• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

An Experiment to Determine the Empirical Formula of Lead Iodide

Extracts from this document...

Introduction

An Experiment to Determine the Empirical Formula of Lead Iodide Aim: The empirical formula shows the simplest whole number of ratio of atoms of each element in a molecule of a compound. In this experiment, we will determine the empirical formula of lead iodide. An insoluble lead iodide compound is formed by dissolving the lead in nitric acid and reacting with potassium iodide. Then, the lead iodide is filtered and dried for a week. In the end of the experiment, the mass of lead, filter papers and filter paper with lead iodide are recorded. The ratio of moles of lead to moles of iodine in the compound can be obtained either by weighing/calculating the mass of each element from the experiment raw data. In order to get the mass of iodide in the lead iodide compound, we can calculate by deducting the initial mass of lead from the mass of the precipitate. By knowing the mass of each element, we can determine the amount of each element (moles) than the ratio of the compound. . Variables: Controlled 30 cm3 3M HNO3 ; distilled water ; 1.2g potassium iodide (KI) Independent Amount of Lead (Pb) Dependent Amount of lead iodide formed Method for controlling variables: Glassware, electronic balance Materials: Equipment: - 3 x 250 cm3 beaker - 1 x 150 cm3 beaker - 3 x watch glass - 1 x Electronic balance (300 g, readability �0.001 g) ...read more.

Middle

21) The precipitate was washed twice with cold distilled water and transferred completely from the beaker to the filter paper. 22) After the water was completely drained through the filter, the filter paper I, II and III were removed from the funnel and were dried by wrapping it in paper towel to absorb the water. 23) Each filter paper was air dried for one week at room temperature. 24) Each filter paper with lead iodide was weighed for each trial on the electronic balance to the nearest 0.001 g and the mass was recorded. Data Collection and Processing Raw Data: Table 1: Mass of lead, filter paper and filter paper with lead iodide Run 1 Run 2 Run 3 Mass of granulated lead /g � 0.001 g 0.151 0.152 0.129 Mass of filter paper /g � 0.001 g 0.837 0.858 0.865 Mass of filter paper with lead iodide /g � 0.001 g 1.129 1.168 1.127 Observations: Color of solution: Potassium iodide and nitric acid containing lead were all colorless. When the potassium iodide was added in the beaker containing lead and nitric acid, the solution turned into bright yellow component. State of solution: Potassium iodide and nitric acid containing lead were liquid. When the potassium iodide was added in the beaker containing lead and nitric acid, an aqueous component is formed (lead iodide). ...read more.

Conclusion

0.129 6.23 1:1.7 Average 0.144 1.13 0.144 6.95 1:1.6 Conclusion: The ratios of the mole of lead to Iodine were calculated and are 1: 1.6. We can assume that the compound is lead iodide PbI2 since the oxidation status of lead is either 2 or 4. Therefore, the theoretical ratio of iodine to lead is 1:2. Compared to the ratio in this experiment, the iodine in the theoretical ratio is greater by 0.4 than that of obtained calculation from our experiment. Moreover, the percentage error of each calculation is below 5%, thus, the results are acceptable. If the percentage error is larger than 5%, the experiment is not valid. Evaluation: However, the result 1:1.6 obtained is slightly different from the theoretical possible ratio 1:2. Hence, the component was not completely gathered. The product may have escaped when the watch glass was lifted, lost during filtration, or left on the wall of the beaker containing lead iodide. Furthermore, the purity of lead will influence the ratio. The lead used might not be pure. If the lead was impure, other product apart from lead iodide would be formed. To further improve the experiment, we must avoid lead iodide to escape, avoid lead iodide on the funnel, and filter all the lead iodide in the beaker. These would give greater ratio of iodine to lead. Reference: - http://en.wikipedia.org/wiki/Lead_iodide ?? ?? ?? ?? European International School Manila Page 1 of 5 ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our International Baccalaureate Chemistry section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Chemistry essays

  1. Experiment - The Empirical Formula of Magnesium Oxide

    Theoretically, the empirical formula for magnesium oxide should be MgO because MgO is an ionic compound. When magnesium atoms and oxygen atoms react with each other, magnesium atoms lose 2 electrons to become Mg2+ (magnesium ions have a valence of 2+ charge)

  2. Preparation and Composition of Tin (IV) Iodide

    1st equation IO3- + 5I- + 6H+ ----------> 3I2 + 3H2O 2nd equation IO3- + 2I2 + 6H+ + 5Cl- ----------> 5ICl + 3H2O Proving that KIO3 is equivalent to 2I- We multiply the 1st equation by 2 and the 2nd equation by 3.

  1. Limiting Reagent Lab. Purpose To ...

    mPbI = 1.41 � 0.10g = E (Table 3) 2 E Percent yield = 100 � T 1.41� 100 1.111 = 127 % (Table 3) The percent yield exceeds 100%, which suggests that more was produced than PbI2 expected. This value is explainable when it is presumed that substance other than PbI2Pb(NO3)2 KNO3 , or was left in the filter paper when it was removed from the funnel.

  2. Lab Experiment : The change in mass when magnesium burns. (Finding the empirical formula ...

    Based on the data processing in trail run, 1,2,3,4 the Mg/O ratio of 1.2 and I would have conclude from my experiment that the empirical formula of magnesium oxide is Mg6O5 . My method for determining this is as follows: Explanation of conclusion: If the Mg/O ratio is 1.2, that means that the empirical formula would be Mg1.2 O.

  1. Investigating Stoichiometry - The table shows the mass of reactants potassium iodide and lead(II) ...

    Pb(NO3)2 x = 0.0038646176mol PbI2 Using KI: Moles of KI = 1.701g KI x = 0.010246988mol KI Moles of PbI2 = 0.010246988mol KI x = 0.005123494 mol PbI2 ?Pb(NO3)2 is the limiting reagent.

  2. Potassium Iodide Lab

    Solution was then transferred to filter paper *Two filter papers were used for double filtration 6. The filter paper was put into a funnel of a beaker to the solution would separate VI. Data: a) mass of filter papers: b)

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work