Question

although the percent yield should always be maximized, fairly low yields are acceptable if the starting materials are cheap and the product is valuable. this is often the case for pharmaceutical chemicals.
in this experiment, gypsum, caso₄·2h₂o, will be prepared by the reaction of alum with calcium chloride. aik(so₄)₂·12h₂o(aq) + 2 cacl₂(aq) → 2caso₄·2h₂o(s) + aici₃(aq) + kci(aq) + 8h₂o(ℓ)
the gypsum is insoluble in hot water, while the starting materials and the other products are water soluble. therefore the gypsum can be purified by repeated washing with hot - water, followed by drying.
materials: top loading balance, filter paper, alum, calcium chloride
procedure:
part a: weigh 1.25 to 1.35 g of aik(so₄)₂·12h₂o and 0.80 to 0.90 g of cacl₂ into separate 100 - ml beakers. record masses to the nearest 0.0001 g. add 30 ml of water to each beaker, heat the contents to above 60 °c. stir until all the solids are dissolved.
pour the calcium chloride solution into the alum solution, rinsing it in with about 3 ml of water. stir to mix and heat the beaker and contents, with stirring, on a hot plate for ten minutes on a lower setting (2 or 3). filter through a preweighed and initialed filter paper using a vacuum filtration aparatus. when all the liquid has passed through, fill the filter paper 1/3 full of hot water. let this water drain through the filter paper. then carry out a second similar rinse using acetone. let the liquid drain thoroughly, then place the filter paper and product on a weighed watch glass and place in the oven.
part b: repeat part a except use about 0.50 g of cacl₂.
after drying, weigh the filter paper with the gypsum and calculate the experimental yield of gypsum by difference. turn in your product in a labeled sample vial.
calculate the theoretical yield of gypsum, experimental yield, and the percent yield you obtained. also determine the limiting reagent, excess reagent, and the amount of excess reagent left.

trial	grams alum	grams calcium chloride	mass of filter paper	mass of watchglass	theo yld of gypsum	exp yld of gypsum	l. r.	ex. r.	% yld
b	1.3446	0.5145	0.2376	17.0031		0.4423

Explanation:

Step1: Write the balanced chemical equation

\[ \text{AlK(SO}_4)_2\cdot12\text{H}_2\text{O(aq)} + 2\text{CaCl}_2\text{(aq)} ightarrow 2\text{CaSO}_4\cdot2\text{H}_2\text{O(s)}+\text{AlCl}_3\text{(aq)}+\text{KCl(aq)} + 8\text{H}_2\text{O(l)} \]

Step2: Calculate molar - masses

The molar mass of \(\text{AlK(SO}_4)_2\cdot12\text{H}_2\text{O}\):
\[ \begin{align*} M_{\text{AlK(SO}_4)_2\cdot12\text{H}_2\text{O}}&=27 + 39+2\times(32 + 4\times16)+12\times(2\times1 + 16)\\ &=27+39 + 2\times(32 + 64)+12\times18\\ &=27+39+2\times96 + 216\\ &=27+39 + 192+216\\ &=474\ g/mol \end{align*} \]
The molar mass of \(\text{CaCl}_2\) is \(M_{\text{CaCl}_2}=40+(2\times35.5)=111\ g/mol\), and the molar mass of \(\text{CaSO}_4\cdot2\text{H}_2\text{O}\) is \(M_{\text{CaSO}_4\cdot2\text{H}_2\text{O}}=40 + 32+4\times16+2\times(2\times1 + 16)=172\ g/mol\)

Step3: Determine the limiting reagent for Trial A

For Trial A, \(m_{\text{AlK(SO}_4)_2\cdot12\text{H}_2\text{O}} = 1.3421\ g\), so \(n_{\text{AlK(SO}_4)_2\cdot12\text{H}_2\text{O}}=\frac{1.3421\ g}{474\ g/mol}=0.002831\ mol\)
\(m_{\text{CaCl}_2}=0.8941\ g\), so \(n_{\text{CaCl}_2}=\frac{0.8941\ g}{111\ g/mol}=0.008055\ mol\)
From the balanced equation, the mole - ratio of \(\text{AlK(SO}_4)_2\cdot12\text{H}_2\text{O}\) to \(\text{CaCl}_2\) is \(1:2\).
The amount of \(\text{CaCl}_2\) required to react completely with \(0.002831\ mol\) of \(\text{AlK(SO}_4)_2\cdot12\text{H}_2\text{O}\) is \(2\times0.002831\ mol = 0.005662\ mol\). Since \(0.008055\ mol>0.005662\ mol\), \(\text{AlK(SO}_4)_2\cdot12\text{H}_2\text{O}\) is the limiting reagent.

Step4: Calculate the theoretical yield for Trial A

From the balanced equation, 1 mole of \(\text{AlK(SO}_4)_2\cdot12\text{H}_2\text{O}\) produces 2 moles of \(\text{CaSO}_4\cdot2\text{H}_2\text{O}\).
So the theoretical yield of \(\text{CaSO}_4\cdot2\text{H}_2\text{O}\) is \(n_{\text{CaSO}_4\cdot2\text{H}_2\text{O, theo}} = 2\times0.002831\ mol = 0.005662\ mol\)
\(m_{\text{CaSO}_4\cdot2\text{H}_2\text{O, theo}}=0.005662\ mol\times172\ g/mol = 0.9739\ g\)
The experimental yield is given as \(m_{\text{CaSO}_4\cdot2\text{H}_2\text{O, exp}} = 0.5279\ g\)
The percent yield is \(\text{Percent Yield}=\frac{0.5279\ g}{0.9739\ g}\times100\% = 54.2\%\)
The amount of excess \(\text{CaCl}_2\) is \(n_{\text{CaCl}_2,\text{excess}}=0.008055\ mol - 0.005662\ mol = 0.002393\ mol\)
\(m_{\text{CaCl}_2,\text{excess}}=0.002393\ mol\times111\ g/mol = 0.2656\ g\)

Step5: Determine the limiting reagent for Trial B

For Trial B, \(m_{\text{AlK(SO}_4)_2\cdot12\text{H}_2\text{O}} = 1.3446\ g\), so \(n_{\text{AlK(SO}_4)_2\cdot12\text{H}_2\text{O}}=\frac{1.3446\ g}{474\ g/mol}=0.002837\ mol\)
\(m_{\text{CaCl}_2}=0.5145\ g\), so \(n_{\text{CaCl}_2}=\frac{0.5145\ g}{111\ g/mol}=0.004635\ mol\)
The amount of \(\text{CaCl}_2\) required to react completely with \(0.002837\ mol\) of \(\text{AlK(SO}_4)_2\cdot12\text{H}_2\text{O}\) is \(2\times0.002837\ mol = 0.005674\ mol\). Since \(0.004635\ mol<0.005674\ mol\), \(\text{CaCl}_2\) is the limiting reagent.

Step6: Calculate the theoretical yield for Trial B

From the balanced equation, 2 moles of \(\text{CaCl}_2\) produce 2 moles of \(\text{CaSO}_4\cdot2\text{H}_2\text{O}\).
So \(n_{\text{CaSO}_4\cdot2\text{H}_2\text{O, theo}} = 0.004635\ mol\)
\(m_{\text{CaSO}_4\cdot2\text{H}_2\text{O, theo}}=0.004635\ mol\times172\ g/mol = 0.7972\ g\)
The experimental yield is given as \(m_{\text{CaSO}_4\cdot2\text{H}_2\text{O, exp}} = 0.0423\ g\)
The percent yield is \(\text{Percent Yield}=\frac{0.0423\ g}{0.7972\ g}\times100\% = 5.31\%\)
The amount of excess \(\text{AlK(SO}_4)_2\cdot12\text{H}_2\text{O}\) is \(n_{\text{AlK(SO}_4)_2\cdot12\text{H}_2\text{O, excess}}=0.002837\ mol-\frac{0.004635\ mol}{2}=0.002837\ mol - 0.0023175\ mol=0.0005195\ mol\)
\(m_{\text{AlK(SO}_4)_2\cdot12\text{H}_2\text{O, excess}}=0.0005195\ mol\times474\ g/mol = 0.2462\ g\)

Answer:

Trial	Limiting Reagent	Theoretical Yield of Gypsum (g)	Experimental Yield of Gypsum (g)	Percent Yield of Gypsum	Amount of Excess Reagent Left (g)
B	\(\text{CaCl}_2\)	\(0.7972\)	\(0.0423\)	\(5.31\%\)	\(\text{AlK(SO}_4)_2\cdot12\text{H}_2\text{O}:0.2462\)