So, in the last lecture if you remember we had started off with discussion of mass
percentage composition in empirical and molecular formula and I had discussed that how do
you determine the mass percentage in various elements in a compound. To do that I had
told that we will take 1 mole of the compound and we determined in that how many moles
are there of the individual elements, and therefore we will see corresponding to them how
many masses are there. Like in glucose we saw 72 gram carbon, 12 gram of hydrogen and 96
grams of oxygen, the sum comes out to be 180 grams which is molar mass of glucose. So,
what is the percentage of carbon? 72 gram divided by 180 gram into 100 which gives you the
percentage as 40%. Overall if we see by generalising it, we had written the formula, mass
percentage is mass of an element in 1 mole of a compound divided by the mass of 1 mole of
compound that is its molar mass into 100, correct. So this we have done and we also have
done some examples for glucose that, sorry. We defined Empirical formula and molecular
formula first. What is empirical formula? It represents the atoms in a molecule in their
simplest whole number ratio. So, what is the ratio of glucose simplest whole number? CH2O
and if we see that of acetic acid, the ratio is same, CH2O, for it the molar mass is 180 gram
and for this the molar mass is 60 gram. We calculate for the empirical formula mass CH2O.
The empirical formula mass comes out as 30 gram. If I divide 180 gram with 30 gram what
will be the answer, 6. And I am telling if I multiply the empirical formula CH2O with 6, do I
get the molecular formula. So, we see a relationship, we will define n factor which is the
molar mass divided by empirical formula mass and what we will do, with the help of this we
will get the molecular formula, okay. This is what we do. Now, we have not seen how to get
the empirical formula, that is the process that we have to start today. But this is basically the
process. I have also told you that many times the data of molecular weight will not be given
directly, you will be given the data of vapour density. So, do you remember the relationship
of vapour density and molecular weight? What is vapour density? So, you can calculate the
molecular weight as two times vapour density, correct. Okay, after that we had discussed
what is, we had done some important points about reaction or stoichiometric calculation
that involve gases. We had told that 1 mole of a gas occupies 4 litres at STP. In this we have
make assumption that temperature is zero degree centigrade and that pressure is 1
atmosphere, if we take 1 bar what will be the volume? And that is the new convention but
for most of the calculations we still follow the old convention that is 22.4 litres. We had
made ideal gas equation PV is equal to nRT and in this we had defined that R is universal gas
constant and we had written two values, 8.314 joules per moles per Kelvin and the second
one 0.0821 litre atmosphere per mole, okay, this is what we had done.
Okay, so today let’s start with the estimation of molecular, sorry empirical formula of a
compound. Now, let’s learn how to derive the empirical formula of a compound if their mass
percentage composition is given to us, okay. I will write it here, if mass percentage
composition is given. Now, before you start on with this particular data, I want to tell you a
very simple logic, very simple, very elementary thing. In C6H12O6, what is the ratio of moles
of carbon to hydrogen, 1 is to 2, is it correct. Okay, suppose I have 0.1 mole of glucose, will
this ratio change. This ratio depends on what, the amount of carbon to the amount of
hydrogen. If you tell that you have C6H12O6, so how many moles of carbon is there, 0.6 and
hydrogen moles are 1.2. Now also, number of moles of carbon to number of moles of
hydrogen, what will be the ratio? 1 is to 2, even if you see in 1 mole or 100 moles or see in x
moles. If x moles were there, what would have written 6 x and 12 x, so what would be ratio
again 6x by 12x which is 1 is to 2, do you understand this thing. This is the crux for
calculating the empirical formula of any compound, okay. Now, we will see how to solve it.
See, what would be given is, if we see here. You will be given mass percent data and by that
we have to find empirical formula, let me take one example in which I will give some data
and we will learn how to calculate it. First of all we will create a table in which we will mark
element, it’s atomic mass, it’s mass percent composition, okay and after this what we will
mark apart from atomic mass and mass percentage composition. With this we will also say
number of moles in 100 grams of compound then relative number of atoms, and if required
we will also mark one more thing which is called as the simplest whole number ratio, okay.
Now see how we will do it. We noted all these things; let me write down the elements.
Carbon, hydrogen, oxygen and here if I write atomic mass data, what will come, 12, 1, 16,
suppose mass percent is given 40, okay, for hydrogen it is given 6.67 and this is 53.33 and we
know why this was done by us, what should the empirical formula finally be CH2, let’s make
sure it is that. See, first of all what we are going to do. To calculate what is number of mole
in 100 gram of compound, these percentages are there, so in 100 gram how much gram
carbon will be – 40 gram, in 100 gram how much will be hydrogen – 6.67 and how much will
be oxygen – 53.33. So, we will say, let’s try to calculate the number of atoms as mass
percentage data divided by the atomic mass. If we write this here then what we will get, 40
divided by 12, 40 divide by 12, very good, you will get it as 3.33, for hydrogen you will get
6.67 and for oxygen 53.33 by 60, how much it will come, check it. You see, you will get 3.33
only. So, these element moles are there, in 100 gram and you take whatever grams you
want, it will come same as this. We have to take its ratio and that is important. And what do
we want, empirical formula. What we are interested in is the simplest ratio and if I tell, how
we will ..