Estimation of Sulphur Phophorous & Oxygen - Organic Chemistry: Some Basic Principals, class 11 Video Lecture

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exam now let's do estimation of sulfur
sulfur also use serious method in this
case sulfur is heated with sodium
peroxide or nitric acid sulfur so the
sulfur whatever is present it will get
oxidized to sulfuric acid correct and it
is precipitated as varium sulfate by
adding vacat in the water and now we
have a BS here to this place it is
filter wash and fight similarly to ACC
n2 and we can find the percentage of
sulfur correct so as I told you my
sulfur this first reacted with sodium
peroxide or hno3 nitric acid right what
you get is sulfuric acid see this is a
sulfur with we have water and oxygen
also here right little liquid actually
right so you have water oxygen we have
sulfur you react with this nitric acid
you get sulphuric acid now you have the
sulfuric acid you react with beige in
play you get barium sulphate and this is
what precipitates out - listen this guy
you can filter again dry and then from
this you can find actually the
composition of circle correct so let us
take one example on this so in the
sulfur estimation this gram of organic
compound yeah this gram of barium
sulphate so I have my organic compound
and this gives billions under I have to
find the percentage on sulfur
we find the molar mass of barium sulfate
is nothing but B M is 137 sulfur is 32
oxygen is 16 84 64 you add this 33 this
is my molecular mass of billions of it
so if I have got 230 and sulfur is 32
right so that the equation if you see is
from sulfur only you get VA suppose this
is my balanced equation actually so if I
have got one mole of sulfur from one
muzaffer I'll get one mole of barium
circle right so one mole of barium
sulfate is what 33 so do 33 grams of
barium sulfate if I god this implies I
have one mole that is 32 gram of sulfur
in organic um Oh hope you understand -
33 grams of barium sulfate implies if I
have called as output this implies 33
grams of software now going upon so if I
have got one gram of barium sulphate
this implies 32 by 33 grams of sulfur in
organic component but I have how much
I've got I've got 0.48 1 3 grams of
barium sulphate oh god this - implies
this value into point four eight one
grams of sulfur inaudible this is this
is what I have sulfide or getting upon
to find the percentage of sulphur what
you do you take the whole value divided
by the weight of warning Omega bonds you
take this is nothing but 32 by 2 33 into
0.48 1 3 divided by this mass of this
calling a bond in 200 you solve this you
get forty two point one zero percentage
and that is the percentage of sulphur in
this organic compound similar to sulphur
we can do quantitative estimation
phosphorus also
so in this what we do we take some
organic compound of known mass we heat
this with nitric acid and you get what
you get phosphoric acid so let me write
the reaction for this phosphorus you
heat with nitric acid you get is three
yes correct
now it is now precipitated as ammonium
phospholipids by adding ammonia or
ammonium molecule that is one way or the
alternative is what you can do is you
have this phosphoric acid right
you want to precipitate it as this by
adding magnesium so as its reaction so I
am is Threepio for Aisle eight add
magnesium sure so I will get mg in this
for PU for this guy right and this if I
heat actually on ignition this guy will
a heat if I heat this so let me write
this reaction here right here only mg n
ash or you whole if I heat this I get mg
2 e 2 O 7 plus I get ammonia when I get
correct this is what I get and from this
again find the concentration of circle
because I have got this compound and
this is how sulfur looks right so there
are two methods and explained so if you
use the first method the possibility of
sulfur will come out to be this same
thing same logic be applied this is the
maha sulfur this was the whole mass of
the complex compound is if again the
same thing here right so using that
logic you'll see that percent of
phosphorous in first case if you are
using the first option where you get
this this compound NH 4 3 pure 4.12 moo3
you have to use this formula or if you
are using if you are precipitating out
mg 2 P do say and because you have to
precipitate something right there only
you can dry it and I played the
percentage of phosphorous so either you
can precipitate this guy or this guy H
if you are going for the magnesium
approachment
you get magna suppose you get this then
you use this form rough if you're going
for this approach yes you use this form
correct let's estimate oxygen so
generally you know how to find all the
other components nitrogen oxygen or a
nitrogen phosphorus chlorine so you
generally we find all the other
compounds and then we subtract them from
hundred whatever you get is that
percentage of oxygen that is the channel
cross but there is a direct method also
because this is a tricky one if we have
we have carbon hydrogen nitrogen
chlorine so you have these things or
sulfur also interpreted compound so you
have to find oxygen also so first
approach is you find conservation of all
these subtract 100 you get oxygen sort
of percentage submission hundred that's
a very cool approach to the ID method is
what you do you take out any compound
right and you heat this in nitrogen gas
right
now you pass this because you look at
oxygen right you hitting the nitrogen
gas whatever oxygen you get another gas
you get this oxygen want to get you
passed with a red-hot coke it in from
carbon monoxide right and now that is
also gas right it's very difficult to
measure so what we do is we you pass
this guy with I - fine I didn't paint
oxide so here you get carbon dioxide and
iodine gas so either from iodine or from
carbon and I said any of these you can
actually find the concentration of
oxygen correct see if you see if my
carbon dioxide let suppose form is M one
gram right and then you have this
relationship by this for T program of
carbon dioxide means four into gram of
oxygen because this guy has how much
this guy is my total carbon dioxide is
44 out of this this was my 2l and this
is 32 16 to do so if you take carbon
dioxide that means try to get 44 grams
carbon dioxide it's implies I have 32
grams of oxygen right so with that if
you find this percentage you get
percentage of oxygen give me this day 32
by 44 into M 1 by M into M where m where
is the mass of carbon dioxide mom
correct thank you
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