Question

1. preparing and mixing the reactants

mass of weighing boat and CuSO4*5h2O, g : 2.66g

Mass of weighing boat, g: .66g

mass of CuSO4*5h2O,g: 2.00g

Molarity of ammonia solution, M: ????

in this part we added 2.00g of CuSO4*5h2O, and added 6.4mL of water. we then heated the mixture and added 6.45mL of NH3. after it was done heating, 8.3mL of room temp ethanol was added. we then made an ice water bath, and put the beaker in the ice water bath for 15 mins.  how do I calculate the molarity of ammonia solution??

III. Processing the product

in this part we prepared EtOH (10mL) in a test tube 1 and put it in an ice water bath. in another test tube 2 was 10mL of EtOH and 10mL of NH3 which was also placed in the water bath. we had to isolate the beaker from part 1 with the solution by a vacuum filtration. the solution was transferred into a filter flask, and was attached to the water aspirator vacuum. the remaining solid that was left in the beaker was rinsed out with the NH3 ethanol solution that was in test tube 2. then we washed the the solid in the funnel with test tube 1 that contained only ethanol. the powder remaining in the funnel was transferred into a weighing boat.

mass of weighing boat and [Cu(NH3)4]SO4*H2O,g: 4.64g

mass of weighing boat, g: 2.28g

mass of product, g: 2.36g

# of moles CuSO4*5h2O added, mol: ??

# moles of NH3 added, mol: ??

formula of the limiting reagent: ??

theoretical yield of [Cu(NH3)4]SO4*H2O, g : ???

percent yield of [Cu(NH3)4]SO4*H2O, % : ???

things you need to know: we used concentration 28% NH3, density is 0.9g/moL, and we used 95% ethanol. please show your work as I am super confused on how to do calculations and what to use when calculating. thank you!

Answer 1

Molarity of ammonia:

28% solution = 28 g in 100 ml

Moles of ammonia = 28 g/17 g/mol = 1.647 mol

Molarity = (1.647/100 ml)*1000 ml = 16.47 M

III) Processing the product:

a) Moles of CuSO4.5H2O :

Molar mass of CuSO4.5H2O = 159.61 g/mol

Number of moles = 2.00 g/(159.61 g/mol) = 0.0125 mol

b) Moles of ammonia added = molarity * volume (in L)

= 16.47 M * (10 ml/1000)

= 0.1647 mol

c) As per balanced equation:

CuSO₄ + 4NH₃ -----> [Cu(NH₃)₄]SO₄

For 0.0125 moles of CuSO4, moles of NH3 required = 0.0400 mol

Given amount I.e. 0.1647 mol is more than required.

Hence, limiting reagent is CuSO4.5H2O

d) Theoretical yield of [Cu(NH3)4]SO4.H2O :

Molar mass = 245.79 g/mol

Moles to be formed = 0.0125 mol

Mass to be formed = 0.0125 mol * 245.79 g/mol

= 3.072 g

e) percent yield:

% yield = (Observed mass/Theoretical mass)*100

= (2.36 g/3.07 g)*100

= 76.9 %

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Ammonia is assumed to be 28 % (w/v)

In case, ammonia given is 28 % (w/w)

Then 28 g of ammonia is present in 100 g

Volume of 100 g solution = mass/density

= 100 g/0.9 g/ml

= 111.11 ml

Molarity = (28/17)*(1000/111.11)

= 14.82 M

Moles of ammonia = (14.28 M)*(10 ml/1000 ml)

= 0.1428 mol

Taking these calculations also, CuSO4.5H2O is limiting reagent.