1. Introduction 2. Dalton’s Atomic Theory · 2015-11-19 · ˜˚˛˝˙ˆ˚ˇ˘˝ ˙ ˇ ˙ ˚ ˘ ˚ ˝ 16 PLANCESS CONCEPTS Many unexplained chemical phenomena were quickly explained

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

Postulates of Daltons theory:

rePresentative samPle Chemistry: mole ConCePt and stoiChiometry

Mole Concept and Stoichiometry

w w w . p l a n c e s s j e e . c o m

16

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”.

Modification to the Law of Conservation of mass

The total sum of mass and energy of the system remains constant.

PLANCESS CONCEPTS

The 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 different 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, The law states that “A pure chemical compound always consists of the same elements combined together in a fixed (definite) proportion by weight.”

Illustration 1: The 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.

Mole Concept and Stoichiometry17

Sol:

In the first 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 = ×Weightofoxygen


= × =(0.54g)

100 20%(2.70g)

Weight of copper oxide = 1.15 g

Weight of oxygen = 0.23 g

% copper = ×Weightof copper


= × =(0.92g)

100 80%(1.15g)

% oxygen = ×Weightofoxygen


= × =(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 definite 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 different 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.



18

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

Sol:

For the first oxide: For the second oxide:

% of carbon in first oxide = 42.8

% of oxygen in first 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 fixed 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. The 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 fix 1g of hydrogen as the fixed weight in the three compounds.

In the first 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 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.


PLANCESS CONCEPTS

Dalton became the first 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 “ The Law States that , “The ratio of the weights of two elements, A and B which combine separately with a fixed 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 “

The 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. Thus, the weight of C and O which combine with fixed 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 first 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 fixed 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

Thus, 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 fix 1g of phosphorus (P) as the fixed 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 = =77.43

g 3.43g22.57

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 = =8.82

g 0.097g91.18

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

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

Weight of hydrogen = 100 – 97.23 = 2.77g

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

Let us compare the two ratios.

These are related to each other as =3.43 97.23

: 35.37:35.100.097 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.”


This 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.

The 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

Thus, 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.

Thus 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 find the relationship between molecular mass and vapour density of a gas

Vapour density (VD) = =Densityofgas Massof acertainvolumeof thegas

Densityofhydrogen Massof thesamevolumeofhydrogenatthesametemp,andpressure

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

VD = =Massofnmoleculesofthegas Massof1moleculeofthegasMassofnmoleculesofhydrogen Massof1moleculeofhydrogen



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= =Massof1moleculeofthegas MolecularmassMassof1moleculeofhydrogen 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 first 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 identified. The 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

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

Thus,

A + 2B ⟶ 4C

Initially — 5 moles 12 moles 0 moles

Finally — 0 moles 2 moles 20 moles

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. The reactant that is completely consumed is called limiting reactant, whereas unreacted reactants are called excess reactants.


Amounts of substances produced are called yields. The amounts calculated according to stoichiometry are called theoretical yields whereas the actual amounts are called actual yields. The 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

(contd.)



24

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 CuSO4.5H2O Washing soda crystals Na2CO3.10H2O

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

× == 90

100 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

× == 180100 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 = l l× =22.4

100 5044.8

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 = 22.4L 3.4 kg or 3400 g will have volume = l× =22.4

3400 448017

(contd.)


Exercise 1— For school Examinations

Fill in the Blanks

Q.1 During a chemical reaction, the sum of the 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 The Avogadro constant is defined as the number of atoms in exactly of carbon-12.

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

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

Q.8 A chemical formula is also known as a

Q.9 Those ions which are formed from single atoms are called

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

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

Q.12 Mole is a link between the and

Q.13 The SI unit of amount of a substance is

True / False

Q.14 Formula mass of Na2O is 62 amu.

True False

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

Q.16 Formula for sulphur dioxide is SO3.

True False

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

True False(contd.)



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Exercise 2 – For Competitive examinations

Multiple choice Questions

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 The 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 The 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) or y 75 5 3 cm= =

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 The 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 The 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

(contd.)


SOLUTIONS

EXERCISE 1

Fill in the Blanks

Sol.1 masses Sol.2 definite Sol.3 polyatomic

Sol.4 chemical formula Sol.5 6.022 x 1023, 12g Sol.6 molar mass

Sol.7 symbol Sol.8 molecular formula. Sol.9 simple ions.

Sol.10 metal and non-metals. Sol.11 equal.

Sol.12 mass of atoms & number of atoms. Sol.13 mole

True/False

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

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

Match the Column

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

Very Short Answer Questions

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

Sol. 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. This number is Avogadro’s number i.e., 6.023 x 1023.

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

− =Number of gram atomMass of the element in grams

Atomic mass

(contd.)



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SOLUTIONS

EXERCISE 2

Multiple Choice Questions

Sol.1 (d) Sol.2 (d)

Sol.3 (b) Percentage of hydrogen = ×Massofhydrogeninwater

100Molarmassofwater

= × =2

100 11.12%18

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

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

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

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

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

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

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

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

Sol.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.

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

0.57: 1 : 12 : 1.7031 : 2 : 3

Sol.38 (a) Molecular weight of C60H122

= 12 x 60 + 122 x 1 = 720 + 122 = 842

6 x 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

(contd.)

1. Introduction 2. Dalton’s Atomic Theory · 2015-11-19 · ˜˚˛˝˙ˆ˚ˇ˘˝ ˙ ˇ ˙ ˚ ˘ ˚ ˝ 16 PLANCESS CONCEPTS Many unexplained chemical phenomena were quickly explained

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