Resolution of Matter into Pure Substances II: Fractional Crystallization Answer Key

Answers

1. Data-sheet calculations

Use the masses recorded on your own report sheet. The arithmetic is:

$$m_{sample}=m_{sample+150\,mL\,beaker}-m_{150\,mL\,beaker}$$ $$m_{SiC}=m_{SiC+filter}-m_{SiC\,filter}$$ $$\%SiC=\frac{m_{SiC}}{m_{sample}}\times100\%$$ $$m_{KNO_3,\ recovered}=m_{beaker+KNO_3}-m_{empty\,beaker}$$ $$\%\text{ sample recovered as }KNO_3=\frac{m_{KNO_3,\ recovered}}{m_{sample}}\times100\%$$

For the recrystallized product:

$$m_{pure\ KNO_3}=m_{filter+pure\ KNO_3}-m_{KNO_3\ filter}$$

The copper(II) sulfate pentahydrate percentage is not obtained from a mass subtraction. Match the blue color of the ammonia complex in the test solution to the prepared standards, then report the standard's corresponding mass percentage. If the sample is lighter than the lowest standard, report it as below that method's detection range rather than exactly zero.

2. What each separation step accomplishes

• Hot-water dissolution: KNO3 and CuSO4·5H2O enter solution while insoluble SiC remains solid.
• First filtration: the SiC fraction is collected. Any solution retained in the filter cake lowers soluble-product recovery.
• Cooling crystallization: KNO3 becomes much less soluble and crystallizes; the smaller amount of copper sulfate is intended to remain mainly in the mother liquor.
• Cold washing: a small amount of cold water removes adhering mother liquor and improves purity, but too much wash water dissolves product and lowers yield.
• Recrystallization: redissolving the crude KNO3 in a minimum of hot water and cooling it again rejects more copper impurity into the mother liquor.

3. Advance Study Assignment calculations

The supplied solubility graph gives approximately 180 g CuSO4·5H2O and 240 g KNO3 per 100 g water at 100 °C.

For 3.9 g CuSO4·5H2O:

$$3.9\ \mathrm{g}\times\frac{100\ \mathrm{g\ H_2O}}{180\ \mathrm{g}} =2.2\ \mathrm{g\ H_2O}$$

For 37 g KNO3:

$$37\ \mathrm{g}\times\frac{100\ \mathrm{g\ H_2O}}{240\ \mathrm{g}} =15\ \mathrm{g\ H_2O}$$

The larger requirement controls, so about 15 g of water can dissolve both substances at 100 °C under the exercise's assumption that each solute does not appreciably change the other's solubility.

After adding 12 g more water, the solution contains 27 g water. At 0 °C, the graph gives about 10. g KNO3 and 20. g CuSO4·5H2O per 100 g water.

$$m_{KNO_3,\ dissolved}=27\times\frac{10.}{100}=2.7\ \mathrm{g}$$ $$m_{KNO_3,\ crystals}=37-2.7\approx34\ \mathrm{g}$$

The 27 g of water could retain about $27\times20./100=5.4$ g CuSO4·5H2O, more than the 3.9 g present, so the idealized calculation predicts no copper sulfate pentahydrate crystallization.

$$\%KNO_3\ recovery=\frac{34}{37}\times100\%\approx92\%$$

4. Yield versus purity

The first crystallization normally gives more product but may trap more copper-containing mother liquor. Recrystallization improves purity, yet some KNO3 remains dissolved or is lost during transfers and washing. A lower mass after recrystallization is therefore not automatically an error; interpret it together with the copper-impurity test.

Conclusion

Complete the report sheet with your measured masses and color comparison. The fixed solubility exercise predicts about 34 g KNO3 crystallized, 92% recovery, and no CuSO4·5H2O crystallized under its idealized assumptions; those figures are not substitutes for the experimental recovery data.