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03_ATOM AND MOLECULES

 

 

CHEMISTRY (CLASS-IX)                                  Chapter-3                         Atoms and Molecules

 

Matter is called ‘PADARTH’ that cannot be divided further. The combined form of atoms is called ‘molecules’. Different kinds of atoms and molecules have different properties.As we know that sugar, salt, fabric, oxygen, copper, silver, water etc., are all different kinds of matter. We have discussed earlier that what is matter and how the matter is classified into solid liquid and gases on the basis of their physical states or into elements, compounds and mixture on the basis of their chemical properties. Now we will study it in detail.

1.1       ATOMS AND MOLECULES : ANCIENT THOUGHTS

Ancient Indian and Greek philosophers have always wondered about the unknown and unseen form of matter or we can say that what matter is ultimately made up of. It was around 500 BC that an Indian philosopher Maharishi Kanad had postulated that “matter is divisible” i.e. if we go on dividing the matter we will get smaller and smaller particles and ultimately we will get the smallest particle of matter which can not be divided any further. These indivisible particles were named by him as “Parmanu”.

Around the same era ancient Greek philosopher Democritus and Leuccipus suggested the same idea. However he called the smallest indivisible particles as “atoms” (Greek : means uncutable)

Another Indian philosopher PakudhaKatyayama said that these particles normally exist in a combined form called molecules which give us various forms of matter.

Thus, we may conclude that matter is made up of small particles which may be atoms or molecules. Different kinds of atoms and molecules have different properties that’s why different kinds of matter also show different properties.

Till eighteenth century there were no experimental work done to validate these philosophical ideas. By the end of eighteenth century, the fact had been established that pure substances can be either elements or compounds. Scientist became interested in finding out why and how elements combine and what happens when they combine?

Experimental studies were carried out to understand the laws according to which the elements combine to form compounds. These laws are called law of chemical combination.

All matter is made up as small particles called atoms and molecules

 

 

 

 

            As mentioned above,whenever reactants (elements) react together to form a compound they do so according to certain laws. These laws are called “laws of chemical combination”. There are two important laws of chemical combination. These are

(i)    Law of conservation of mass

(ii)   Law of constant proportions

The laws of chemical combination are the experimental laws which have been established by scientists after performing a large number of experiments.

2.1       LAW OF CONSERVATION OF MASS

This law was put forward by Antoine Lavoisier in 1774. It deals with the masses of reactants and products involved in a chemical reaction.

It was observed by the scientists that if they carried out a chemical reaction in a closed container then there was no change in mass. This preservation of mass in chemical reaction lead to the formulation of law of conservation of mass (or matter).

 

Matter is neither created nor destroyed in a chemical reaction.

 

Or

 

In any chemical reaction the total mass of the reactants is equal to the total mass of the products.

Or

 

There is no change in mass during a chemical reaction.

The following experiment illustrate the validity of this law

 

2.1.1    Experiment :  When matter undergoes a physical change

A piece of ice (water) is taken in a small conical flask. It is well corked and weighed. This flask is then heated gently to melt the ice (solid) into water (liquid)

            Observation :After heating, the ice changes into water. When we weighed the flask the mass does not change though a physical change has taken place

            Conclusion :From the above experiment, we lead to the conclusion that during a physical change mass does not change or in other words mass is conserved during the physical change.

2.1.2    When matter undergoes a chemical change

When barium chloride reacts with sodium sulphate, barium sulphate and sodium chloride is formed.

Barium chloride + Sodium sulphate Barium sulphate + sodium chloride

(solution)                    (solution)                     (whiteppt)                    (solution)

Experiment :Take a clean conical flask fitted with a cork and a small test tube having a long thread tied to its neck. Weigh the flask, cork and tube together to find the initial mass of the apparatus. Now prepare the two solutions.

(i)    Approximately 5% solution of barium chloride is prepared by dissolving 5g of barium chloride in 100 ml of water. Take a small amount of barium chloride solution in the conical flask. Put some sodium sulphate solution in the small test tube. Suspend the test tube in the flask with the thread as shown in figure. Weigh the complete system. Subtract the mass of the empty apparatus.

The difference will give the mass of the reactants taken. Let the mass of the reactants be x.

Now loosen the cork so that the thread is loosened and the test tube fall into the flask due to which sodium sulphate solution mixes with barium chloride solution to form a white precipitate. of barium sulphate and sodium chloride solution (figure). Weigh the complete apparatus again along with the contents. If we subtract the initial mass of the apparatus from this mass, we will get the mass of the products. Suppose the mass of the product is y.

            Observation :It was observed that the mass of the reactants (x) comes out to be same as that of the products (y). This is in accordance with the law of conservation of mass.

            Conclusion :From the above experiment we lead to the conclusion that during a chemical change, mass remains the same or unchanged or in other words mass is conserved during a chemical change.

Example: 1

Question:       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 g sodium ethanoate, 2.2 g of carbon dioxide and 0.9 g of water. Show that this data verifies the law of conservation of mass.

Solution:        The above reaction can be written as

Sodium carbonate + Ethanoic acid  ® Sodium ehtanoate + carbon dioxide + water

In the above reaction sodium carbonate and ethanoic acid are reactants while sodium ethanoate, carbon dioxide and water are products. So

(i) Mass of reactants      = Mass of sodium carbonate + Mass of ethanoic acid

= 5.3 g + 6.0 g

= 11.3 g

(ii) Mass of products     = Mass of sodium ethanoate + Mass of CO2 + Mass of water

= 8.2 g + 2.2 g + 0.9 g

= 11.3 g

After the calculation we find that the mass of the reactants and that of the products is same i.e. 11.3 g, thus the given data verifies the law of conservation of mass.

2.2       LAW OF CONSTANT PROPORTIONS

Lavoisier, along with other scientists found that many compounds were composed of two or more elements and each such compound had the same elements in the same proportions, irrespective of from where the compound came.

These observations led to the ‘law of constant proportions’ which is also known as the ‘law of definite proportions’. This law deals with the composition of elements present in a given compound. This law was given by J.L. Proust. According to this law

In a chemical substance the elements are always present in definite proportions by mass.

Or

A chemical compound is always made up of the same elements combined together in the same fixed proportion by mass.

            For example:In water, hydrogen and oxygen combined together in the same fixed proportion of 1 : 8 by mass, irrespective of the source of water (like river, rain or tap water).

If we decompose 9 g of pure water by electrolysis i.e. passing electricity through it, then 1 gm of hydrogen and 8 gm of oxygen are obtained. Now, if we repeat this experiment by taking pure water from different sources the same masses of hydrogen and oxygen elements are obtained.

This experiment shows that water always consists of hydrogen and oxygen combined together in the same constant proportion of 1 : 8 by mass.

Law of constant proportions also help us to calculate the percentage of any element in the given compound using the following expression.

% of an element in the compound =

Example: 2

Question:        Calculate the mass of carbon present in 4g of carbon dioxide.

Solution:         Carbon dioxide (CO2) contains carbon and oxygen in the fixed proportion by mass,.which is 12 : 32 i.e. 3 : 8. This means that 3 g of carbon combine with 8 g of oxygen to form 11 g of CO2 or we can say that

11 g of CO2 contain = 3g of carbon

1 g of CO2 contain =  g of carbon

so 4 g of CO2 will contain =  g = 1.090 g of carbon

Example: 3

Question:        A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percent composition of the compound by mass. 

Solution:         (i)        Mass of boron in compound      = 0.096 g

Mass of compound                   = 0.24 g

So, percentage of boron in the compound

 

= 40%

(ii)        Mass of oxygen in compound = 0.144 g

Mass of compound                   =0.24 g

So percentage of oxygen in compound =

=

= 60%

 

After the above two laws of chemical combination were put forward, the next problem faced by scientist was to give appropriate explanation of these laws. This led John Dalton to put forward a theory in 1808 about the nature of matter. Dalton picked up the idea of divisibility of matter, which was till then just a theoretical idea. The theory is based on certain postulates called postulates (or assumptions) of Dalton’s atomic theory. His theory provided an explanation for the law of conservation of mass and the law of definite proportion. According to Dalton’s atomic theory all matter, whether an element, a compound or a mixture is composed of small particles called atoms.

3.1       POSTULATES OF DALTON’S ATOMIC THEORY

The main postulates of Dalton’s atomic theory are as follows :

(i)   All matter is made up of very small particles called atoms.

(ii)  Atoms are indivisible particles, which can not be created or destroyed in a chemical reaction.

(iii)  Atoms of a given element are identical in all respects i.e. size, shape, mass and chemical properties.

(iv) Atoms of different elements have different size and masses and also posses different properties.

(v)  Atoms of the same or different elements combine in the ratio of small whole numbers to form compounds.

(vi) The relative number and kinds of atoms are constant in a given compound.

(vii)      Atoms of the same elements or two different elements may combine in different ratios to form more than one compound.

3.2       EXPLANATION OF THE LAWS OF CHEMICAL COMBINATION

Dalton’s atomic theory was the first modern attempt to describe the properties of matter in terms of atoms. This theory provides a simple explanation for the laws of chemical combination.

3.2.1    Explanation of law of conservation of mass

According to Dalton’s atomic theory matter is made up of atoms and, number of various types of atoms in the products of a chemical reaction is the same as that of the reactants. As the same number of various atoms in products and reactants will have the same mass, so the total mass of products is equal to the total mass of reactants or the mass remain unchanged during a chemical reaction and this is the law of conservation of mass.

3.2.2    Explanation of law of constant proportion

According to one of the postulates of Dalton’s atomic theory the number and kind of atoms in a compound is fixed. From this we can infer that a compound is always made up of the same elements combined together in the same proportion by mass and this is the law of constant proportion.

3.3       LIMITATIONS OR DRAWBACKS OF DALTON’S ATOMIC THEORY

With the advancement in scientific studies, Dalton’s atomic theory suffered from the following drawbacks :

(i)   Atom is no longer considered as the smallest indivisible particle.

(ii)  According to Dalton’s atomic theory, all the atoms of an element have exactly the same mass. Though, it is now known that atoms of the same elements may have different masses.

(iii)  According to Dalton’s atomic theory, atoms of different elements have different masses. However it is now known that even atoms of different elements can have the same mass.

(iv) Substances made up of the same kind of atoms may have different properties. For example, charcoal, graphite and diamond are all made up of carbon atoms but have different physical properties.

 

 

As all the houses are made up of bricks similarly all matter whether an element, or a compound or a mixture is made up of the smallest indivisible particles called atoms. In other word atoms are the building blocks of all matter. In chemistry atom is defined as the smallest particle of an element that can take part in a chemical reaction and which may or may not be capable of free existence. Atoms of most of the elements are very reactive and do not exist in the free state, instead, they exist in combination with the atoms of the same element or another element. For example, H2, N2, O2, HCl, NH3, etc.

4.1       HOW BIG ARE THE ATOMS ?

Atoms are very-very small in size. They are so small that they can not be seen even under a microscope. To imagine about their size, it is very much interesting to note that if millions of atoms are stacked one above the other, the thickness produced may not be equal to the thickness of the sheet of a paper.

The size of an atom is indicated by its radius which is called ‘atomic radius”. Atomic radius is measured in nanometers, which is represented by the symbol ‘nm’.

1 nm = 109 m

1m = 109 nm

 

Atomic radii of some common elements

S. No. Element Atomic radius
1.

2.

3.

4.

5.

6.

Hydrogen

Carbon

Nitrogen

Oxygen

Chloride

Sulphate

0.037 nm

0.077 nm

0.074 nm

0.073 nm

0.099 nm

0.104 nm

 

Atoms are so small that they can not be seen even under the most powerful optical microscope. However it is only recently that a highly sophisticated microscope known as scanning tunneling microscope (STM), has magnified image of surface of elements showing atoms.

Relative size  
Radii (in meter) Example
10–10 molecule of water
10–9 atom of hydrogen
10–8 molecule of Haemoglobin
10–4 Grain of sand
10–3 ant

4.2       WHAT ARE THE MODERN DAY SYMBOLS OF ATOMS OF DIFFERENT ELEMENTS ?

Symbol is a short method of representing anything. In case of elements a short method of representing the full name of an element is known as symbol.

4.2.1    Dalton’s symbols of element

            Dalton was the first scientist to suggest the symbols for elements in a very specific way. Dalton’s symbol for an element represent the “element’ as well as one atom of that element. Thus, we can say that the symbol used by him also represent the quantity of the element. A few of these symbols as proposed by Dalton are as follows :

Dalton’s symbols were difficult to draw and inconvenient to use so they could not become popular.

4.2.2    Berzelius suggestion for symbols of elements

J.J. Berzelius, a Swedish chemist, suggested a more scientific method for representing an element. He suggested that the first one or two letter of the name of an element can be used as its symbols. This idea led to the development of modern symbols of elements.

4.2.3    Modern symbols of elements

In the beginning, the names of elements were derived from the name of the place where they were first found, discovered or after the name of the scientist who discovered it. For example the name copper was taken from Cyprus, Rutherfordrium after Rutherford etc. Some names were taken from specific colours. For example gold was taken from the English word meaning yellow. However as more and more elements were discovered an international committee was set up called ‘International Union of Pure and Applied Chemistry’ (IUPAC) which approved the names of the different elements.

Though the names of most of the elements have been taken from English there are some elements which have been taken from Latin, German or Greek. In all cases the symbol of an element is the “first letter” or “the first letter and another letter” of the English or Latin name of the element. For example :

The symbol of Hydrogen is H

The symbol of oxygen is O

So, in the case of hydrogen and oxygen the first letter of their English names are taken as their symbols.

It should be noted that in a “two letter” symbol, the first letter is the capital letter but the second letter is the small letter. The necessity of adding another letter arises only in case of elements whose names start with the same letter. For example the name of the elements viz. carbon, chlorine calcium and copper starts with the common letter C.

Hence,

Carbon is represented by the symbol C

Chlorine is represented by the symbol Cl

Calcium is represented by the symbol Ca

Copper is represented by the symbol Cu

Symbols of elements derived from English names

English name of the element Symbol English name of the element Symbol
1.   Argon Ar 15.   Iodine I
2.   Arsenic As 16.   Lithium Li
3.   Aluminium Al 17.   Magnesium Mg
4.   Boron B 18.   Manganese Mn
5.   Barium Ba 19.   Nitrogen N
6.   Bromine Br 20.   Neon Ne
7.   Carbon C 21.   Nickel Ni
8.   Calcium Ca 22.   Oxygen O
9.   Chlorine Cl 23.   Phosphorus P
10. Chromium Cr 24.   Platinum Pt
11. Cobalt Co 25.   Sulphur S
12. Fluorine F 26.   Uranium U
13. Hydrogen H 27.   Zinc Zn
14. Helium He    

Symbols derived from Latin names

English name of the element Latin name of the element Symbol
1.   Antimony Stibium Sb
2.   Copper Cuprum Cu
3.   Gold Aurum Au
4.   Iron Ferrum Fe
5.   Lead Plumbum Pb
6.   Mercury Hydragyrum Hg
7.   Potassium Kalium K
8.   Silver Argentum Ag
9.   Sodium Natrium Na
10. Tin Stannum Sn

4.3       ATOMIC MASS       

The most remarkable concept that Dalton’s atomic theory proposed was that of the atomic mass. According to Dalton, each element had a characteristic atomic mass. We know that atoms are extremely small particles. Hence determining the mass of an individual atom was a relatively difficult task. For example, actual mass of an atom of hydrogen is found to be 1.673 ´ 1024 g which is extremely small. However, it was found easy to compare the masses of atoms of different elements with some reference atom. The masses thus obtained are called relative atomic mass.

The reference which was chosen earlier was hydrogen atom as it is the lightest element. Its mass was taken as 1. However, by using hydrogen as the reference element the masses of atoms of other elements came out to be fractional. Due to this the reference was changed to oxygen (taken as 16) or in other words 1/16th of the mass of an atom of naturally occurring oxygen was taken as one unit. This was considered relevent due to the following two reasons :

  • Oxygen reacted with a large number of elements and formed compounds.
  • This atomic mass unit gave masses of most of the elements as whole numbers.

However, in 1961, a universally accepted atomic mass unit carbon-12 isotope was chosen as the standard reference for measuring atomic masses. It is called carbon twelve (C – 12) and is represented as 12C. The atomic mass can be defined as :

“One atomic mass unit is a mass unit equal to exactly one twelfth (1/12th) the mass of one atom of carbon-12. The relative atomic masses of all elements have been found with respect to an atom of carbon -12”.

Thus, 1/12th of the mass of an atom of carbon -12 isotope represents one unit of mass on the atomic scale. This is called one atomic mass unit (amu). Now it is represented simply by ‘u’ which stands for unified mass.

Atomic mass of an element may therefore also be defined as “the number of times an atom of that element is heavier than 1/12th of the mass of an atom of C-12 isotope”.

For example, an atom of magnesium is found to be two times heavier than an atom of
C-12 i.e. 24 times heavier than 1/12th of the mass of C-12 atom. Hence, atomic mass of magnesium = 24 amu.

 

 

Atomic masses of some common elements

Element Symbol Atomic mass Element Symbol Atomic mass
1.   Hydrogen H 1 14.  Sulphur S 32
2.   Helium He 4 15.  Chlorine Cl 35.5
3.   Lithium Li 7 16.  Argon Ar 40
4.   Boron B 11 17.  Potassium K 39
5.   Carbon C 12 18.  Calcium Ca 40
6.   Nitrogen N 14 19.  Iron Fe 56
7.   Oxygen O 16 20. Copper Cu 63.5
8.   Fluorine F 19 21. Zinc Zn 65
9.   Neon Ne 20 22. Silver Ag 108
10. Sodium Na 23 23. Platinum Pt 195
11. Magnesium Mg 24 24. Gold Au 197
12. Aluminium Al 27 25. Lead Pb 207
13. Phosphorus P 31 26. Uranium U 238

 

 

            Atoms of most elements are not able to exist independently. Only the atoms of a few elements exist in free state.

Atoms usually exist in two ways :

(i)    In the form of molecules

(ii)   In the form of ions

When they form molecules or ions they become stable. Molecules and ions aggregate in large numbers to form matter which we see around us.

Though we can not see individual atoms, molecules or ions but we can see the matter. For example, we cannot see the Na+ and Cl ions but we can see the sodium chloride compound (common salt).

 

 

 

 

A molecule is in general a group of two or more atoms that are chemically bonded together i.e. tightly held together by attractive forces.

Or

A molecule is the smallest particle of an element or a compound which can exist freely and possesses all the properties of that substance.

Atoms of the same element or of different elements can join together to form molecules.

6.1       MOLECULE OF AN ELEMENT

The molecules of an element are constituted by the same type of atoms.

Or

The molecules of an element contain two (or more) similar atoms chemically combined together. Molecules of many elements such as Argon (Ar), Helium (He), etc. are made up of only one atom of that element. But this is not the case with most of the elements. Depending upon whether the molecule contains one, two, three or four atoms they are called monoatomic, diatomic triatomic, tetra atomic or polyatomic. A few examples of molecules of different types are as follows:

            (i)    Monoatomic molecules :Noble gases like Helium, Neon, etc. exist as single atoms i.e. He, Ne etc. Hence they are called monoatomic.

 

            (ii)   Diatomic molecules :Molecules of Hydrogen, Oxygen, Nitrogen contain two atoms of each element respectively and are represented by H2, O2, N2 etc.

 

        œA molecule of hydrogen or oxygen cannot be represented as 2H or 2O. This will represent two atoms of hydrogen and oxygen and not a molecule of hydrogen. It applies to other molecules as well.

 

 

            (iii)  Triatomic molecules :Molecules containing 3 atoms are called triatomic molecules. For example, ozone contains 3 atoms of oxygen element combined together.

 

            (iv)  Tetratomic molecules :Molecules containing 4 atoms of an element are called tetratomic molecules. Most common example is that of phosphorus represented by P4.

            (v)   Polyatomic molecules :Molecules containing more than four atoms of particular element are called polyatomic molecules. For example, a molecule of sulphur contains 8 atoms of sulphur and is represented by S8.

6.1.1    Atomicity      

The number of atoms present in one molecule of a substance is known as its atomicity.

Let us look at the atomicity of some non-metals.

Atomicity of some elements
Type of Element Name Atomicity
Non-metal Argon

Helium

Oxygen

Hydrogen

Nitrogen

Chlorine

Phosphorus

Sulphur

   Monoatomic

Monoatomic

Diatomic

Diatomic

Diatomic

Diatomic

Tetra-atomic

Poly-atomic

!

        œAll metals (except mercury) are represented by their atomic symbols and are taken as monoatomic.

 

6.2       Molecules of compounds

The molecules of a compound consist of two or more atoms of different elements combined together in a definite proportion by mass to form a compound that can exist freely.

For example, carbon dioxide contains one atom of carbon and two atoms of oxygen combined together in a fixed ratio of 3 : 8 by mass.

Molecules of some compounds
Compound Combining Elements Ratio by Mass
Water

Ammonia

Carbon

dioxide

Hydrogen, Oxygen

Nitrogen, Hydrogen

 

Carbon, Oxygen

1:8

14:3

 

3:8

In the above table we have given the ratio by mass of atoms present in one molecule of that particular compound. The atomic masses of different elements are H = 1.0u, O = 16.0u, N = 14.0u, C = 12.0u. By comparing the data we can find out the ratio by number of atoms of elements in the molecule of the particular compound as follows :

 

 

The molecular compounds composed of metals and non-metals contain charged species. The charged species are known as ions. Ions can be defined as a positively or negatively charged atom (or group of atoms). Depending upon the charge they carry, ions can be of two types :

            (i)    Cation :A positively charged ion is known as the cation. For example, Sodium ion (Na+), Magnesium ion (Mg++) etc. A cation is formed by the loss of one or more electrons by an atoms. This can be represented as follows :

 

sodium ion (cation)

 

        œSince a cation is formed by the removal of electrons from an atom, therefore a cation contain less electrons than a normal atom, and hence it is positively charged.

 

            (ii)   Anion : A negatively charged ion is known as anion. For example chloride ion (Cl), oxide ion (O) are anions as they are negatively charged.

An anion is formed by the gain of one or more electrons.

 

Chlorine atom             chloride ion (anion)

        œSince an anion is formed by the gain of one or more electrons by an atom therefore it contains more electrons than a normal atom.

 

It is important to know that in any compound the total positive charge of cation is equal to the negative charge of anion, so that the ionic compound as a whole is electrically neutral.

7.1       SIMPLE IONS AND COMPOUND IONS (POLYATOMIC IONS)

7.1.1    Simple ions

Those ions which are formed from single atoms are called simple ions.

For example, etc.

7.1.2    Polyatomic ion

Ions formed from a group of atoms carrying a charge (either negative or positive) is known as a polyatomic ion or compound ion.

For example .

Some of the common ionic compound their formula and the ions present in them are given below :

 

Some Ionic compounds

Name Formula Ions present
1. sodium chloride             NaCl             Na+      and      Cl
2. Potassium chloride             KCl             K+        and      Cl
3. Ammonium chloride             NH4Cl             NH4+   and      Cl
4. Magnesium             MgCl2             Mg2+    and      Cl
5. Calcium chloride             CaCl2             Ca2+    and      Cl
6. Magnesium oxide             MgO             Mg2+    and      O2
7. Calcium oxide             CaO             Ca2+    and      O2
8. Aluminium oxide             Al2O3             Al3+      and      O2
9. Sodium hydroxide             NaOH             Na+      and      OH
10. Copper sulphate             CuSO4             Cu2+    and      SO
11. Calcium nitrate             Ca(NO2)2             Ca2+    and      NO3

 

In what form atoms can exist in aqueous solution? Give example.

 

 

Before we learn how to write a chemical formula we should know what is a chemical formula? A chemical formula may be defined as the “Symbolic representation of its composition” or in other words.

“A chemical formula of a molecular compound represents the actual number of atoms present in one molecule of the compound”.

For example: H2O is the chemical formula of water, NH3 is the chemical formula of ammonia.

In other words, the formula of a compound tells us the “kind of atoms” as well as “the number of atoms” of various elements present in one molecule of the compound.

“Chemical formula of an ionic compound represents the cations and anions present in the structure of the compound”.

For example: Na+Cl represents that sodium chloride contains Na+ and Cl ions in the ratio of 1:1.

To understand how to write the chemical formula of different compounds, we need to learn the symbols and combining capacity of the elements.

 

8.1       CONCEPT OF VALENCY

            Valency can be defined as the combining capacity of that particular element.

To find out how the atoms of an element will combine with the atom (s) of another element to form a compound, valency can be used.

For example valency of oxygen is 2, this means that one atom of oxygen can combine with 2 atoms of hydrogen or in other words we can say that valency of hydrogen is one so 2 atoms of hydrogen can combine with one atom of oxygen to form water molecule.

The concept of valency can be understood more easily by the following examples. The valency of the atom of an element can be considered as hands of that atom.

We know that human beings have two hands and an octopus has eight. If one octopus has to catch hold of a few people in such a way that both arms of all the human and all the eight arms of octopus are locked. How many humans according to you an octopus can hold?

If we represent the octopus with O and humans with H than we can write a formula for this combination as OH4 where 4 is the subscript which indicates the number of humans held by the octopus.

As molecular compounds are formed by the combination of non-metal atoms, the valencies of some of them are as follows:

Valencies of some common non-metal elements
Element Symbol Valency Element Symbol Valency
Hydrogen H 1 Oxygen O 2
Fluorine F 1 Sulphur S 2, 4, 6
Chlorine Cl 1 Nitrogen N 3, 5
Bromine Br 1 Phosphorus P 3, 5
Iodine I 1 Carbon C 4

 

For writing the chemical formula of an ionic compound,valency of an ion can be defined as – “the units of positive or negative charge present on the ion.

For example:  Na+ ion has one unit positive charge

Ca2+ ion has two unit positive charge

Cl ion has one unit negative charge

SO ion has two unit negative charge

Depending upon whether the ions has 1, 2, 3 or 4 unit charge (positive or negative), they are called monovalent, divalent, trivalent and tetravalent ions respectively.

Below is given a table showing names and symbols of some ions.

 

Names and symbols of some ions
Valency Name of ion Symbol Non-metallic element Symbol Polyatomic ions Symbol
1. Sodium Na+ Hydrogen H+ Ammonium NH4+
Potassium K+ Hydride H Hydroxide OH
Silver Ag+ Chloride Cl Nitrate NO
Copper (I)* Cu+ Bromide Br Hydrogen

carbonate

HCO
Iodide I
2. Magnesium Mg2+ Oxide O2 Carbonate CO
Calcium Ca2+ Sulphide S2 Sulphite SO
Zinc Zn2+      

Sulphate

 

SO

Iron (II)* Fe2+
Copper (II)* Cu2+
3. Aluminium Al3+  

Nitride

 

N3

 

Phosphate

 

PO

Iron (III)* Fe3+

 

From the above table, it is seen that some elements show a number of valencies (called variable valencies). For example, copper shows a valency of 1 as well as 2, iron shows a valency of 2 as well as 3. These valencies are sometimes shown by Roman numeral in a bracket after the symbol of the element, e.g., Cu (I), Cu (II), Fe (II), Fe (III) etc. Some non-metal atoms also show variable valency.

 

8.2       RULES FOR WRITING THE CHEMICAL FORMULAE

While writing a chemical formulae of molecular or ionic compounds the following steps are to be followed:

(i)    In case of simple molecular compounds (compounds made up of only two elements). The symbols of the two elements are written side by side and their respective valencies are written below their symbols.

(ii)   In case of simple ionic compounds, the symbol of the cation or metal atom is written first followed by the symbol of the anion or non-metal atom and their respective valencies are written below their symbols. For example in CaO, symbol of calcium (Ca, a metal) must be written first followed by symbol of oxygen (which is a non-metal).

(iii)  The valencies or charges on the ion must be balanced.

(iv)  In case of compounds containing polyatomic ions, the formula of the polyatomic ion is written in brackets and the valencies are written below.

(v)   In any of the above cases, if there is a common factor between the valencies of the cation and anion, the valencies are divided by the common factor.

(vi)  Finally we apply cross-over of the valencies so that they appear on the lower right hand side of the symbols. However, 1 appearing on the lower right hand side of the symbol is omitted. Similarly we also omit the + and – signs of the charges of the ions.

8.2.1    Formulae of Simple Compounds

The simplest compounds, which are made up of two different elements are called binary compounds.

By applying the above mentioned rules, the chemical formulae of some simple molecular compounds can be written as follow:

Example 1: Steps for writing the formula of Hydrogen chloride:

(i)    Elements present are:     hydrogen and chlorine

(ii)   Symbols of the elements:            H         Cl

(iii)  Valency of the elements:             1          1

(iv)  Cross-over of the valency:

 

(v)   We stated that one appearing on the lower right hand side of the symbol is omitted. So the formula of the compound would be HCl.

Example 2:Steps for writing the formula of Methane

(i)    Elements present are:           Carbon    Hydrogen

(ii)   Symbols of the elements:            C         H

(iii)  Valency of the elements:             4          1

(iv)  Cross-over of the valency:

 

(v)   So the chemical formula of the methane can be written as CH4 (As we omit one appearing on the lower right hand side of carbon element).

 

 

Example 3:Steps for writing the chemical formula of Carbon dioxide

(i)    Elements present are:           Carbon    Oxygen

(ii)   Symbols of the elements:            C         O

(iii)  Valency of the elements:             4          2

(iv)  Dividing by common factor:        = 2   = 1

(v)   Cross-over of the valency:

 

(vi)  So the chemical formula of carbon dioxide can be written as CO2 (as we omit one appearing on the lower right hand side of carbon element).

 

8.2.2    Chemical formulae of some simple ionic compound

Example 1:Steps for writing the chemical formula of Sodium chloride

(i)    Elements present in the compound:        Sodium            Chlorine

(ii)   Symbols of the elements:                        Na                   Cl

(iii)  Charge on the ions                                  +1                    -1

(iv)  Valency of the elements:                         1                      1

(v)   Cross-over of the valency:

(vi)  So the chemical formula of sodium chloride can be written as NaCl (As we omit 1 appearing on the lower right hand side of Na and Cl atoms and + and – sign of the charges of ions).

Example 2:Steps for writing the chemical formula of Aluminium oxide

(i)    Elements present in the compound:        Aluminium       Oxygen

(ii)   Symbols of the elements:                        Al                     O

(iii)  Charge on the ions                                  +3                    -2

(iv)  Valency of the elements:                         3                      2

(v)   Cross-over of the valency:

 

(vi)  So the chemical formula of aluminium oxide can be written as Al2O3.

 

8.2.3    Chemical formulae of compounds containing polyatomic ions

Whilewriting the chemical formulae of compounds containing polyatomic ions, same rules will apply except that the formula of polyatomic ion is written in brackets.

Example 1:Steps for writing the formula of potassium nitrate

(i)    Symbols of the ions:                                K                      (NO3)

(ii)   Charge on the ions:                                 1+                    1-

(iii)  Valency of the ions:                                 1                      1

(iv)  Cross-over of the valency:

 

(v)   So the chemical formula of Potassium nitrate can be written as KNO3 (As we omit one appearing on the right hand side of the ions).

Example 2:Steps for writing the chemical formula of Calcium nitrate

(i)    Symbols of the ions:                                Ca                   (NO3)

(ii)   Charge on the ions:                                 2+                    1-

(iii)  Valency of the ions:                                 2                      1

(iv)  Cross-over of the valency:

 

(v)   So the chemical formula of Calcium nitrate can be written as Ca (NO3)2 (As we omit one appearing on the right hand side of the Caions).

 

            Example 2:     An element M forms the oxide M2O3. What will be the formula of its phosphate?

Solution: (i)   In M2O3total charge on 2 oxide ion = 3 x (-2) = -6

(ii)   As the compound M2O3 is neutral so the charge on two metal atoms should be = +6

\ Charge on one metal atom = + 6/2 = + 3

i.e.valency of metal atom = 3

Steps for writing the formula of metal phosphate.

(i)    Symbols of the elements/ion present:     M                     (PO4)

(ii)   Charge:                                                    3+                    3-

(iii)  Valency:                                                   3                      3

(iv)  Dividing by the common factor               = 1               = 1

(iv)  Cross-over

 

(v)   Thus we can write the chemical formula as MPO4

 

 

The molecular mass of a substance is the sum of the atomic masses of all the atoms in a molecule of the substance. It is therefore the relative mass of a molecule expressed in atomic mass unit.

Molecular mass of a substance can be defined as follows:

“Molecular mass of a substance (element or compound) is the average relative mass of its molecule as compared with that of an atom of C-12 isotope taken as 12”.

OR

“Molecular mass of a substance represents the number of times the molecules of that substance is heavier than 1/12th of the mass of an atom of C-12 isotope”.

For example the molecular mass of hydrogen is 2, which means that a molecule of hydrogen is two times heavier than the 1/12th of the mass of an atom of C-12 isotope.

9.1       CALCULATION OF MOLECULAR MASS

As we know that molecules are made up of two or more atoms of same or different elements and each element/atom has a definite atomic mass, therefore, molecular mass of a molecule of a substance can be calculated by adding atomic masses of all the atoms present in one molecule of the substance. For example,

(a)   A molecule of water has the formula H2O. Hence molecular mass of

H2O = (2 x atomic mass of hydrogen) + (1 x atomic mass of oxygen)

= (2 x 1.0 u + 1 x 16.04) – 18 u

(b)   A molecule of sulphuric acid has the formula H2SO4. Hence molecular mass of

H2SO4 = (2 x atomic mass of hydrogen) + (1 x atomic mass of sulphur)

+ (4 x atomic mass of oxygen)

= (2 x 1.0 u) + (1 x 32.0 u) + (4 x 16.0 u)

= (2.0 u + 32.0 u + 64.0 u)

= 98.0 u

 

Example: 1

Question:         Calculate the molar mass of the following substances :

                        (a) Ethyne, C2H2                        (b) Sulphur molecule, S8

                        (c) Phosphorus molecule, P4(Atomic mass of phosphorus = 31)         

                        (d) Hydrochloric acid, HCl        (e) nitric acid, HNO3

Solution:         (a) Molecular mass of C2H2 = 2 x Atomic mass of C + 2 x Atomic mass of H

= 2 x 12 u + 2 x 1 u = 26 u

\ Molar mass of C2H2 = 26 g.

(b) Molecular mass of S8 = 8 x Atomic mass of sulphur = 8 x 32 u = 256 u

\ Molar mass of S8 = 256 g.

(c) Molecular mass of P4 = 4 x Atomic mass of P = 4 x 31 u = 124 u

\         Molar mass of P4 = 124.0 g.

(d) Molecular mass of HCl = Atomic mass of H + Atomic mass of Cl

= 1.0 u + 35.5 u = 36.5 u

\         Molar mass of HCl = 36.5 g.

(e) Molecular mass of HNO3 = Atomic mass of H + Atomic mass of N + 3 x Atomic

mass of O

= 1.0 u + 14.0 u + 3 x 16.0 u = 63 u

\       Molar mass of HNO3 = 63.0 g.

 

Calculate the molecular mass of potassium permanganate (KMnO4). Atomic mass K = 39.0 U, Mn = 55.0 U; O = 16.0 U.

 

 

9.2       FORMULA UNIT MASS

Before describing the formula of unit mass, we should be aware of the meaning of ‘Formula Unit’ of an ionic compound.

An ionic compound consists of a very large but equal number of positively charged and negative charged ions (as an ionic compound is electrically neutral) arranged in a definite order in a crystal lattice. Thus the actual formula of an ionic compound should be written as (C+)n (a)n or (C+a)n, where ‘n’ is a very large number. Thus formula unit can be defined as follows:

The simplest combination of ions that produces an electrically neutral unit, is called a ‘formula unit’ of the ionic compound. It is thought to be the smallest unit of that compound. For example the formula unit of sodium chloride is NaCl.

As we have discussed that ionic compounds do not exist in molecular form, so we can not give its molecular mass. In such case, the term “Formula unit mass” is used.

The formula unit mass of an ionic compound can be defined as follows:

“The formula unit mass or formula mass of an ionic compound is the sum of the atomic masses of all the atoms present in one formula unit of the compound”.

Or

“The formula mass of an ionic compound is the relative mass of its “formula unit” as compared with the mass of a carbon-12 atom taken as 12 unit”.

In order to calculate the formula unit mass of an ionic compound, we must be aware about the formula of the compound as well as the atomic masses of its constituent atoms.

Example: 2

Question:       Calculate the formula unit mass of Na2SO4.10H2O Atomic masses Na = 23.0u, S = 32.0u, O = 16.0u, H = 1.0u.

Solution:        Formula unit mass  = (2 x atomic mass of Na) + (2 x atomic mass of S) + (4 x atomic mass of oxygen) + 10 (2 x atomic mass of H + atomic mass of oxygen)

= [2 x 23.0u] + [32.0u] + [4 x 16.0u] + 10 [ 2 x 1.0u + 16.0u]

= [46.0u + 32.0u + 64.0u] + 10 [2.0u + 16.0u]

= [142.0u] + 10 [18.0u]

= 142.0u + 180.0u]

= [322.0u]

Thus the formula mass of Na2SO4.10H2O is 322.0u

 

Calculate the formula masses of the following compounds: (a) calcium chloride (b) sodium carbonate. Give: Atomic masses: Ca = 40.0u; Cl = 35.5u; Na = 23.0u; C = 12.0u; O = 16.0u.

 

9.3       GRAM ATOMIC MASS AND GRAM MOLECULAR MASS

In order to understand the “Mole concept” we should first know the meaning of the terms “gram atomic mass” and “gram molecular mass”.

9.3.1    Gram Atomic Mass

Atomic mass expressed in grams is called gram atomic mass of that element.

Or

The amount of a substance whose mass in grams is numerically equal to its atomic mass, is called the gram atomic mass of that substance.

In other words, we can say that atomic mass and gram atomic mass of a substance is numerically equal, but their units are different.

For example:

(a)                  Atomic mass of Na = 23.0u

Gram atomic mass of Na = 23.0g

(b)                  Atomic mass of Cl = 35.5u

Gram atomic mass of Cl = 35.5g and so on.

 

        œ“Gram atomic mass of an element” and “molar mass of an element” are just the same”.

 

9.3.2    Gram Molecular Mass

Like gram atomic mass, gram molecular mass can also be defined as follows:

            “Molecular mass expressed in grams is called gram molecular mass of that element”

Or

            “The amount of substance whose mass in grams is numerically equal to its molecular mass is called gram molecular mass of that substance”.

In other words, molecular mass and gram molecular mass of a substance is numerically equal, only their units are changed.

For example:

(a)          Molecular mass of H2 = 2.0u

Gram molecular mass of H2 = 2.0g

(b)        Molecular mass of H2O = 18.0u

Gram molecular mass of H2O = 18.0g.

Thus to convert molecular mass or atomic mass into gram molecular mass or gram atomic mass, we have to replace ‘u’ by ‘g’.

        œ“Gram molecular mass of a substance” and “Molar mass of a substance” are just the same.

 

9.3.3    Gram Formula Unit Mass

Similarly we can define the gram formula unit mass as follows:

            “Formula unit mass expressed in grams is called as gram formula unit mass”.

For example:

Formula unit mass of NaCl = 23.0u + 35.5u

= 58.5u

Gram formula unit mass of Nacl = 58.5u

        œThe amount of substance having mass equal to its gram atomic mass or gram molecular mass or gram formula unit mass is called “one gram atom” or “one gram molecule” or “one gram formula” unit respectively.

 

9.4       MOLE CONCEPT

In our daily life we buy some products in terms of numbers such as dozens (for 12), or gross (for 144) or in terms of mass such as kilograms (1000g) or quintals (100kg).

For example: We buy bananas, oranges, mangoes etc. in dozens where as rice, wheat, pulses etc. in kilograms.

1 dozen banana          = 12 banana

1 dozen mango           = 12 mangoes

1 kilogram wheat        = 1000g wheat

1 quintal rice                = 100 kilogram rice and so on.

We can not buy rice, pulses, sugar etc. in terms of numbers as they are very small in size. Similarly atoms and molecules are also very small in size. They are so small that we can not see them through our eyes.

So to express a definite amount of a chemical substance a new bigger unit “mole” was introduced. In Latin ‘mole’ means ‘heap’ or ‘collection’ or ‘pile’. The mole may be expressed in terms of mass or in terms of number.

9.4.1    Mole in Terms of Mass

“A mole of atoms is defined as” the amount of the substance i.e. element which has mass equal to its gram atomic mass”.

Or

“A mole of atoms is equal to one gram atom of that particular element”.

For example:

1 mole of Hydrogen (H) atom                        = 1g atom of H = 1.0g

1 mole of Oxygen (O) atom               = 1g atom of O = 16.0g

1 mole of Nitrogen (N) atom               = 1g atom of N  = 14.0g

1 mole of Sodium (Na) atom              = 1g atom of Na = 23.0g

similarly

“A mole of molecule is defined as that amount of the substance (element or compound) which has mass equal to gram molecular mass”.

Or

            “A mole of molecule is equal to one gram molecule of the substance”.

For example,

1 mole of Hydrogen (H2) molecule    = 1 g molecule of H2    = 2.0g

1 mole of Oxygen (O2) molecule       = 1 g molecule of O2   = 32.0g

1 mole of Ammonia (NH3) molecule  = 1 g molecule of NH3 = 17.0g

1 mole of Carbon dioxide molecule    = 1 g molecule of CO2 = 44.0g

 

 

Why is the statement “1 mole of hydrogen” not correct? What should be the correct statements ?

 

9.4.2    Mole in Terms of Number

As we have discussed earlier that one dozen contain 12 number of a particular substance or one gross contain 144 number of the substance in consideration. Similarly the number of particles (atoms, molecules or ions) present in 1 mole of any substance is fixed. The concept of mole in terms of number was arrived as follows:

Consider any two chemical substance, suppose carbon and oxygen.

Atomic mass of carbon                      = 12.0u

Molecular mass of oxygen (O2)         = 32.0u

Thus the ratio of mass of 1 atom of C to 1 molecule of oxygen (O2) = 12:32. Now if we take a million atoms of carbon and a million molecules of oxygen (O2), the ratio of their masses will remain the same. Conversely, it implies that the masses of carbon and oxygen taken in the ratio of 12:32 will contain the same number of atoms or oxygen molecules. Thus we can say that if we take 32.0g of oxygen (i.e. 1 mole of oxygen), it will contain the same number of oxygen molecules as the number of atoms in 12g of carbon (i.e. 1mole of carbon atoms).

Thus is general mole can be defined as follows:

            “A mole of particles (atoms, molecules or ions) is defined as that amount of the substance which contains the same number of particles as there are C-12 atoms in 12g of carbon”.

Experimentally, it has been found that 12g of C-12 isotope contain 6.022 x 1223 atoms. This number is called Avogadro’s number or Avogadro’s constant and is represented by the symbol N0. Thus

Avogadro’s number (N0) = 6.022 x 1023.

Thus a mole of particles can also be defined as follows:

“A mole of particles (atoms, molecules or ions) is that amount of the substance which contain 6.022 x 1023 particles”.

For example:

1 mole of C atoms                  = 6.022 x 1023 C atoms

1 mole of O atoms                  = 6.022 x 1023 O atoms

1 mole of H atoms                  = 6.022 x 1023 H atoms

1 mole of H2O molecules       = 6.022 x 1023H2O molecules

1 mole of CO2 molecules       = 6.022 x 1023 CO2 molecules

1 mole of Na+ ions                  = 6.022 x 1023 Na+ ions

 

9.4.3    Relationship Between Mole, Avogadro Number and Mass

 

9.4.4    Significance of the mole

A mole represents the following:

(i)         It represents 6.022 x 1023 particles of the substance

(ii)        The mass of one mole of an element is equal to the mass of 6.022 x 1023 atoms of that element.

(iii)       One mole of a substance represents one gram, formula mass of that substance.

9.4.5    Formulae for calculation

            (a)        Number of moles =

i.e.

(b)       Number of moles =

                                                          

 

 

 

Question:       Calcium salt of a hypothetical anion Z has the molecular formula Ca3Z2. What is the valency of Z and what would be the molecular formula of aluminium salt of Z?

Solution:         Valency of Z is – 3. So, formula of salt is AlZ.

 

Question:        The valency of potassium is +1. Write the molecular formulae for potassium dichromate and potassium permanganate.

Solution:         Potassium dichromate                          Potassium permanganate

                        

 

Question:        Calculate the formula unit mass of CaCl2.

Solution:         Formula unit mass of CaCl2 = Atomic mass of Ca + 2 ´ Atomic mass of Cl.

= 40.0 u + 2 ´ 35.5= 40.0 u + 71.0 u  = 111.0 u

 

Question:       What is the mass of :

                        (a) 1 mole of nitrogen atoms.

                        (b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27) ?

Solution:        (a) Mass (m) = No. of moles (n) x Molar mass (M)

= 1 mole x 14 g mol1 = 14 g. (Molar mass of N = 14 g mol1)

(b) Mass (m) = 4 mol x 27 g mol1 = 108 g. (Molar mass of Al = 27 g mol1)

 

 

 

 

Question:       Calculate the number of moles in 5.75 g of sodium (Atomic mass of sodium   = 23.0 u)

Solution:        1 mole of sodium atoms = gram atomic mass of sodium

= 23.0 g

As we know that

Given :                          m = 5.75 g, M = 23.0 g

moles

 

Question:       Calculate the mass of 5 mol of ammonia ?

Solution:        we know that

 

or         m = n ´ M

Given,   n = 5 mole, M = 17 g

So,       m = 5 ´ 17

= 85 g

 

Question:       The mass of single atom of an element M is 3.15 ´ 1023 g. What is its atomic mass ? What could the element be ?

Solution:        Gram atomic mass        = mass of 6.022 ´ 1023 atoms

= mass of 1 atom ´ (6.022 ´ 1023)

= (3.15 ´ 1023 g) ´ (6.022 ´ 1023)

= 3.15 ´ 6.022

= 18.97 g

Atomic mass of the element = 18.97

Thus, the element is most likely to be fluorine.

 

Question:       If one mole of carbon atoms weight 12 gram, what is the mass in gram of 1 atom of carbon ?

Solution:        1 mole of C atoms = 6.022 ´ 1023 atoms of C = 12 g (given)

Thus, 6.022 ´ 1023 atoms of carbon have mass = 12 g

\         1 atom of carbon will have mass g

= 1.993 ´ 1023 g.

 

Question:       Write down the formulae of (i) aluminium oxide (ii) aluminium chloride (iii) hydrogen sulphide(iv) calcium hydroxide.

Solution:        (i)         Aluninium oxide :

 

Symbol

 

Charge

Valency

   

= Al2O3

                        (ii)         Aluminium chloride :

 

Symbol

 

Charge

Valency

   

= Alcl3

                        (iii)        Hydrogen sulphide :

 

Symbol

 

Charge

Valency

   

= H2S

                        (iv)        Calcium hydroxide :

 

Symbol

 

Charge

Valency

   

= Ca(OH)2

 

Question:       (a) Calculate the relative molecular mass of water (H2O).

                        (b) Calculate the molecular mass of HNO3.

Solution:        (a) Relative molecular mass of H2O = 2 x Atomic mass of hydrogen + Atomic mass of oxygen

= 2 x 1 u + 16 u = 18 u.

(b) Molecular mass of HNO3 = Atomic mass of H + Atomic mass of N + 3 x Atomic mass

of O

= 1 u + 14 u + 3 x 16 u = 63 u.

 

Question:       What is the mass of :

                        (a) 0.2 mole of oxygen atoms?

                        (b) 0.5 mole of water molecules?

Solution:        (a) Mass (m) = No. of mole (n) x Molar mass (M)

                                                            Molar mass of O-atoms = 16 g mol1

\ Mass of 0.2 mole of O-atoms = 0.2 mol x 16 g mol1 = 3.2 g.

(b) Molar mass of H2O = 2 x 1.0 + 16 g mol1 = 18 g mol1

                                    \ Mass of 0.5 mole of water molecules = 0.5 mol x 18 g mol1 = 9.0 g.

 

 

Question:       Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur.

Solution:                                No. of moles (n) =

No. of molecules (N) = No. of mole (n) x Avogadro’s No. (N0)

=             (on substituting the value of n)

Molar mass of S8 = 8 x 32 g mol1 = 256 g mol1

  • of molecules in 16 g sulphur = x 6.022 x 1023 molecules mol

= 3.76 x 1022 molecules.

 

Question:       Calculate the formula unit mass of ZnO, Na2O, K2CO3. Given atomic mass of Zn = 65 u, Na = 23 u, K = 39 u, C 12 u, O = 16 u, S = 32 u.

Solution:        Formula unit mass of ZnO    = Atomic mass of Zn + Atomic mass of O

= 65 u + 16 u = 81 u

Formula unit mass of Na2O  = 2 x Atomic mass of Na + Atomic mass of O

= 2 x 23 u + 16 u = 46 u + 16 u = 62 u

Formula unit mass of K2CO3 = 2 x Atomic mass of K + Atomic mass of C + 3 Atomic            mass of O

= 2 x 39 u + 12 u + 3 x 16 u

= 78 u + 12 u + 48 u = 138 u.

 

Question:       Calculate the number of particles in each of the following :

                        (i)         46 g of Na atoms                      (ii)        8 g of O2 molecules

                        (iii)       0.1 mole of carbon atoms

Solution:        (i)          moles

 

or          atoms.

(ii)         mole

molecules.

(iii)        atoms

 



KEY POINTS

  • 1 mole of particles = 6.022 × 1023 particles (atoms/molecules/ions/electrons/protons/neutrons /nucleons).
  • The weight of 1 mole atoms of an element = gram atomic weight of the element.
  • The weight of 6.022 ´ 1023 atoms of an element = gram atomic weight of the element.
  • The weight of 1 mole molecules of a compound = gram molecular weight of a compound.
  • The weight of 6.022 ´1023 molecules of a compound = gram molecular weight of the compound.
  • The weight of 1 mole of formula units of a salt = gram formula weight of the salt.
  • 1 mole of any gas at S.T.P. occupies 22.4 litres of volume.
  • 022×1023molecules of any gas at S.T.P. occupies 22.4 litres of volume.
  • Gram molecular weight of any gas occupies 22.4 litres of volume at S.T.P.

 

UNDERSTANDING OF MOLE

 

 

 

CBSE ASSIGNMENT

Dalton Atomic Theory

  1. Which postulate of Dalton’s atomic theory is a result of the law of conservation of mass?
  2. Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
  3. Define the atomic mass unit?
  4. Write the postulates of Dalton’s atomic theory.
  5. Write the relation between nanometer and meter.

Law of Chemical Combinations

  1. Who put forward the law of conservation of mass?
  2. When 5 g of calcium is burnt in 2 g of oxygen, then 7 g of calcium oxide is produced. What mass of calcium oxide will be produced when 5 g of calcium is burnt in 20 g of oxygen? Which law of chemical combination will govern your answer?
  3. When 5 g of calcium is burnt in 2 g of oxygen, then 7 g of calcium oxide is produced. What mass of calcium oxide will be produced when 5 g of calcium is burnt in 20 g of oxygen? Which law of chemical combination will govern your answer?
  4. When 4.2 g of NaHCO3 is added to a solution of acetic acid (CH3COOH) weighing 10.0g, it is observed that 2.2 g of CO2 is released into the atmosphere. The residue left behind is found to weigh 12.0 g. Show that these observations are in agreement with the law of conservation of mass.
  5. Define law of conservation of mass.
  6. When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?

Law of Definite Proportion and Constant Composition

  1. State “Law of definite proportion”. Explain it with one suitable example.
  2. Which molecules have hydrogen and oxygen in the ratio 1 : 8 by mass. Find the ratio of number of elements in water molecules.
  3. Magnesium and oxygen combine in the ratio of 3 : 2 by mass to form magnesium oxide. What mass of oxygen gas would be required to react completely with 24 g of magnesium?
  4. Define law of definite proportions.
  5. 6.488 of lead combine directly with 1.002 g of oxygen to form lead peroxide (PbO2). Lead peroxide is also produced by heating lead nitrate and it was found that the percentage of oxygen present in lead peroxide is 13.38 percent. Use these data to illustrate the law of constant composition.
  6. 2.16 g of copper metal when treated with nitric acid followed by ignition of the nitrate gave 2.70 g of copper oxide. In another experiment 1.15 g of copper oxide upon reduction with hydrogen gave 0.92 g of copper. Show that the above data illustrate the Law of Definite Proportions.

Valency Related Questions

  1. Calcium salt of a hypothetical anion Z has the molecular formula Ca3Z2. What is the valency of Z and what would be the molecular formula of aluminium salt of Z?
  2. The valency of potassium is +1. Write the molecular formulae for potassium dichromate and potassium permanganate.
  3. Write a brief note on variable valency.
  4. State and explain the following.

(i) Atom (ii) Molecule (iii) Cation (iv) Anion (v) Atomic Mass unit.

  1. The valency of potassium is +1. Write the molecular formulae for potassium dichromate and potassium permanganate.
  2. The mass of a single atom of an element X is 2.65 x 1023 g. What is its atomic mass? What could this element be?
  3. The valencies (or charges) of some of the ions are given below :
Ion Valency (Charge) Ion Valency (Charge)
Sodium ion 1+ Nitrate ion 1-
Copper ion 2+ Sulphide ion 2-

Using this information, write down the formulae of :

(i)  Sodium sulphide

(ii) Copper nitrate

  1. An element X forms the following compounds with hydrogen, carbon and oxygen :

H2X,    CX2,    XO2,    XO3

State the three valencies of element X which are illustrated by these compounds.

  1. Valencies or charges of some ions are given below :

Hydrogen ion, H+ = + 1                                Oxide ion, O2 = – 2

Aluminium ion, Al3+ = + 3                              Nitride ion, N3 = – 3

Calcium ion, Ca2+ = + 2                         Hydroxide ion, OH = – 1

Sodium ion, Na+ = + 1                      Carbonate ion,  = – 2

Using the above information, write down the chemical formulae of the following :

(i)   Aluminium oxide                                      (ii) Calcium nitride

(iii) Calcium hydroxide                                    (iv) Sodium carbonate

(v)  Carbonic acid                                            (vi) Water

  1. If the aluminium salt of an anion X is Al2X3, What is the valency of X ? What will be the formula of the magnesium salt of X ?

Element, Atom and Molecule

  1. What is Latin name of Mercury? What is its symbol?
  2. What is meant by atomicity? Explain with example.
  3. Explain the difference between 2N and N2.
  4. What is the order of size of atoms?
  5. What is the atomicity of the following?

(a) Oxygen                        (b) Ozone                    (c) Neon

(d) Sulphur                        (e) Phosphorus (f) Sodium

  1. Calculate the mass of 5 mol of ammonia ?
  2. State and explain the following.

(i) Atom (ii) Molecule (iii) Cation (iv) Anion (v) Atomic Mass unit.

MOLE CONCEPT

LEVEL-1

  1. Calculate the number of moles in 5.75 g of sodium (Atomic mass of sodium = 23.0 u)
  2. Calculate the mass of 5 mol of ammonia ?
  3. The mass of single atom of an element M is 3.15 ´ 1023 g. What is its atomic mass ? What could the element be ?
  4. If one mole of carbon atoms weight 12 gram, what is the mass in gram of 1 atom of carbon ?
  5. (a) Calculate the relative molecular mass of water (H2O).

(b) Calculate the molecular mass of HNO3.

  1. What is the mass of :

(a) 0.2 mole of oxygen atoms?

(b) 0.5 mole of water molecules?

  1. Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur.
  2. Calculate the formula unit mass of ZnO, Na2O, K2CO3. Given atomic mass of Zn = 65 u,

Na = 23 u, K = 39 u, C 12 u, O = 16 u, S = 32 u.

  1. Calculate the number of particles in each of the following :

(i)         46 g of Na atoms                     (ii)        8 g of O2 molecules

(iii)       0.1 mole of carbon atoms

  1. Calculate the mass of the following :

(i) 0.5 mole of N2 gas                          (ii) 0.5 mole of N atoms

(iii) 3.011 x 1023 N atoms                    (iv) 6.022 x 1023 N2 molecules.

  1. Calculate the number of aluminium ions present in 0.051 g of aluminium oxide (Atomic mass of Al = 27 u).
  2. Calculate the number of moles for the following :

(i)         52 g of He                   (ii)        12.044 ´ 1023 atoms of He

  1. Which has more atoms, 100 gram of sodium or 100 gram of iron (given, atomic mass of

Na = 23 u, Fe = 56 u).

LEVEL-2

  1. Define a “formula unit mass”.
  2. What is the difference between atomic mass and gram atomic mass.
  3. How many atoms are present in 1 mole of hydrogen gas?
  4. Write the relation between 1 mole of an element and Avogadro number.
  5. How many moles are there in 34.5 g of sodium?
  6. Define mole in terms of mass.
  7. Calculate the mass of 3.011 x 1024 atoms of carbon.
  8. Convert into mole:

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

  1. Calculate the mass of the following:

(i) 0.5 mole of N2 gas (mass from mole of molecule)

(ii) 0.5 mole of N atoms (mass from mole of atom)

(iii) 3.011 x 1023 number of N atoms (mass for number)

(iv) 6.022 x 1023 number of N2 molecules (mass from number)

  1. Which of the following weigh the most? (Atomic masses : O = 16, N = 14, Fe = 56, C = 12)

(i) 23 g of oxygen                                (ii) 2.0 g atoms of nitrogen

(iii) 0.5 mole of iron                            (iv) 3.011 x 1023 atoms of C

  1. Calculate the number of atoms in each of the following:

(i) 0.08 g of hydrogen              (ii) 0.008 g of sulphur              (iii) 0.8 g of iron

(Atomic masses : H = 1, S = 32, Fe = 56)

  1. (a) Give the relationship between molar mass and Avogadro number.

(b) Write the chemical formulae of the following compounds:

(i) Hydrogen oxide      (ii) Hydrogen nitride

(iii) Hydrogen peroxide                                       (iv) Hydrogen carbide

  1. If 1 g of sulphur dioxide contains x molecules, what will be the number of molecules in 1 g of methane?(S = 32 u, O = 16 u, C = 12 u, H = 1 u)
  2. 10 g silver nitrate solution are added to 10 g of sodium chloride solution? What change in mass do you expect after the reaction and why?
  3. How many atoms are present in 1 mole of hydrogen gas?
  4. How are mass, molar mass and number of moles related to each other?
  5. How are number of molecules of a substance related to its number of moles?
  6. The mass of a single atom of an element X is 2.65 x 1023 g. What is its atomic mass? What could this element be?
  7. How many moles are 9.033 x 1024 atoms of helium (He)?
  8. Calculate the number of moles for the following :

(i)   52 g of He (finding mole from mass).

(ii)  12.044 x 1023 number of He atoms (finding mole from number of particles).

 

LEVEL-3

  1. Calculate the number of atoms in each of the following :

(i)   0.08 g of hydrogen

(ii)  0.8 g of iron.

(Atomic masses : H = 1, S = 32, Fe = 56)

  1. 1022 atoms of an element ‘X’ are found to have a mass of 930 mg. Calculate the molar mass of the element ‘X’.
  2. Calculate the number of atoms in each of the following :

(i)   16 moles of He

(ii)  16 u of He

(iii) 16 g of He (Atomic mass of He = 4.0 u)

  1. Arrange the following in order of increasing masses : (i) 0.1 g atom of silver (ii) 0.1 mole of H2SO4 (iii) 1023 molecules of CO2 gas (iv) 1 gram of carbon (v) 1023 atoms of calcium.

(Atomic masses : Ag = 108 u, S = 32, N = 14 u, Ca = 40 u)

  1. Calculate the formula unit mass of CaCl2.

IONS

  1. What do we call those particles which have:

(a)  More electrons than the normal atoms?

(b)  Less electrons than the normal atoms?

  1. Write the symbols / formulae of two simple ions and two compound ions (or polyatomic ions).
  2. What is an ion ? Give two example each of bivalent cation and bivalent anions.
  3. What are polyatomic ions? Give examples.

 

 

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