2. 0.9192 g of Cu(NO) 2.5H20 was dissolved completely in approximately 40-mL of

Question

2. 0.9192 g of Cu(NO) 2.5H20 was dissolved completely in approximately 40-mL of deionized water. Then, the solution was poured into in a 100.00-mL volumetric flask and water was added to the flask to the calibration mark. The solution is mixed thoroughly and labelled as "Stock Cu solution". Calculate the molarity of the stock Cu2 solution. (Hint: The recitation guide provides instructions on how to calculate the molar mass of Cu(NO) 2.5H20.) 3. Assume that you are preparing standard Cu(NHs) solution no. 4 (see lab manual Table 8.1). 8.00 mL of 0.0400 M stock Cu solution was mixed with 2.00 mL 15 M NHs(a?) Then, the solution was further diluted to 25.00 mL. by adding deionized water and mixed thoroughly. Note that because NHs(ag) is an excessive reagent in this reaction, all Cu2 in the solution is converted into Cu(NH)2 a. Determine the stoichiometric ratio of Cu2 to Cu(NHs)2 b. Calculate the moles of Cu(NHs) produced. c. Calculate the final molarity of Cu(NH,)2 in standard solution no. 4. 68, l mg of copper was found in a 2.534-g penny. Calculate the %mass of copper in the penny. (Hint: Make the units match!) 4,

Explanation / Answer

Solution:-(2) The given hydrate formula is Cu(NO3)2*2.5H2O

Molar mass = 63.546 + 2(14.01) + 6(16.00) + 2.5[2(1.01) + 16.00)]

= 63.546 + 28.02 + 96.00 + 2.5(18.02)

= 63.546 + 28.02 + 96.00 + 45.05

= 232.616 g/mol

Let's calculate the moles of the hydrate present in given sample.

0.9192 g x (1mol/232.616g) = 0.00395 mol

Volume of the solution is 100.00 mL that is 0.100 L.

So, the molarity of the solution = 0.00395 mol/0.100L = 0.0395 M

Since the hydrate has just one Cu in it, so the concentration of Cu2+ is also 0.0395 M that could be round to 0.0400 M.

(3) Cu2+(aq) + 4NH3(aq) -----> [Cu(NH3)4]2+(aq)

Moles of Cu2+ = 8.00 mL x (1L/1000mL) x (0.0400 mol/L) = 0.00032 mol

moles of NH3 = 2.00 mL x (1L/1000mL) x (15mol/L) = 0.0300 mol

(a) Cu2+ is limiting and there is 1:1 mol ratio between Cu2+ and [Cu(NH3)4]2+,

(b) As the mol ratio is 1:1, the moles of [Cu(NH3)4]2+ formed would 0.00032.

(c) Total volume = 8.00 mL + 2.00 mL + 25.00 mL = 35.00 mL = 0.03500 L

Molarity of [Cu(NH3)4]2+ = 0.00032 mol/0.03500L = 0.00914 M

(4) Mass of Cu = 68.1 mg = 0.0681 g

mass of penny = 2.534 g

mass percent of Cu = (0.0681/2.534)*100

mass percent of Cu = 2.69%

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