Question

A 1.000-g sample of XI₂, is dissolved in water, and excess silver nitrate is added to precipitate all of the iodide as Agl. The mass of the dry Agl is found to be 1.375 g. Calculate the atomic weight (mass) of X.

Short answer

The atomic weight (mass) of X is determined as 87.8 amu.

Step-by-step solution

Definition

The molar mass of a materialis defined as the mass of the substancecontained in one moleof the substance, and it is equal to the total massof the component atoms. It may be stated mathematically as follows:

Molar mass =$\frac{Weighting}{numberofmoles}$

Determination of number of moles of iodine

Here, the mass of Agl = 1.375g

And the mass of Agl = (1(107.87)+(126.90))g/mol

= 234.8g/mol

So, the number of moles in iodine is 1.375gAgl

$$\begin{gathered} =1.375\mathrm{g}\mathrm{Agl}\times \frac{1\mathrm{mol}\mathrm{Agl}}{234.8\mathrm{g}\mathrm{Agl}}\times \frac{1\mathrm{mol}\mathrm{I}}{1\mathrm{mol}\mathrm{Agl}} \\ =5.856\times {10}^{-3}\mathrm{mol}\mathrm{I} \end{gathered}$$

Thus the number of moles of iodine in 1.375g Agl is $5.856\times {10}^{-3}molI$.

Determination mass of iodine

Let’s find the mass of I using its molar mass

Here, the mass of $5.856\times {10}^{-3}molI\times \frac{126.90gI}{1molI}$

= 0.7431 g I

Determination of atomic mass of X

From the given, it is given as 1.000-g sample of XI₂, and then we found that rest of the gram is calculated as 0.7431 g I.

Therefore the gram of X is (1.000 g - 0.7431 g).

Let’s compute the moles of X using molar ratio,

Moles of $X=5.856\times {10}^{-3}\times \frac{1molX}{2molH}$

$=2.928\times {10}^{-3}molX$

Lastly, let’s calculate the atomic weight of X,

Molar mass $=\frac{0.257gX}{2.928\times {10}^{-3}molX}$

$$\begin{gathered} =87.8\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right. \\ =87.8amu \end{gathered}$$

Therefore the atomic mass is87.8 amu.