IB DIPLOMA CHEMISTRY LAB REPORTPRACTICAL 5

Title: Determining the Percentage of Yield of a salt from a reaction involving two substances.

Aim: To determine percentage yield of lead (II) iodide, PbI2, formed from the reaction of lead (II) nitrate, Pb (NO3)2 and potassium iodide, KI.

Research Question: What is the percentage yield of lead (II) iodide, PbI2 formed from the reaction of lead (II) nitrate, Pb (NO3)2) and potassium iodide, KI?

Hypothesis: The experiment mass or the amount of product actually obtained from a reaction is almost less than the theoretical mass.

Variables:Independent – Limiting ReagentDependent – The Experimental Mass

Constant – Temperature, pressure

Introduction: It is relatively easy to calculate the mass of the product that should be produced from a

known mass of a reactant. This is the theoretical yield. However, the mass of the product actually obtained, the experimental yield, is usually less than the theoretical yield. The relationship between these two yields is expressed in a quantity called the percentage yield.

Apparatus and Material:

1) 1 Measuring Cylinders (10 mL)2) 2 Pipets (5 mL)3) 1 Beaker (250 mL)4) 2 Volumetric flask (250 mL)5) Filter paper6) Filter funnel7) Analytical balance (+ 0.0001 g)8) Drying Oven9) Distilled water10) 50 mL of 1.0 M lead(II) nitrate, Pb(NO3)2

11) 50 mL of 1.0 M potassium iodide, KI

Methods:1) 5 mL of each of the two solutions carefully measured out using the different

pipette and the two solutions are added into the beaker. The mixed solution is swirled gently.

2) The mass of the filter paper is measured.

3) The filter paper is folded into quatres and it is placed in the filter funnel supported by a flask.

4) The mixture is poured into the funnel slowly. Be careful not to overflow the filter paper. A wash bottle of distilled water is used to wash any remaining solid out of the beaker and onto the filter paper.

5) After filtering process, the filter paper is dried in a drying oven set at the low temperature. When it is completely dry, its mass is determined. Be careful not to remove any of the solid products on the filter paper.

6) The entire procedure is repeated for two times, and average your three trial masses.

Results:Data Collecting

A. Quantitative Data

The table of the lead (II) iodide, PbI2 mass

First Reading Second Reading Third ReadingMass of the filter

paper (+ 0.0001 g)0.8279 0.8254 0.8233

Mass of the filter paper and the lead (11) iodide, PbI2

(+ 0.0001)

1.5570 1.6110 1.7478

Mass of the lead (II) iodide, PbI2 (+ 0.0001)

0.7291 0.7876 0.9245

B. Qualitative Data

1. T he volume of 1.0M lead(II) nitrate, Pb(NO3)2 solution used is 5.0 ±0.5 cm³2. The volume of 1.0M potassium iodide, KI solution used is 5.0 ±0.5 cm³3. The salt which produced from the mixture of the two solutions is yellow in colour

Data Processing

1. The equations of the reactions a. The Chemical equation is

Pb (NO3)2 + 2KI PbI2 + 2KNO3

b. The ionic equation isPb2+ + I- PbI2

2. The average amount of the lead (II) iodide

From the table above, the average amount of the lead (II) iodide is

Average of PbI2 = 0.7291 + 0.7876 + 0.9245 3

= 0.8137 g

The uncertainty of PbI2 = (0.8137-0.7291) + (0.8137-0.7876) + (0.8137-0.9245)

3 = 0.073 x 100% = 7.38 %

Hence the average amount of lead (II) iodide is 0.8137 ± 7.38% g

3. The number of moles of lead (II) nitrate, Pb (NO 3)2 and potassium iodide, KI

From the equation above, we can determine the number of moles of lead (II) nitrate, Pb (NO3)2 and potassium iodide, KI,

The no. of moles of Pb (NO3)2 = MV M = The molarity of Pb (NO3)2, 1.0 1000 moldm-3

= (1.0 x 5) V = 5 mL of Pb (NO3)2

1000= 0.005 mole

The no. of moles of KI = MV M = The molarity of KI, 1.0 1000 moldm-3

= (1.0 x 5) V = 5 mL of KI 1000

= 0.005 mole

4. The limited reagent of the reaction

Pb (NO3)2 + 2KI PbI2 + 2KNO3

According to the equation, 1 mole of Pb (NO3)2 reacted with 2 moles of KI.

There is 0.005 mole of Pb (NO3)2 used, so number of mole of KI should be used is 0.010 mole. But there is only 0.005 mole of KI present.

Hence the Limited Reagent is potassium iodide which is 0.005 mole.

5. The exact number of mole of PbI 2 obtained

In order to determine the exact number of mole, we must use the no. moles of the limited reagent, potassium iodide to determine the theoretical mass of lead (II) iodide, PbI2,

From the above chemical equation, 2 moles of KI will produce 1 mole of PbI2

So 0.005 mole of KI will produce 0.0025 mole of PbI2

Hence the exact number of mole of PbI2 is 0.0025 mole.

6. The percentage yield of PbI 2

The theoretical mass of PbI2 = No. of mole x JMR= 0.0025 x (207.19 + 2(126.90))= 1.1525 g

Therefore the percentage of yield of salt, lead (II) iodide, PbI2 is

The percentage of yield = experimental mass (average) x 100%theoretical mass

= 0.8137 g x 100% 1.1525 g = 70.6 %

Hence the percentage yield of lead (II) iodide, PbI2 obtained is 70.6%.

Conclusion: The percentage of yield of salt, lead (II) iodide, PbI2 is 70.6%.The result showed that the yield is below 100% and the hypothesis is accepted.

Limitation and Recommendation

1) The two solutions did not mix well so the mass of salt is lower than the theoretical mass. The mixture must swirl gently to make sure the mixture mix well.

2) The left remaining solids still in the beaker after the mixture was pouring into the filter funnel. We must wash the beaker with the distilled water so any remaining solids not left out. The step must be done because the remaining solids can decrease the mass of the product.

3) All of the mixture not filtering, so the product maybe contained water. My suggestion is we must filter all the mixture by measure the drop of water dropped into the beaker. The less the drop of water, the less water in the product.

4) The solids not completely dry in the drying oven, so when we measured the product using analytical balance, it maybe contained water. My suggestion is the solids must put in the drying oven more than 45 minutes and less than 1 hour to get the approximate measurement.

5) When we measure the solids using analytical balance, some of the product not measure because it was outside the rounded mass. To get the result which has the less uncertainly, we must take and put all the solids in the centre of the rounded mass.