Question

Suppose you synthesized a Salt with formula Ni(en)2(H2O)2SO4•5H2O. In your synthesis you used 0.037mol of NiSO4.6H2O and 0.007mol of en. The actual yield (salt weighed out after synthesis) is 0.824g. Calculate the percent yield. Report your answer in % but don't report the units (e.g. 58 for 58%)

Answer 1

Answer on Question#75716 – Chemistry – General chemistry

Question: Suppose you synthesized a Salt with formula $\mathrm{Ni(en)_2(H_2O)_2SO_4\cdot 5H_2O}$. In your synthesis you used $0.037\mathrm{mol}$ of $\mathrm{NiSO_4.6H_2O}$ and $0.007\mathrm{mol}$ of en. The actual yield (salt weighed out after synthesis) is $0.824\mathrm{g}$. Calculate the percent yield. Report your answer in % but don't report the units (e.g. 58 for 58%)

Solution:

en is ligand with chemical formula $\mathrm{C_2H_8N_2}$

$$\mathrm{Ni(en)_2(H_2O)_2SO_4\cdot 5H_2O} = \mathrm{Ni(C_2H_8N_2)_2(H_2O)_2SO_4\cdot 5H_2O}$$

A reaction of synthesis the salt:

$$\mathrm{NiSO_4\cdot 6H_2O} + 2\mathrm{C_2H_8N_2} \rightarrow \mathrm{Ni(C_2H_8N_2)_2(H_2O)_2SO_4\cdot 5H_2O}$$ $$0.037 \mathrm{~mol} \text{ of } \mathrm{NiSO_4\cdot 6H_2O} > \frac{0.007 \mathrm{~mol} \text{ of } \mathrm{en}}{2} = 0.0035 \mathrm{~mol} \text{ of } \mathrm{en}$$

Therefore, en is limiting reagent.

Theoretical yield of the salt in moles = $\frac{0.007 \mathrm{~mol}}{2} = 0.0035 \mathrm{~mol}$

$$\mathrm{M(C_2H_8N_2)} = 60.10 \mathrm{~g/mol}$$ $$\mathrm{M(Ni(C_2H_8N_2)_2(H_2O)_2SO_4\cdot 5H_2O)} = 401.06 \mathrm{~g/mol}$$ $$n(\text{salt}) = \frac{m(\text{salt})}{M(\text{salt})} = \frac{0.824 \mathrm{~g}}{401.06 \mathrm{~g/mol}} = 0.00205 \mathrm{~mol} - \text{actual yield in moles}$$ $$\% \text{yield} = \frac{0.00205 \mathrm{~mol}}{0.0035 \mathrm{~mol}} \times 100\% = 59\%$$

Answer: 59.

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