MDCAT Lec#1,UHS Topic 1A,Fundamental Concepts

ATOM

The smallest particle of an element which takes part in the chemical reaction is called an atom.

May exist independently

 Examples 

Monatomic gases

    •   Helium   (He)

    •   Neon       (Ne)

    •   Argone   (Ar)

 May not exist independently

 Example

    •  Hydrogen  (H2)

    •  Oxygen       (O2)

    •  Nitrogen     (N2)

a) RELATIVE MASS

 Relative mass is the mass of a given substance scaled with carbon- 12.

C-12 is used as standard in this scale because

1. It is highly stable isotope

2. Its mass is exactly in whole numbers that is 12.000

3. It can be handled easily

Relative atomic mass

The mass of an atom of an element as compared to the mass of an atom of carbon- 12 

Atomic mass unit (amu)

The unit is used to express relative atomic mass is called atomic mass unit (a.m.u) and it is 1/12 of the mass of one atom of C -12.

          1 a.m.u   =  1.661 × 10^- 27 kg 

          1 a.m.u   =  1.661 × 10^-24 g

 Examples 

i) Chlorine

 Isotopes                             Cl-35            Cl-37

Relative abundance          75%             25% i) Relative isotopic mass 17Cl-35 = 35 amu ii) Relative isotopic mass 17Cl-37 = 37 amu

iii) Relative atomic mass = 

Relative isotopic mass × R A + relative isotopic mass × RA/100

iv) Relative atomic mass of Cl = 35 × 75 + 37 × 25 /100 

= 26.25 + 9.25 = 35.5 amu

ii) Neon

Isotopes      10Ne-20       10Ne-21     10Ne-22

Relative 

abundance  90.92%        0.26%          8.82%

Relative atomic mass of Neon

= 20 × 90.92 + 21 × 0.26 + 22 × 8.82 /100 

= 20.18 amu

Relative isotopic mass

 The mass of an isotope of an element as compared to the mass of an atom of carbon - 12

 Examples

i)  Relative isotopic mass of chlorine-35 is 35 amu

 ii) Relative isotopic mass of chlorine-37 is 37 amu 

Relative molecular mass

 The mass of a molecule as compared to the mass of an atom of carbon-12 Examples

i) Relative molecular mass of water is 18 amu

ii) Relative molecular mass of carbon dioxide is 44 amu

Relative formula mass

 The mass of a formula unit as compared to the mass of an atom of carbon-12 Examples 

i) Relative formula mass of NaCl is 58.5 amu

ii) Relative formula mass of Na2SO4 is 142 amu

b) CONCEPT OF MOLE

When relative mass of any substance represent in grams (g) is called one mole of substance or molar mass.

• The mole is related to mass,volume and number of particles of the given substance.

Examples:

There are four different chemical substances can be represent in moles.

i) Atom

Relative mass of Na atom = 23 amu

When it represent in grams it becomes

23 amu ----> 23 grams = One Mole or simply Mole = Molar atomic mass or Molar mass = One gram mole

ii) Molecule

Relative mass of H2O molecule = 18 amu

When it represent in grams it becomes

18 amu ----> 18 grams = One Mole or simply Mole = Molar molecular mass or Molar mass = One gram mole

iii) Formula Unit

Relative formula mass of NaCl = 58.5 amu

When it represent in grams it becomes

58.5 amu ----> 58.5 grams = One Mole or simply Mole = Formula unit mass or Molar mass = One gram mole

iv) Ion

Relative mass of H+ ion = 1.008 amu

When it represent in grams it becomes

1.008 amu ----> 1.008 grams = One Mole or simply Mole = Molar ionic mass or Molar mass = One gram mole

We know according to Avogadro's 

One mole of any substance = NA = 6.02 × 10^23 particles (atoms or molecules or formula units or ions)

In case of ideal gases one mole of ideal gase at STP (standard temperature and pressure) =Molar volume (Vm) = 22.414 dm3 = 6.02 × 10^23 atoms or molecules

Interconversion of different Quantities into Mole

i. Mass into Moles and vice versa

Moles (n) = Mass (m) / Molar mass (M)

ii. Number of particles into Moles

Moles(n) = Number of particles(N)/NA

iii. Volume of ideal gas into Moles

Moles (n) = Vol.(V)/Molar Vol.(Vm)

Relationship between different formulae

• m / Mr = N / NA
• m / Mr = V / Vm
• N / NA  = V / Vm

Multiple Choice Questions

Q.1  The 4.4 g CO2 occupies ___________ volume at STP

A)  22.414 dm3

B)  11.207 dm3

C)  2.2414 dm3

D)  4.4000 dm3

Q.2 How many total number of atoms and electrons are present in 1.8 g of NH4- ions

A) 3.01 × 10^23 and 1.2 NA

B) 6.01 × 10^22 and 12 NA

C) 1.20 × 10^23 and 1.1 NA

D) 6.01 × 10^23 and 11

Q.3 The total number of molecules are present in 0.5 mole N2 gas?

A) 6.02 × 10^23

B) 3.01 × 10^23

C) 1.2 × 10^24

D) 1.5 × 10^ 23

Q.4 How many number of atoms are present in 0.5 mole of nitrogen gas?

A) 6.02 × 10^23

B) 3.01 × 10^23

C) 1.2 × 10^24

D) 1.5 × 10^ 23

Q.5 Thee mass of 1/2 mole of nitrogen gas is

A) 7 g

B) 14 g

C) 28 g

D) 56 g

Q.6 The largest number of moles are present in

A) 3.6g H2O

B) 2.8g CO

C) 4.8g C2H5OH

D) 5.4g N2O5

Q.7 Gram atoms of He in 8.0 g He gas

A) 1 gram

B) 2 gram

C) 4 gram

D) 8 gram

Q.8 22414 cm3 volume of Hydrogen gas at STP contains number of hydrogen atoms

A) 1 NA

B) 1.5 NA

C) 2 NA

D) 4 NA

Q.9 All of the followings are relative isotopic mass of Ne except

A) 20 amu

B) 21 amu

C) 22 amu 

D) 20.18 amu

Q.10  1 a.m.u is equal to

A) 1.661 × 10^-27 g

B) 1.661 × 10^ -24 g

C) 1.661 × 10^ -19 g

D) 1.661 × 10^ -30 g 

Answers:

1. C

2. A

3. B

4. A

5. B

6. A

7. B

8. C

9. D

10. B

Explanation:

1.

Use formula:

m / Mr = V / Vm

Or

V = m / Mr × Vm

V = 4.4 g / 44 g × 22.414 dm3

V = 0.1 × 22.414 = 2.2414 dm3

2.

Use formula for calculating total no.of atoms

N / NA = m / Mr × atomicity

Or

N = m / Mr × NA × 5

N = 1.8 g / 18 gmol-1 × 6.02 × 10^23 × 5

N = 0.1 × 6.02 × 10^23 × 5

N = 3.01 × 10^23 (Total atoms)

Use formula for calculating total no.of atoms

N / NA = m / Mr × No.of es- in one ion

Or

N = m / Mr × NA × 12

N = 1.8 g / 18 gmol-1 × 6.02 × 10^23 × 12

N = 0.1 × 6.02 × 10^23 × 12

N = 1.2 ×  6.02 × 10^23 or 1.2 NA (Total electrons)

Note: Explanation of remaining MCQs uploaded soon...