Table of contents  
Atom and Atomic Weight  
Average Atomic Mass  
The Molecule & Molecular Mass  
Molar Mass  
Formula Mass  
Mole Concept  
Methods of Calculations of Mole  
Density 
Structure of Atom
Note: This picture is just for explanatory purpose , the structure of an atom is much more complex and includes, shells, subshells, orbitals and more.
AMU versus GRAMS
Relationship between gram and amu:
⇒ 1 amu = 1/12 wt of one C12 atom
⇒ For C, 1 mole C = 12 gm = 6.023 × 10^{23} atoms or wt. of 6.023 × 10^{23} atoms = 12 gm
⇒ wt. of 1 atom of C = 12/N_{A} gm (NA → Avogadro's number = 6.23 × 10^{23})
⇒ 1 amu = 1/12 wt of one C12 atom
⇒ 1 amu = 1/12 × 12/N_{A }gm
⇒ 1 amu = 1/N_{A} g
So, 1 amu = 1.66 × 10^{24} g
1 amu = 1.66 × 10^{27} kg
Today, ‘amu’ has been replaced by ‘u’ which is known as unified mass.
When we use atomic masses of elements in calculations, we actually use average atomic masses of elements which are explained below.
Mass of an atom of hydrogen = 1.6736 × 10^{}^{24} g
Thus, in terms of amu, the mass of a hydrogen atom is:
Example 1. Use the date given in the following table to calculate the molar mass of naturally occurring argon.
Solution. Molar mass of Ar = 35.96755 × 0.071 + 37.96272 × 0.163 + 39.96924 × 0.766
= 39.352 g mol^{−1}
Example 2. Carbon occurs in nature as a mixture of carbon 12 and carbon 13. The average atomic mass of carbon is 12.011. What is the percentage abundance of carbon 12 in nature?
Solution. Average atomic mass is:
1201.1 = 12x + 1300 – 13x
x = 1300 – 1201.1 = 98.9%
Example 3. Nitrogen is made up of two isotopes, N14 and N15. Given nitrogen's atomic weight of 14.007, what is the percent abundance of each isotope?
Solution. (14.003074) (x) + (15.000108) (1  x) = 14.007
Notice that the abundance of N14 is assigned 'x' and the N15 is 'one minus x.' This is the "trick" refered to above. The two abundances always add up to one (or, if you prefer, 100%)
For example, I might have this:
(14) (x) + (15) (1  x) = 14.007
14x + 15  15x = 14.007
x = 15  14.007 = 0.993 & 1  x = 0.007
Example 4. Copper is made up of two isotopes, Cu63 (62.9296 amu) and Cu65 (64.9278 amu). Given copper's atomic weight of 63.546, what is the percent abundance of each isotope?
Solution. (62.9296) (x) + (64.9278) (1  x) = 63.546
Once again, notice that 'x' and 'one minus x' add up to one.
Now solve for x:
x = 0.6915 (the decimal abundance for Cu63)
Relative Molar Mass is defined as the smallest mass unit of a compound with onetwelfth of the mass of one carbon – 12 atom.
Molecular Mass of Sugar (Sucrose)
Molecular mass C_{12}H_{22}O_{11} = 12(mass of C) + 22(mass of H) + 11(mass of O)
Molecular mass C_{12}H_{22}O_{11} = 12(12) + 22(1) + 11(16)
Molecular mass C_{12}H_{22}O_{11} = 342g/mol
One can use the following methods to find the Molecular weight:
(a) Vapour Density Method
Molecular mass = 2 × V.D.
(b) Diffusion Method: According to Graham’s law of diffusion, rate of diffusion of a gas is inversely proportional to the square root of its molecular mass.
(c) Victor Meyer Method: This method applies to volatile organic liquids. Suppose vapour of an organic liquid having mass W g occupies a volume of V mL at STP. Then its molecular mass is:
(a) If no. of some species is given, then no. of moles is:
(b) If the weight of a given species is given, then no of moles is:
or
(c) If the volume of a gas is given along with its temperature (T) and pressure (P)
use:
where R = 0.0821 literatm/molK (when P is in atmosphere and V is in litre.)
Note:
1 mole of any gas at STP (0°C & 1 bar) occupies 22.7 litres.
1 mole of any gas at STP (0°C & 1 atm) occupies 22.4 litres.
A mole of any substance is related to:
Density is of two types:
(a) Absolute density
(b) Relative density
Example 1. Find the weight of water present in 1.61 g of Na_{2}SO_{4}. 10H_{2}O
Solution. Moles of Na_{2}SO_{4}.10H_{2}O = 0.005
Moles of water = 10 × moles of Na_{2}SO_{4}. 10H_{2}O = 10 × 0.05 = 0.05
wt. of water = 0.5 × 18 = 0.9 gm
Example 2. A molecule of a compound has 13 carbon atoms, 12 hydrogen atom, 3 oxygen atoms and 3.02 × 10^{23} gm of other element. Find the molecular wt. of compound.
Solution. Wt. of the 1 molecule of a compound = 13 × 12 × 3 × 3.02 × 10^{23}
= 234.18 / N_{A} = 234 amu.
Example 3. Calculate the number of moles of electron present in 1 kg of it. (me = 9.1 × 10^{–31} kg)
Solution. Number of electrons in 1 kg = 1 kg/9.1 × 10^{31} kg
Number of moles = Number of particles/6.023 × 10^{23}
Example 4. What will be the number of moles in 13.5 g of SO_{2}Cl_{2}?
Solution. Molecular mass of SO_{2}Cl_{2} = 32 + 32 + 71 = 135 g mol^{–1}
Example 5. What will be the number of moles of oxygen in one litre of air containing 21% oxygen by volume under STP conditions?
Solution. 100 ml of air at STP contains 21 ml of O_{2}
1000 ml of air at STP contains = 21/100 × 1000 = 210 ml O_{2}
∴ No. of moles = Volume in ml under STP conditions/22400 litre
= 210/22400 = 9.38 × 10^{3 }mole.
Example 6. Calculate the mass of (i) an atom of silver (ii) a molecule of carbon dioxide.
Solution. (i) 1 mole of Ag atoms = 108 g
(∴ Atomic mass of silver = 108u)
= 6.022 × 10^{23} atoms
6.022 × 10^{23} atoms of silver have mass = 108g
∴ Mass of one atom of silver
(ii) 1 mole of CO_{2} = 44 g
(Molecular mass of CO_{2} = 1 x 12 + 2 x 16 = 44u)
= 6.022 × 10^{23} molecules
Thus, 6.022 × 10^{23} molecules of CO_{2} has mass = 44 g
∴ 1 molecule of CO_{2} has mass
Example 7. Calculate the number of molecules present
(i) In 34.20 grams of cane sugar (C_{12}H_{22}O_{11}).
(ii) In one litre of water assuming that the density of water is 1 g/cm^{3}.
(iii) In one drop of water having mass 0.05 g.
Solution. (i) 1 mole of C_{12}H_{22}O_{11} = 342 g
[Molecular mass of cane sugar C_{12}H_{22}O_{11}]
= 12 x 12 + 22 × 1 + 11 × 16 = 342 amu
= 6.022 × 10^{23} molecules
Now 342 g of cane sugar contain 6.022 × 10^{23} molecules.
∴ 34.2 g of cane sugar will contain
(ii) 1 mole of water = 18 g = 6.022 × 10^{23} molecules.
Mass of 1 litre of water = Volume × density = 1000 × 1 = 1000 g
Now 18 g of water contains = 6.022 x 10^{23} molecules.
∴ 1000 g of water will contain
= 3.346 x 10^{25} molecules
(iii) 1 mole of H_{2}O = 18 g = 6.022 × 10^{23} molecules.
Mass of 1 drop of water = 0.05 g
Now 18 g of H_{2}O contain = 6.022 × 10^{23} molecules.
∴ 0.05 g of H_{2}O will contain =
Example 8. Calculate the number of moles in each of the following:
(i) 392 grams of sulphuric acid
(ii) 44.8 litres of carbon dioxide at STP
(iii) 6.022 × 10^{23} molecules of oxygen
(iv) 9.0 grams of aluminium
(v) 1 metric ton of iron (1 metric ton = 10^{3} kg )
(vi) 7.9 mg of Ca
(vii) 65.5 mg of carbon
Solution.
(i) 1 mole of H_{2}SO_{4} = 98 g
∵ Molecular mass of H_{2}SO_{4} = 2 × 1 + 32 + 4 × 16 = 98u)
Thus 98 g of H_{2}SO_{4} = 1 mole of H_{2}SO_{4}
∴ 392 g of H_{2}SO_{4}
(ii) 1 mole of CO_{2} = 22.4 litres at STP
i.e. 22.4 litres of CO_{2} at STP = 1 mole
∴ 44.8 litres of CO_{2} at STP
= 1/22.4 × 44.8
= 2 moles CO_{2}
(iii) 1 mole of O_{2} molecules = 6.022 × 10^{23} molecules
6.022 × 10^{23} molecules = 1 mole of oxygen molecules
(iv) 1 mole of Al = 27 g of Al
(∴ Atomic mass of aluminium = 27 g)
i.e. 27 g of aluminium = 1 mole of Al
(v) 1 metric ton of Fe = 10^{3} kg = 10^{6} g
1 mole of Fe = 56 g of Fe
∴ 10^{6} g of Fe = 10^{6}/56 molecules
= 1.786 × 10^{4} moles
(vi) 7.9 mg of Ca = 7.9 × 10^{−3} g of Ca
(vii) 65.5 µg of C = 65.5 10^{6} g of C
Q.1. Find the weight of 6023 molecules of CO_{2} in grams.
Ans. 4.4 × 10^{–19} gm
Q.2. Calculate the total number of electrons present in 1.6 g of methane. (Molecular wt. of methane = 16 a.m.u.)
Ans. 6.023 × 10^{23} electrons
Q.3. 0.24 gm of a volatile substance upon vaporization gives 45 ml vapours at STP. What will be the vapour density of the substance?
Ans. 59.73
Q.4. What is the mass of carbon present in 0.5 moles of K_{4}[Fe(CN)_{6}]?
Ans. 36 gm
Q.5. Calculate how many methane molecules and how many hydrogen and carbon atoms are there in 25.0 g of methane.
Ans. 9.41 × 10^{23} CH_{4} molecules, 9.41 × 10^{23} carbon atoms, and 37.64 × 10^{23} hydrogen atoms.
Q.6. One litre of gas at NTP weighs 1.97g. Find the molecular mass of gas.
Ans. 44.128 g
172 videos306 docs152 tests

1. What is the difference between atomic weight and average atomic mass? 
2. How is molar mass different from molecular mass? 
3. What is the formula mass and how is it calculated? 
4. How is the concept of mole used in chemistry calculations? 
5. What are some common methods used to calculate the number of moles in a sample? 

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