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Chemical Reactions - Part V |
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5.
Concepts of mole and calculations The
amount of substance in grams equal to its atomic mass in grams is called the
gram atomic mass. For example, atomic mass of Na is 23, then gram atomic
mass of Na is 23 gm. Atomic mass of O is 16, gram atomic mass of O is 16 gm.
Gram atomic mass is also known as the gram atomic weight of a
substance.
It has been discovered by Avogadro, that gram atomic mass of substance contains 6.23 x 1023 atoms. This means that if you take 23 gm of Na, there will be 6.23 x 1023 atoms of Na. Similarly, if you take 16 gm of O, you will get 6.23 x 1023 atoms of O. In earlier chapters, we have studied that as the mass of atoms increase, their sizes also increases. Thus Avogadro's number 6.23 x 1023 atoms remains constant for different atomic masses. The mass of substance that contains 6.23 x 1023 atoms, is also known as the mole mass. Thus 1 mole of Na will be 23 gm of Na, 1 mole of O will be 16 gm of O. The amount of substance in grams equal to the molecular mass in grams
is called the gram molecular mass. For example, molecular mass of NaCl is
60. Thus gram molecular mass of NaCl = 60 gm. Molecular mass of O2
is 32. The gram molecular mass of O2 is 32 gm. Molecular mass of
water is 18, gram molecular mass of water is 18 gm. As discussed above, 1 mole of NaCl will contain 6.23 x 1023 molecules
of NaCl. Similarly 1 mole of O2 molecule will contain 6.23 x 1023
molecules of O2. Example 1 :What
is the mass in grams of 5 moles of Fe ? The atomic mass of Fe = 56gm 5 moles of Fe = 5 x 56 gm = 280 gm Example 2
: Which has more atoms: 10 gm of
Ag (silver) or 10 gm of Au (gold) The atomic mass of Ag = 108 gm 1 mole of Ag = 108 gm = 6.23 x 1023 atoms
of Ag 1 mole of Au = 197 gm of Au - 6.23 x 1023 atoms
of Au Thus 10 gm of Ag will have more number of atoms than
10 gm of Au. Example 3
:
How many moles of NaCl are present in 20 gm of the substance? 1 mole of NaCl = molecular mass of NaCl in grams = 60
gm 60 gm of NaCl = 1 mole Thus
20 gm of NaCl will have
0.33
mole of NaCl. Example 4
: Calculate the mass of 0.5 mole of methane (CH4). How many
molecules does the 0.5 mole of methane contain? Molecular mass of
CH4
= 16 gm 1
mole of CH4 = 16 gm
= 6.23 x 1023 molecules of CH4 Thus
0.5 mole of CH4 will
weigh 8 gm and it will have 3.12 x 1023 molecules of CH4. Example 5
: In which of the following, the
number of hydrogen atoms is more a) 3 moles of H2O or b) 10 moles
of HCl ? 1 mole of H2O molecules has 2 moles of H
atoms. Thus
3 moles of H2O molecules will have 6 moles of H atoms. 1 mole of HCl molecules will have 1 mole of H atoms. Thus
10 moles of HCL will have 10 moles of H atoms. Thus 10 moles of HCL will have more atoms than 3
moles of H2O. Example 6
: How many grams of O is present in 50 gm of CaCO3? Molecular mass of CaCO3
is 100 gm. This has 48 gm of O. Thus
50 gm of CaCO3 will have 24 gm of O. Example 7
: Consider the following reaction How
many moles of KClO3 are needed to release 0.4 moles of O2 ? 2
moles of KClO3 release
3 moles of O2. To
have 0.4 moles of O2 released, the amount of KClO3 needed
is 0.6 moles. Example
8: Consider the equation
CaCO3
+ 2HNO3 How
many moles of CO2 will
be released from 6 moles of HNO3 acid ? 2
moles of HNO3 gives off 1 mole of CO2 gas. Thus
6 moles of HNO3 will release 3 moles of CO2 gas. Example
9 : A gaseous compound of nitrogen and hydrogen contains 75 %
nitrogen by mass. Find the empirical formula of the compound. (Atomic masses
N = 14, H = 1) Nitrogen
(%) = 75 % (by mass) In
terms of moles, number of moles of N = 75/14 = 5.35 moles The
number of moles of H = 25/1 = 25 moles Thus
in 100 gms of the compound, the ratio of N : H is 5.35 : 25. This comes out
to be 1 : 4.7. This
shows that for every N atom there are 4.7 H atoms. So the empirical formula
for the compound comes out to be NH4.7. Example
10 :
What is the chemical formula of ferric oxide which contains 65 % Fe and 35 %
O? (Atomic masses of Fe = 56, O=16) The number of moles of Fe = 65/56 = 1.1 moles The
number of moles of O = 35/16 = 2.18 moles The
ration of the atoms will give the empirical formula of the Fe-O compound to
be FeO2. Similarly we can calculate the molecular formula of a
compound, percentage of composition in a compound, etc. It is important to
remember that the calculations become very convenient when we consider
moles. |
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