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 C-12 atom
⇒ For C, 1 mole C = 12 gm = 6.023 × 1023 atoms or wt. of 6.023 × 1023 atoms = 12 gm
⇒ wt. of 1 atom of C = 12/NA gm (NA → Avogadro's number = 6.23 × 1023)
⇒ 1 amu = 1/12 wt of one C-12 atom
⇒ 1 amu = 1/12 × 12/NA gm
⇒ 1 amu = 1/NA 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Today, ‘amu’ has been replaced by ‘u’ which is known as unified mass.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, N-14 and N-15. 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 N-14 is assigned 'x' and the N-15 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, Cu-63 (62.9296 amu) and Cu-65 (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 Cu-63)
Relative Molar Mass is defined as the smallest mass unit of a compound with one-twelfth of the mass of one carbon – 12 atom.
Molecular Mass of Sugar (Sucrose)
Molecular mass C12H22O11 = 12(mass of C) + 22(mass of H) + 11(mass of O)
Molecular mass C12H22O11 = 12(12) + 22(1) + 11(16)
Molecular mass C12H22O11 = 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 liter-atm/mol-K (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
Solved Examples
Example 1. Find the weight of water present in 1.61 g of Na2SO4. 10H2O
Solution. Moles of Na2SO4.10H2O = 0.005
Moles of water = 10 × moles of Na2SO4. 10H2O = 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 / NA = 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 × 1023
Example 4. What will be the number of moles in 13.5 g of SO2Cl2?
Solution. Molecular mass of SO2Cl2 = 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 O2
1000 ml of air at STP contains = 21/100 × 1000 = 210 ml O2
∴ 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 × 1023 atoms
6.022 × 1023 atoms of silver have mass = 108g
∴ Mass of one atom of silver
(ii) 1 mole of CO2 = 44 g
(Molecular mass of CO2 = 1 x 12 + 2 x 16 = 44u)
= 6.022 × 1023 molecules
Thus, 6.022 × 1023 molecules of CO2 has mass = 44 g
∴ 1 molecule of CO2 has mass
Example 7. Calculate the number of molecules present
(i) In 34.20 grams of cane sugar (C12H22O11).
(ii) In one litre of water assuming that the density of water is 1 g/cm3.
(iii) In one drop of water having mass 0.05 g.
Solution. (i) 1 mole of C12H22O11 = 342 g
[Molecular mass of cane sugar C12H22O11]
= 12 x 12 + 22 × 1 + 11 × 16 = 342 amu
= 6.022 × 1023 molecules
Now 342 g of cane sugar contain 6.022 × 1023 molecules.
∴ 34.2 g of cane sugar will contain
(ii) 1 mole of water = 18 g = 6.022 × 1023 molecules.
Mass of 1 litre of water = Volume × density = 1000 × 1 = 1000 g
Now 18 g of water contains = 6.022 x 1023 molecules.
∴ 1000 g of water will contain
= 3.346 x 1025 molecules
(iii) 1 mole of H2O = 18 g = 6.022 × 1023 molecules.
Mass of 1 drop of water = 0.05 g
Now 18 g of H2O contain = 6.022 × 1023 molecules.
∴ 0.05 g of H2O 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 × 1023 molecules of oxygen
(iv) 9.0 grams of aluminium
(v) 1 metric ton of iron (1 metric ton = 103 kg )
(vi) 7.9 mg of Ca
(vii) 65.5 mg of carbon
Solution.
(i) 1 mole of H2SO4 = 98 g
∵ Molecular mass of H2SO4 = 2 × 1 + 32 + 4 × 16 = 98u)
Thus 98 g of H2SO4 = 1 mole of H2SO4
∴ 392 g of H2SO4
(ii) 1 mole of CO2 = 22.4 litres at STP
i.e. 22.4 litres of CO2 at STP = 1 mole
∴ 44.8 litres of CO2 at STP
= 1/22.4 × 44.8
= 2 moles CO2
(iii) 1 mole of O2 molecules = 6.022 × 1023 molecules
6.022 × 1023 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 = 103 kg = 106 g
1 mole of Fe = 56 g of Fe
∴ 106 g of Fe = 106/56 molecules
= 1.786 × 104 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 CO2 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 × 1023 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 K4[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 × 1023 CH4 molecules, 9.41 × 1023 carbon atoms, and 37.64 × 1023 hydrogen atoms.
Q.6. One litre of gas at NTP weighs 1.97g. Find the molecular mass of gas.
Ans. 44.128 g
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