Question

a student has 480.0 ml of a 0.1374 m aqueous solution of mnso₄ to use in an experiment. he accidentally leaves the container uncovered and comes back the next week to find only a solid residue. the mass of the residue is 12.34 g. determine the chemical formula of this residue.

Explanation:

Step1: Calculate moles of MnSO₄ initially

First, convert volume to liters ($V = 480.0\ mL=0.4800\ L$) and use the formula $n = M\times V$. Given $M = 0.1374\ M$, then $n = 0.1374\ mol/L\times0.4800\ L = 0.065952\ mol$.

Step2: Calculate molar mass of the residue

The molar mass $M_{residue}=\frac{m}{n}$, where $m = 12.34\ g$ and $n = 0.065952\ mol$. So $M_{residue}=\frac{12.34\ g}{0.065952\ mol}\approx187.1\ g/mol$.
The molar mass of $MnSO_{4}$ is $M_{MnSO_{4}}=54.94 + 32.07+4\times16.00=151.01\ g/mol$. The difference in molar - mass might be due to water of hydration. The molar mass of water is $18.02\ g/mol$. $(187.1 - 151.01)\div18.02\approx2$.

Answer:

$MnSO_{4}\cdot2H_{2}O$