Test Papers – ICSE – Class – X
Chemistry 11,Empirical and Molecular Formula Laws of Chemical Combinations POAC
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 ..
Chemistry 09, Stoichiometric Calculations involving gases
So, in the last lecture if you remember we had discussed the effect of dilution and mixing of
solution and also we had discussed these calculations which we do in stoichiometry based
on the strength terms like on basis of molarity, you had seen acid based neutralization or
double displacement example in which you had seen precipitation of barium sulphate and
you had found that how much molarity of barium fluoride solution or how much volume of
barium flouride solution was require, this kind of question, we had seen. Now, today what
we are going to do is, we are first going to review one important aspect in dilution which is
effect of temperature on molarity and molality. What is the effect of temperature on
molarity and molality? Can you tell me what is the effect here. Okay, first tell me the
definition of molality, molality. It is represented by small m, number of moles of solute
divided by the mass of solvent in kgs, absolutely right. And what is the expression for
molarity? Number of moles of salute …in all these terms which is the term dependent on
temperature. So, we say that, this term volume is temperature dependent, since this is
temperature dependent term, we observe that generally as temperature increases, what
happens to volume? It increases, you read expansion, right, thermal expansion, and if we
heat then the volume of the solution is going to expand. So what happens to morality? You
can write, molarity of a solution decreases with increase in temperature. Correct, can
everyone understand this. Molarity of a solution decreases with increase in temperature.
What happens to molality? You can write it here, molality is independent of temperature.
Till here it is okay, everyone understands. Everybody wrote it.
Today what we will do, is first we will note down some important information, you can write
it down. Stoichiometric calculations involving gases. When we have to do stoichiometric
calculations involving gases that time we require some data like, vapour density of a gas, we
will abbreviate it as VD is equal to the molecular weight of the gas divided 2. Remember this
and we will see its proof in the gaseous state. For now just memorise it, okay. Second is you
must be aware of one very important data which you have learnt in 9th, when you studied
the properties of gas. That 1 mole of a gas at STP occupies how much, very good, 22.4 litres.
Now, in this particular case there is very important thing to note down, what is the definition
of STP? STP stands for Standard Temperature and Pressure. Now, in this particular case we
will say that temperature is taken to be zero degrees centigrade or which is 273 Kelvin and
pressure is taken to be 1 atmosphere. Please, understand, when you are writing 22.4 litres it
is corresponding to 1 atmosphere and this is old convention which is still followed because
most of the questions that are present in the books follow this convention. That is why now
also we remember the 22.4 litre data. But if you were to see the new convention, according
to new convention we write down that 1 mole of a gas at STP occupies 22.7 litres. And the
change that occurs in this, that is STP which is again standard temperature and pressure but
this time what will be the temperature. 273 Kelvin just like last time but pressure is 1 bar.
For now you have to assume that atmosphere and bar are units of pressure. There is no
need to understand the relationship between two of these. You can explore that in the
chapter of gaseous state. That time you will go into detail. You will learn more units like
millilitre of mercury, centimetre of mercury, okay, all these things we will discuss at that
time. We will do all the relations, for now we will just keep this much in mind. Now the third
aspect which you have to remember PV is equal to nRT which is called as the Ideal Gas
Equation, very good. Now, in this case we say P is representing what, T representing, V
representing, n represents and we will write all these things. P represents pressure, V
represents volume, n represents number of moles and T represents temperature, keep in
mind about temperature, as temperature is written always as Kelvin, temperature is always
marked K, okay and R is the, have we done this before, universal gas constant, do you
remember its value. Right, everybody copied till here, everybody mark it. The notations
whatever I am marking please mark it first and then we will discuss how to do it. Everybody
wrote it? Now, the value of R comes out as 8.314 joules per mole per Kelvin. Now this is the
SI unit which is more relevant in thermodynamics but for gaseous state or for calculations in
Stoichiometry you can typically use this unit 0.0821 litre atmosphere per mole per Kelvin,
okay, this is the unit that you will us. Trust me, this unit looks very bad and to study also is
bad but to solve it is very easy.
