Some Basic Concepts of Chemistry - Notes, Topics, Formula, Books, FAQs

Edited By Team Careers360 | Updated on Sep 17, 2024 07:23 PM IST

'Some basic concepts of chemistry' is the most fundamental chapter of complete chemistry. It gives information about the atomic number and mass number of elements. In any chemical reaction, it is important for us to know about the amount and number of reactants that will consume and the products that will produce, thus for estimating all these calculations, we use the laws of chemical combinations.

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This Story also Contains

Overview of the Chapter
Dalton's Atomic Theory
Mole Concept
Empirical Formula and Molecular formula
Stoichiometry and Stoichiometric Calculations
Some important Concentration Terms
How to prepare for States of matter?
Real life application based on Some Basic Concepts in Chemistry
Some Solved Examples
Conclusion

Important topics in some basic concepts of chemistry are mole concept, stoichiometry, molarity, normality, molality, etc. These topics are explained in detail along with some important solved examples later in this article.

Overview of the Chapter

Some basic concepts of chemistry are the most basic chapter of chemistry. It has various important concepts that you need to have greater insights for a better understanding of the whole of the chemistry. This article will help you to give important insights into this chapter and will also guide you in some important tips and guidelines.

Nature of Matter

Everything in the universe that has some mass and occupy some space is known as matter. Matter exists in three different physical forms i.e, solid, liquid and gas.

Property	Solid	Liquid	Gas
Tightness	Very tightly packed	Tightly packed	Loosely packed
Intermolecular space	Minimum	Intermediate	Maximum
Force of attraction	Maximum	Intermediate	Minimum
Kinetic Energy	Minimum	Intermediate	Maximum
Density	Maximum	Intermediate	Minimum
Volume	Fixed	Fixed	Variable
Shape	Fixed	Variable	Variable
Compressibility factor	Minimum	Intermediate	Maximum

At the macroscopic level, the matter can be classified into two categories i.e, mixtures and pure compounds as shown in the figure.

matter

Mixtures are those substances in which two or more components are mixed. Mixtures are further classified as homogeneous and heterogeneous mixtures. Homogeneous mixtures are the one in which components are present in a fixed ratio and the properties of this kind of mixture are the same throughout, for example, solution of sugar in water. But heterogeneous mixtures are those in which the components are not mixed in a definite ratio and properties of the mixture vary at different positions of the mixture, for example, sand in water.

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Laws of Chemical Combination

The combination of elements to form some new product follows three basic laws as shown in the figure

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Law of conservation of mass: This law states that "matter can neither be created nor destroyed". This simply means that the number of reactants that are used in the reaction will be equal to the number of products formed.
Law of definite proportions: This law states that any compound always contains its components in the fixed ratio by mass irrespective of its source and method of preparation.
Law of multiple proportions: This law states that when two elements combine to form one or more than one compound, then the masses of one element that combine with the other element are in the ratio of small whole numbers.

Dalton's Atomic Theory

John Dalton proposed a theory about the matter and atoms in which he proposed the following postulates:

(i) Matter consists of small individual particles known as 'atoms'.
(ii) All atoms of any particular element have the same properties and same mass but atoms of different elements have different properties and mass.
(iii) Compounds are formed when atoms of different elements are combined in the fixed ratio.
(iv) Chemical reactions involve the reorganization of atoms.

Mole Concept

Mole is the standard unit to measure the number of particles like atoms or molecules in a given sample. Mathematically, one mole is equal to 6.022 x 10²³. In other words, it can be said that it is equal to as many particles as there are atoms in 12g of the carbon-12 isotope. For example, if we have one mole of oxygen gas, that means we have 6.022 x 10²³ molecules of O₂.

Empirical Formula and Molecular formula

Empirical formula is the simplest whole-number ratio of atoms present in any particular molecule and the molecular formula is the actual representation of the number of atoms present in the molecule.

For example glucose i.e C₆H₁₂O₆, its molecular formula is C₆H₁₂O₆ but its empirical formula is CH₂O.
Mathematically, the relation between empirical formula mass and molecular formula mass is given as follows:

$\mathrm{n\, =\, \frac{Molecula\: mass}{\: empirical\: formula\: mass}}$

where n is the simplest ratio.

Limiting Reagent

The reactant is consumed first in the reaction. When we are dealing with the balanced chemical equation, if the number of moles of reactants is not in the ratio of the stoichiometric coefficient of the balanced chemical equation, then there should be one reactant that should be limiting reactant.

% yield

Sometimes, experimentally, the reaction does not undergo 100% completion because of many factors which are involved in the actual industrial processes. So in such cases, we need the concept of % yield.

It is defined as the ratio of actual moles of product(s) formed to the number of moles that should have been theoretically formed assuming 100% completion of the reaction.

$\mathrm{\% \:yield = \frac{\:Actual \:number \:of \:moles \:formed}{\:Theoretical \:moles \:that \:should \:have \:formed}}$

Stoichiometry and Stoichiometric Calculations

This concept helps us to calculate the mass or amount of reactants and products in the given chemical reaction. For calculations, first we must have a balanced chemical equation, only then we can predict the mass of reactants and products. For example
The chemical reaction is given as follows:
$\mathrm{2H_{2}\,+O_{2}\rightarrow 2H_{2}O}$
Now, this chemical equation is a balanced equation, thus we are able to predict that one mole oxygen will combine with 2 moles of hydrogen and form 2 moles of water and thus accordingly, we can calculate the masses of the respective elements.

In this concept, we also study about the "limiting reagent". It is the substance that is present in reactants in a smaller amount. In other words, limiting reagent is a substance which will be completely finished or react in the given chemical reaction.

Some important Concentration Terms

Mole fraction: It is the ratio of the moles of any substance present in the solution to the total moles of the solution. Mathematically, it is given as follows:
$\mathrm{Mole\, fraction\, of\, A=\, \frac{n_{A}}{n_{A}\, +\, n_{B}}}$
where n_A is the moles of component A and n_B is the moles of component B.
Molarity: It is the concentration of any substance present in the solution. In other words, it is moles of solute per unit volume of solution in litres. Mathematically, it can be given as follows:
$\mathrm{Molarity\, =\, \frac{Moles\, of\, solute}{Volume\, of\, solution\, in\, litre}}$
Molality: It is the moles of solute dissolved in the given amount of solvent. Mathematically, it can be represented as follows:
$\mathrm{Molality\, =\, \frac{Moles\, of\, solute}{Mass\, of\, solvent(Kg)}}$
Normality: It is defined as the number of equivalents of solute dissolved in one litre volume of solution. Mathematically, it can be represented as follows:
$\mathrm{Normality\, =\, \frac{Number\, of\, equivalents\, of\, solute}{Volume\, of\, solution\, in\, litre}}$

Related topics

How to prepare for States of matter?

This chapter is the beginning of chemistry. This chapter is one of the most important chapters of the complete chemistry syllabus. Its concepts, laws, numerical and graphs all are important both for basic foundation of chemistry and for scoring good marks in the examination.
Read this chapter carefully, as it has all the basic concepts like the mole concept, stoichiometry, molarity, normality, etc.
This chapter is the foundation stone of the whole of the chemistry syllabus.
Rest this chapter is very simple, just be regular and be consistent in your numerical practice.

Real life application based on Some Basic Concepts in Chemistry

In cooking, it is always necessary to know the exact amount of ingredients to be add in the food to get the best and tasty dish. This calculation of ingredients in stoichiometry.
Saturn planet is much larger than earth but still, it is less dense. The specific gravity of Saturn is even less than 1.
The GPS system that we use for finding out the way is related to stoichiometry. All these signals come from satellites. The stoichiometry helps to calculate the fuel and its components of reactions in the complete journey of satellites.

Prescribed Books for the States of Matter

First, you must finish the class XI NCERT textbook and solve each and every example and unsolved question given in it. Then for advanced level preparation like JEE and NEET, you must follow R.C. Mukherjee and O.P. Tandon. You must definitely solve the previous year papers. Meanwhile, in the preparation, you must continuously write the mock tests for the depth of knowledge. Our platform will help you to provide with the variety of questions for deeper knowledge with the help of videos, articles and mock tests.

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Some Solved Examples

Example 1:

A sample of KCl is placed in 50 ml of solvent. What should be the mass (in gm) of the sample for the molarity to be 2M ?

1) (correct) 7.45

2) 7.81

3) 6.81

4) 7

Solution

Number of Moles = molarity x volume

= 2 x 0.05 = 0.1

So, mass = (39 + 35.5) x 0.01g = 7.45 g

Hence, the answer is an option (1).

Example 2:

The amount (in g) of sugar $\left ( C_{12}H_{22}O_{11} \right )$ required to prepare 2L of its 0.1 M aqueous solution is:

1) 17.1

2) (correct) 68.4

3) 136.8

4) 34.2

Solution

Molarity -Molarity (M) = (Number of moles of solute)/(volume of solution in litres)

It is defined as the number of moles of the solute in 1 litre of the solution.

As we have learned in the mole concept.

The formula of molarity = $\frac{(n)_{solute}}{V_{solution} (in\ lit)}$

$0.1 = \frac{\frac{wt}{342}}{2}$

wt(C₁₂H₂₂O₁₁) = 68.4 gram

Hence, the answer is the option (2).

Example 3:

Calculate the molality of a solution containing of Acetic acid in Ethanol if the mass of solute = 10g and the density of Ethanol = 0.789 gmL^-1.

1) (correct) 0.2112

2) 0.2012

3) 0.1992

4) 0.2002

Solution

We know.

Molality (m) = (number of moles of solute)/(mass of solvent in kg)

Now,

Moles of solute Acetic acid = mass / molar mass

$\mathrm{= \frac{10}{60}\ moles}$

If Volume of ethanol = 1 L

Formula, Density = mass/volume

So,

Weight of ethanol = 1000 x 0.789 = 789g = 0.789 Kg

$\mathrm{Molality = \frac{mass\;of\;solute}{mass\;of\;solution\ in\ kg } }$

$\mathrm{Molality = \frac{10}{60\times\frac{789}{1000}} = 0.2112\ m/kg}$

Hence, the answer is (0.2112 m/kg).

Conclusion

Basic concepts of chemistry tells about various important concepts like the Mole concept and stoichiometry which play an important role in industry-based setups to find the required amount of reagents. Various chemical reactions and their limiting agents can be identified using these concepts. Students are suggested to read the prescribed books above in this article to get a good command on the entire subject.