Important Basic Concepts of Chemistry Definitions:

Stoichiometry :

Stoichiometry is a quantitative chemistry. “Calculations based on law of conservation of mass, law of constant composition and balanced chemical equations are known as stoichiometric calculations.” 

Atom:

Atom is derived from Greek word “Atomos” meaning indivisible.

“smallest particle of matter which possesses all the properties

of that matter but which cannot exist independently”

is known as an atom e.g. O, H, N, Cl, Br etc.

Fundamental Sub-atomic particles of an atom:

Fundamental sub-atomic particles of atoms are as follow :

• Electrons
• Protons
• Neutrons

Molecule:

It is defined as “smallest particle of matter which possesses all the properties of that matter and which can exist independently.” e.g. O₂, Cl₂, C₆H₁₂O₆, S₈, H₂O, He etc.

Ions:

Charge carrying species is called ion.

Types of Ions:

Cation:

Ions bearing positive charge are called cations

Anion:

Ions bearing negative charge are called anions.

Formation of cations is endothermic process while formation of anions is exothermic process. Ions are of three types.

(i) simple ion: If a single atom bears +ve or –ve charge then it is called simple ion e.g. H⁺, O^-2. (ii) compound ion: When a group of atoms bear +ve or –ve charge then it is called compound ion e.g. NH₄⁺, SO₄^-2 etc.

(iii) molecular ion: When a molecule gains or loses electrons, the resulting species is called molecular ion e.g. CH₄⁺, O₂⁺, CO⁺, N₂⁺ etc. Cationic molecular ions are more common than anionic molecular ions.

Recommended video:

Element:

A substance which is made up of only one kind of atoms is called element e.g. O₂, O₃, Cl₂, S₈, He etc. There are 92 natural elements while total elements are more than 110.

Compound:

A substance which is made up of more than one kind of atoms chemically combined together, is called compound e.g. CO₂, H₂SO₄, C₆H₁₂O₆ etc.

Isotope:

Atoms having same atomic number but different atomic masses are called isotopes. e.g. Hydrogen has three isotopes protium ₁H¹, deuterium ₁H², tritium ₁H³, Carbon has three isotopes

₆C¹², ₆C¹³, ₆C¹⁴, oxygen also has three isotopes ₈O¹⁶, ₈O¹⁷, ₈O¹⁸ etc. More than 300 isotopes occur in nature they include 40 radioactive isotopes. Often, elements with even atomic number have more number of isotopes and are abundant.

Atomic Number (Z):   

“Number of protons in the nucleus of an atoms” or “Number of electrons in an atom” is called atomic number.

Number of electrons in an atom          = Z

Number of protons in an atom            = Z

Number of neutrons in an atom          = A – Z

Atomic masses:

Atomic masses are expressed in three ways:

(i)         Mass Number (A):   

    Sum of protons and neutrons in an atom is called its mass number or nucleon number.

                          A = Protons + Neutrons

(ii)        Relative atomic mass (Ar):    

  The atomic mass of an atom as compared to mass of Carbon-12, is called relative atomic mass e.g. relative atomic mass of Hydrogen (protium i.e. ₁H¹) is 1.0078 amu.

(iii)       Average atomic mass (Ar):  

    The weight average of isotopic masses of an element is called average atomic mass of the element.   

e.g. (i)                          Relative isotopic mass             Relative abundance

            Carbon-12                   12.00000 amu                         98.89%

            Carbon-13                   13.00335 amu                         1.11%

Average atomic mass =      12.00000 x 98.89   +   13.00335 x 1.11 / 100

                                                                               =  12.01 amu        

   amu: This is a unit to measure atomic masses. It is one-twelfth of the mass of carbon-12.

                                 1 a.m.u = 1.66 x 10^-24g            or 1.66 x 10^-27kg

Relative molecular masses (Mr): Sum of relative atomic masses of atoms present in a molecule gives relative molecular masses

Chemical Formulae:

            The symbolic representation of a compound that shows elements in the compound and also the ratio of their atoms is called chemical formula.

Types of Chemical formulae

Chemical formulae are of three types:

(i)   Empirical formula or Simple formula:            

            The formula of a compound that shows elements in the compound and their simple atomic ratio is called empirical or simple formula.

(ii)   Molecular formula:      

            The formula of a compound that shows elements in the compound and actual number of their atoms is called molecular formula.

(iii)  Structural formula:       

            The formula of a compound that shows elements in the compound, actual number of their atoms and arrangement of the atoms in the molecule is called structural formula.

Important MCQ’s question answers with brief explanation:

• A species that contains one proton, one neutron and two electron is
  1. H^-1
  2. D^-1
  3. He
  4. He⁺²

Answer ) D^-1 ( Deuterium )

Brief Explanation :

Deuterium atom has one electron, one proton and one neutron. Thus D^-1 ion wills one contains one proton, one neutron and two electrons.

---

• A substance that can be separated into components only by chemical method?
  1. Air
  2. Petroleum
  3. Bleaching Powder
  4. Molten NaCl

Answer ) Molten NaCl

Brief Explanation :

Molten NaCl is a chemical compound and chemical compound can be separated into its components by chemical methods like electrolysis

---

• Atoms which are chemically same and differ only in mass-dependent properties are
  1. ₆C¹² , ₆C¹³ , ₆C¹⁴
  2. ₁H¹ , ₂He⁴ , ₆C¹²
  3.  ₁₄Si³⁰ , ₁₅P³¹ , ₁₆S³²
  4. ₁₈Ar⁴⁰ , ₁₉K⁴⁰ , ₂₀Ca⁴⁰

Answer ) ₆C¹² , ₆C¹³ , ₆C¹⁴

Brief Explanation :

₆C¹² , ₆C¹³ , ₆C¹⁴ are isotopes having the same atomic numbers but different mass numbers. Isotopes have the same chemical property due to same atomic number but different mas dependent physical properties and rates of reactions due to different mass number

---

• Elements are arranged in periodic table, in increasing order of their
  1. Valency
  2. Mass Number
  3. Oxidation State
  4. Charge on nucleus

Answer ) Charge on nucleus

Brief Explanation :

Atomic number (charge on Nucleus) is the fundamental property and elements in the periodic table have been arranged in order of their increasing atomic numbers

---

• H₂O, CH₄, HF and Na⁺¹ have the same
  1. Number of protons
  2. Number of electrons
  3. Number of neutrons
  4. Average mass

Answer ) Number of electrons

Brief Explanation :

H₂O, CH₄, HF and Na⁺¹ ions, all have the same number of electrons (10)

---

• One atomic mass unit (a.m.u) stands for
  1. One C¹² atom
  2. One H-atom
  3. 1/12^th of the mass of H-atom
  4. 1/12^th of the mass of C¹²-atom

Answer ) 1/12^th of the mass of C¹²-atom

Brief Explanation :

1 a.m.u = 1.66 x 10^-24 g

Mass of 6.023 x 10²³ C¹² atoms = 12 g

Mass of one C¹² atom = 12 x 1 /6.023 x 10²³

                                        = 1.9927 x 10^-23 g

We have one atomic mass unit is 1/12^th of the mass of C¹² atoms

1 amu    = 1 /12 x mass of C¹² atom

1 amu    = 1 /12 x 1.9927 x 10^-23 g

               = 1.66 x 10^-24g

---

• One mole of NaCl contains6.02×10²³
  1. Ions
  2. Atoms
  3. Molecules
  4. Formula Unit

Answer ) Formula Unit

Brief Explanation :

• One mole of Sodium Chloride contains 6.023 x 10²³ NaCl Formula units

---

• Mass of one mole of oxygen is
  1. 32 g
  2. 16 x N_A g
  3. 32 x N_A g
  4. 32 / N_A g

Answer ) 32 / N_A g

Brief Explanation :

Mass of N_A (6.023 x 10²³) O₂ – molecules = 32g

Mass of one O₂ – molecule = 32/N_A g

---

• Number of atoms in 64 a. m. u. of SO₂ is
  1. 3
  2. 64
  3. 3 x N_A
  4. 64 x N_A

Answer ) 3

Brief Explanation :

64 a.m.u. is the mass of one molecule of SO2 and thus contains three (3) atoms. However 64 g is the mass of one mole of SO2 molecule and contains 3 x 6.023 x 10²³atoms.

---

• A balanced chemical equation may not obey
  1. Law of conservation of mass
  2. Law of conservation of volume
  3. Law of constant composition
  4. All of these

Answer ) Law of conservation of volume

Brief Explanation :

In a balanced chemical equation, each compound has definite composition with correct chemical formula. The total mass of reactants is always equal to total mass of products in balanced equation. However, volumes of reactants and products may not be equal. Thus balanced equation obeys the law of conservation of mass and law of definite proportion but may or may not obey the law of conservation of volume.

---

• The limiting reactant in a chemical reaction is
  1. Is used up earlier and completely
  2. Controls the yield of reaction
  3. Gives the least amount of product
  4. All of these

Answer ) All of these

Brief Explanation :

A limiting reactant in a chemical reaction is always used up earlier, gives the least amount of product and controls the yield of reaction. It is present in lesser amount than the required stoichiometric amount.

---

• When equal moles of the reactant react together, the limiting reactant is

1. K₂Cr₂O₇
2. HCl
3. Cl₂
4. None

Answer ) HCl

Brief Explanation :

K₂Cr₂O₇ + 14 HCl to form 3Cl₂ + other products. Let’s one mole of each reactant takes part in a chemical reaction.

One mole of K₂Cr₂O₇ = 3 mole Cl₂

Similarly, 14 moles of HCl = 3 mole Cl₂

1 mole of HCl = 3/14 mole of Cl₂

As 1 mole of HCl produces the least amount of Cl₂ product so HCl is a limiting reactant.

---

• A yield of reaction that is independent of their reaction conditions is

---

• The yield of reaction is becomes 100% when
  1. Actual yield < Theoretical yield
  2. Actual yield > Theoretical yield
  3. Actual yield ≤ Theoretical yield
  4. Actual yield  = Theoretical yield

Answer ) Actual yield  = Theoretical yield

Brief Explanation :

The yield of reaction is  become 100% when actual yield equal to theoretical yield of reaction

---

• Actual yield is always lesser than theoretical yield due to
  1. Lack of optimum reaction conditions
  2. Mechanical loss of reactants and / or products
  3. Reversibility of reaction
  4. All of these

Answer ) All of these

Brief Explanation :

Actual yield is always lesser than theoretical yield due to one or more reason like not availability of reagent and the most suitable reaction conditions; loss of reactants / products during experimental work, reversibility of reaction of by-products formation etc.

---

---

• Mass of Cl^-1 ions in a sample of AlCl₃ that contains 6.023 x 10²³ Al⁺³ ions
  1. 3
  2. 3 x 10²¹
  3. 6.023 x 10²³ / 3
  4. 3 x 6.023 x 10²³

Answer ) 3

Brief Explanation :

From the formula AlCl₃ we have

1 Al⁺³ ions = 3Cl^-1 ion

6.023 x 10²³ Al⁺³ion = 3 x 6.023 x 10²³ Cl^-1 ion

                                         = 3 mole of Cl^-1 ion

---

• The simplest ration b/w atoms (Pb:O) of a binary compound of oxygen containing 90.66% lead ₈₂Pb²⁰⁷, is
  1. 1:1
  2. 1:2
  3. 3:4
  4. 9:1

Answer ) 3:4

Brief Explanation :

We have that

Lead (Pb) = 90.66% = 90.66/207 mol = 0.4379 mol

Oxygen (O2) = 100 – 90.66 = 9.34% = 9.34/16 mol = 0.5838 mol

Relative number of atoms of Pb = 0.4379/0.4379 = 1

Relative number of atoms of Oxygen = 0.5838/0.4379 = 1.333

To change number of atoms into whole number, multiplying both the values with 3 i.e.

Atoms of Pb = 1 x 3 = 3

Atoms of Oxygen = 1.333 x 3 = 3.999 ~ 4 Hence the simple ration between Pb and Oxygen is (3:4)

---

• When 60cm³ each of H₂ and O₂ reacts together, the theoretical decrease in volume of the reacting gasses will be
  1. 120 cm³
  2. 90 cm³
  3. 60 cm³
  4. 30 cm³

Answer ) 90 cm³

Brief Explanation :

From the balanced equation  2H₂ +O₂ to form 2H₂O ,

 We have that 2cm³ of H₂ reacts with = 1 cm³ of O₂

60 cm³ of H₂ reacts with = 60/2 = 30 cm³ of O₂

Total volume of reacting gases = 60+60= 120 cm³

Volume of gases used = 60 cm³ of H₂ + 30 cm³ of O₂ = 90 cm³

---

• Moles of fluorine required to reach with 0.6 mole of metal M to form46.8 grams of MF₂
  1. 0.12 mole
  2. 1.00 mole
  3. 1.20 mole
  4. 2.00 mole

Answer ) 1.20 mole

Brief Explanation :

• From the formula  MF₂ we have that

1 mole of metal M reacts with = 2 mole Fluorine

 0.6 mole of metal M will reacts with= 2 x 0.6 = 1.2 mole of Fluorine

Notes of all the Topics of Stoichiometry with Important Explanation in detail:

Mole and Avogadro’s Number | Important Questions

How do you calculate the percent composition of an element in a compound

Mole and Chemical Equation

Law of conservation of mass

Law of Constant Composition(law of definite proportions)

How to find Excess Reactant and Limiting Reactant

How do you calculate the yield of a reaction

---

Related Important Questions for Exams !!

Q1.What is the mass of 0.5 mole of CaCO3 ?

Moles of CaCO3 = 0.5 mol

Mass of CaCO3 =?

SOLUTION:

Molar mass of CaCO3 = 100 g/ mol

 FORMULA:

No of moles= mass / molar mass

Mass of CaCO3 = moles. X   molar mass

Mass of CaCO3 =  0.5   X100

                       = 50g CaCO3

---

Q2.Calculate the Formula mass of the following:

(i) MgSO₄ (ii) C₃H₆O (iii) C₃H₈ (iv) C₂H₅OH (v) Al₂O₃

Solution:

(i) formula mass of MgSO₄ = 24 + 32 + 16 x 4 = 24 + 32 + 64 = 120

(ii) formula mass of C₃H₆O = 12 x 3 + 1 x 6 + 16 = 36 + 6 + 16 = 58

(iii) formula mass of C₃H₈ = 12 x 3 + 1 x 8 = 36 + 8 = 44

(iv) formula mass of C₂H₅OH = 12 x 2 + 1 x 6 + 16 = 24 + 6 + 16 = 46

 (v) formula mass of Al₂O₃ = 27 x 2 + 16 x 3 = 54 + 48 = 102

  (vi) formula mass of K₂Cr₂O₇ = 39 x 2 + 52 x 2 + 16 x 7 = 78 + 104 + 112 = 294

Q3.What is the mass of one mole Carbon atom?

Ans. The mass of one mole of carbon atom is 12g.

---

Q4.How many moles are there in 60g of NaCl  ?

Solution:

Given:

Mass of NaCl= 60g

SOLUTION:

Moles of NaCl = ?

SOLUTION:

Molar mass of NaCl= 23+35.5= 58.5g/mol

 FORMULA:

No of moles= mass / molar mass

No of moles= 60 / 58.5

No of moles(n) =1.02 mol

---

---

Q5.What is formula mass?

Ans) Sum of relative atomic masses of atoms in a formula unit of a compound is called formula mass.

---

---

Q6.Calculate the number of molecules in 12g of ice?

Given:

Mass of ice (water) = 12g

Molar mass of water = 18g/mol

Solution:

Number of molecules of ice (water) = Mass of ice (water) x N_A

Molar mass of water

Number of molecules of ice (water) = 12g x 6.022 x 10²³

18g/mol

= 4.01 x 10²³ molecules

---

Q7.Define molar volume of a gas. What will be the volume of 2.5 moles of H₂ gas and 60 g of NH₃ at STP.

• Volume of one mole of a gas at a given temperature and pressure is called molar volume of the gas at the given condition e.g. molar volume of a gas at STP 22.4 dm³ while molar volume of the gas at RTP is 24dm³.

Volume of H₂ gas at STP = moles x 22.4dm³ = 2.5 x 22.4dm³ = 56dm³

Volume of NH₃ gas at STP = ?

Moles of NH₃ = mass / molar mass = 60g / 17g mol^-1 = 3.53 mol

Volume of NH₃ gas at STP = moles x 22.4 = 3.53 x 22.4 = 79.07 dm³

---

Given the equation

CH_4(g) + 2O_2(g) → CO_2(g) + 2H₂O_(g) + Heat

How can this chemical equation be read in terms of particles , moles and masses?

Ans) CH_4(g) + 2O_2(g) → CO_2(g) + 2H₂O_(g) + Heat

The equation shows 1 molecule of methane reacts with 2 molecules of oxygen to produce 1 molecule of carbon dioxide and 2 molecules of steam along with liberation of heat.           

Or        the equation shows 1 mole of methane reacts with 2 moles of oxygen to produce 1 mole of carbon dioxide and 2 moles of steam along with liberation of heat.

Or        the equation shows 1 dm³ of methane reacts with 2 dm³ of oxygen to produce 1 dm³ of carbon dioxide and 2 dm³ of steam along with liberation of heat.

Or        the equation shows 16 g of methane reacts with 64 g of oxygen to produce 44 g  of carbon dioxide and 36 g of steam along with liberation of heat.

---

Q8.Calculate the percentage composition of a compound. ( MgSO₄ )

Molar mass of MgSO₄ = 1 x 24 + 1 x 32 + 4 x 16 = 24 + 32 + 64 = 120

               % age of Mg =         Mass of Mg             x 100       =  24  x 100    = 20 %

                                           Molar mass of MgSO₄     120 

               % age of S =         Mass of S            x 100       =  32  x 100    = 26.67 %

                                           Molar mass of MgSO₄     120 

               % age of O =         Mass of O           x 100       =  64  x 100    = 53.33 %

                                           Molar mass of MgSO₄     120 

---

Q9.Aluminium sulphate hydrate [Al₂(SO₄)₃.x H₂O] contains 8.20% of Al by mass, calculate x.

Solution:

            Formula mass of Al₂(SO₄)₃.x H₂O      = 27×2 + 32×2 + 16×12 + x(1×2 + 16)

or         Formula mass of Al₂(SO₄)₃.x H₂O      = 54 + 96 + 192 + x(18)

or         Formula mass of Al₂(SO₄)₃.x H₂O      = 342 + 18x

Percentage of Al in Al₂(SO₄)₃.x H₂O = 8.20%

     But      Percentage of Al in Al₂(SO₄)₃.x H₂O =     Amount of Al in Al₂(SO₄)₃.x H₂O x 100

                                                                                     Formula mass of Al₂(SO₄)₃.x H₂O

8.20     =          54                  x 100

                                                                                     342 + 18x

Or                                            8.20 (342 + 18x)    =   5400

Or                                            2804.40 + 147.60x = 5400

Or                                            147.60x = 5400- 2804.40

Or                                             147.60x = 2595.60

Or                                            x = 2595.60 / 147.60 =   17.59   ≈ 18

Also Read !!

Stoichiometry Important mcq with answers and Explanation

Visit Digital Kemistry YouTube Channel for Free Chemistry Tutorials

Get in touch !!