Subject: Chemistry

The term ‘stoichiometry’ comes from the Greek word which means “to measure an element”. The calculation which deals with mass and volume relationship among reactant and product is known as stoichiometry. A chemical equation is a basis for such calculations. The combination of the element to form the compound and chemical reactions are mainly governed by following five laws known as laws of stoichiometry or law of chemical combination.

- Law of conservation of mass
- Law the of constant composition (definite proportion)
- Law of multiple proportions
- Law of reciprocal proportion ( equivalent proportion)
- Gay – Lussace’s law of gaseous volume

**1.** Law** of conservation of mass**

** Statement:** The total mass of substance taking part in a chemical reaction remains constant through out the change. In other words, the total mass of reactants is always equal to the total mass of product in the given chemical reaction.

i.e. Total mass of reactant = Total mass of product

This law if also known as the Law of the indestructibility of matter. According to Dalton’s atomic theory, the atom can neither be created nor be destroyed. During the chemical reaction, simply the rearrangement of atoms takes place. Since an atom has definite mass, therefore, total mass during reaction remains conserved. So, the law of conservation of mass is in accordance with Dalton’s atomic theory.

**Illustration **

The law of conservation of mass can be illustrated by Landolt's experiment. Landolt took AgNO_{3} solution and NaCl solution separately in two limbs of an H-shaped tube. The tube was weighed initially with reactants and the solutions were allowed to mix by shaking the tube whereby reaction occurred and white ppt. of AgCl was formed. The tube was again weighed after reaction until the constant mass was observed. It was found that there was no change in total mass i.e. Mass of (AgNO_{3} + HCl) = Mass of (AgCl + NaNO_{3})

AgNO_{3} (aq) + NaCl (aq) → AgCl ↓ + NaNO_{3}

(white ppt.)

**Modification**

According to Einstein, mass – energy relationship, mass and energy are interconvert able to each other. Therefore, mass during the chemical reaction may not be conserved particularly for those reactions which involve liberation of the large quantity of energy in the form of heat or light. Therefore, this law can be restated as

“ Total mass and energy during chemical reaction remain constant”

**2. Law of constant composition**

**Statement:** A chemically pure substance ( compound ) always contains the same element combined together in definite proportion by mass irrespective of sources or mode of formation of the compound. In other words, the percentage composition of an element in a compound is always fixed.

**Illustration **

**a) Composition of water : **Pure water obtained from different sources such as springs, well, rivers, sea, etc. always found to contain only 2 elements; hydrogen and oxygen combined together in the ratio 1:8 by mass regardless of the source of water.

**b) Composition of copper oxide**

Copper oxide can be prepared by the following method

i) By heating copper hydroxide

$$Cu(OH)_2\xrightarrow{\Delta} CuO+H_2O$$

ii) By heating copper carbonate

$$CuCo_3\xrightarrow{\Delta}CuO+Co_2$$

iii) By heating copper nitrate

$$2Cu(No_3)_2\xrightarrow{\Delta}2CuO+4No_2+O_2$$

To analyse copper oxide sample, a known mass each of all three sample is heated in current of hydrogen whereby the following reaction occurred :

$$CuO+H_2\longrightarrow Cu+H_2O$$

The mass of copper thus formed is then measured in all three samples and it is found that the ratio of Cu:O in all samples of CuO is same and nearly equal to 4:1 which illustrates the law of constant composition.

**Limitations**

i) This law of definite proportion does not hold good for non-stoichiometric compounds like ZnO, FeO and also in some compounds which are formed by the different isotopes of the same element.

Eg: H_{2}O, H: O = 1:8

D_{2}O , D:O = 1: 4

**3. Law of multiple proportions**

**Statement:** “ When two elements combine to form one or more compounds, the mass of one element that combines with the constant mass of another element bears simple whole number ratio”.

The simple whole number ratio indicates the existence of individual particle as the smallest ultimate particle ( i.e. element combine in simple whole number ratio of atoms). Due to this, it encouraged Dalton to propose the theory about the structure of atoms known as Dalton’s atomic theory. According to Dalton’s atomic theory, atoms of two or more elements combine in simple whole number ratio to give compound atom ( nowadays called molecule).

**Illustration**

1) Sulphur and oxygen combine with each other to give sulphur trioxide and sulphur dioxide.

Sulphur + Oxygen

In SO_{2}, 32 P/M of sulphur combines with 32 P/M of oxygen

In SO_{3}, 32 P/M of sulphur combines with 48 P/M of oxygen

In both compounds, the mass of oxygen that combines with fixed mass of sulphur is in the ratio of 32:48 or 2:3, which is simple whole number ratio. It illustrates the law of multiple proportions.

2) Nitrogen combines with oxygen to give five oxides of nitrogen. The masses of oxygen which combines with 14 P/M of nitrogen in five oxides of nitrogen is given below in the following table:

Oxides of Nitrogen | Mass of nitrogen | Mass of oxygen | Mass of oxygen that combines with 14P/M of nitrogen |

Nitrous oxide(N_{2}O) | 28 | 16 | 8 |

Nitric oxide (NO) | 14 | 16 | 16 |

Nitrogen trioxide(N_{2}O_{3}) | 28 | 48 | 24 |

Nitrogen peroxide (N_{2}O_{4}) | 28 | 64 | 32 |

Nitrogen pentoxide (N_{2}O_{5}) | 28 | 80 | 40 |

Thus the mass ratio of oxygen that combines with the constant mass of nitrogen [14P/M] in five oxides of nitrogen is 8:16:24:32:40 or 1:2:3:4:5 which is simple whole number ratio. This illustrates the law of multiple proportions.

**4.Law of reciprocal proportion ( Equivalent proportion)**

**Statement:** When two elements combine separately with the constant mass of the third element, the ratio in which they will do so will be either same of the simple multiple of the ratio in which they combine with each other.

**Illustration**

a) Let us consider three elements sulphur, hydrogen and oxygen which combine to give compounds H_{2}S, H_{2}O and SO_{2}

In H_{2}S, 32 P/M of sulphur combines with 2P/M of hydrogen

In SO_{2}, 32 P/M of sulphur combines with 32 P/M of oxygen

Now, the mass ratio of hydrogen and oxygen that combines with the constant mass of sulphur [32 P/M] is 2:32 or 1:16. This implies that if hydrogen and oxygen combine, according to the law of reciprocal proportion, the ratio in which they will do so will be either 1: 16 or simple multiple of it.

In H_{2}O, the mass ratio of hydrogen and oxygen is 1:8 which is a simple multiple of 1:16.

b) Three elements C, H and O combine to give CH_{4}, CO_{2} and H_{2}O

In CH_{4}, 12 P/M of carbon combines with 4P/M of hydrogen

In CO_{2}, 12 P/M of carbon combines with 32 P/M of oxygen

Now, the mass ratio of hydrogen to oxygen that combines with the constant mass of carbon ( 12 P/M ) is 4:32 or 1:8. This implies that, if hydrogen and oxygen combined, according to the law of reciprocal proportion, the ratio will be either 1:8 or simple multiple of it.

In H_{2}O, the mass ratio of hydrogen and oxygen is 1:8 which is same. This illustrates the law of reciprocal proportion.

**5. Gay – Lussace’s law of gaseous volume**

**Statement**: When two different gases combine with each other under similar condition of temperature and pressure to give gaseous product, the volume of reactant, as well as product, bear simple whole number ratio”

Let a gas ‘A’ combines with gas ‘B’ to give gaseous product ‘AB’. If ‘ x ‘ cc of A combines with ‘ y’ cc of B to give ‘z’ cc of AB, then according to this law, x : y : z must be the simple whole number.

**Example: **

H_{2} (g) + Cl_{2}(g) → 2HCl (g)

1vol 1 vol 2 vol

N_{2} + 3H_{2}→ 2NH_{3}

1vol 3vol 2 vol

In the first reaction, the ratio of the volume of reactants and products os 1:2 and in the second reaction, the ratio of the volume of reactants and products is 1:3:2. Both of the ratios are simple whole number ratio.

**AVOGADRO's HYPOTHESIS AND ITS APPLICATIONS**

**Origin of Avogadro's Hypothesis **

According to Dalton's atomic theory, atoms combine in simple whole number ratio to give compound atom ( nowadays called molecule).

According to Gay Lussace's law of gaseous volume, gases combine in simple whole number ratio by volume to give the gaseous product.

Gay Lussace's law of gaseous volume and Dalton's atomic theory clearly indicates that there must be some sort of relationship between gaseous volume and number of atoms. This relationship was proposed by Berzelius and was further developed by Avogadro.

**Berzelius hypothesis**

According to Berzelius hypothesis ' under the similar condition of temperature and pressure, the equal volume of all gases contains the same number of atoms'. Let us consider the following fact

Hydrogen + Chlorine → Hydrogen Chloride

\( 1 vol\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 1 vol\:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 2vol (experimentally determined)\)

Let 1 volume of hydrogen contains 'n' atoms of it and applying Berzelius hypothesis

\(n \:atoms\:\:\: n\: atoms\:\:\: 2n\: compound\: atoms\)

\(1\: atom\:\:\: 1 \:atom\:\:\: 2 \:compound \:atoms\)

\( \frac {1}{2} atom\:\:\:\) \( \frac {1}{2} atom\:\:\:\) \(1 \:compound\: atom\)

Hence, \( \frac {1}{2}\) atom of hydrogen combines with \( \frac {1}{2}\) atom of chlorine to give 1 compound atom of hydrogen chloride. According to Dalton's atomic theory, atom is indivisible. This means \( \frac {1}{2}\) atom or fraction of atom is not possible.

So, some flaw was left on Berzelius hypothesis.

**Avogadro's hypothesis **

Avogadro explained that number of molecules (n) is directly proportional to volume of gases (V)

i.e. V∝n

He replaced the term 'atoms' in Berzelius hypothesis by the term 'molecules'. According to Avogadro's hypothesis, " Under the similar condition of temperature and pressure, an equal volume of all gases contains the same number of molecules".

Let us consider above reaction and apply Avogadro's hypothesis

Hydrogen + Chlorine → Hydrogen Chloride

\( 1 vol\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 1 vol\:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 2vol (experimentally determined)\)

Let 1 volume of hydrogen contains 'n' molecules of it and applying Avogadro's hypothesis

\(n \:molecules\:\:\: n\: molecules\:\:\: 2n\: molecules\)

\(1\: molecule\:\:\: 1 \:molecule\:\:\: 2 \:molecules\)

\( \frac {1}{2}molecule\:\:\:\) \( \frac {1}{2} molecule \:\:\:\) \(1 \:molecule\)

Hence, \( \frac {1}{2}\) molecule of hydrogen combines with \( \frac {1}{2}\)molecule of chlorine to give 1 molecule atom of hydrogen chloride.Since a molecule is composed of atoms, hence half-molecule or fraction of the molecule is possible. This shows the validity of Avogadro's hypothesis because it is in accordance with Dalton's atomic theory.

**Applications of Avogadro's hypothesis**

**i) Deduction of atomicity of elementary gases**

Atomicity can be defined as the number of atoms present in a molecule. Most of the elementary gases are diatomic molecules. Example: H_{2}, O_{2},N_{2}, etc.

#**Prove that hydrogen is a diatomic molecule.**

Consider,

Hydrogen + Chlorine → Hydrogen Chloride

\( 1 vol\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 1 vol\:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 2vol (experimentally determined)\)

Let 1 volume of hydrogen contains 'n' molecules of it and applying Avogadro's hypothesis

\(n \:molecules\:\:\: n\: molecules\:\:\: 2n\: molecules\)

\(1\: molecule\:\:\: 1 \:molecule\:\:\: 2 \:molecules\)

\( \frac {1}{2}molecule\:\:\:\) \( \frac {1}{2} molecule \:\:\:\) \(1 \:molecule\)

Hydrogen chloride reacts with sodium hydroxide to give only one type of salt which is common salt. This means 1 molecule of hydrogen chloride contains only 1 displaceable hydrogen atom which comes from half molecule of hydrogen.

Therefore,

\( \frac {1}{2}\) molecule of hydrogen = 1 atom of hydrogen

1 molecule of hydrogen = 2 atoms of hydrogen

Hence, hydrogen is a diatomic molecule. Similar to this, chlorine can also be proved to be diatomic.

#**Prove that hydrogen is a diatomic molecule.**

Consider,

Hydrogen + Chlorine → Hydrogen Chloride

Let 1 volume of hydrogen contains 'n' molecules of it and applying Avogadro's hypothesis

\(n \:molecules\:\:\: n\: molecules\:\:\: 2n\: molecules\)

\(1\: molecule\:\:\: 1 \:molecule\:\:\: 2 \:molecules\)

\( \frac {1}{2}molecule\:\:\:\) \( \frac {1}{2} molecule \:\:\:\) \(1 \:molecule\)

Hydrogen chloride reacts with sodium hydroxide to give only one type of salt which is common salt. This means 1 molecule of hydrogen chloride contains only 1 displaceable hydrogen atom which comes from half molecule of hydrogen.

Therefore,

\( \frac {1}{2}\) molecule of hydrogen = 1 atom of hydrogen

1 molecule of hydrogen = 2 atoms of hydrogen

Hence, hydrogen is a diatomic molecule. Similar to this, chlorine can also be proved to be diatomic.

#**Prove that oxygen is a diatomic molecule.**

Consider,

Hydrogen + Oxygen → Water(Vapor)

\(2 vol\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 1 vol\:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 2vol (experimentally determined)\)

Applying Avogadro's hypothesis

\(2n \:molecules\:\:\: n\: molecules\:\:\: 2n\: molecules\)

\(2\: molecule\:\:\: 1 \:molecule\:\:\: 2 \:molecules\)

\( 1molecule\:\:\:\) \( \frac {1}{2} molecule \:\:\:\) \(1 \:molecule\)

Here, 1 molecule of hydrogen and \( \frac {1}{2}\) molecule of oxygen combine together to give water molecule.

The molecular mass of water has been found to be 18 amu and the molecular mass if hydrogen is 2 amu.

Here, the mass of oxygen in 1 molecule of water = (18 - 2)amu = 16 amu which is atomic mass of oxygen.

This means 1 molecule of water contains 1 atom of oxygen coming from \( \frac {1}{2}\) molecule of oxygen.

ie.\( \frac {1}{2}\) molecule of oxygen = 1 atom of oxygen

ie. 1 molecule of oxygen = 2 atoms of oxygen

Hence, oxygen is diatomic.

**ii) Deduction of relationship between molecular mass and vapour density ( Molecular mass = 2 × vapour density) **

Vapour density of substance may be defied as the ratio of the mass of given volume of gas or vapour to the mass of the same volume of hydrogen gas under the similar condition of temperature and pressure.

Vapor density = \( \frac {Mass \:of\: V\: cc\: of\: gas }{Mass \:of\: same\: volume\: of\: hydrogen\: gas\: at\: same\: T\: and\: P }\)

= \( \frac {Mass \:of\: n\: molecules\: of\: gas }{Mass \:of\: n\: molecules\: of\: hydrogen\: gas\: }\)

= \( \frac {Mass \:of\: 1\: molecule\: of\: gas }{Mass \:of\: 1\: molecule\: of\: hydrogen\: gas\: }\)

= \( \frac {1}{2}\) \( \frac {Mass \:of\: 1\: molecule\: of\: gas }{Mass \:of\: 1\: atom\: of\: hydrogen\: gas\: }\)

[Because hydrogen is diatomic gas]

or, 2 × Vapor density =\( \frac {Mass \:of\: 1\: molecule\: of\: gas }{Mass \:of\: 1\: atom\: of\: hydrogen\: gas\: }\) ---------------(i)

Since molecular mass is the number of times a molecule is a heaver than mass of 1 atom of H_{2} i.e.

Molecular mass = \( \frac {Mass \:of\: 1\: molecule\: of\: gas }{Mass \:of\: 1\: atom\: of\: hydrogen\: gas\: }\)----------(ii)

Combining equation (i) and (ii)

**Molecular mass = 2 × vapor density**

Hence, molecular mass of a substance is twice the vapor density.

**iii) Deduction of molar volume of gas ( i.e. GMV = 22.4 lit at NTP) **

We know,

Molecular mass = 2 × vapor density

= \(2 × \frac {Mass \:of\: certain\: volume\: of\: gas }{Mass \:of\: same\: volume\: of\: hydrogen\: gas\: at\: same\: T\: and\: P }\)

Let us consider 1 lit. of gas at NTP

Molecular mass = \(2 × \frac {Mass \:of\: 1\: liter\: of\: gas \: at\:NTP }{Mass \:of\: 1\: liter\: of\: hydrogen\: gas\: at\: NTP }\)

= \(2 × \frac {Mass \:of\: 1\: liter\: of\: gas \: at\:NTP }{0.089 }\)[Mass of 1 liter of hydrogen gas at NTP = 0.089 gm]

Molecular mass in gram = 22.4 × Mass of 1 liter of gas at NTP

= Mass of 22.4 liter of gas at NTP

Hence, mass of 22.4 liter of any gas at NTP has mass equal to its molecular mass in gram. Molecular mass expressed in gram is called gram molecular mass or gram mole or simply mole. Therefore, 1 mole of any gaseous substance at NTP occupies 22.4 colume which is known as molar volume or gram molecular volume (GMV).

**iv) Deduction of molecular formula of compound from volumetric composition **

Avogadro's hypothesis can be applied to determine simple formula (empirical formula) as wellas molecular formula of compound from its volumetric composition. For example, formula of ammonia can be established as follows:

Hydrogen + Nitrogen → Ammonia

\( 3 vol\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 1 vol\:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 2vol (experimentally determined)\)

Let 1 volume of hydrogen contains 'n' molecules of it and applying Avogadro's hypothesis

\(3n \:molecules\:\:\: n\: molecules\:\:\: 2n\: molecules\)

\(3\: molecule\:\:\: 1 \:molecule\:\:\: 2 \:molecules\)

\( \frac {3}{2}molecule\:\:\:\) \( \frac {1}{2} molecule \:\:\:\) \(1 \:molecule\)

\(3\: atoms\:\:\: 1 atom\:\:\: 1 \:molecule\)

Hence, empirical formula of Ammonia = NH_{3}

Empirical formula mass of ammonia = ( 14 + 1 x 3 ) = 17

It is experimentally observed that vapor density of ammonia = 8.5

Hence, the molecular mass of ammonia = 2 x 8.5 = 17

Now,

Value of n =\( \frac {Molecular\:formula\:mass}{Empirical\:formula\:mass}\)

= \( \frac {17}{17}\) = 1

Now,

Molecular formula of ammonia = Empirical formula x n = {NH_{3}} x 1 = NH_{3}

- The law of conservation of mass can be illustrated by Dandolt's experiment
- A chemically pure substance ( compound ) always contains the same element combined together in definite proportion by mass
- The simple whole number ratio indicates the existence of individual particle as the smallest ultimate particle
- According to Dalton's atomic theory, atoms combine in simple whole number ratio to give compound atom
- According to Gay Lussace's law of gaseous volume, gases combine in simple whole number ratio by volume to give the gaseous product.

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