Explain the estimation of phosphorus and sulphur present in the organic compound

Explanation

Estimation of phosphorus : Known mass of organic compound (.a. gm) is heated with fuming `HNO_(3)` in a carius tube. .p. is oxidised to `H_(2)PO_(4)` acid. This acid is precipitated as ammonium phosphomolybdate (.b. gm) by adding ammonia and ammonium molybdate solutions.
`% g ` of phosphorus `= (100)/(a) = (b)/(1877) xx 31 gm`
Molecular mass of `(NH _(4)) _(3) PO _(4) 12 MO O _(2)` (ammonium phosphomoly bdate) = 1877
Estimation of sulphur : Known mass of organic compound (.a. gm) is heated with sodium peroxide in a carius tube. If sulphur is present it is oxidised to sulphuric acid. The acid is precipitated as barium sulphate (big) by adding excess of aq. `BaCl_(2)` solution. The ppt. is filtered, washed, dried and weighed.
`%g` of sulphur `= (100)/(a) xx ( b xx 32)/(233) g `
Molecular mass of `Ba SO _(4) = 233`
Estimation of Oxygen : The percentage of oxygen in. an organic compound is usually found by difference between the total percentage composition (100) and the sum of the percentages of all other elements. However, oxygen can also be estimated directly as follows:
A definite mass of an organic compound is decomposed by heating in a stream of nitrogen gas. The mixture of gaseous products containing oxygen is passed over .red-hot coke when all the oxygen is converted to carbon monoxide. This mixture is passed through warm iodine. pentoxide `(I _(2) O _(2))` when carbon monoxide is oxidised to carbon dioxide producing iodine.
Compound `overset("heat") to O _(2) + ` other gaseous products
`2 C + O _(2) overset( 1372 K) to 2 CO] xx 5 (A)`
` I _(2) O _(5) + 5 CO to I _(2) + 5 CO _(2) ] xx 2(B)`
On making the amount of CO produced in equation (A) equal to the amount of co used in equation (B). by multiplying the equations (A) and (B) by 5 and 2 respectively, we find that each mole of oxygen liberated from the compound will produce two moles of carbondioxide. Thus 88 g carbon dioxide is obtained if 32 g oxygen is liberated.
Let the mass of organic compound taken be mg
Mass of carbon dioxide produced be `m _(1) g `
`therefore m _(1) g` carbon dioxide is obtained from `(32 xx m _(1))/( 88) g O _(2)`
`therefore ` Percentage of oxygen` = (32 xx m_(1) xx 100 )/( 88 xx m ) %.`