Chapter 6, Problem 146MP

Three students were asked to find the identity of the metal in a particular sulfate salt. They dissolved a 0.14 72-g sample of the salt in water and treated it with excess barium chloride, resulting in the precipitation of barium sulfate. After the precipitate had been filtered and dried, it weighed 0.2327 g.

Each student analyzed the data independently and came to different conclusions. Pat decided that the metal was titanium. Chris thought it was sodium. Randy reported that it was gallium. What formula did each student assign to the sulfate salt?

Look for information on the sulfates of gallium, sodium, and titanium in this text and reference books such as the CRC Handbook of Chemistry and Physics. What further tests would you suggest to determine which student is most likely correct?

Interpretation Introduction

Interpretation: The formula of the sulphate salt assigned by each student has to be identified.

Concept Introduction: When two soluble solutions are mixed together, an insoluble salt formation occurs called as precipitate. These precipitates fall out of the solution and the reactions are called as precipitation reaction.

Answer to Problem 146MP

The formula of sulphate salt assigned by Pat is Titanium sulphate ${\text{(TiSO}}_{\text{4}}\text{)}$

           The formula of sulphate salt assigned by Chris is Sodium sulphate ${\text{(Na}}_{\text{2}}{\text{SO}}_{\text{4}}\text{)}$

           The formula of sulphate salt assigned by Randy is Gallium sulphate ${\text{(Ga}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}\text{)}$

Explanation of Solution

Given:

Record the given info

Weight of the sample = $\text{0}\text{.1742}\text{g}$

Weight of the precipitate = $\text{0}\text{.2327}\text{g}$

The weight of the sample and weight of the precipitate are recorded as shown above.

To calculate the moles of ${\text{BaSO}}_{\text{4}}$

Molecular mass of ${\text{BaSO}}_{\text{4}}$ = $\text{233}\text{.4}\text{g}$

Moles of ${\text{BaSO}}_{\text{4}}$= $\text{0}\text{.2327}\text{g×}\frac{\text{1}\text{mol}}{\text{233}\text{.4}\text{g}}\text{=9}{\text{.970×10}}^{\text{-4}}\text{mol}{\text{BaSO}}_{\text{4}}$

The moles of ${\text{BaSO}}_{\text{4}}$ depends on the formula of the sulphate salt

The general equation can be given as,

${\text{M}}_{\text{x}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{y(aq)}}{\text{+yBa}}^{\text{2+}}{}_{\left(\text{aq}\right)}\to {\text{yBaSO}}_{\text{4(s)}}{\text{+xM}}^{\text{z+}}$

The mole ratio between the unknown sulphates salt and ${\text{BaSO}}_{\text{4}}$ shows variation depending upon the value of $\text{y}$ .

The moles of ${\text{BaSO}}_{\text{4}}$ is calculated by plugging in the values of molar mass of ${\text{BaSO}}_{\text{4}}$ to the weight of the precipitate. The moles of ${\text{BaSO}}_{\text{4}}$ is found to be $\text{9}{\text{.970×10}}^{\text{-4}}\text{mol}$

To write the formula of sulphate salt assigned by Pat

Pat thinks that formula of sulphate salt is ${\text{TiSO}}_{\text{4}}$

Then the equation becomes,

${\text{TiSO}}_{\text{4(aq)}}{\text{+Ba}}^{\text{2+}}{}_{\left(\text{aq}\right)}\to {\text{BaSO}}_{\text{4(s)}}{\text{+Ti}}^{\text{2+}}{}_{\left(\text{aq}\right)}$

There is $\text{1:1}$ mole ratio between ${\text{BaSO}}_{\text{4}}$ and ${\text{TiSO}}_{\text{4}}$, therefore $\text{9}{\text{.970×10}}^{\text{-4}}\text{mol}$ of ${\text{TiSO}}_{\text{4}}$ is needed.

Assuming the molar mass of ${\text{TiSO}}_{\text{4}}$,

$\text{Molar}\text{mass}\text{of}{\text{TiSO}}_{\text{4}}\text{=}\frac{\text{0}\text{.1472}\text{g}}{\text{9}{\text{.970×10}}^{\text{-4}}\text{mol}}\text{=147}\text{.6}\text{g/mol}$

The value of calculated molar mass of ${\text{TiSO}}_{\text{4}}$ is compared with the standard molar mass of ${\text{TiSO}}_{\text{4}}$ using the atomic masses in the periodic table.

The standard molar mass of ${\text{TiSO}}_{\text{4}}$= $\text{143}\text{.95}\text{g/mol}$

On comparing the standard molar mass and calculated molar mass of ${\text{TiSO}}_{\text{4}}$, the Titanium sulphate assigned by Pat seems to be reasonable.

The formula assigned by Pat is Titanium sulphate. The molar mass of Titanium sulphate is calculated using the weight of the sample and the moles of ${\text{BaSO}}_{\text{4}}$. The standard molar mass of ${\text{TiSO}}_{\text{4}}$ is calculated using the standard atomic masses from the periodic table. On comparing both the standard and calculated molar mass of ${\text{TiSO}}_{\text{4}}$ the formula of sulphate salt given by Pat seems to be reasonable.

To write the formula of sulphate salt assigned by Chris

Chris thinks that formula of sulphate salt is ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$

The equation becomes,

${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}{\text{+Ba}}^{\text{2+}}{}_{\left(\text{aq}\right)}\to {\text{BaSO}}_{\text{(s)}}{\text{+2Na}}^{\text{+}}{}_{\left(\text{aq}\right)}$

There is $\text{1:1}$ mole ratio between ${\text{BaSO}}_{\text{4}}$ and ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$, therefore $\text{9}{\text{.970×10}}^{\text{-4}}\text{mol}$ of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ is needed.

Then the molar mass of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ can be given as,

$\text{Molar}\text{mass}\text{of}{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\text{=}\frac{\text{0}\text{.1472}\text{g}}{\text{9}{\text{.970×10}}^{\text{-4}}\text{mol}}\text{=147}\text{.6}\text{g/mol}$

The value of calculated molar mass of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ is compared with the standard molar mass of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ using the atomic masses in the periodic table.

The standard molar mass of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$= $\text{142}\text{.05}\text{g/mol}$

On comparing the standard molar mass and calculated molar mass of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$, the Sodium sulphate assigned by Chris also seems to be reasonable.

The formula of sulphate salt assigned by Chris is Sodium sulphate. The molar mass of Sodium sulphate is calculated using the weight of the sample and the moles of ${\text{BaSO}}_{\text{4}}$. The standard molar mass of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ is calculated using the standard atomic masses from the periodic table. On comparing both the standard and calculated molar mass of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ the formula of sulphate salt given by Chris seems to be reasonable.

To write the formula of sulphate salt assigned by Randy

The formula of sulphate salt assigned by Randy is ${\text{Ga}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}$

The equation becomes,

${\text{Ga}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3(aq)}}{\text{+3Ba}}^{\text{2+}}{}_{\left(\text{aq}\right)}\to {\text{3BaSO}}_{\text{4(s)}}{\text{+2Ga}}^{\text{3+}}{}_{\left(\text{aq}\right)}$

The molar mass of ${\text{Ga}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}$ can be given by,

$\text{Molar}\text{mass}\text{of}{\text{Ga}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}\text{=}\frac{\text{0}\text{.1742}\text{g}{\text{Ga}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}}{\text{9}{\text{.970×10}}^{\text{-4}}\text{mol}{\text{BaSO}}_{\text{4}}}\text{×}\frac{\text{3}\text{mol}{\text{BaSO}}_{\text{4}}}{\text{mol}{\text{Ga}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}}\text{=442}\text{.9}\text{g/mol}$

The value of calculated molar mass of ${\text{Ga}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}$ is compared with the standard molar mass of ${\text{Ga}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}$ using the atomic masses in the periodic table.

The standard molar mass of ${\text{Ga}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}$= $\text{427}\text{.65}\text{g/mol}$

On comparing the standard molar mass and calculated molar mass of ${\text{Ga}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}$, the Gallium sulphate assigned by Randy also seems to be reasonable.

The formula of sulphate salt assigned by Randy is Gallium sulphate. The molar mass of Gallium sulphate is calculated using the weight of the sample and the moles of ${\text{BaSO}}_{\text{4}}$. The standard molar mass of ${\text{Ga}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}$ is calculated using the standard atomic masses from the periodic table. On comparing both the standard and calculated molar mass of ${\text{Ga}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}$ the formula of sulphate salt given by Randy also seems to be reasonable.

Interpretation Introduction

Interpretation: the most likely correct salt of sulphate has to be given.

Concept Introduction: When two soluble solutions are mixed together, an insoluble salt formation occurs called as precipitate. These precipitates fall out of the solution and the reactions are called as precipitation reaction.

Answer to Problem 146MP

The sulphate salts identified by the students are tested with aqueous Sodium hydroxide.

Explanation of Solution

To identify which salt is likely correct and what test.

By references,

Sodium sulphate ( ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$) occurs as white solid with orthorhombic crystals

Gallium sulphate ( ${\text{Ga}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}$) occurs as white powder, whereas

Titanium sulphate occurs as green powder and has a formula of ${\text{Ti}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}$

The calculated molar mass of ${\text{Ti}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}$ should be $\text{389}\text{g/mol}$ since the formula of ${\text{Ti}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}$ is similar to the formula of ${\text{Ga}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}$ whose calculated molar mass is $\text{442}\text{.9}\text{g/mol}$.

Therefore, the salt is unlikely to be Titanium sulphate.

In order, to distinguish the like salt between ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ and ${\text{Ga}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}$. The sulphate salts of ${\text{Na}}^{\text{+}}\text{and}{\text{Ga}}^{\text{3+}}$ are treated with aqueous Sodium hydroxide.

The sulphate salts are made to dissolve in water and later are treated with Sodium hydroxide.

Gallium ( ${\text{Ga}}^{\text{3+}}$) forms a precipitate with hydroxide where Sodium ( ${\text{Na}}^{\text{+}}$) doesn’t give any precipitate.

Based on the solubility rules and the references, Gallium hydroxide is insoluble

The salt that is likely to be correct is Gallium sulphate.

The most unlike salt is to be Titanium sulphate because Titanium sulphate takes a formula of ${\text{Ti}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}$ and not ${\text{TiSO}}_{\text{4}}$ . The formula of ${\text{Ti}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}$ has a similar form of ${\text{Ga}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}$ whereas the calculated molar masses completely differ from one other. Therefore, the most likely salts are either Sodium sulphate or Gallium Sulphate. In order to distinguish, the sodium sulphate and gallium sulphate, the sulphate salts assigned/identified by each student is treated with aqueous solution of Sodium hydroxide. It is seen that Gallium sulphate would precipitate Gallium hydroxide whereas no precipitate formation with Sodium sulphate.

Hence, the most likely correct salt predicted is by Randy as Gallium sulphate.