Chemistry Notes : Mole concept, mass percentage, & elemmental analysis
Chemistry Notes : Mole concept, mass percentage, & elemmental analysis CHEM 116
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This 2 page Class Notes was uploaded by Jamisha Evans on Friday March 18, 2016. The Class Notes belongs to CHEM 116 at Western Kentucky University taught by Bangbo Yan in Spring 2016. Since its upload, it has received 10 views. For similar materials see INTRO TO COLLEGE CHEMISTRY in Chemistry at Western Kentucky University.
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Date Created: 03/18/16
Chapter 3 notes Bangbo yan Lecture 2 Mole concept Molar mass To find molar mass, convert grams of substances to moles EXAMPLE: 646 grams of PbCrO is how4many moles? Divide amount of grams of PbCrO (644g) by the mass of PbCrO (323g) 4 646g/323g=2 mol Mass percentage from the formula Mass percentage To find mass percentage, divide mass of the whole compound by mass of A in compound. EXAMPLE: What is the mass percentage of Na in NaCl? (Hint: A is Na) Divide mass of Na (22.99g) by mass of NaCl (58.44). Then multiply by 100 ***If you were trying to find the mass percentage of Na t2en you would multiply the mass of Na by 2 then divide by the mass of the whole compound*** 22.99g/58.44g x 100=39.34 is the mass % of A Percentage composition from the formula EXAMPLE: What is the mass percentage of each element in CH O (forma2dehyde). Keep 3 sigfigs. C: 12g/30.0 x 100 = 40% H 2: 1.01 (2.02) /30.0 x 100 = 6.73% O: 16/ 30.0 x 100 = 53.3 % Empirical formula (simplest formula) Empirical formula: The formula of a substance written with the smallest integer subscripts. EXAMPLE: What is the empirical formula of hydrogen peroxide? H 2 (2 lecular formula) HO (empirical formula) What is the empirical formula of benzene? C 6 6molecular formula) CH (empirical formula) Elemental analysis Used when grams of each element is unknown/ not given. EXAMPLE: combust 11.5 g ethanol Collect 22.0g C2 and 13.5 g H2O CO : 2 22.0 g CO 2 1 mol CO 2 12.01 g C 1 mol C 44.01 g CO 2 1 mol C 12.01 g C =.500 mol C .500 x 12 = 6.00 g C H 2: 13.5 g H2O 1 mol H 2 1.01 g H 1 mol H 32.32 g H 2 1 mol H 1.01 g H =1.50 mol H 1.50 mol H x 1.01 g H = 1.51 g H O: g of O = g of sample (11.5) – g of C (6.00) + g of H (1.51) = 4.00 g O 4.00g O 1 mol O 16 g O = .25 mol O C 0.5 1.5 0.25 Divide by the smallest subscript (0.25) Answer: C H O 2 6
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