Yield Essay Sample

Output of CuCl2. 2DMSO
Formula weight ( Mr ) of CuCl2 = 63. 55 + ( 35. 45 ten 2 ) =134. 45g/mol Formula weight of merchandise CuCl2. 2DMSO = 134. 45 + 2 [ 16 + 32. 06 + ( 12. 01 ten 2 ) + ( 1. 0079 x 6 ) ] = 290. 704g/mol

Mass of CuCl2= 0. 850g
Equation for reaction
CuCl2 + 2DMSO – & gt ; CuCl22DMSO
Mole ratio between CuCl2 and CuCl22DMSO = 1:1
Mole of CuCl2 = Mass/ Mr = 0. 850/134. 45 = 0. 00632 moles

Since the ratio between CuCl2 and CuCl22DMSO = 1:1. mole of CuCl2DMSO is besides 0. 0063 moles. To happen theoretical output of CuCl2. 2DMSO
Mass = Mr x mole
=290. 704 ten 0. 00632
= 1. 837g
The existent output gained is 1. 573g
The per centum output = ( existent yield/ thereotical output ) x 100 % = ( 1. 573/1. 837 ) x 100 % = 85. 63 %

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Output of Cu ( gly- ) 2. H2O ( Isomer 1 )
Formula weight ( Mr ) of ( CH3CO2 ) 2Cu. H2O = [ [ 12. 01 + ( 1. 0079 x 3 ) + 12. 01 + ( 16 x 2 ) ] x 2 ] + 63. 55 + ( 1. 0079 x2 ) + 16 = 199. 65g/mol Formula weight ( Mr ) of Cu ( gly- ) 2. H2O = 63. 55 + [ [ 14. 01 + ( 1. 0079 x 2 ) + 12. 01 + ( 1. 0079 x 2 ) + 12. 01 + ( 16 x 2 ) ] x 2 ] + ( 1. 0079 x 2 ) ] + 16 = 229. 68g/mol

Mass of ( CH3CO2 ) 2Cu. H2O = 2. 0g
Equation for reaction
( CH3CO2 ) 2Cu. H2O + 2NH2CH2COOH  Cu ( NH2CH2COO ) 2. H2O + 2CH3CO2H

Mole ratio between ( CH3CO2 ) 2Cu. H2O and Cu ( NH2CH2COO ) 2. H2O is 1:1. Mass of ( CH3CO2 ) 2Cu. H2O = 2. 0g
Gram molecules of ( CH3CO2 ) 2Cu. H2O = Mass / Mr = 2/199. 65 = 0. 01 moles Since Mole ratio between ( CH3CO2 ) 2Cu. H2O and Cu ( NH2CH2COO ) 2. H2O is 1:1. mole of Cu ( NH2CH2COO ) 2. H2O = 0. 01 mole Theoretical output of Cu ( NH2CH2COO ) 2. H2O = Mr x mole = 229. 68 ten 0. 01 = 2. 296 g Actual output addition = 2. 587g

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Percentage of output = ( Actual yield/ Thereotical Yield ) x 100 % = ( 2. 587/2. 296 ) X 100 % = 112. 67 %

Output of Cu ( gly- ) 2. H2O ( Isomer 2B and 2A )
Formula weight ( Mr ) of Cu ( gly- ) 2. H2O = 63. 55 + [ [ 14. 01 + ( 1. 0079 x 2 ) + 12. 01 + ( 1. 0079 x 2 ) + 12. 01 + ( 16 x 2 ) ] x 2 ] + ( 1. 0079 x 2 ) ] + 16 = 229. 68g/mol Mass of ( CH3CO2 ) 2Cu. H2O = 0. 2g ( method B )

Mass of ( CH3CO2 ) 2Cu. H2O = 1. 5 ( method a )
Equation for reaction
Cu ( NH2CH2COO ) 2. H2O ( isomer 1 )  Cu ( NH2CH2COO ) 2. H2O ( isomer 2 ) + H2O Mole ratio between isomer 1 and 2 = 1:1

Isomer 2B
Mole of ( CH3CO2 ) 2Cu. H2O ( isomer 1 ) = Mass/ Mr = 0. 2/229. 68 = 0. 0008707 mole Since mole ratio between isomer 1 and 2 = 1:1. mole of isomer 2 is besides 0. 0008707 mole. Thereotical output of isomer 2 = Mr x mole = 229. 68 ten 0. 0008707 = 0. 199g Actual output addition = 0. 188

Percentage of output = ( Actual yield/ Thereotical Yield ) x 100 % = ( 0. 188/0. 199 ) x 100 % = 94. 47 %

Isomer 2A
Mole of ( CH3CO2 ) 2Cu. H2O ( isomer 1 ) = Mass/ Mr = 1. 5/229. 68 = 0. 00653mole Since mole ratio between isomer 1 and 2 = 1:1. mole of isomer 2 is besides 0. 00653mole mole. Thereotical output of isomer 2 = Mr x mole = 229. 68 ten 0. 00653mole = 1. 499g Actual output addition = 0. 883g

Percentage of output = ( Actual yield/ Thereotical Yield ) x 100 % = ( 0. 883/1. 499 ) x 100 % = 58. 9 %