CHEM 113
GENERAL CHEMISTRY LABORATORY
YIELD OF A CHEMICAL REACTION
Submitted by
PARTNER’S NAME
EXPERIMENT DATE
SUBMISSION DATE
Yeditepe University
İstanbul
2
1. AIM OF THE EXPERIMENT
The aim of the experiment is to determine the experimental (actual) and theoretical
yields, and the percent yield of a reaction.
2. INTRODUCTION
In the present experiment, equal volumes of aqueous
(Pb(CH3COO)2) and 0.3 M sodium phosphate (Na3PO4)
to react to completion.
solutions were mixed and allowed
The experimental and theoretical yields of the precipitate, lead
phosphate, were determined and percent yield was calculated.
3. EXPERIMENTAL SET-UP
3.1. Equipment
- 2 Volumetric flasks (50 mL)
- Beaker (250 mL)
- Glass rod
- Buchner funnel
- Filtering flask
- Filtering adaptor
- Filter paper
- Watch glass
- Dropper
- Spatula
- Laboratory scale
3.2. Chemicals
- 0.2 M Pb(CH3COO)2 (Lead(II) acetate) solution
- 0.3 M Na3PO4 (Sodium phosphate) solution
- Distilled water (H20)
0.2 M lead(II) acetate
3
4. OBSERVATIONS, DATA and CALCULATIONS
4.1. Data:
Table 5.1. Experimental Data
Mass of filter paper
0.76 g
Mass of filter paper + precipitate
3.00 g
Mass of dry precipitate
2.24 g
(experimental/actual yield)
4.2. Observations: A white precipitate, assumed to be pure lead(II) phosphate-Pb3(PO4)2- was
observed.
Reaction equation
(molecular): 3Pb(CH3COO)2(aq) + 2Na3PO4(aq) → Pb3(PO4)2(s) + 6NaCH3COO(aq)
(ionic) :
3Pb(aq)2++6CH3COO-(aq)+ 6Na(aq)++ PO4(aq)3-→
Pb3(PO4)2(s)+ 6Na(aq)+6CH3COO(aq)
(net ionic) : 3Pb(aq)2++2 PO4(aq)3-→ Pb3(PO4)2(s)
4.3. Calculations:
Theoretical yield:
nPb(CH3COO)2 =0.2
mol
×0.05 L=0.01=nPb moles
L
mol
×0.05 L=0.015=nPO4 3- moles
L
1 mol ppt
nPb3 (PO4 ) =0.01 mole Pb×
=0.0033 moles
2
3 mol Pb
nNa3 PO4 =0.3
MW(Pb3 (PO4 ) =811.54 g/mol
2
m(Pb3 (PO4 ) =3.3×10-3 ×
2
811.54g
mol
=2.68 g (Theoretical yield)
% yield:
% yield=
experimental yield
×100
theoretical yield
4
=
2.24 g
×100=83.6 %
2.68 g
5. DISCUSSION and CONCLUSIONS
The experiment was about determining the yield of the precipitate lead (II) phosphate in
the reaction between aqueous lead (II) acetate and aqueous sodium phosphate.
As seen in the above part of the report, the percent yield of the product lead(II) phosphate
(Pb3(PO4)2, the precipitate), was calculated to be 83.6 %.
Although according to the internet survey conducted, [2] an experimental yield above 80
% is considered to be “very good”, it has been noted that the removal of the filter paper with
the precipitate from the Buchner funnel in the experiment was not perfect. Some of the
precipitate was stuck to the sides of the funnel and could not be thoroughly recovered. This
loss could have been avoided if the mixture was more slowly poured onto the filter paper in the
Buchner funnel. On the other hand, the fact that drying of the precipitate took place in the
laboratory hood, not in a drying oven in which ideal conditions exist for moisture removal,
sheds some doubts on the measured dry mass of the precipitate. It could be that the mass lost
during the removal of precipitate from the funnel was compensated to some degree by moisture
that could not be removed from the precipitate in the laboratory hood.
Since the solubility of Pb3(PO4)2 in water is so very low [1] (1.37x10-5 g in 100 mL
water; molar solubility of 1.69x10-7 mol/L) the fact that some of the precipitate has dissolved
in the solution cannot possibly be shown as a source of loss in yield.
The present experiment has resulted in a percent yield of 83.6 % for the precipitation
reaction between aqueous lead(II) acetate and aqueous sodium phosphate solutions, using the
experimental and the theoretical yields of the product lead(II) phosphate (Pb3(PO4)2 .