Gravimetric Analysis of Chloride in Solution Lab PDF

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Gravimetric Analysis Lab Report...

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Gravimetric Analysis of Chloride in Solution Lab Report Introduction: The purpose of this experiment is to determine the identity of a chloride-containing solute by reacting it with silver nitrate and producing some quantity of silver chloride to determine the amount of chloride in the sample. Experimental: We will measure some amount of the unknown chloride-containing solute and react it with silver nitrate in nitric acid to produce silver chloride. The nitric acid prevents the production of unwanted solutes. We will then dry out the AgCl precipitate and determine its mass to calculate the amount of chloride initially in the unknown sample. Data Analysis: Sample #1 (g)

Sample #2 (g)

Mass sample (g):

.2541

.2568

Mass of crucible:

30.6923

24.9161

Mass of crucible and precipitate:

31.3145

25.5608

Mass of precipitate only:

0.6222

0.6447

Percentage Cl -- in Sample #1: 0.6222 g AgCl x (1 mol AgCl / 143.32 g) = 0.004341 mol AgCl = 0.004341 mol Cl-- .004341 mol Cl-- x (35.453 g / 1 mol Cl) = .1539 g Cl-- (.1539 g Cl-- / .2541 g sample) * 100 = 60.57% Cl-- Percentage Cl -- in Sample #2: 0.6447 g AgCl x (1 mol AgCl / 143.32 g) = 0.004498 mol AgCl = 0.004498 mol Cl-- .004341 mol Cl-- x (35.453 g / 1 mol Cl) = .1595 g Cl-- (.1595 g Cl-- / .2541 g sample) * 100 = 62.10% Cl-- Average Percentage Cl -- in Samples #1-2: (60.57 + 62.10) / 2 = 61.34% Identity of Sample: Assume sample is in form XCl: Therefore, mass of X = 35.453 g / .6134 * .3866 = 22.34 g/mol with charge +1. The closest element is Na, with mass 22.99 with charge +1. Assume sample is in form XCl2: Therefore, mass of X = ((35.453 g * 2) / .6134 * .3866) = 44.69 g/mol with charge +2. Such an element does not exist in group 2. Thus, the sample must be in form XCl.

Discussion: In this lab, we were able to form an average of .63345 grams of silver chloride precipitate. Our two trials were within 0.0225 grams of each other, suggesting that our data was precise. From these obtained masses of AgCl precipitate, we were able to calculate the amount of chloride produced to be .1539 g and .1595 g in samples 1 and 2, respectively. These values were within 0.0056 g of each other, once again indicating the precision of our data. We were then able to use these values with the initial masses to determine the percent chloride by mass to be, on average, 61.34% in the unknown solute. Through this number, we were able to determine the identity of the solute to be NaCl. This makes sense because NaCl is a very common chloride that is relatively inexpensive and white in color, as was the unknown solute. 1. a. Gravimetric analysis is not used for determining trace amounts of ions because the method of gravimetric analysis is dependent on the assumption that the reactions being measured have gone to completion. This will likely not be the case with trace amounts of ions because reactions proceed with random collisions, and there won’t be enough ions in solution to complete the reaction (in a timely manner). b. Assuming Na+ is the correct cation, the percent error between our obtained molar mass for the cation and that of sodium is given by: ((22.34-22.99) / 22.99) * 100 = 2.827% Despite our error being very small, possible sources of error could have arisen from a few sources. Primarily, error could have arisen from the light sensitivity of silver chloride. We exposed this sample to light during the beginning of this experiment, and it could have caused some dissociation of AgCl into Ag+ and Cl- ions, which would cause us to underestimate the amount of chlorine. Another source of error could have been the slight solubility of AgCl in water. This could have also caused us to underestimate our amount of Cl- , as some of the solute may have dissolved (admittedly negligible amounts, but our error was also nearly negligible so this must be considered). 2. 0.523 g AgCl x (1 mol AgCl / 143.32 g) = .00365 mol AgCl = .00365 mol Cl- x (35.453 g / 1 mol Cl) = .129 g Cl-- → (.129 / .323) * 100 = 40.1% Cl-- ; The theoretical molar weight of the metal in a compound MxCly a. for x=1, y=1: 35.453 = .401(MMT) → MMT* .599 = 53.0 → M = 53.0 g/mol b. for x=1, y=2: (35.453*2) = .401(MMT) → MMT*.599 = 106 → M = 106 g/mol c. for x=2, y=4: (35.453*4) = .401(MMT) → MMT*.599 = 212 → M = 212 / 2 = 106 g/mol Conclusion: From this experiment, we determined the identity of a chloride-containing solute to be sodium chloride (NaCl). We obtained this identity through reacting the unknown chloride-containing solute with silver nitrate in acidic conditions to form a silver chloride precipitate. By determining the mass of the silver chloride, we were able to determine the amount of chloride precipitated, which is assumed to be equal to the amount of chloride in the initial sample. By using the percent mass of chloride in the compound, we were able to determine the cation in the unknown solute to be sodium, given that it expected the cation to have an atomic mass of 22.34 g/mol and a charge of +1. This atomic mass is within 2.837% of

the actual atomic mass of sodium (22.99 g/mol). This small error could have arisen from the light sensitivity of silver chloride and the solubility of silver chloride in water....