Chemistry 116: Study guide
Chemistry 116: Study guide CHEM 116
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This page Study Guide was uploaded by Jamisha Evans on Wednesday March 23, 2016. The Study Guide belongs to CHEM 116 at Western Kentucky University taught by Bangbo Yan in Spring 2016. Since its upload, it has received 50 views. For similar materials see INTRO TO COLLEGE CHEMISTRY in Chemistry at Western Kentucky University.
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Date Created: 03/23/16
Chemistry 116 Test 3 Study guide Bangbo Yan Lecture 10 0 v Molecular Mass Sum of atomic masses of all atoms in molecule of substance Unit measured in amu EXAMPLE Molecular mass of H20 is 1802 amu H2 2 X 1008 2016 amu O 1 X1600 1600 amu 1802 amu add 2016 and 1600 round to correct amt of sigfigs 339 Formula Mass 0 Sum of atomic masses of all atoms in formula unit of the compound Whether molecular or not Unit measure in amu EXAMPLE Formula mass of CaC12 is 11098 Ca 1 X 4008 7040 amu C12 2 x 3545 4008 amu 11098 amu add and round v The mole Concept Mole quantity of a given amount of substance that contains as many molecules or formula units as the number of atoms in exactly 12g of carbon12 0 Symbol mol no unit for mole similar to When we say 1 pair of shoes or 12 dozen eggs they are not measured in units Avogadro s number number of atoms in exactly 12g of carbon12 Avogadro s number is 6022 X 1023 Molar mass mass of one mole of substance The molar mass in grams per mol is numerical equal to the formula mass in amu 0 Unit grams g EXAMPLE 1 mol 0 1600 g 0 1600 amu same but different unit 0 How to calculate the mass of one atom 0 Take the mass of the atom and divide it by the Avogadro s number 6022 X1023 EXAMPLE Mass in grams of one oxygen atom is 2657 X 10 1 mol 0 6022X 1023 O atoms Mass of O 1600g 1600g O6022 X 1023 O atoms 2657 X 103923g 0 How to calculate the mass of one molecule 0 Take the mass of the entire the molecule each atom in the molecule and divide it by Avogadro s number EXAMPLE Mass in grams of a hydrogen bromide molecule is 1343 X 10 3922g 1 mol of HBr 6022X 1023 Mass of HBr 8091g 8091 g HBr 6022X 10 23 HBr atoms 1343X 103922g 0 How to convert moles of substances to grams 0 Take the amount of moles given and multiply it by the molar mass of the substance EXAMPLE Convert 0721 mol of C6H1206 glucose to grams Molar mass of C6H1206 18016g 0721 mol X 18016g12990 Lecture 1 I 339 Mole concept 0 Molar mass 0 To find molar mass convert grams of substances to moles EXAMPLE 646 grams of PbCrO4 is how many moles Divide amount of grams of PbCrO4 646g by the mass of PbCrO4 323g 646g323g2 mol 339 Mass percentage from the formula 0 Mass percentage 0 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 2299g by mass of NaCl 5844 Then multiply by 100 If you were trying to find the mass percentage of Na2 then you would multiply the mass of Na by 2 then divide by the mass of the whole compound 2299g5844g x 1003934 is the mass of A 0 Percentage composition from the formula EXAMPLE What is the mass percentage of each element in CH2O formaldehyde Keep 3 sigfigs C 12g300 x 100 40 H2 2 x 101 202 300 x 100 673 O 16 300 X 100 533 0 Empirical formula simplest formula 0 Empirical formula The formula of a substance written with the smallest integer subscripts EXAMPLE What is the empirical formula of hydrogen peroxide H202 molecular formula HO empirical formula What is the empirical formula of benzene C6H6 molecular formula CH empirical formula 339 Elemental analysis 0 Used when grams of each element is unknown not given EXAMPLE combust 115 g ethanol Collect 220g C02 and 135 g H2O Q 220 g C02 1 mol CO2 1201 g C 1 mol C 4401 g C02 1 mol C 1201 g C 500 mol C 500molCX12gC600gC H20 135 g H20 1mol H20 101 g H 1mol H 3232 g H20 1 mol H 101 g H 150 mol H 150 mol H X 101 g H 151 g H Q g of O g of sample 115 g of C 600 g ofH 151 400 g 0 400g O 1 m010 16 g 0 25 mol 0 C05H150025 Divide by the smallest subscript 025 ANSWER C2H60 Lecture 12 339 Empirical formula from percentage composition 0 An analysis of sodium dichromate give the following mass percentages 175 Na 397 Cr and 428 0 What is the empirical formula of this compound 0 In 100 g 1 0 Na 100g X 175100 175 g Na 0 Cr 100g X 397100397g Cr 0 0100g X 428100428g 0 gt quot The above was done to change the percentages to be eXpressed as a decimal 2 o 175 Na X 1 mol230g Na0761 mol Na 0 397g Cr X 1 mol520g Cr 0763 mol Cr 0 428 g 0 X 1 mol16g 0 268 m010 gt quot The above is used to convert grams to moles 3 o 268 mol 0 0761 mol Na352 0 0763 mol Cr 0761 mol Na1 0 0761 mol Na 0761 mol Na1 DiVide each by the smallest amount of moles to find out how many moles are in each 4 0 NaCrO352 352 is not a whole number so multiply each by the simplest whole number which in this case is 2 ANSWER NazCr207 v Molecular formula from empirical formula 0 Molecular mass 11 X empirical formula mass 0 n molecular massempirical formula mass 0 Acetic acid is said to be 399 C 67 H and 5345 0 What is its empirical formula What is its molecular fomula with the molecular mass of acetic acid being 600 amu 0 Do steps 13 for C H amp O 0 You should get the following answers 1 0 C 399 g o H 67 g 0 O 5345 g 2 0 C 332 mol 0 H 663 mol 0 O 334 mol 3 0 C 100 mol 0 H 200 mol 0 O 100 mol 0 Based on the answers above the empirical formula is CH20 0 Multiply by the smallest whole number to find the molecular formula which is C2H402 0 What is the formula mass of a 050 mol sample of a molecular compound weighing 940g 0 Us algebra to create new equation empirical formula mass molecular mass moles 0 Empirical formula mass 940g050 mol 0 188 g 339 Stoichiometry quantitative relations 0 the quantitative study of reactants and products in a chemical reaction 339 Molar interpretation of a chemical equation 0 In terms of 0 Number of moles 0 Number of molecules 0 Mass of substance 0 Mole method 0 Use moles to calculate the amount of product in a reaction 0 How many moles of NH3 can be produced from 27 mol of H2 N2 g 3 H2 g NH3 g 0 Balance the equation N2 g 3 H2 g 2NH3 g ANSWER 2 moles o How many gram of HCL react with 500 g of manganese dioxide accoriding to the following equation 0 4HCl aq MnO2 2H2O l MnCl aq C12 500 g MnO2 4 mol HCl 36 46 g HCl 8694 g 1 mol HCl MnO2 839 g HCL Lecture 12 0 v Limiting reactant Theoretical and Percentage yields 0 Limiting reactant the reactant that us entirely consumed when a reaction goes to completion EXMPLE 2H2 O2gt 2H20 02 is the limiting reactant 9 Think of the process of making a sandwich If you have 4 pieces of bread 2H2 but only 2 pieces of cheese 02 will you have extra bread left over or extra cheese left over The answer is extra Bread H2 Which is why H2 is on the reactant and the product side while 02 is only on the reactant side because it was entirely consumed when the reaction went to completion 0 Once the reactant has been completely consumed the reaction stops 0 Any problem giving the starting amount for more than one reactant is the limiting reactant problem 0 The moles of product are always determined by the staring moles of limiting reactant EXAMPLE Zns 2HCl aq ZnCl2 aq H2 g If 030 mol of Zn is added to hydrochloric acid containg 052 mol of HCl how many moles of H2 are produced Both reactants are given so it s a limited reactant problem Find out which reactant produces les product 030 mol Zn 1 mol H2 1 mol Zn 030 mol H2 052 mol HCl 1 mol H2 2 mol HCl 026 mol H2 gt HCL is the limiting Iquotecietantgt ltgt lt EXAMPLE limiting reactant involving masses 2CH3CHO l 02 g 392HC2H302 1 In a laboratory test of this reaction 200g CH3CHO and 100 g 02 were put into a reduction vessel AHow many grams of acetic acid HC2H302 can be produced by this reaction from these amounts of reactants B How many grams of this excess reactant remain after the reaction is complete This is a limiting reactant problem because both reactants are given 339 Percentage yield 200g CH3CHO 1 mol CH3CHO 2 mol HC2H302 44 lg CH3CHO 2 mol CH3CH3CHO 0454 mol HC2H302 CH3CHO is the limiting reactant 100g 02 1 mol 02 2 mol HC2H302 32g 02 1 mol 02 0625 mol HC2H302 0454 mol HC2H302 1 mol HC2H302 2 mol HC2H302 273 g HC2H302 0454 mol HC2H302 601g HC2H302 1 mol HC2H302 7 26g 02 Percentage yield the actual yield expressed as a percentage of the theoretical yield Actual yieldtheoretical yield x 100 0 Theoretical yield maximum amount of product that can be obtained by a reaction from given amount of reactants Based on limiting reactant 0 Actual yield amount of product actually obtained from reaction Normally less than theoretical yield EXAMPLE CH3OH l CO g gtHC2H302 1 In an experiment 150 G of methanol and 100 g of carbon monoxide were placed in a reaction vessel A What is the theoretical yield of the acetic acid B If the actual yield is 191 g What is the percentage yield 150 g MeOH 1 mol MeOH 1 mol HC2H302 320 g MeOH 1 mol MeOH 0469 mol HC2H302 100 g MeOH 1 mol MeOH 1 mol HC2H302 280 g MeOH 1 mol CO 0357 mol HC2H302 MeOH is the limiting reactant 0357 mol CO 601 g HC2H302 1 mol HC2H302 Percentage yield 191g215g x 100 888 215 g HC2H302 theoretical yield
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