Question

QUESTIONS 1. Using the MSDS for copper(II) phosphate, is there any danger with this compound? Iuitatimoy te eyes, espirateuy aystem and sKimt May Lerg-term Altre, mawsra, vemitimy and burmime foelima's and othus 2. If your filtrate is blue, why is it this color? caue aaveve elfects un thu bquatic enirom mend

Answer 1

Q#2:

The presence of blue color in the filtrate indicates the presence of copper(II) salt in the filtrate. The product copper(II) phosphate is insoluble in water (insoluble in filtrate). However, CuCl₂ is soluble in water (soluble in filtrate). Therefore, the blue color in the filtrate is due to the presence of CuCl₂. Since, CuCl₂ is the limiting reagent therefore, this fact indicates that the reaction between Na₃PO₄ and CuCl₂ is not complete.

Q#3:

Since, CuCl₂ is the limiting reagent therefore, if the filtrate is colorless that means the reaction between Na₃PO₄ and CuCl₂ is complete. In the filtrate excess sodium phosphate and generated sodium chloride would be present. Both these sodium salts are colorless.

Q#4:

Since, CuCl₂ is the limiting reagent therefore the yield of the products should be calculated based on the number of moles of CuCl₂ used in the reaction.

Theoretically, 3 mmol of CuCl₂ and 2 mmol of Na₃PO₄ produce 1 mmol of copper(II) phosphate and 6 mmol of sodium chloride. Thus, theoretically, 5.98$\times$10^-3 mol = 5.98 mmol of CuCl₂ will give 5.98$\times$2 = 11.96 mmol of NaCl. The molecular weight of NaCl = 58.44 g/mol. The weight of NaCl = 11.96$\times$10^-3 mol $\times$ 58.44 g/mol =0.698 g (answer)

The percent yield of copper phosphate will be higher than 100 % if the NaCl is not washed. Because, the obtained product would be mixture of copper phosphate 0.867 g (theoretical value for 100 % yield) and NaCl. Thus, the total weight of the product would be more than 0.867 g. Hence, during yield calculation, the percent yield would be more than 100 %.

Q#5a:

The yield would be inaccurate (more than 100 %) if the copper phosphate is not dried. Because, the yield calculation would be based on the molecular formula of copper phosphate. However, the wet copper phosphate would contain some water molecules. Hence, the molecular weight of wet copper phosphate would be more than water free copper phosphate. Therefore the obtained copper phosphate would be more than 0.867 g and the yield would be more than 100 %.

Q#5b:

If residue is not washed than the product would contain copper phosphate and sodium chloride. Therefore, the yield would be more than 100 % (see answer to Q#4).

Q#5c:

If copper(II) phosphate is not completely transferred from the beaker to the filtration apparatus then theoretically the yield would be less than 100 % as the yield is calculated based on weight of the obtained product in the filtration apparatus.