Question

a student has 460.0 ml of a 0.1239 m aqueous solution of mgi₂ to use in an experiment. she accidentally leaves the container uncovered and comes back the next week to find only a solid residue. the mass of the residue is 24.07 g. determine the chemical formula of this residue.

Explanation:

Step1: Calculate moles of MgI₂ initially

Use the formula $n = M\times V$. Given $M = 0.1239\ M$ and $V=460.0\ mL = 0.4600\ L$.
$n = 0.1239\ mol/L\times0.4600\ L=0.0570\ mol$

Step2: Calculate molar mass of MgI₂

The molar mass of Mg is $24.31\ g/mol$ and of I is $126.90\ g/mol$. So molar mass of $MgI_{2}=24.31 + 2\times126.90=278.11\ g/mol$

Step3: Calculate mass of MgI₂ if no water - loss and no reaction

$m = n\times M = 0.0570\ mol\times278.11\ g/mol = 15.85\ g$

Step4: Analyze the situation

The mass of the residue ($24.07\ g$) is greater than the mass of pure $MgI_{2}$ we calculated. This indicates that the residue is a hydrate.
Let the formula of the hydrate be $MgI_{2}\cdot xH_{2}O$.
The mass of water in the hydrate is $m_{water}=24.07 - 15.85=8.22\ g$
The moles of water $n_{water}=\frac{8.22\ g}{18.02\ g/mol}=0.456\ mol$
The ratio of moles of water to moles of $MgI_{2}$ is $\frac{n_{water}}{n_{MgI_{2}}}=\frac{0.456\ mol}{0.0570\ mol}=8$

Answer:

$MgI_{2}\cdot8H_{2}O$