Jump to content

  • You cannot start a new topic
  • You cannot reply to this topic

2005 - Written Paper - Q3(b)(ii) Question 3(b)(ii)

AmAnDa

Posted 23 May 2006 - 02:45 PM

Can someone please show me how to do this question,

Thank you

Michael

Posted 23 May 2006 - 03:13 PM

Ok here it goes:

1 mole ethanol -----> 1 mole ethyl etanoate
46g -------->88g
1g--------> 88/46 g
5g ------> 88/46 x 5 = 9.565g

% yield = actual yield / theoretical yield x 100
% yield = 5.8/9.565 x 100
% yield = 60.6% (note the answers round it up to 61%)

Nathan

Posted 23 May 2006 - 03:16 PM

why does everyone have such a problem with yield??

i thought it was one of the easier calculations in higher chemistry unsure.gif

AmAnDa

Posted 24 May 2006 - 11:07 AM

QUOTE(Dhesi @ May 23 2006, 04:13 PM) View Post

Ok here it goes:

1 mole ethanol -----> 1 mole ethyl etanoate
46g -------->88g
1g--------> 88/46 g
5g ------> 88/46 x 5 = 9.565g

% yield = actual yield / theoretical yield x 100
% yield = 5.8/9.565 x 100
% yield = 60.6% (note the answers round it up to 61%)


Thank you, much appreciated. biggrin.gif

  • You cannot start a new topic
  • You cannot reply to this topic