Mole Concept - Some Basic Concepts - Physical Chemistry - Chemistry - LOGiota

This concept will clear the relationship between a mole, molar mass and the number of atoms.

By Sakshi Goel | 28 Oct'18 | 2 K Views |

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Theory

A mole is defined as the amount of substance that contains as many atoms as there are atoms present in exacyly 12 g of carbon.
1 mole = 6.023 $Ã—$ 10²³ particles
6.023 $Ã—$ 10²³is called the Avogadro constant or Avogadro number.
One mole of anything is 6.023 $Ã—$ 10²³units of that thing.
For example: one mole of oxygen atoms = 6.023 $Ã—$ 10²³atoms
one mole of oxygen molecules = 6.023 $Ã—$ 10²³molecules
one mole of sodium ions = 6.023 $Ã—$ 10²³ions

Converting between number of moles and number of atoms:

1 mol = 6.023 $Ã—$ 10²³atoms
So, if we have 3.4 mol of Fe and we want to calculate number of Fe atoms in it. Avogadro number will be used as a conversion factor.
1 mol Fe = 6.023 $Ã—$ 10²³Fe atoms
$3.4 m o l F e = \frac{6.023 Ã— 10^{23} F e a t o m s}{1 m o l F e} Ã— 3.4 m o l F e = 20.48 Ã— 10^{23} F e a t o m s$

Converting between mass and number of moles:

The mass of one mole of a substance in grams is called molar mass.
For example: 1 mol of C = 12.01 g C
Suppose we want to calculate number of moles of aluminium in 2.50 g of aluminium sheet, given the molar mass of Al is 26.98 g/mol.
26.98 g of Al = 1 mol of Al
$2.50 g o f A l = \frac{1 m o l o f A l}{26.98 g o f A l} Ã— 2.50 g o f A l = 0.09 m o l o f A l$

Converting between mass and number of atoms:

For this, we need to use both the above concepts.
1 mol of C = 6.023 $Ã—$ 10²³C atoms = 12.01 g C
For example: Calculate number of atoms in 3 g of Cu. Molar mass of Cu = 63.55 g/mol
63.55 g of Cu = 1 mol
$3 g o f C u = \frac{1 m o l o f C u}{63.55 g o f C u} Ã— 3 g o f C u = 0.047 m o l o f C u$
1 mol of Cu = 6.023 $Ã—$ 10²³Cu atoms
$0.047 m o l o f C u = \frac{6.023 Ã— 10^{23} C u a t o m s}{1 m o l o f C u} Ã— 0.047 m o l o f C u = 2.8 Ã— 10^{22} C u a t o m s$

MCQ

Problem:
The maximum number of molecules is present in
(1) 15 L of H₂ gas at STP
(2) 5 L of N₂ gas at STP
(3) 0.5 g of H₂ gas
(4) 10 g of O₂ gas

Solution:

For each of the options, we will calculate the number of moles.

At STP, 22.4 L = 1 mol

Moles of H₂(g) = $\frac{15}{22.4}$ = 0.67 mol

moles of N₂ (g)= $\frac{5}{22.4}$ = 0.223 mol
Moles of 0.5 g H₂ gas = $\frac{0.5}{2}$ = 0.25 mol

moles of 10 g O₂ gas = $\frac{10}{32}$ = 0.1325 mol

Maximum number of moles = maximum number of molecules = option (a)

Problem:
Which has maximum molecules?
(a) 7 g N₂O
(b) 20 g H₂
(c) 16 g NO₂
(d) 16 g SO₂

Solution:

Maximum number of moles have maximum number of molecules.
Moles of N₂O = $\frac{7}{44}$ = 0.16 mol

Moles of H₂ = $\frac{20}{2}$ = 10 mol

Moles of NO₂ = $\frac{16}{46}$ = 0.35 mol
Moles of SO₂ = $\frac{16}{64}$ = 0.25 mol

H₂ has maximum number moles and thus maximum number of molecules. Correct option is (b).

Problem:
The total number of electrons in 1.6 g of CH₄ to that in 1.8 g of H₂O
(a) Double
(b) Same
(c) Triple
(d) One fourth

Solution:

Avogadro number = N_o

_{Total number of e}^__{in CH₄ = 10}

Number of e^â€“ in 1.6 g of CH₄ = $\frac{1.6}{16} Ã— 10 Ã— N_{o}$ = N_o

_{Total number of e}^__{in H₂O = 10}

Number of e^â€“ in 1.8 g of H₂O = $\frac{1.8}{18} Ã— 10 Ã— N_{o}$ = N_o

So, correct option is (b).

Problem:
Which has the maximum number of molecules among the following?
(a) 8 g H₂
(b) 64 g SO₂
(c) 44 g CO₂
(d) 48 g O₃

Solution:

Maximum number of moles have maximum number of molecules
So, we will calculate number of moles for each of these.
8 g H₂ = $\frac{8}{2}$ = 4 moles

44 g of CO₂= $\frac{44}{44}$ = 1 mole
64 g SO₂ = $\frac{64}{64}$ = 1 mole

48 g of O₃ = $\frac{48}{48}$ = 1 mole

Option (a) is correct

Problem:
Weight of 1 x 10²² molecules of CuSO₄.5H₂O is
(a) 41.4 g
(b) 4.14 g
(c) 414.9 g
(d) 24.6 g

Solution:

molar mass of CuSO₄.5H₂O = 249.5 g/mol
1 mole = 6.023 x 10²³ molecules
Mass of 6.023 x 10²³ molecules = 249.5 g
mass of 1 x 10²² molecules = $249.5 Ã— \frac{1 Ã— 10^{22}}{6.023 Ã— 10^{23}} = 4.14 g$
Correct option is (b).

Problem:
Which has the maximum number of atoms?
(a) 10.8 g of Ag
(b) 2.4 g of C
(c) 5.6 g of Fe
(d) 54 g of Al

Solution:

The best way is to convert all the options into the moles.
Option (a) 10.8 g of Ag: 108 g of Ag = 1 mole
10.8 g of Ag = 10.8/108 = 0.1 mol
Option (b) 2.4 of C: 12 g of C = 1 mole
2.4 g of C = 2.4/12 = 0.2 mol
Option (c) 5.6 g of Fe: 56 g of Fe = 1 mole
5.6 g of Fe = 5.6/56 = 0.1 mol
Option (d) 54 g of Al: 27 g of Al = 1 mole
54 g of Al = 54/27 = 2.0 mol
More is the number of moles, more are the number of atoms.
So, correct option is (d).

Problem:
Number of electrons present in 1.6 g of methane is
(a) 6.02 x 10²⁴
(b) 6.02 x 10²³
(c) 6.02 x 10²²
(d) 6.02 x 10²¹

Solution:

1 mole of methane = 16 g = 6.02 x 10²³molecules of CH₄
1.6 g = (1.6/16) x 6.02 x 10²³ = 6.02 x 10²²molecules of CH₄
C has 6 electrons and H has 1 electron.
In total, 1 molecule of CH₄ has 10 electrons.
So, 6.02 x 10²²molecules of CH₄will have 6.02 x 10²² x 10electrons = 6.02 x 10²³ electrons
Correct option is (b).

Problem:
Which of the following possess highest mass?
(a) 0.2 mol of CO₂ gas
(b) 2.24 L SO₂
(c) 3.01 x 10²³molecules of water
(d) 20 g calcium

Solution:

For option (a),
1 mole of CO₂ = 44 g
then, 0.2 mol of CO₂ = 0.2 x 44 = 8.8 g
For option (b),
22.4 L of SO₂ = 1 mole = 64 g
2.24 L of SO₂ = (2.24/22.4) x 64 = 6.4 g
For option (c),
6.02 x 10²³molecules of water = 1 mole = 18 g
3.01 x 10²³molecules of water = (3.01/6.02) x 18 = 9 g
For option (d),
mass given = 20 g calcium
So, correct option is (d).

Problem:
Volume of oxygen at NTP, required to completely burn 1 kg of coal (100 % carbon) is
(a) 22.4 L
(b) 1.86 x 10³ L
(c) 22.4 x 10³ L
(d) 1000 L

Solution:

The equation for reaction of coal with oxygen during burning is
$C + O_{2} â†’ {CO}_{2}$
1 mole of C reacts with 1 mole of O₂.
Mass of coal = 1 kg = 1000 g
Molar mass of C = 12 g/mol
Moles of C = mass/molar mass = 1000/12 = 83.3 mol
83.3 moles of carbon will react with 83.3 moles of oxygen.
At NTP, 1 mol = 22.4 L
83.3 moles = 22.4 x 83.3 = 1866 L = 1.86 x 10³ L
Option (b) is correct.

Problem:
The density of a solution of 3.69 g KI in 21.70 g H₂O is 1.11 g/mL. Calculate the % w/v of KI in the solution.
(a) 14.5 % (w/v)
(b) 16.1 % (w/v)
(c) 18.9 % (w/v)
(d) 20.8 % (w/v)

Solution:

Mass of KI = 3.69 g
Mass of water = 21.70 g
Total mass of solution = 3.69+21.70 = 25.39 g
Density of solution = 1.11 g/ml
Total volume of solution = mass of solution/density = 25.39/1.11 = 22.87 ml
So, 22.87 ml of solution contains 3.69 g of KI
100 ml of solution will contain mass of KI = 3.69 x 100/22.87 = 16.13 g
So, % w/v of KI is 16.1 % (w/v)
Option (b) is correct.

Short Answer

Problem:
Haemoglobin contains 0.334% of iron by weight. The molecular weight of haemoglobin is approximately 67200.
Calculate the number of iron atoms (Atomic weight of Fe is 56) present in one molecule of haemoglobin.

Solution:

Weight of Fe in haemoglobin = $\frac{0.334}{100} Ã— 67200$ = 224.48 g
Mass of one Fe atom = 56 g
Total number of Fe atom = $\frac{224.48}{56}$ = 4

Problem:
Calculate the number of atoms in 0.1 mol of a tetraatomic gas.

Solution:

Tetraatomic means 4 atoms in one molecule.

So, total number of atoms = 0.1 Ã— 4 Ã— 6.022 Ã— 10²³ = 2.4 Ã— 10²³ atoms

Problem:
Calculate the number of molecules in 4.25 g of NH₃.

Solution:

Moles of NH₃ = $\frac{4.25}{17}$ = 0.25 moles
Number of molecules = 0.25 Ã— 6.022 Ã— 10²³ molecule = 1.50 Ã— 10²³ molecules

Problem:
What is the volume occupied by one molecule of water (density = 1 g cm^-³)?

Solution:

As water is liquid its density = 1 g/mL
i.e., 1 g of H₂O have volume = 1 mL
Mass of one molecule = $\frac{18}{6.02 Ã— 10^{23}} g$
$\frac{18}{6.02 Ã— 10^{23}} g$ of H₂O have volume = $\frac{18}{6.02 Ã— 10^{23}}$ mL = 3.0 Ã— 10^-²³ mL

Problem:
What volume of oxygen gas (O₂) measured at 0Â°C and 1 atm, is needed to burn completely 1 L of propane gas (C₃H₈) measured under the same conditions?

Solution:

$C_{3} H_{8} + 5 O_{2} â†’ 3 {CO}_{2} + 4 H_{2} O$

For 1 mol propane, 5 mol O₂ gas is needed.
22.4 L propane = 5 Ã— 22.4 L of O₂ gas needed
So, 1 L propane = 5 L of O₂ gas is required

Problem:
A sample of ammonium phosphate (NH₄)₃PO₄ contains 3.18 moles of hydrogen atoms. Calculate the number of moles of oxygen atoms in the sample?

Solution:

In (NH₄)₃PO₄, moles of 'H' are present with 4 moles of oxygen atom.

So, 3.18 moles of 'H' are present with = $\frac{4}{12} Ã— 3.18$

= 1.06 moles of oxygen atom

Long Answer

Problem:
4 g of hydrogen reacts with 20 g of oxygen to form water. What is the mass of water formed?

Solution:

$2 H_{2} + O_{2} â†’ 2 H_{2} O$
4 g 32 g 36 g
When 4 g of H₂ reacts with 32 g of O₂ gives 36 g of H₂O.
Now present oxygen is 20 g
So, O₂ will be the limiting reagent and H₂O will be calculated from O₂
32 g of O₂ given = 36 g of H₂O
20 g of O₂ given = $\frac{36}{32} Ã— 20$ = 22.5 g H₂O

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Physical Chemistry molar mass Avogadro number

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