Report

Determining Mr. Lawson’s Wealth ($) 1. At the front of the room is a Silver coin 2. With your phone take a close-up picture of BOTH sides of a coin 3. Examine the picture CLOSELY and determine how many TOTAL atoms of silver make up Mr. Lawson’s net worth! 4. DO NOT OPEN THE CASE, Mr. Lawson is a new teacher and does not get paid very much… • Not all materials are pure (entirely made of JUST the chemical we are interested in) • In order to find the purity of a substance we use the following formula… percent purity = mass of pure chemical x100% = mass of impure chemical pure chemical = 0.9995 x 1.000oz = 0.9995oz Percent Purity 31.10348g 1 mol 6.02x10 23 atoms ? atoms Ag = 0.9995oz Ag x x x = 1.73x10 23 atoms 1oz 107.87g 1 mol Answer: Percent Purity Percent Yield Chemicals don’t always exist in pure form. • The purity of a chemical is indicated as the percentage purity. • The impure substance contains another substance to make the mass higher than a pure substance Percent Purity V. Percent Purity An analogy What is the percent purity of gold in gold quartz? When refined the gold weighs 6.0 g Gold Quartz Weighs 28.0 g V. Percent Purity Percent Purity = Mass of pure chemical Mass of impure chemical x 100 % What is the percent purity of gold in gold quartz? When refined the gold weighs 6.0 g Gold Quartz Weighs 28.0 g V. Percent Purity Percent Purity = 6.0 g pure gold 28.0 g impure gold x 100% = 21% What is the percent purity of gold in gold quartz? When refined the gold weighs 6.0 g Gold Quartz Weighs 28.0 g Did you know? Gold Karat is a measure of percentage purity, not weight. Karats are often mistaken to be a weight measure used by jewelers. Instead karats are a measure of gold's purity. Pure gold is too malleable to be used as-is by jewelers: a ring made of pure gold would bend and loose its shape and be impossible to wear. So jewelers "dilute" gold by mixing it with other stronger metals. One Karat measures the fineness of gold in 1/24 part which is 4.2%. Hence a 18 karat gold ring is made of 18/24 parts of gold, or 75% gold and 25% other metals. Metals added to gold are of cheaper value, so for equals weight, the price of a jewelry item goes up when the karat count goes up too. Gold Karat Percent Purity = Mass of pure chemical Mass of impure chemical Percent Purity x 100 % If 100.0g of FeO produces 12.0g of pure Fe according to the reaction FeO + C + O2 Fe + CO2 What is the percentage purity of FeO used? Example 1 2FeO + 2C + O2 ® 2Fe + 2CO2 Use 12.0g of pure Fe to find pure FeO. 1 mol 2FeO 71.85g 12.0g Fe x x x = 15.4 g FeO 55.85g 2Fe 1mol 15.4g % Purity = x100% = 15.4% 100.0g Example 1 What mass of impure zinc metal having a purity of 89.5% is required to produce 975 mL of hydrogen gas at STP according to the reaction: Zn + HCl ZnCl2 + H2 Example 2 Zn + 2HCl ® ZnCl2 + H 2 1L 1 mol 1 Zn 65.39g 975ml x x x x = 2.85g PURE Zn 1000mL 22.4 L 1 H 2 1 mol % Purity = mass pure x100% mass impure mass impure = mass pure x 100% 100% = 2.85g x = 3.18g impure Zn % Purity 89.5% Example 2 1. Please collect a % Purity worksheet from the front of the class 2. Work through the Quick Check and then attempt the practice problem on the back. 3. Make sure you TRY the practice problem before viewing the answer Purity Problems Sometimes 100% of the expected amount of products cannot be attained from a reaction. This can occur because: 1. The reactants may not all react • They may not be 100% pure 2. Some of the products are lost due to the experiment procedures Percent Yield Percent Yield = Mass of product attained Mass of product expected Percent Yield x 100 % V. Percent Yield An analogy What is the percent yield? 30 kernels What is the percent yield? Actual yield Percent yield = Theoretical yield 30 kernels What is the percent yield? Actual yield Percent yield = Theoretical yield = 30 30 kernels What is the percent yield? Actual yield = 24 Percent yield = Theoretical yield = 30 30 kernels 24 popped corn What is the percent yield? Actual yield = 24 Percent yield = Theoretical yield = 30 30 kernels = 80% 24 popped corn Given the following equation: _____ K2PtCl4 + _____ NH3 a. b. c. _____ Pt(NH3)2Cl2 + _____ KCl Balance the equation. Determine the theoretical yield of KCl if you start with 34.5 grams of NH3. Starting with 34.5 g of NH3, and you isolate 76.4 g of Pt(NH3)2Cl2, what is the percent yield? Example 1 – The process Given the following equation: _____ K2PtCl4 + __2___ NH3 a. b. c. _____ Pt(NH3)2Cl2 + __2___ KCl Balance the equation. Determine the theoretical yield of KCl if you start with 34.5 grams of NH3. Starting with 34.5 g of NH3, and you isolate 76.4 g of Pt(NH3)2Cl2, what is the percent yield? Example 1 – The process Given the following equation: _____ K2PtCl4 + __2___ NH3 a. b. c. _____ Pt(NH3)2Cl2 + __2___ KCl Balance the equation. Determine the theoretical yield of KCl if you start with 34.5 grams of NH3. 151 g KCl Starting with 34.5 g of NH3, and you isolate 76.4 g of Pt(NH3)2Cl2, what is the percent yield? Example 1 – The process Given the following equation: _____ K2PtCl4 + __2___ NH3 a. b. c. _____ Pt(NH3)2Cl2 + __2___ KCl Balance the equation. Determine the theoretical yield of KCl if you start with 34.5 grams of NH3. 151 g KCl Starting with 34.5 g of NH3, and you isolate 76.4 g of Pt(NH3)2Cl2, what is the percent yield? % yield = 25.1% Example 1 – The process K 2 PtCl4 + 2NH 3 ® Pt(NH 3 )2 Cl2 + 2KCl 34.5g NH 3 x 1 mol 2 KCl 74.55g x x = 151g KCl 17.04g 2 NH 3 1 mol 1 mol 1 Pt(NH 3 )2 Cl2 300.06g 34.5g NH 3 x x x = 304g Pt(NH 3 )2 Cl2 17.04g 2 NH 3 1 mol Actual Yield 76.4g % Yield = x100% = x100% = 25.1% Theoretical Yield 304g Example 1 Given the following reaction: CH4 + Cl2 CH3Cl + HCl When 15.0g of CH4 is reacted with Cl2, a total of 29.7g of CH3Cl is formed. What is the percentage yield of the reaction? Example 2 CH 4 + Cl2 ® CH 3Cl + HCl ? theoretical g CH 3Cl formed 1 mol 1 CH 3Cl 50.49g 15.0g CH 4 x x x = 47.19g CH 3Cl 16.05g 1 CH 4 1 mol actual 29.7g % Yield = x100% = x100% = 62.9% theoretical 47.19g Example 2 What mass of K2CO3 is produced when 1.50g of KO2 is reacted according to the reaction, KO2 + CO2 K2CO3 + O2 if the reaction has a 76.0% yield? Example 3 4KO2 + 2CO2 ® 2K 2CO3 + 3O2 ? theoretical g K 2CO3 formed 1 mol 2 K 2CO3 138.21g x x = 1.46g K 2CO3 71.10g 4KO2 1 mol actual theoretical x % Yield 1.46g x 76.0% % Yield = x100% ; = actual = = 1.11g K 2CO3 theoretical 100% 100% 1.50g KO2 x Example 3 What mass of CuO is required to make 10.0g of Cu according to the reaction NH3 + CuO N2 + Cu + H2O if the reaction has a 58.0% yield? Example 4 2NH 3 + 3CuO ® N 2 + 3Cu + 3H 2O actual actual 10.0g x100% ; theoretical = x100% = x100% = 17.2g Cu theoretical % Yield 58.0% 1 mol 3 CuO 79.55g 17.2g Cu x x x = 21.6g CuO 63.55g 3 Cu 1 mol % Yield = Example 4