composition of gaseous mixture under the above mentioned conditions at the end?

Sol: N2 (g) + 3H2 (g) ⟶ 2NH3 (g)

1 vol 3 vol 2 vol

or 10 l 30 l 20 l

�e actual yield of the product = 10 L (50% of the expected yield)

Vol. of N2 used for formation of 10 L NH3 = 5 l

Vol. of H2 used for formation of 10 L NH3 = 15 l

Vol. of N2 left unreacted = 10 – 5 = 5 l

and Vol. of H2 left unreacted = 30 – 15 = 15 l

Vol. of NH3 formed = 10 l

Example 21: What volume of oxygen gas at NTP is necessary for complete combustion of 20.0 litres of propane (CH3CH2CH3) measured at 27°C and 760 mm pressure?

Sol: Conversion of 20 litres of propane at 27°C and 760 mm pressure to NTP conditions.

Given conditions NTP conditions

P1 = 760 mm P2 = 760 mm

V1 = 20.0 litres V2 =?

T1 = 27 + 273 = 300 K T2 = 273K

Applying ideal gas equation,


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1 1 2 2

1 2

P V P V

T T ; ×× 2760 V760 20

300 273 ;

× ××2

760 20 273V

300 760 = 18.2 litres

�e relevant chemical equation is

C3H8 + 5O2 ⟶ 3CO2 + 4H2O

1 mole 5 moles

1 vol. 5 vol.

1 litre 5 litres

�us, 1 litre of propane requires = 5 litres of O2

18.2 litres of propane requires 51

×18.2 =91.0 litres of O2

Example 22: 15cc of a gaseous sample believed to contain ethylene (C2H4) and acetylene (C2H2), required 45cc of oxygen for complete combustion. Determine the percentage of the mixture.

Sol: Total vol. of the sample = 15 cc

Let the vol. of ethylene in the sample = cc.

Vol. of acetylene in the sample = (15–x)cc

Writing the relevant balanced chemical equations.

(i) C2H4 + 3O2 ⟶ 2CO2 + 2H2O

1 vol. 3 vol.

1 cc. 3 cc.

(ii) 2C2H2 + 5O2 ⟶ 4CO2 + 2H2O

2 cc 5 cc

From equation,

1cc. of C2H4 requires = 3 cc. of O2

cc. of C2H4 requires = 31

× cc.

= 3 cc. of O2

From equation,

2 cc. of C2H2 requires 5 cc. of O2

(15–) cc. of C2 H2 requires 52

(15–) cc.

Total vol. of O2 consumed = 3 + (15–) cc.

Vol. of O2 given in question = 45 cc

3+ (15–) = 45; 6 + 75 – 5 = 90; = 90 – 75 = 15

Vol. of C2H4 in the sample = 15 m Vol. of C2H2 in the sample = 15–15= 0 mL


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Example 23: A gas mixture of 3.0 litres of propane and butane on complete combustion at 25°C produced 10 litres of CO2. Find out the composition of the gas mixture.

Sol:

Let the volume of propane = V litres Let the volume of butane = (3 –V) litres

C3H8 + 5O2 ⟶ 3CO2 + 4H2O

(V) L (5V) L (3V) L (4V) l

C4H10 + O2 ⟶ 4CO2 + 5H2O

(3 –V) L (3–V) L 4(3 –V) L 5(3 –V) l

Given Volume of CO2 produced = 10 l

3V + 4 (3 –V) = 10; 3V – 4V = 10 – 12 or V = 2l

Volume of propane = 2 l

Composition of propane = 23

×100 = 66.66%

Volume of butane = 1 l

Composition of butane = 13

×100 = 33.33%

Example 24: 1 litre of a mixture of CO and CO2 is taken. �is is passed through a tube containing red hot charcoal. �e volume now becomes 1.6 litre. �e volumes are measured under the same conditions. Find the composition of the mixture by volume. When the mixture of CO and CO2 is passed through a tube containing charcoal, CO2 present in the mixture is reduced by charcoal (C) to CO.

Sol: Total volume of the mixture = 1 litre = 1000ml

Let the vol. of CO in the mixture = x ml

Vol. of CO2 in the mixture = (1000 –x) ml

When the mixture of CO and CO2 is passed through a tube containing charcoal, CO2 present in the mixture is reduced by charcoal (C) to CO with the result, the product (1.6 litres) will consist of only CO. �e relevant chemical equation is

CO2 + C ⟶ 2CO

1 mole 2 moles

1 vol. 2 vol.

1 ml 2 ml

�us,1 ml of CO2 gives = 2 ml of CO

(1000–x) ml of CO2 gives 21

×(1000–x)ml of CO = 2 (1000–x) ml of CO

Total vol. of CO = Vol. of CO in mix + Vol. of formed from CO2 = x + 2 (1000–x)

But, total volume of the product = 1.6 litres = 1600 ml (given)


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x + 2 (1000–x) = 1600 ; x + 2000 – 2x = 1600 ; x = 400

Vol. of CO in the mixture = 400 ml

Vol. of CO2 in the mixture = 1000 – 400 =600 ml

Example 25: Nine volumes of a mixture of a gaseous organic compound A and just su�cient amount of oxygen required for complete combustion yielded on burning four volumes of CO2, six volumes of H2O vapour and two volumes of N2, all being measured at the same temperature and pressure. If compound A contained C, H and N only, what is the molecular formula of A?

Sol: According to the given statement, the relevant chemical equation is

→

2 2 2 2A O CO H O N

9 vol. 4 vol. 6 vol. 2 vol.

9 moles 4 moles 6 moles 2 moles

A + O2 ⟶ 4CO2 + 6H2O + 2N2

�e total oxygen atoms present in product are 14, therefore, 7 moles of oxygen must have been present in the reactant. Hence, A + 7O2 ⟶ 4CO2 + 6H2O + 2N2

�e total no. of moles of the reactants = 9(given) ∴ No. of moles of A = 9 – 7 = 2

Hence, the fully balanced equation will be as below 2A + 7O2 ⟶ 4CO2 + 6H2O + 2N2

Since, all oxygen atoms of the products are balanced by O2 of reactant. A must have other atoms of products, i.e., 2C, 6H and 2N (atoms are calculated in one mole of A).

Hence, the molecular formula of A is C2H6N2


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EXERCISE 1 – For School Examinations

Fill in the Blanks

Directions: Complete the following statements with an appropriate word/term to be �lled in the blank space(s).

Q.1 During a chemical reaction, the sum of of the reactants and products remains unchanged.

Q.2 In a pure chemical compound, elements are always present in a proportion by mass.

Q.3 Clusters of atoms that act as an ion are called ions.

Q.4 In ionic compounds, the charge on each ion is used to determine the of the compound.

Q.5 �e Avogadro constant is de�ned as the number of atoms in exactly of carbon-12.

Q.6 Mass of 1 mole of a substance is called its

Q.7 �e abbreviation used for lengthy names of elements are termed as their

Q.8 A chemical formula is also known as a

Q.9 �ose ions which are formed from single atoms are called

Q.10 Ionic compounds are formed by the combination between and

Q.11 �e valency of an ion is to the charge on the ion.

Q.12 Mole is a link between the and

Q.13 �e SI unit of amount of a substance is

True / False

Directions: Read the following statements and write your answer as true or false.

Q.14 Formula mass of Na2O is 62 amu.

True False

Q.15 �ose particles which have more or less electrons than the normal atoms are called ions True False


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Q.16 Formula for sulphur dioxide is SO3.

True False

Q.17 Molar mass of ethyne (C2H2) is 26 g/mol.

True False

Q.18 22gm of CO2 consists of 1 mole.

True False

Q.19 Number of molecules in 32 gram of oxygen is 6.02 x 1023.

True False

Q.20 Water is an atom.

True False

Q.21 Formula for sulphur dioxide is SO2.

True False

Q.22 Clusters of atoms that act as an ion is called polyatomic ion.

True False

Q.23 Mass of 1 mole of a substance is called its formula mass.

True False

Q.24 In a pure chemical compound, elements are always present in a de�nite proportion by Mass.

True False

Match the Following Columns

Directions: Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in column I have to be matched with statements (p, q, r, s) in Column II

Q.25 Column II gives molecular mass in amu for substance in Column I. Match them correctly.

Column I Column II

(A) H2O (p) 58.5

(B) HNO 3 (q) 111

(C) NaC1 (r) 18

(D) CaCl2 ( s ) 63


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Q.26 Column II gives no. of moles for description given in Column I. Match them correctly.

Column I Column II

(A) 52 g of He (p) 2

(B) 12.044 x 1023 He atoms (q) 13

(C) 8 of gm O2 (r) ¼

(D) 28 of gm N2 (s) 1

Very Short Answer Questions

Directions: Give answer in one word or one sentence.

Q.27 De�ne term mole.

Q.28 What is Avogadro’s number?

Q.29 What is gram atomic weight?

Q.30 What is the di�erence between CO2, and 2CO2?

Q.31 Name the following compounds PCl3 and SO2.

Q.32 In what smallest whole-number ratio must N and O atoms combine to make dinitrogen to tetroxide N2O4? What is the mole ratio of the elements in this compound?

Q.33 How many moles of sodium atoms correspond to 1.56 x 1021 atoms of sodium?

Q.34 What is the relationship between the formula weight of a substance and its molar mass?

Q.35 How many grams of silver are in 0.263 mol of Ag?

Q.36 How many atoms are 1.00 x 10-9 g of lead?

Q.37 How many grams of iron are needed to combine with 25.6g of O to make Fe2O3?

Q.38 What is the mass of 4 moles of aluminium atoms? (Atomic mass of Al = 27u)

Q.39 Calculate the mass of 6.022 x 1022 atoms of He.

Q.40 Calculate the number of moles in 3.011 × 1022 molecules of carbon dioxide.

Q.41 A sample of 45.8g of H2SO4 contains how many moles of H2SO4?

Q.42 What is the mass in grams of 5 moles of Fe?

Q.43 How many moles of NaCl are present in 20 gm of the substance?

Q.44 How many grams of O is present in 50 gm of CaCO3?

Q.45 How many grams of CO2 are present in 0.1 mole CO2?


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Q.46 Describe the di�erence between the mass of a mole of oxygen atoms (O) and the mass of a mole of oxygen molecules (O2).

Q.47 What do you understand by the term Formula unit?

Short Answer Questions

Directions: Give answer in two to three sentences.

Q.48 Convert into moles:

(a) 12 g of oxygen gas (b) 22 g of carbon dioxide

Q.49 What is molecular weight? Explain with example.

Q.50 Write formulas for ionic compounds formed from

(a) Na and F, (b) Na and O

(c) Mg and F (d) Al and C

Q.51 Write formulas for

(a) Aluminium sul�de (b) Strontium �uoride

(c) Titanium (IV) oxide (d) Calcium bromide

Q.52 How many moles of Al atoms are needed to combine with 1.58 mole of O atoms to make aluminium oxide, Al2O3?

Q.53 What is the volume of 32g of sulphur dioxide measured at STP?

Q.54 How many grams of sodium will have the same number of atoms as 6 grams of magnesium?

(Na = 23, Mg = 24).

Q.55 How many grams of Chromium (molar mass Cr = 52 g/mol) are there in 85 g of Cr2S3?

Q.56 Which has more number of atoms: 100 grams of sodium or 100 grams of iron? (Given atomic mass of Na = 23 u and Fe = 56 u)

Q.57 A compound of nitrogen and oxygen has the formula NO. In this compound, there are 1.143g of oxygen for each 1.000g of nitrogen. A di�erent compound of nitrogen and oxygen has the formula NO2.How many grams of oxygen would be combined with each 1.000g of nitrogen in NO2?

Q.58 How many moles of nitrogen atoms are combined with 8.60 mol of oxygen atoms in dinitrogen pentoxide, N2O5?

Q.59 Calculate the volume occupied by 2.8 g of N2 at STP.

Q.60 How much weight of carbon dioxide is present in same volume which is occupied by 4.0 grams of oxygen? Volumes are measured at normal temperature and pressure. (NTP)

Q.61 What is the mass of 0.5 mole of water (H2O)? (Atomic masses : H = 1u, O = 16u)


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Q.62 How many moles of sulphur atoms and Oxygen atoms are present in one mole each of H2SO4, H2SO3 and SO2?

Q.63 What is the percentage composition of the elements in ammonia, NH3 ? (at. mass: H = 1, N = 14)

Q.64 Calculate the theoretical percentage composition of N2O3.

Q.65 10 grams of CaCO3 on heating gave 4.4g of CO2 and 5.6 of CaO. Show that these observations are in agreement with the law of conservation.

Q.66 In which of the following, the number of hydrogen atoms is more (a) 3 moles of H2O or (b) 10 moles of HCl?

Q.67 If we decompose 50g of water by passing electricity through it, how many grams of oxygen and hydrogen we obtain?

Q.68 Hydrogen and oxygen combine in the ratio of 1 : 8 by mass to form water. What mass of oxygen gas would be required to react completely with 3g of hydrogen gas?

Q.69 Calculate the number of molecules in 32 grams of oxygen gas and 14 grams of nitrogen.

Q.70 Calculate the mass of a sample of iron-metal that would contain 0.250 moles of iron atoms.

Q.71 Calculate the number of atoms in a 0.123 gram sample of aluminium foil.

Long Answer Questions

Directions: Give answer in four to �ve sentences.

Q.72 Calculate the formula mass of each of the following and round your answer to the nearest 0.1u.

(a) NaHCO3 (b) K2Cr2O7

(c) (NH4)2CO3 (d) Al2(SO4)3

(e) CuSO4.5H2O

Q.73 Calculate the molar mass of the following substances.

(a) Ethyne, C2H2

(b) Sulphur molecule, S8

(c) Phosphorus molecule, P4 (atomic mass of phosphorous = 31)

(d) Hydrochloric acid, HCl

(e) Nitric acid, HNO3

Q.74 Calculate the number of aluminium ions present in 0.051g of aluminium oxide (Al2O3). (Atomic mass : Al = 27 u, 0 = 16u)


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Q.75 (I) Write the chemical formulas for the following compounds:

(a) Copper (1) oxide, (b) Potassium peroxide

(c) Mercury (I) bromide (d) Iron (III) carbonate,

(e) Sodium hypobromite

(II) Give the name for each of the following acids

(a) HBrO3, (b) HBr,

(c) H3PO4, (d) HClO

(e) HIO3

Q.76 (I) Calculate both the average mass of a single molecule of carbon dioxide and glucose and the molecular weight of these compounds.

(II) Determine the number of carbon atoms in 0.500 grams of carbon dioxide, CO2.

Q.77 (I) 1.375g of cupric oxide was reduced by heating in a current of hydrogen and the weight of copper that remained was 1.098g. In another experiment, 1.179g of copper was dissolved in nitric acid and resulting copper nitrate converted into cupric oxide by ignition. �e weight of cupric oxide formed was 1.476g. Show that these results illustrate the law of constant proportion.

(II) In an experiment, 1.288 g of copper oxide was obtained from 1.03g of copper. In another experiment, 3.672g of copper oxide gave, on reduction, 2.938 g of copper. Show that these �gures verify the law of constant proportions.

Q.78 Calculate the following quantities :

(a) Mass, in grams, of 1.73 mol CaH2

(b) Moles of Mg (NO3)2 in 3.25g of this substance

(c) Number of molecules in 0.245 mol CH3OH

(d) Number of H atoms in 0.585 mol C4H1o.

(e) Number of moles of Al in 2.16 mol of Al2O3

Q.79 (i) Distinguish between an atom and a molecule.

(ii) Di�erentiate between atom and ion.


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EXERCISE 2 – For Competitive Examinations

Multiple Choice Questions

Directions: �is section contains multiple choice questions. Each question has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct.

Q.1 Which of the following has largest number of particles?

(a) 8g of CH4 (b) 4.4g of CO2

(c) 34.2g of C12H22O11 (d) 2g of H2

Q.2 �e number of molecules in 16.0g of oxygen is

(a) 6.02 x 1023 (b) 6.02 x 10-23

(c) 3.01 x 10-23 (d) 3.01 x 1023

Q.3 �e percentage of hydrogen in H2O is -

(a) 8.88 (b) 11.12

(c) 20.60 (d) 80.0

Q.4 Find the mass of oxygen contained in 1 kg of potassium nitrate (KNO3)

(a) 475.5 g (b) 485.5 g

(c) 475.24 g (d) 485.2 g

Q.5 How many moles of electron weight one kilogram?

(a) 6.023 x 1023 (b) × 31110

9.108

(c) × 546.02310

9.108 (d) ×

×81

109.108 6.023

Q.6 25.4g of iodine and 14.2g of chlorine are made to react completely to yield a mixture of ICl and ICl3. Calculate the ratio of moles of ICI and ICl3(a) 1 : 1 (b) 2 : 1 (c) 3 : 1 (d) 1 : 2

Q.7 �e mass of sodium in 11.7 g of sodium chloride is

(a) 2.3 g (b) 4.32 g (c) 6.9 g (d) 7.1 g

Q.8 �e formula of a chloride of a metal M is MCl3, the formula of the phosphate of metal M will be

(a) MPO4 (b) M2PO4 (c) M3PO4 (d) M2(PO4)3


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Q.9 Which of the following contains the largest number of molecules ?

(a) 0.2 mol H2 (b) 8.0 g H2 (c) 17 g of H2O (d) 6.0 g of CO2

Q.10 One gram of which of the following contains largest number of oxygen atoms?

(a) O (b) O2 (c) O3 (d) All contains same

Q.11 �e percentage by weight of O2 in CaSO4 (O= 16, S = 32, Ca = 40) is –

(a) 64 (b) 28.2 (c) 47.2 (d) 16.2

Q.12 �e percentage by weight of Zn in white vitriol, ZnSO4.7H2O (Zn = 65, S = 32, O = 16, H = 1), is approximately

(a) 23 (b) 33 (c) 43 (d) 13

Q.13 �e formation of SO2 and SO3 explain

(a) �e law of conservation of mass

(b) �e law of multiple proportions

(c) �e law of de�nite properties

(d) Boyle’s law

Q.14 �e law of de�nite proportions was given by -

(a) John Dalton (b) Humphry Davy (c) Proust (d) Michael Faraday

Q.15 Molecular mass is de�ned as the

(a) Mass of one atom compared with the mass of one molecule

(b) Mass of one atom compared with the mass of one atom of hydrogen

(c) Mass of one molecule of any substance compared with the mass of one atom of C-12

(d) None of the above

Q.16 0.001 g of C is required to write a letter with a graphite pencil. �e total number of C atoms used in writing the letter is

(a) 5.00 x 1012 (b) 5 x 1019

(c) 5.0 x 1024 (d) 6.023 x 1023

Q.17 One mole of a gas occupies a volume of 22.4 l �is is derived from

(a) Berzelius’ hypothesis (b) Gay-Lussac’s law

(c) Avogadro’s law (d) Dalton’s law

Q.18 �e mass of one C atom is -

(a) 6.023 x 1023 g (b) 1.99 x 10-23 g

(c) 2.00 g (d) 12g


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Q19 �e chemical symbol for barium is

(a) B (b) Ba (c) Be (d) Bi

Q.20 �e chemical symbol P stands for

(a) Phosphorus (b) Potassium

(c) Polonium (d) Promethium

Q.21 A group of atoms chemically bonded together is a (an)

(a) Molecule (b) Ion

(c) Salt (d) Element

Q.22 Adding electrons to an atom will result in a (a

(a) Molecule (b) Anion (c) Cation (d) Salt

Q.23 When an atom loses electrons, it is called a (an) and has a charge.

(a) Anion, positive (b) Cation, positive (c) Anion, negative (d) Cation, positive

Q.24 �e molecular formula P2O5 means that

(a) A molecule contains 2 atoms of P and 5 atoms of 0

(b) �e ratio of the mass of P to the mass of O in the molecule is 2:5

(c) �ere are twice as many P atoms in the molecule as there are O atoms

(d) �e ratio of the mass of P to the mass of O in the molecule is 5 : 2

Q.25 �e correct symbol for silver is

(a) Ag (b) Si (c) Ar (d) Al

Q.26 Aspartame, an arti�cial sweetener, has the molecular formula C 14H18N2O5. What is the mass in grams of one molecule? (Atomic weights: C = 12.01, H = 1.008, N = 14.01, O = 16.00).

(a) 4.89 x 10-21 (b) 2.24 x 10-21 (C) 3.85 x 10-22 (d) 4.74 x 10-22

Q.27 Morphine, an addictive drug, has the molecular formula C17H19NO3. What is the mass in grams of one molecule? (Atomic weights; C = 12.01, H = 1.008, N = 14.01, O=16.00)

(a) 4.89 x 10-21 (b) 2.24 x10-21 (c) 3.85 x10-22 (d) 4.89 x10-22

Q.28 �e controversial arti�cial sweetener saccharin has the molecular formula C3H5O3NS. What is the mass in grams of one molecule? Atomic weights: C = 12.01, H =. 1.008. O= 16.00, N = 14.01, S =32.06).

(a) 5.55 x 10-21 (b) 3.85 x 10-22 (c) 2.24 x 10-22 (d) 2.24 x 10-21


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Q.29 �e Statue of Liberty is made of 2.0 × 105 lbs of copper sheets bolted to a framework. ( 1 lb = 454 g) How many atoms of copper are on the statue? (Atomic weight: Cu =63.5).

(a) 2.1 x 1027 (b) 8.6 x 1029 (c) 4.3 x 1026 (d) 8.6 x 1026

Q.30 Selenium ingested in the amount of 90 micrograms per day causes loss of hair. How many selenium atoms are in this size sample? (Atomic weight: Se =78.96).

(a) 6.9 x 1023 (b) 8.8 x 1017 (c) 8.8 x1022 (d) 6.9 x1017

Q.31 Novocain is C13H16N2O2. What is the total number of moles of atoms in 0.020 moles of novocain? (Atomic weights: C = 12.01, 0 = 16.00, N = 14.01, H = 1.008).

(a) 0.033 (b) 3.3 (c) 0.66 (d) 0.33

Q.32 Methoxychlor, a garden insecticide, has the molecular formula C16H15C13O2. What is the total number of moles of atoms in a 3.0 mg sample? (Atomic weights: C = 12.01, H= 1.008, Cl=35.45, O =16.00).

(a) 8.7x 10-6 (b) 3.0 x 10-5 (c) 3.1 x 10-1 (d) 3.5 x 10-4

Q.33 What are the total number of moles of atoms in 4.32 g of Sc(NO3)3 ? (Atomic weights: Sc = 45.0.0 = 16.00. N = 14.01).

(a) 0.0132 (b) 0.324 (c) 0.0187 (d) 0.243

Q.34 Two samples of lead oxide were separately reduced to metallic lead by heating in a current of hydrogen. �e weight of lead from one oxide was half the weight of lead obtained from the other oxide. �e data illustrates –

(a) Law of reciprocal proportions (b) Law of constant proportions

(c) Law of multiple proportions (d) Law of equivalent proportions

Q.35 �e percentage of copper and oxygen in samples of CuO obtained by di�erent methods were found to be the same. �e illustrate the law of

(a) Constant proportion (b) Conservation of mass

(c) Multiple proportion (d) Reciprocal proportion

Q.36 �e total number of atoms represented by the CuSO4.5H2O is

(a) 27 (b) 271 (c) 5 (d) 8

Q.37 In compound A, 1.00g of nitrogen unites with 0.57g of oxygen. In compound B, 2.00g of nitrogen combines with 2.24g of oxygen. In compound C, 3.00g of nitrogen combines with 5.11g of oxygen. �ese results obey the following law

(a) law of constant proportion (b) law of multiple proportion

(c) law of reciprocal proportion (d) Dalton’s law of partial pressure


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Q.38 �e weight of a molecule of the compound C60H122 is

(a) 1.4 × 10-21 g (b) 1.09 × 10-21 g (c) 5.025 × 1023 g (d) 16.023 × 1023 g

Q.39 �e mass of a molecule of water is

(a) 3 × 10-26 kg (b) 3 × 10-25 kg (c) 1.5 × 10-26kg (d) 2.5 × 10-26kg

Q.40 �e number of atoms in 4.25g of NH3 is approximately

(a) 1 × 1023 (b) 2 × 1023 (c) 4 × 1023 (d) 6 × 1023

Q.41 Volume of a gas at STP is 1.12 x 10-7 cc. Calculate the number of molecules in it -

(a) 3.01 × 1020 (b) 3.01 × 1012 (c) 3.01 × 1023 (d) 3.01 × 1024

Q.42 �e number of molecules of CO2 present in 44g of CO2 is

(a) 6.02 × 1 023 (b) 3 × 1023 (c) 12 × 1023 (d) 3 × 1010

Q.43 �e volume occupied by 4.4g of CO2 at STP is

(a) 22.4 l (b) 2.24 l (c) 0.224 l (d) 0.1 l

Q.44 How many molecules are present in one gram of hydrogen?

(a) 6.02 × 1023 (b) 3.01 × 1023 (c) 2.5 × 1023 (d) 1.5 × 1023

More than One Correct

Directions: �is section contains multiple choice questions. Each question has 4 choices (a), (b), (c) and (d) out of which ONE OR MORE may be correct.

Q.45 Which of the following has same mass?

(a) 4g of He (b) 6.023 × 1023 atoms of He

(c) 1 atom of He (d) 1 mole atoms of He

Q.46 Which of the following pairs of substances illustrate the law of multiple proportions?

(a) CO and CO2 (b) H2O and D2O

(c) SO2 and SO3 (d) MgO and Mg(OH)2

Q.47 Which of the following is/are the best example of law of conservation of mass?

(a) 12g of carbon combines with 32g of oxygen to form 44g of CO2.

(b) When 12g of carbon is heated in a vacuum, there is no change in mass.

(c) A sample of air increases in volume when heated at constant pressure but its mass remains unaltered.

(d) 2 g of hydrogen combines with 16 g of oxygen to form 18 g of water.


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Q.48 Which one of the following pairs of gases contains the same number of molecules?

(a) 16 g of O2 and 14 g of N2 (b) 8 g of O2 and 22 g of CO2

(c) 28 g of N2 and 22 g of CO2 (d) 8 g of O2 and 7 g of N2

Q.49 Which of the following represents a polyatomic ion?

(a) Sulphite (b) Chloride (c) Sulphate (d) Phosphate

Q.50 Which of the following statements is /are correct?

(a) An atom is the smallest particle of matter according to Dalton’s theory(b) An atom.is the smallest particle of an element(c) An atom is the smallest indivisible particle of an element that can take part in a chemical change(d) An atom is the radioactive emission

Q.51 Which of the following symbols does not represent an element?

(a) CO (b) Ar (c) K (d) NO

Q.52 Which of the following element(s) has a symbol having two letters?

(a) Tin (b) Uranium (c) Carbon (d) Aluminium

Q.53 Which of the following is / are not a correct symbol for an element(s)?

(a) Ng (b) Fi (c) Bk (d) Zc

Q.54 0.220 g of a gas occupies a volume of 112 ml at a pressure of 1 atm and temperature of 273 K. �e gas can be

(a) Nitrogen dioxide (b) Nitrous oxide (c) Carbon dioxide (d) Propane

Q.55 Which of the following contains the same number of molecules?

(a) 1g of O2, 2 g of SO2,

(b) 1 g of CO2, 1g of N2O

(c) 112 ml of O2 at STP, 224 ml of He at 0.5 atm and 273K

(d) 1 g of oxygen, 1 g of ozone

Q.56 SO2 gas is slowly passed through an aqueous suspension containing 12 g CaSO3 till the milkiness just disappears. What amount of SO2 would be required?

CaSO3(s) + H2O(l) + SO2(g) → Ca(HSO3)2

(soluble)

(a) 12.8 g (b) 6.4 g (c) 0.1 mole (d) 0.2 mole

Q.57 8g of O2 has the same number of molecules as

(a) 7g CO (b) 14g N2 (c) 11g CO2 (d) 1 6g SO2


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Q.58 A vessel contains 4.4 g of CO2. It means that it contains

(a) 0.1 mole of CO2 (b) 6.02 × 1022 molecules of CO2

(c) 8.8g atoms of oxygen (d) 1120 m l of CO2 S.T.P.

Q.59 1g Mg was burnt in a closed vessel containing 2 g oxygen. Which of the following are not correct?

(a) 0.25 g of Mg will be left unburnt (b) 1.33 of O2 will be left unreacted

(c) 2.5 g of MgO will be formed (d) �e mixture at the end will weigh 3 g.

Fill in the Blanks

Directions: Complete the following statements with an appropriate word/term to be �lled in the passage.

Q.60 Laws to explain the chemical behaviour of matter are called law of (1) . Law of conservation of mass is given by (2) Suppose xg and yg of oxygen combines with �x zg of sulphur then x/z and y/z are in the ratio of small (3) numbers. �is data illustrates the law of (4) Gay lussac’s law is applicable when reactant and product are in (5) state.

Q.61 In one gram atomic weight of any element (1) atoms of that element are present. �is quantity of the element is known as (2) mole. One mole of any element is the value of (3) in grams. For any compound it is the value of (4) in gram.

Passage Based Questions

Directions: Study the given passage (s) and answer the following questions.

Ashok is conducting an experiment in laboratory involving compound x. One mole of compound x weighs 35 g. If he consumed 15 g of x in his experiment �en

Q.62 �e no. of moles of x used in above experiment is

(a) 1/7 (b) 2/7 (c) 3/7 (d) 4/7

Q.63 �e name of compound x is

(a) NH4OH (b) NaOH (c) NaCl (d) Na2O2

Q.64 If in another experiment, only 70g of those substances are required. �en what will be the number of moles now?

(a) 3 (b) 2 (c) 4 (d) 1


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Assertion and Reason

Directions: Each of these questions contains an assertion followed by reason. Read them carefully and answer the questions on the basis of following options. You have to select from following the one that best describes the two statements.

(a) Both assertion and reason are correct and reason is the correct explanation of assertion

(b) Both assertion and reason are correct, but reason is not the correct explanation of assertion.

(c) Assertion is correct but reason is incorrect.

(d) Assertion is incorrect but reason is correct.

Q.65 Assertion: Molecular weight of oxygen is 16.

Reason: Atomic weight of oxygen is 16.

Q.66 Assertion: One mole of SO2 contains double the number of molecules present in one mole of O.

Reason: Molecular weight of SO2 is double to that of O2.

Q.67 Assertion: Pure water obtained from di�erent sources such as, river, well, spring, sea etc. always contains hydrogen and oxygen combined in the ratio : 8 by mass.

Reason: A chemical compound always contains elements combined together in same proportion by mass, it was discovered by French chemist, Joseph Proust (1799).

Q.68 Assertion: 1 amu equals to 1.66 x 10-24 g

Reason: 1.66 x 10-24 g equal to 112

the of mass of a C12 atom.

Q.69 Assertion : 1 mol of H2 and O2 each occupy 22.4 L of volume at 0°C and 1 bar pressure.

Reason : Molar volume for all gases at the same temperature and pressure has the same volume.

Multiple Matching Questions

Directions: Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in column I have to be matched with statements (p, q, r, s) in Column II

Q70.

Column I Column II

(A) 1.5 mole of CO2(g (p) 33600 ml at STP

(B) 3.0 g of H2 (q) Total number of atoms=4.5 x NA

(C) 1.5 mole of O3 (g) (r) Weighs 72 g

(D) 1 mole of oxygen (s) Weighs 32 g


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Subjective Questions

Directions: Answer the following questions.

Q.71 Write formulas for the chlorides and oxides formed by (a) chromium and (b) copper.

Q.72 �e formula for arsenic acid is H3AsO4. What is the name of the salt Na3AsO4?

Q.73 Sucrose (table sugar) has the formula C12H22O11. In this compound, what is the

(a) Atom ratio of C to H? (b) Atom ratio of H to O ?

(c) Mole ratio of C to O? (d) Mole ratio of H to O?

Q.74 A sample of ascorbic acid (vitamin C) is synthesized in the laboratory. It contains 1.50g of carbon and 2.00g of oxygen. Another sample of ascorbic acid isolated from citrus fruits contains 6.35g of carbon. How many grams of oxygen does it contain ? Which law are you assuming in answering this question?

Q.75 Calculate the weight of (i) one atom of oxygen, and (ii)one molecule of oxygen.

G.A.W. of O2 = 16.0g

Q.76 Why can’t we say, ‘an atom of water’?

Q.77 How does Dalton’s atomic theory account for the fact that when 1.000g of water is decomposed into its elements, 0.111 g of hydrogen and 0.889g of oxygen are obtained regardless of the source of the water?

Q.78 A chemist �nds that 30.82g of nitrogen will react with 17.60g, 35.20g, 70.40g or 88.00g of oxygen to form four di�erent compounds. (a) Calculate the mass of oxygen per gram of nitrogen in each compound. (b) How do the numbers in part (a) support Dalton’s atomic theory?

Q.79 Calculate gram molecular weights of the following gases: N2 (if 360 cm3 at STP weighs 0.45g)

Q.80 A silicon chip used in an integrated circuit of a microcomputer has a mass of 5.68 mg. How many silicon (Si) atoms are present in this chip?

Q.81 How many atoms of oxygen are present in 300g limestone?

Q.82 Elements X and Y form the compound XY4. When these elements react, it is found that 1.00g of X combines with 5.07g of Y. When X combines with oxygen, it forms the compound XO2 in which 1.00 g of X combines with 1.14g of 0. What is the atomic mass of Y ?

Q.83 Aluminium sulfate, Al2(SO4)3, is a compound used in sewage treatment plants.

(a) Construct a pair of conversion factors that relate moles of aluminium to moles of sulfur for this compound.

(b) Construct a pair of conversion factors that relate moles of sulfur to moles of Al2(SO4)3.

(c) How many moles of Al are in a sample of this compound if the sample also contains 0.900 mol S ?

(d) Calculate the number of moles of sulphur if number of aluminium sulphate present is 1.16 mol.


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Q.84 How many moles of UF6 would have to be decomposed to provide enough �uorine to prepare 1.25 mol of CF4 ? (Assume su�cient carbon is available.)

Q.85 One sample of CaC2 contains 0.150 mol of carbon. How many moles and how many grams of calcium are also in the sample ?

Q.86 An atom is 200 times heavier than 112

of the mass of an atom of carbon (C-12), what is its mass in amu (or U)?

Q.87 Calculate the number of moles, molecules and atoms in 11.2 L of H2 at NTP

Q.88 Find the percentage of nitrogen in urea (H2NCONH2)

Q.89 How many molecules of water of crystallisation are present in 24.95 mg of hydrated copper sulphate (CuSO4.5H2O)?

Q.90 Sodium carbonate reacts with ethanoic acid to form sodium ethanoate, carbon dioxide and water. In an experiment, 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid to form 8.2 of sodium ethanoate, 2.2 g of carbon dioxide and 0.9g of water. Show that this data veri�es the

law of conservation of mass. �e given reaction is

→

Reac tants

Pr oducts

Sodium Carbonate Ethanoic acid

Sodium ethanoate Carbon dioxide Water

Q.91 375 gm of pure cupric oxide was reduced by heating in a current of pure dry hydrogen and the mass of copper that remained 1.0980 gms. In another experiment, 1.179 g of pure copper was dissolved in pure HNO3 and the resulting copper nitrate converted into cupric oxide by ignition. �e mass of copper oxide formed was 1.476 g. Show that the results illustrate the law of constant composition with in the limits of experimental error.

Q.92 What is the formula of carbon dioxide if 2.73 grams of carbon combine with 7.27 grams of oxygen molecules (O2) when the carbon burns?

Q.93 Write the chemical formula for each substance mentioned in the following word descriptions (a) Zinc carbonate can be heated to form zinc oxide and carbon dioxide. (b) On treatment with hydro�uoric acid, silicon dioxide forms silicon tetra�uoride and water. (c) Sulfur dioxide reacts with water to form sulfurous acid. (d) �e substance hydrogen phosphide, commonly called phosphine, is a toxic gas. (e) Perchloric acid reacts with cadmium to form cadmium (II) perchlorate.


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SOLUTIONS

Exercise 1 – For School Examinations

Fill in the Blanks

1. Masses 2. De�nite 3. Polyatomic

4. Chemical formula 5. 6.022 x 1023 6. Molar mass

7. Symbol 8. Molecular formula 9. Simple ions

10. Metal and non-metals. 11. Equal

12. Mass of atoms and number of atoms. 13. Mole

True / False

14. True 15. True 16. False 17. True 18. False 19. True

20. False 21. True 22. True 23. False 24. True

Match the Column

25. A ->(r), B -> (s), C -> (p), D->(q), 26. A->(q), B-> (p), C -> (r), D -> (s)

Very Short Answer Questions

27. In one gram atomic weight of any element 6.02 x 1023 atoms of that element are present. �is quantity of the element is known as one mole.

28. One gram atomic mass of any element contains the same number of atom of that element as there are carbon atoms in exactly 12 g of carbon-12. �is number is Avogadro’s number i.e., 6.023 x 1023.

29. �e atomic mass of an element when expressed in grams is known as gram atomic mass or simply as gram atom (g-atom). �us, one g-atom of carbon (C-12) weighs 12.0 g. �e number of gram-atoms of any element in its certain weight is given by:

30. CO2 represents one molecule of triatomic carbon dioxide, while the notation 2CO2 represents two molecules of carbon dioxide.

Number of gram atom Mass of the element in gramsAtomic mass


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31. Phosphorus trichloride and sulfur dioxide.

32. 1:2, 2:4

33. 2.59 x 10-3, mole of Na atoms.

34. It’s the same.�e formula weight of a substance in amu has the same numerical value as the molar mass expressed in grams.

35. No. of grams of Ag (0.263 mol of Ag) = 107.9g of Ag1mol of Ag

× 0.263 = 28.4 g of Ag

36. No. of atoms of Pb= No. of moles of Pb × NA

923 1210

6.022 10 2.906 10 atoms207.2

× × ×

37. No. of g of Fe

1molof O 2mol of Fe 55.8g of Fe(25.6 g of O)

16.0 g mol of O 3 mol of O 1mol of Fe× =59.5 g of Fe

38. �e atomic mass of aluminium is given to be 27u. �is means that mole of aluminium atoms has a mass of grams.

Now, 10 mole of aluminium atoms=27g

So, 4 moles of aluminium atoms=27 x 4g=108g

39. 6.022 x 1023 atom of He weights 4 g

∴ 6.022 x 1022 atoms of He will weighs × ×

×

22

23

4 6.022 100.4g

6.022 10

40. 6.022 x 1023molecules of CO2 =1 mole of CO2

∴ 3.011 × 1022 molecules of CO2

×

×

22

223

3.011 100.05 moles of CO

6.022 10

41. No. of mol H2SO4

=(45.8 g of H2SO4)

2 42 4

2 4

1mol of0.467 mol of

98.1 g o

H SOH SO

H SOf

42. �e atomic mass of Fe=56 gm

1 mole of Fe=56 gm

5 moles of Fe=5 × 56 gm =280gm


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43. 1 mole of NaCl = Molecular mass of NaCl in grams = 23 + 35.5 = 58.5

60 gm of NaCl =1 mole

20 gm of NaCl will have 0.341 mole of NaCl.

44. Molecular mass of CaCO3 is 100 gm. �is has 48 gm of O. �us, 50 gm of CaCO3 will have 24 gm of O.

45. Molecular weight of CO2=12+32=44

∴ Quantity of CO2 present in 1 mole CO2 is 44 grams

∴ Quantity present in 0.1 mole of CO2 would be ×44 0.14.4gm

1

46. Because the atomic weight of oxygen is 15.999 amu, a mole of oxygen atoms has a mass of 15.999 grams.

Each O2 molecule has two atoms, so the molecular weight of O2 molecules is twice as large as the atomic weight of the atom.

1 mol O = 15.999 g,

1 mol O2 = 31.998g

47. It is the smallest repeating formula in the structure of ionic compound.

Short Answer Questions

48. (a) 32 g O2 = 1 mol

∴ 12 g O2 = 1232

= 0.375 mol

(b) 44 g CO2 = 1 mol

∴ 22 g CO2 = 0.5 mol

49. �e molecular mass of a substance (element or compound) is the number of times, the molecule of the substance is heavier than 1/12th the mass of an atom of carbon-12 isotope.

Molecular weight is calculated by adding the atomic weights of all the constituent atoms present in a molecule.

For example: Molecular weight of a molecule of hydrogen

(H2) = 2 × atomic weight of hydrogen = 2 × 1 = 2 amu

50. (a) NaF (b) Na2O

(c) MgF2 (d) Al4C3

51. (a) Al2S3 (b) SrF2

(c) TiO2 (d) CaBr2


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52. According to formula of Al2O3, In one mole of aluminium oxide,

3 moles of oxygen combines with 2 moles of aluminium.

1 mole of oxygen combines with 23

moles of aluminium.

1.58 mole of oxygen combines with 23

x 1.58 moles or

1.58 mole of oxygen combines with 1.053 moles of aluminium.

53. Molecular weight of SO2 = 32 + 2 × 16 = 64g

∴ 64g of SO2 at STP occupies 22.4 litres

Hence, 32g of SO2 at STP will occupy

×22.4

32 11.2litres64

54. Number of moles of magnesium = 2.3 12.8 4

mol

∴ Number of moles of sodium should also be 14

and 1 mol of Na = 23 g

∴ 14

mole of Na will be 23 × 14

=5.57 g

55. Molar mass of Cr2S3 =52 × 2 + 32 × 3 = 200 g

∴ 200 g of Cr2S3 contains 2 mol of Cr or 2 × 52 = 104 g of Cr

∴ 85 g of Cr2S3 contain 2.3 1042.8 200

× 85 = 44.2 g of Cr.

56. Number of atoms of Na

= ×Given mass of sodium

Avogadro numberMolar mass of sodium

× × ×23 241006.022 10 2.618 10

23

Number of atoms of Fe × × ×23 241006.022 10 1.075 10

56

We �nd that 100 g of Na contains more number of atoms than 100 g of Fe.

57. �e amount of oxygen per gram of nitrogen in NO2 should be exactly twice as that of NO, as required by the formulas of the two substances. �erefore, 2.285g oxygen would combine with 1.000g nitrogen.

58. No. of moles of N

=(8.60 mol of O) 2 mol of N3.44 mol of N

5 mol of O


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59. Molecular weight of N2 = 2 x 14 = 28 g

28 g of N2 at STP occupies = 22.4L

2.8 g of N2 at STP =?

×22.4 2.8g2.24L

28g

2.8 g of N2 at STP occupies a volume of 2.24 L.

60. At NTP, weight of 22.4 litres of one mole of gas is equal to its molecular weight. So volume of 32 grams of Oxygen at NTP = 22.4 litre.

∴ Volume of 4.0 gram of oxygen at NTP is = ×22.4

4.0 2.8litres32

2.8 liter

Weight of 22.4 litres of CO2 at NTP = 44 grams.

∴ Weight of 2.8 litres of CO2 at NTP ×44 2.85.5gram

22.4 gram

61. In order to solve this problem, we should know the mass of 1 mole of water. �is can be obtained by using the given values of the atomic masses of hydrogen and oxygen as follows:

1 mole of water (H2O) = Molecular mass of H2O in grams

= Mass of 2H atoms + Mass of O atom = 2 x 1 + 16 = 2 + 16 = 18 grams

�us, the mass of 1 mole of water is 18 grams.

So, Mass of 0.5 mole of water = 18 x 0.5g = 9g

�us, the mass of 0.5 mole of water (H2O) is 9 grams

62. One moles of H2SO4 contains 1 mole of sulphur atoms and 4 mole of oxygen atoms.

One mole of H2SO3 has 1 mole of sulphur atoms and 3 moles of oxygen atoms.

One mole of SO2 has 1 mole of sulphur atoms and 2 moles of oxygen atoms.

63. Molar mass of NH3 = 14 × 1 + 1 × 3 = 17 g/mol

Percentage of nitrogen ×3

3

Molar mass of N in NHMolar mass of NH

100

×100 82

1417

.35%

Percentage of hydrogen ×100 173

17.65%


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64. We �rst determine the number of grams of each element that are present in one mole of sample:

2 mol of N × 14.01 g/mol = 28.02 g of N

3 mol of O × 16.00 g/mol = 48.00 g of O

�e percentages by mass are then obtained by using the formula mass of the compound (76.02 g) :

×

×

28.02%N 100 36.86%of N

76.0248.00

%O 100 63.14% of O76.02

65. Mass of the reactants = 10g

Mass of the products = 4.6 + 5.4g = 10g

Since the mass of the reactants is equal to the mass of the products, the observations are in agreement with the law of conservation of mass.

66. 1 mole of H2O molecules has 2 moles of H atoms.

3 moles of H2O molecules will have 6 moles of H atoms.

1 mole of HCl molecules will have 1 mole of H atoms.

10 moles of HCl will have 10 moles of H-atoms.

�us 10 moles of HCl will have more atoms than 3 moles of H2O.

67. Water (H2O) contains hydrogen and oxygen in the ratio of 1 : 8 (by weight). It means if 9g of water is decomposed it will produce 1 g of hydrogen and 8g of oxygen. �us, on decomposing 50g of water.

Weight of hydrogen obtained = ×1

50g 5.56g9

Weight of oxygen obtained = ×8 400

50g or 44.44g9 9

68. Here we have been given that hydrogen and oxygen always combine in the �xed ratio of 1 : 8 by mass. �is means that

1 g of hydrogen gas requires = 8g of oxygen gas

So, 3g of hydrogen gas requires

= 8 x 3g of oxygen gas = 24 g of oxygen gas

�us, 24 grams of oxygen gas would be required to react completely with 3 grams of hydrogen gas.

69. As number of molecules of a substance present in one gram molecular is 6.02 X 1023.

So, number of molecules present in 32 grams of oxygen is 6.02 × 1023.


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Number of molecules present in 28 grams of nitrogen is 6.02 × 1023.

So number of molecules present in 14 grams of nitrogen is × ××

23236.02 10 14

3.01 1028

70. �e molar mass of iron is 55.85 g/mol. �is can be represented in terms of either of the following unit factors.

1 mol Fe 55.85Feor

55.85 g Fe 1 mol Fe

In order to convert from moles to grams, we need the unit factor that tells us how many grams of iron can be found in one mole of this metal.

×

55.85Fe0.250mol Fe 14.0g

1 mol Fe

71. Before we can do anything else, we need to know the number of moles of aluminium metal in the sample. �is can be calculated from the mass of the sample and the molar mass of aluminium, which is 26.982 g/mol.

× × 31 mol Al0.123g Al 4.56 10 mol Al

26.982g Al

Now, we can use Avogadro’s number to calculate the number of atoms in the sample.

4.56 × 10-3 mol Al × × 236.02 10 Al atoms26.982g Al

=2.75 × 1021 Al atoms

Long Answer Questions

72. (a) NaHCO3 = 1 Na + 1H+ 1C + 3 O

= (22.99) + (1.01) + (12.01) + (3 × 16.00) = 84.0 g/mole

(b) K2Cr2O7 = 2K + 2Cr + 7O

= (2 × 39.10) + (2 × 52.00) + (7 × 16.00) = 294.2 g/mole

(c) (NH4)2CO3 = 2N + 8H + C + 3O

= (2 × 14.01) (8 × 1.01) + (12.01) + (3 × 16.00) = 96.1 g/mole

(d) Al2(SO4)3=2Al+3S+12O

= (2 × 26.98)+(3 × 32.07) (12 × 16.00) = 342.2 g/mole

(e) CuSO4.5H2O=1Cu+1S+90+10H

= 63.55+32.07+(9 × 16.00)+(10 x1.01) = 249.7 g/mole


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73. (i) Molar mass of Ethylene

= 2 x Molar mass of C atom + 2 x Molar mass of H atom

= 2 x 12.011 + 2 x 1 g = 26.022 g

(ii) Molar mass of S8 molecule

= 8 x Molar mass of S atom

= 8 x 32.066 = 256.53 g

(iii) Molar mass of P4 molecule

= 4 x molar mass of P atom = 4 x 31 g = 124 g

(iv) Molar mass of HCl

= Molar mass of H atom + Molar mass of Cl atom

= 1 + 35.453 g = 36.453 g

(v) Molar mass of HNO 3

= Molar mass of H atom + Molar mass of N atom + 3 × Molar mass of 0 atom

= 1 + 14.007 + 3 × 15.999g

= 1 + 14.007 + 47.997 2

= 63.004 g

74. This problem involves aluminium ions. Please note that the mass of an aluminium ion is the same as that of an aluminium atom. In order to solve this problem, �rst of all we have to �nd out the mass of aluminium atoms in 0.051 g of aluminium oxide (which will give us the mass of aluminium ions) �is can be done as follows:

1 mole of Al2O3 = Formula mass of Al2O3 in grams = Mass of Al × 2 + Mass of O x 3

27 × 2 + 16 × 3 = 54 + 48 = 102 grams

Now, 1 mole of Al2O3 contains 2 moles of Al.

So, Mass of Al in 1 mole of

Al2O3 = Mass of Al × 2 = 27 × 2 = 54 grams

Now. 102 g of aluminium oxide contains = 54 g ofAl

Now, 0.051 g aluminium oxide contains

×54

0.051gAl 0.027gof Al102

�e atomic mass of aluminium is given to be 27u. �is means that 1 mole of aluminium atoms (or aluminium ions) has a mass of 27 grams, and it contains 6.022 × 1023 aluminium ions.

Now, 27g of aluminium has ions = 6.022 × 1023


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So, 0.027g of aluminium has ions

×× ×

23206.022 10

0.027 6.022 1027

�us, the number of aluminium ions (A13+) in 0.051 gram of aluminium oxide is 6.022 × 1023

75. (I) (a) Cu2O (b) K2O2

(c) Hg2Br2 (d) Fe2(CO3)3

(e) NaBrO

(II) (a) Bromic acid (b) Hydrobromic acid

(c) Phosphoric acid (d) Hypochlorous acid

(e) Iodic acid

76. (I) �e average mass of a molecule of carbon dioxide would be equal to the sum of the atomic weights of the three atoms in a CO2 molecule.

Mass of a single CO2 molecule :

1 C atom 1 12.011 amu 12.011 amu

2 O atom 2

___

15.9

____

99 a

____

mu 31.998 amu

44.009___

amu

�e mass of a mole of carbon dioxide would be 44.009 grams.

�e average mass of a molecule of glucose is equal to the sum of the atomic weights of the 24 atoms in a C6H12O6 molecule.

Mass of a single C6H12O6 molecule :

6 C atom 6 12.011 amu 72.066 amu

12 H atoms 12 1.0079 amu 12.095 amu

6O atoms 6 15.999 amu 95.994 amu

________________180.155amu

�e molecular weight of the compound is therefore 180.155 grams/mol.

(II) �e �rst in this calculation involves converting the mass of the sample into the number of moles of CO2.

× 22

2

1mol of CO0.500gof CO

44.009g of CO

=1.14 × 10-2 moles of CO2


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Once we know the number of moles, we can use Avogadro’s number to calculate the number of CO2 molecules in the sample.

1 mole of CO2 = 6.023 × 1023 molecules

1.14 × 10-2 mol of CO2

= 6.023 × 1023 × 1.14 × 10-2 CO2 molecules

= 6.86 × 1021 CO2 molecules

We can know use the chemical formula for carbon dioxide to determine the number of carbon atoms in the sample. �e formula suggests that there is a single carbon atom for each CO2 molecule.

6.86 × 1021 CO2 molecules = 6.86 × 1021C atoms

77. (I) First experiment:

Copper oxide = 1.375g

Copper left = 1.098g

Oxygen present = 1.375 - 1.098 = 0.277g

Percentage of oxygen in

×0.277 100CuO 20.14%

1.375

Second experiment:

Copper taken =1.179g

Copper oxide formed = I .476g

Oxygen present = 1.476 - 1.179 g = 0.297g

Percentage of oxygen = ×0.297

100 20.14%1.476

Percentage of oxygen is same in both the above cases so the law of constant composition is illustrated.

(II) In order to solve this problem, we have to calculate the ratio (or (proportion) of copper and oxygen in the two samples of copper oxide compound. Now:

In the �rst experiment:

Mass of copper = 1.03 g ………………….(1)

And. Mass of copper oxide = 1.288 g

So, Mass of oxygen

= Mass of copper oxide - Mass of copper


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= 1.288 - 1.03 = 0.258 g ..…..…………… (2)

Now, in the �rst sample of copper oxide,

compound : Mass of copper : Mass of oxygen

= 1.03 : 0.258= 3.99 : 1

= 4 : 1 ……………………..(3)

In the second experiment:

Mass of copper = 2.938 g ……..…………….(4)

And, Mass of copper oxide = 3.672 g

So, Mass of oxygen

= Mass of copper oxide - Mass of copper

= 3.672 - 2.938 = 0.734 g ………………………(5)

Now, in the second sample of copper compound:

Mass of copper : Mass of oxygen 2.938 : 0.734

= 4: 1 .…………………. (6)

From the above calculations we can see that the ratio (or proportion) of copper and oxygen elements in the two samples of copper oxide compound is the same 4: 1. So, the law of constant proportions is illustrated.

78. (a) 72.8 g of CaH2

(b) 0.0219 mol of Mg(NO3)2

(c) 1.48 x 1023 CH3OH molecules

(d) 3.52 x 1024 H atoms

(e) No. of mol Al in 2.16 mol of Al2O3 = 4.32 mol of Al

79.

(i)

Atom Molecule

It is the smallest unit of an element It is the smallest unit of an element or a compound.

It may or may not be capable of free existence but it is the smallest unit that takes part in a chemical reaction.

It is capable of free existence.


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(ii)

Atom Ion

It is neutral and carries no charge.

It is a charged particle and carries a positive or negative charge.

�e number of electrons in an atom is equal to its atomic number.

In a cation, (ion with positive charge) the number of electrons are less than that present in corresponding atom. In an anion, (ion with negative charge) the number of electrons are more than that present in corresponding atom.

Exercise 2 – For Competitive Examinations

Multiple Choice Questions

1. (d) 2. (d)

3. (b) Percentage of hydrogen ×Mass of hydrogen in water

100Molar mass of water

×2

100 11.12%18

4. (c) 5. (d) 6. (a) 7. (b)

8. (a) →→ 3 3 343 4MCl M 3Cl M PO MPO;

9. (b) 10. (d) 11. (c) 12. (a) 13. (b) 14. (c)

15. (c) 16. (b) 17. (c) 18. (b) 19. (b) 20. (a)

21. (a) 22. (b) 23. (b) 24. (a) 25. (a) 26. (d)

27. (d) 28. (c) 29. (b) 30. (d) 31. (c) 32. (d)

33. (d) 34. (c) Law of multiple proportions.

35. (a) Constant proportions according to which a pure chemical compound always contains same elements combined together in the same de�nite proportion of weight.

36. (b) 1 atom of Cu + 1 atom of sulphur + 9 atoms of oxygen + 10 atoms of hydrogen. Total number of atoms in compound is 21.

37. (b) Law of multiple proportion. As the ratio of oxygen which combine with �x weights of 1 g of nitrogen bears a simple whole number ratio

0.57: 1 : 12 : 1.7031 : 2 : 3


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38. (a) Molecular weight of C60H122

= 12 × 60 + 122 × 1 = 720 + 122 = 842

6 × 1023 molecule C60H122 has mass = 842 gm.

∴ 1 molecule C60H 122 has mass

× ×

×23 21

23

842140.333 10 gm 1.4 10 gm

6 10

39. (a) 69× 1023 molecule has mass=18gm

1 molecule has mass × ××

23 2623

183 10 gm 3 10 kg

6 10

40. (d) 17 gm NH3 contains 6× 1023 molecules of NH3

∴4.25 gm NH3 contains × 236 104.25

17 molecules of NH3

∴ No. of atoms × ×× ×

23236 10 4.25

4 6 1017

41. (b) 22400cc of gas at STP has 6× 1023 molecules

∴ 1.12× 10-7 of gas at STP has × × ×× ×

23 714 126 10 1.12 10

0.03 10 3 1022400

42. (a) wt. of CO2= 44 mol wt. of CO2 = 44

No. of molecule= × × 232

2

wt.of CO6.02 10

mol wt. of CO× × ×23 2344

6.02 10 6.02 1044

43. (b) Mole of CO2 = 4.40.1

44

Volume = 0.1 × 22.4 = 2.24 l.

44. (b) 2 gm of hydrogen = 6.02× 1023 molecules

∴ 1 gm of hydrogen ××

23236.02 10

3.01 102

molecule


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More than One Correct

45. (b, d)

46. (a, c) Here the amount of oxygen which combines with �xed amount of C and S in their oxides will be in a simple whole number ratio.

47. (a, d)

48. (a, d) 16 g O2 has no. of moles = 16 132 2

14 g N2 has no. of moles = 14 128 2

No. of moles are same, so no. of molecules are same.

8 g of O2 has no. of moles = 8 132 4

7 g of 02 has no. of moles = 7 128 4

49. (a, c, d) Sulphite 23SO Sulphate 2

4SO and phosphate 34PO ions are polyatomic as they

contain more than one ion.

50. (a, b, c)

51. (a, d)

52. (a, d) Symbol of tin and aluminium are Sn and Al respectively.

53. (a, b, d)

54. (b, c, d) Molar mass of gas ×0.220

22400 44112

55. (a, b, c)

56. (b,c) CaSO3(s) + H2O() + SO2(g) → Ca(HSO3)2

(soluble)

mol of SO2 required = mole of CaSO3

12

0.1;120

mass of SO2 = 0.1 × 64 = 6.4g


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57. (a, c, d)

8g O2 =832

mole = 0.25 mole

7g CO = 728

mole = 0.25 mole

14g N2 =1428

mole = 0.5 mole

16g SO2 = 1664

= 0.25 mole

Equal moles contain equal number of molecules.

58. (a, b) 4.4g of CO2 = 4.4

44 mole = 0.1 mole = 0.1 x 6.02 x 1023 molecules = 6.02 x 1022 molecules.

59. (a, b, c)

→

× ×22Mg O 2MgO

2 24g 32g 2 4048g 80g

48g Mg reacts with = 32g O2

∴ 1g Mg will react with 232

O g 0.67g48

O2

�us, whole of Mg will react

∴ O2 left unreacted =2-0.67 =1.33g

MgO formed= 80g 1.67g

48

Total=1.67+1.33=3.0g

Also, by law of conservation of mass, mixture at the end should weigh 1g+2g=3g

Fill In the Blanks60. (1) Chemical combination

(2) Antoine Lavoisier

(3) Whole


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(4) Multiple proportion

(5) Gaseous

61. (1) 6.02 × 1023

(2) One

(3) Atomic mass

(4) Molecular mass

Passage Based Questions

62. (c) 63. (a) 64. (b)

Assertion and Reason

65. (d) Molecular weight of oxygen is 32

66. (d) Molecular weight of SO2 is 64 and of O2 is 32.

67. (a)

68. (a) 6.023 x 1023 atom of C12 = 12g

1 atom of C12 = × 23

12g

6.023 10

112

of one atom of C12=× 23

1g

6.023 10

[1 amu =th

1

12 of one atom of C12]

= 0.166 × 10-23 g = 1.66 × 10-24 g

69. (a)

Multiple Matching Questions

70. A - (p, q); B - (p); C - (p, q, r); D (s)

1. 5 mole of CO2 (g) = 1.5 × 22400 mL at NTP = 33600 ml at NTP

Total number of atoms in one molecule of CO2 = 1 + 2 = 3

Total number of atoms in 1.5 mole of CO2 =1.5 x 3 × NA = 4.5 × NA

3.0 g of H2 = 32

× 22400 mL of H2 at NTP = 33600 ml at NTP


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Number of mole of H2 in 3.0 g H2 32

or 1.5 moles

Number of atoms in 1 molecule of H2 = 2

Number of atoms in 1.5 mole H2 = 2 × 1.5 × NA = 3.0 × NA

Volume of 1.5 moles of ozone at NTP = 22400 × 1.5 ml = 33600 ml

Number of atoms in one molecule of O3 = 3

Total number of atoms in 1.5 mole of O3 = 3 × 1.5 × NA = 4.5 × NA

Weight oil mole of O3 = 48g

Weight of 1.5 mole of O3 = 48 × 1.5 g = 72g

Weight of 1 mole of oxygen (O2) = 32g

Subjective Questions

71. (a) CrCl3 and CrCl2, Cr2O3 and CrO

(b) CuCl, CuCl2, Cu2O and CuO

72. Sodium arsenate

73. (a) 6 : 11 (b) 2 : 1

(c) 12 : 11 (d) 2 : 1

74. In �rst sample of ascorbic acid,

1.50 g of carbon combines with 2.00 g of oxygen

1 g of carbon combines with 2g

1.5 of oxygen.

In second sample, 6.35 g of carbon combines with ×2 6.35g

1.5 of oxygen

∴ 6.35 g of carbon combine with 8.47 g of oxygen.

Sample I Sample II

2 : 1.5 8.47: 6.35

2 : 1.5

�us, the above calculations supports the law of constant composition.

75. (i) Number of oxygen atoms in 16.0g of oxygen

= 6.023 x 1023 atoms

∴ Weight of one atom of oxygen ××

2323

16.02.657 10 g

6.023 10


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(ii) G.M.W. of oxygen = 32.0g

Number of O2 molecules in 32.0 g of O2

= 6.023 x 1023 molecules

∴ Weight of one molecule of oxygen ××

2323

32.05.314 10 g

6.023 10

76. Water molecule consists of three interconnected particles or atoms. Each water molecule divides into two hydrogen atoms and one oxygen atom. Molecules of a substance can be destroyed by a chemical reaction, whereas atoms remain unchanged. For this reason, we cannot say ‘an atom of water’ but we say ‘a molecule of water’.

77. �e atomic theory states that the relative number and kinds of atoms in a compound are constant, regardless of the source. �erefore, 1.0g of pure water should always contain the same relative amounts of hydrogen and oxygen, no matter where or how the sample is obtained.

78. (a) 0.5711g of O/1g of N, 1.142g of O/1g of N, 2.284 g of O/1g of N, 2.855g of 0/1g of N

(b) The numbers in part (a) obey the law of multiple proportions. Multiple proportions arise because atoms are the indivisible entities combining, as stated in Dalton’s atomic theory.

79. 22.4L of gas = 1 gram molecular weight

22.4L = 22.400 cm3. (1 L = 1000 cm3)

360 cm3 of N2 = 0.45g

22,400 cm3 of N2 = ?

×0.45g 22400L

28g360L

Gram molecular weight of N2 is 28 g.

80. �e strategy for doing this problem is to convert from milligrams of silicon to grams of silicon, then to moles of silicon, and �nally to atoms of silicon:

31g of Si

5.68 mg of Si 5.68 10 g of Si1000mg of Si

× ×

33

23

4 23 4

19 20

Given wt. of element 5.68 10 gNo.of moles 0.202 10 mol

Molar mass of element 28.08g

1mole of Si 6.023 10 atom

2.02 10 mol of Si 6.023 10 2.02 10 atom

12.16 10 atoms 1.22 10 atoms

××

∴ ×

× × × ×

× ×

81. �e molecular formula of limestone is CaCO3

Its molar mass is 100 g/mol


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100 g CaCO3 = 1 mol.

300 g CaCO3 = 3 mol.

1 mol CaCO3 contains 3 mol of 0 or 3 x 6.022 x 1023 O atom.

∴ 3 mol CaCO3 would contain

x 3 x 6.022 x 1023 = 5.42 x 1024 of O atom.

82. First calculate the number of moles of oxygen that are combined with X:

(1.14 g of oxygen) 1mole of O16.00g of O

=0.0713 moles of oxygen

�en calculate the moles of X in the oxygen compound:

(0.0713 moles of O) 1mole of X2moles of O

=0.0356 moles of X

Finally, calculate the molar mass of

X: 1.00g of X28.1 g / mole

0.0356 moles of X Note that X is Si, atomic number 14 and that its oxygen

compound is SiO2

�e same number of moles of X are also combined with Y:

4 moles of Y0.0356 moles of X 0.142 moles of Y

1 moles of X�e atomic mass of Y is thus:

5.07g of Y35.6g / mole

0.142 moles of Y

Note that Y is Cl, atomic number 17, and that its compounds with X is SiCl4.

83.

2 3

2 4 32 4 3

2(a) mole of Al mole of S

3(b) mole of S 3 mole of Al O

2 mol of Al(c)No.of mol Al (0.900 mol of S) 0.600 mol of Al

3 mol of S

3 mol of S(d) No.of mol S 1.16 mol of Al (SO ) 3.48 mol of S

1mol of Al (SO )

×

×

84. No. of moles of UF6=(1.25 mol of CF4)4

4 mol of F1 mol of CF


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66

1 mol of UF0.833moles of UF

6 mol of F

85. �e formula CaC2 indicates that there us 1 mole of Ca for every 2 moles of C. �erefore, if there are 0.150 moles of C there must be 0.0750 moles of Ca.

No. of moles of calcium x molar mass of Ca=grams of calcium in the sample

=0 .0750 x 40.078 of Ca=3.01 g of Ca.

86. Its mass is 200 amu (or U).

87. We know, 22.4 L of H2 at NTP=1 mole of H2

2 2 2

11.211.2Lof H at NTP mole of H 0.5 mole ofH

22.4∴

Number of molecules present in one mole of H2

∴ Number of molecules present in 0.5 mole of H2 (or 11.2L of Hydrogen at NTP) =0.5 x 6.023 x 1023

=3.011 x1023molecules

Number of hydrogen atoms in one molecules of H2 = 2

∴ Number of hydrogen atoms in 3.011 x 1023 molecules of H, = 2 x 3.011 x 1023 = 6.022 x 1023

atoms

88. Molecular mass of urea (H2NCONH2)

=2 x 1+1 x 14+1 x 12 +1 x16 + 1 x 14 + 2 x 1 =2+ 14 +12 +16 +14 +2 =60 U

Mass of nitrogen in 1 molecule of urea

= 1 x 14 + 1 x 14 = 14 + 14 = 28U

∴ % age of nitrogen in urea ×28

100 46.6%60

89. Molecular mass of CuSO4.5H2O

=63.5+32+64+5x18 =63.5+96+90=249.5

Number of moles of CuSO4.5H2O in 24.95 mg × 3424.95 10

10 moles249.5

Molecules of water of crystallization present in 1 mole of CuSO4.5H2O=5x6.023x1023

∴10-4moles of CuSO4.5H2O contains molecules of water of crystallization

=5 x6.023 x1023 x 10-4 =30.115 x1019 molecules =3.01 x 1020molecules.


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90. All that we have to do in this problem is to calculate the mass of reactants and products

separately, and then compare the two. If the two masses are equal, then the law of conservation of

mass gets veri�ed. �e given reaction can be written as :

→

Reac tants

Pr oducts

Sodium Carbonate Ethanoic acid

Sodium ethanoate Carbon dioxide Water

(i) Sodium carbonate and ethanoic acid are reactants. So, mass of reactants

= Mass of sodium carbonate + Mass of ethanoic acid = 5.3 + 6 = 11.3 g

(ii) Sodium ethanoate, carbon dioxide and water are products

So, mass of products

= Mass of sodium ethanoate + Mass of carbon dioxide+ Mass of water =8.2 + 2.2 + 0.9 = 11.3g

We �nd that the mass of reactants is 11.3 g and the mass of products is also 11.3 g. Since the mass of products is equal to the mass of reactants, the given data veri�es the law of conservation of mass.

91. 1.375 gm of pure cupric oxide gave 1.098 om ofCu and hence.

percentage of Cu in the oxide = ×1.098

100 79.85%1.375

in another experiment, 1.179 gm of pure copper gave 1.476 gm of the oxide and hence

percentage of Cu is the oxide = ×1.179

100 79.87%1.476

Since, both the oxides have almost the same percentage of. Cu and hence, of oxygen, the result is obeying the law of constant composition.

92. �e atomic weight of carbon is 12.011 amu, which means that a mole of carbon has a mass of 12.011 grams. We can use this information to construct two unit factors.

1 mol C 12.011g Cand

12.011 g C 1 mol C

Converting grams of carbon into moles requires a unit factor that has units of moles in the numerator and grams in the denominator.

×1 mol C

2.73gC 0.227 mol C12.011 g C


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�e same format can be used to convert grams of oxygen into moles of oxygen atoms.

22 2

2

22

1mol O7.27g O 0.227 mol O molecules

31.998 g O

2 O atom0.227 mol O molecules 0.454 mol O atoms

1O molecules

×

×

Ratio between the number of moles of carbon atoms and moles of oxygen atoms in our sample is

0.454mol O2.00

0.227 mol C

�ere are twice as many moles of oxygen atoms as there are moles of carbon atoms in this sample. Because a mole of atoms always contains the same number of atoms, the only possible conclusion is that there are twice as many oxygen atoms as carbon atoms in the compound. In other words, the formula for carbon dioxide must be CO2.

93. (a) ZnCO3. ZnO,

(b) HF, SiO2, SiF4., H2O

(c) SO2, H2O. H2SO3

(d) H3P (or PH 3)

(e) HClO4. Cd. Cd (ClO4)2

1. introduction 2. dalton’s atomic theory

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