Question
Double salts are substances with two cations and one anion. A hydrated sulfate containing two cations has this percentage composition.
(a) (i) Draw one Lewis (electron dot) structure of the sulfate ion.
▶️Answer/Explanation
Ans:
Question
(ii) Calculate the percentage of oxygen present in the double salt.
▶️Answer/Explanation
Ans:
To calculate the percentage of oxygen \((O)\) present in the double salt, we need to determine the remaining percentage after accounting for the percentages of other elements (nitrogen, hydrogen, sulfur, and cobalt). Since oxygen is the only remaining element, its percentage can be calculated as the difference from \(100 \%\) :
Percentage of oxygen \(=100 \%-\) Sum of percentages of other elements Percentage of oxygen \(=100 \%-(7.09 \%+5.11 \%+16.22 \%+14.91 \%)\)
Percentage of oxygen \(=56.67 \%\)(iii) Determine the empirical formula of the double salt. Use section 6 of the data booklet.
Question
(iii) Determine the empirical formula of the double salt. Use section 6 of the data booklet.
▶️Answer/Explanation
Ans:
To determine the empirical formula of the double salt, we need to find the simplest whole number ratio of the elements present in the compound based on their percentage compositions.
Given:
Percentage of nitrogen (\(N\)) = 7.09%
Percentage of hydrogen (\(H\)) = 5.11%
Percentage of sulfur (\(S\)) = 16.22%
Percentage of cobalt (\(Co\)) = 14.91%
Then, we calculate the number of moles of each element:
\[ \text{Moles of nitrogen} = \frac{7.09 \, \text{g}}{14.01 \, \text{g/mol}} \approx 0.506 \, \text{mol} \]
\[ \text{Moles of hydrogen} = \frac{5.11 \, \text{g}}{1.008 \, \text{g/mol}} \approx 5.07 \, \text{mol} \]
\[ \text{Moles of sulfur} = \frac{16.22 \, \text{g}}{32.07 \, \text{g/mol}} \approx 0.506 \, \text{mol} \]
\[ \text{Moles of cobalt} = \frac{14.91 \, \text{g}}{58.93 \, \text{g/mol}} \approx 0.253 \, \text{mol} \]
\[ \text{Moles of oxygen} = \frac{56.67 \, \text{g}}{16.00 \, \text{g/mol}} \approx 3.542 \, \text{mol} \]
Now, we find the simplest whole number ratio of these moles, which is approximately:
\[
\frac{0.506}{ 0.253},\frac{5.06}{ 0.253},\frac{0.506}{ 0.253},\frac{0.253}{ 0.253},\frac{3.54}{ 0.253}
\]
OR
\[
\text { 2.00, 20.0, 2.00, } 1.00 , 14 .00
\]
\[
\mathrm{N}_2 \mathrm{H}_{20} \mathrm{~S}_2 \mathrm{CoO}_{14}
\]
Question
(iv) The molar mass of the empirical formula is the same as the molar mass of the formula unit. Deduce the formula unit of the hydrated double salt.
▶️Answer/Explanation
Ans:
\(\left(\mathrm{NH}_4\right)_2 \mathrm{Co}\left(\mathrm{SO}_4\right)_2 \cdot 6 \mathrm{H}_2 \mathrm{O}\)
Question
(b) 1.20g of the double salt was dissolved in water and an excess of aqueous barium chloride was added, precipitating all the sulfate ions as barium sulfate.
(i) Formulate an ionic equation, including state symbols, for the reaction of barium ions with sulfate ions.
▶️Answer/Explanation
Ans:
The reaction of barium ions (\( \text{Ba}^{2+} \)) with sulfate ions (\( \text{SO}_4^{2-} \)) in aqueous solution to form barium sulfate (\( \text{BaSO}_4 \)) can be represented by the following ionic equation, including state symbols:
\[ \text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{BaSO}_4(s) \]
This equation represents the precipitation reaction where aqueous barium ions react with aqueous sulfate ions to form solid barium sulfate, which precipitates out of solution.
Question
(ii) Calculate the mass of barium sulfate precipitate. Use your answer to part (a)(iii) and section 6 of the data booklet. (If you did not obtain an answer for part (a)(iii),use 400.0g \(mol^{-1}\) as \(M_r\) for the double salt, but this is not the correct value.)
▶️Answer/Explanation
Ans:
\(\begin{aligned} & \frac{\text {1.20 } \mathrm{g} }{400\mathrm{~g} \mathrm{~mol}^{-1}} \\&\text { salt }=2 \times 3.00 \times 10^{-3} \text {mol} \\&\mathrm{SO}_4{ }^{2-}= 6.00 \times 10^{-3} \text { mol } \\ & 233.40 \mathrm{~g} \mathrm{~mol}^{-1} \times 6.00 \times 10^{-3}=1.40 \mathrm{~g}\\ & \end{aligned}\)