91Ó°ÊÓ

• The mole concept is a simple way to express the amount of a substance. Any measurement is divided into two parts: the numerical magnitude and the units in which the magnitude is expressed. For example, if the mass of a ball is 2 kilogrammes, the magnitude is '2' and the unit is 'kilogramme.'
• When dealing with particles at the atomic (or molecular) level, even one gramme of a pure element is known to contain a large number of atoms. This is a common application of the mole concept. It primarily focuses on the unit known as a'mole,' which is a count of a very large number of particles.

In Kjeldahl analysis, $1{\mathrm{molNH}}_3$of is equal to$1\mathrm{molofH}$:

$\mathrm{n}\left({\mathrm{NH}}_3\right)=\mathrm{n}\left({\mathrm{H}}^{+}\right)$

Since there are 2 moles of ${\mathrm{H}}^{+}$in${\mathrm{H}}_2{\mathrm{SO}}_4$they moles are doubled:

$\mathrm{n}\left({\mathrm{NH}}_3\right)=\mathrm{n}\left({\mathrm{H}}^{+}\right)$

After the initial role="math" localid="1654863625697" ${\mathrm{H}}_2{\mathrm{SO}}_4$reacts with ${\mathrm{NH}}_3$ the excess ${\mathrm{H}}_2{\mathrm{SO}}_4$reacts with KI and KI03 and it is titrated with thiosulfate.To determine the ratio of moles of ${\mathrm{H}}_2{\mathrm{SO}}_4$and thiosulfate we have to use reactions 16-18 and 16-19:

$$\begin{gathered} I{O}_3^{-}+6H^{+}+8I^{-}⇌3I_3^{-}+3H_{2}O\left(1\right) \\ {I}_3^{-}+2S_2{O_3}^{2-}⇌S_4{O_6}^{2-}+3I^{-}\left(2\right) \end{gathered}$$

By using reaction (1) we can determine the ratio of moles of sulfuric acid and triiodide:

$$\begin{gathered} 6\mathrm{n}\left({\mathrm{H}}^{+}\right)=3\mathrm{n}(\mathrm{excess}{\mathrm{H}}_2{\mathrm{SO}}_4)=3\mathrm{n}\left({\mathrm{l}}_3^{-}\right)\mathrm{l}:3 \\ 6\mathrm{n}\left({\mathrm{H}}^{+}\right)=\mathrm{n}(\mathrm{excess}{\mathrm{H}}_2{\mathrm{SO}}_4)=\mathrm{n}\left({\mathrm{l}}_3^{-}\right) \end{gathered}$$

By using reaction (2) we can determine the ratio of moles of triiodide and thiosulfate:

$\mathrm{n}\left({\mathrm{l}}_3^{-}\right)=\frac{1}{2}\mathrm{n}\left({\mathrm{S}}_2{{\mathrm{O}}_3}^{2-}\right)$

Therefore, the ratio of moles of sulfuric acid and thiosulfate are:

role="math" localid="1654864150133" $\mathrm{n}(\mathrm{excess}{\mathrm{H}}_2{\mathrm{SO}}_4)-\frac{1}{2}\mathrm{n}\left({\mathrm{S}}_2{{\mathrm{O}}_3}^{2-}\right)$

Moles of${\mathrm{NH}}_3$are:

$$\begin{gathered} \mathrm{n}({\mathrm{NH}}_3)=2\mathrm{n}\left({\mathrm{H}}_2{\mathrm{SO}}_4\right) \\ \mathrm{n}\left({\mathrm{NH}}_3\right)=2(\mathrm{n}(\mathrm{excess}{\mathrm{H}}_2{\mathrm{SO}}_4)-\mathrm{n}(\mathrm{excess}{\mathrm{H}}_2{\mathrm{SO}}_4\left)\right) \end{gathered}$$

By replacing excess moles of sulfuric acid with moles of thiosulfate we get:

$n\left(N{H}_3\right)=2\left(\begin{array}{c}n(initial{H}_{2}S{O}_4)-\frac{1}{2}n\left(S_2{O_3}^{2-}\right)\end{array}\right)$

Hence the relationship between moles of ${\mathrm{NH}}_3$liberated in the digestion and moles of thiosulfate required for titration of iodine is derived. And how thiosulfate titration is related to the nitrogen content of the unknown is explained.