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A 0.132-mol sample of an unknown semiconducting material

Chemistry | 8th Edition | ISBN: 9780547125329 | Authors: Steven S. Zumdahl ISBN: 9780547125329 153

Solution for problem 141 Chapter 10

Chemistry | 8th Edition

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Chemistry | 8th Edition | ISBN: 9780547125329 | Authors: Steven S. Zumdahl

Chemistry | 8th Edition

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Problem 141

A 0.132-mol sample of an unknown semiconducting material with the formula XY has a mass of 19.0 g. The element X has an electron configuration of [Kr]5s 2 4d10. What is this semiconducting material? A small amount of the Y atoms in the semiconductor is replaced with an equivalent amount of atoms with an electron configuration of [Ar]4s 2 3d104p5 . Does this correspond to n-type or p-type doping?

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Week 3 Notes Recrystalization- Single Solvent Goal: Purify an impure or crude experimentally synthesized product Basic Principle: Solubility of a compound usually increases with increasing temperature of the solvent (solubility depends on concentration, polarity, and temperature) Major Difficulty: Most impurities and reaction by-products have similar solubility properties as the desired product. 1. Dissolve the impure product in a minimum amount of a soluble hot solvent. 2. Filter the hot solution to remove any insoluble impuriies. 3. Let the solution cool slowly to allow the pure crystal form without trapping impuities. 4. Filter the purified product Solubility in Recrystalization Example 2: Suppose we take 25 g of sugar and dissolve it in 10 mL of boiling water. We then allow the solution to slowly cool down to room temperature. Will we observe any sugar crystals in the cooled solution Example 3: Suppose we take 25 g of sugar and dissolve it in 12.5 mL of boiling water. We then allow the solution to slowly cool down to room temperature. Will we observe any sugar crystals in the cooled solution Solubility of Sugar in Water Temperature 1.80 g/mL 25 C Using excess solvent in a recrystallization will reduce 2.50 g/mL 100 C the amount of recovered purified product (lower yield!) Example 2: At 100 C: 2.50 gml×10ml = 25g [Can dissolve all sugar] o 1.80 ×10ml=18g [Cannot dissolve all sugar] At25 C: ml 25g initially dissolved - 18 g remaining in solution = 7 g or sugar cyrstals formed Example 3: At 100 C: 2.50 gml×12.5ml = 31.25g [Can dissolve all sugar] At 25 C: 1.80 g ×12.5ml = 22.5g [Cannot dissolve all sugar] ml 25g initially dissolved - 22.5 g remaining in solution = ONLY 2.5 g of sugar crystals formed Using Recrystalization to Remove Impurities Consider the example system where we have 10 g of desired product (A) and 2 g of impurity (B) Temp = 20°C Temp =100°C solubility (0.1 g/mL 1 g/mL solubility (B0.1 g/mL 1 g/mL Dissolve all crude product (10 gAand 2 g B) in 12 mL of solvent at 100oC. Cooling to 20oC, 1.2 g ofAand and 1.2 g of B remain in solution while 8.8 g ofAand 0.8 g of B crystalize out. Before recrystallization 10g / 12g ×100% = 83.3% 12g After recrystallization 8.8g / 9.6g ×100% = 91.7% Choosing a Good Crystallization Solvent A good crystallization solvent should: Be soluble for (dissolve) a large quantity of the product at high temperature Be insoluble for (not dissolve) the product at low temperature Be soluble for (dissolve) the impurities at all temperatures [Optional, becuase very difficult to achieve] Not react with the product Considering Crystalization of Glycine Example 4: Is water or ethanol a better solvent for glycine Solubility of Glycine in Water Temperature Solubility of Glycine in Ethanol Temperature 14.0 g/100 mL 0 C 0.01 g/ 100 mL 0 C 25 g/ 100 mL 25 C 39 g/ 100 mL 50 C 45 g/ 100 mL 25 C 54 g/ 100 mL 75 C o o 65 g/ 100 mL 65 C 67 g/ 100 mL 100 C Considering Crystalization of Glycine Example 4: Is water or ethanol a better solvent for glycine Use percent recovery to decide (amount dissolved at high T - amount at low T / amount at high T) x 100% = Percent Recovery Using water as the solvent: (67 g -14 g) / 67 g % recovery = ×100% = 79% Using ethanol as the solvent: [Preferred Solution] % recovery =(65 g - 0.01g) / 6×100% = 99.98% Synthesis of Aspirn Lab salicylic acid (limiting reagent) + acetic anhydride ——> aspirin + acetic acid Example 5: When performing the aspirin lab, you weigh out exactly 3 g of salicylic acid (SA). Given that you made 3.21 g of aspirin (A) after recrystallization, what is the % yield 3gSA 1mol SA / 138.1g SA) 1mol A / 1 mol SA)180.2g A / 1 mol A=3.91g % yield = actual amount / theoretical amount ×100% = 3.21 g / 3.91 g×100% = 82.1% Recrystalization of Aspirin What solvent do we use to recrystalize aspirin Water: not very soluble Ethanol: way too soluble Solubility ofAspirin in Water Temperature Solubility of Aspirin in Temperature 0.025 g / 100 mL of wat0 C Ethanol 0.30 g / 100 mL of wate25 C 20.0 g / 100 mL 25 C of ethanol 1.0 g / 100 mL of water37 C 35.0 g / 100 mL o of ethanol 70 C Decomposes at high T o 100 C Mixed Solvent System for Recrystalization • Made with two mutually miscible (mixable) liquids • One of the two solvents will dissolve the product more readily than the other solvent • Usually works better than single solvent system We will use a mixed solvent recrystallization for purifying aspirin – use a mixture of water and ethanol! Procedurte for Recrystalization with Mixed Solvent System • Dissolve the impure product in the high-solubility solvent (ethanol) at high temperature. • Add the low-solubility solvent (water) drop-wise until solution turns cloudy. • Heat solution to re-dissolve all crystals. Add a few drops of high- solubility solvent (ethanol) if required to force all the crystals back into solution • Allow solution to cool slowly to allow the pure products to form without trapping impurities. Graphs •Online in powerpoint provided and in notes

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Chapter 10, Problem 141 is Solved
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Textbook: Chemistry
Edition: 8
Author: Steven S. Zumdahl
ISBN: 9780547125329

The full step-by-step solution to problem: 141 from chapter: 10 was answered by , our top Chemistry solution expert on 11/15/17, 04:25PM. This textbook survival guide was created for the textbook: Chemistry, edition: 8. The answer to “A 0.132-mol sample of an unknown semiconducting material with the formula XY has a mass of 19.0 g. The element X has an electron configuration of [Kr]5s 2 4d10. What is this semiconducting material? A small amount of the Y atoms in the semiconductor is replaced with an equivalent amount of atoms with an electron configuration of [Ar]4s 2 3d104p5 . Does this correspond to n-type or p-type doping?” is broken down into a number of easy to follow steps, and 69 words. Since the solution to 141 from 10 chapter was answered, more than 274 students have viewed the full step-by-step answer. This full solution covers the following key subjects: amount, atoms, configuration, type, semiconducting. This expansive textbook survival guide covers 22 chapters, and 2897 solutions. Chemistry was written by and is associated to the ISBN: 9780547125329.

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