Stoichiometry, Part II (and Percentage Yield)

  • #1
Please (pretty please) check all of my work!

1) In an oxidation reaction, ethanol (C2H5OH) is converted to a ethanoic acid (C2H4O2) with a percentage yield of 70%. If you start the reaction with 360g of ethanol, how many grams of ethanoic acid would you expect to produce?
Stoichiometry is 1:1 between the compounds.

% yield = (actual value/expected value) *100

Given: % yield = 70

Find: expected value C2H4O2 in g

Actual value C2H4O2 in g:
360 g C2H5OH (1 mol C2H5OH / 46.068 g C2H5OH) (1 mol C2H4O2 / 1 mol C2H5OH) = 7.81454 mol C2H4O2

7.81454 mol C2H4O2 (60.052 g C2H4O2 / 1 mol C2H4O2) = 469.2785 g C2H4O2 = actual value ??

70 = (469.2785/expected) * 100
expected = 469.2785/.70 = 670 g C2H4O2 ??

Did I muddle this problem big time? I'm confused with what is the actual value and what is the expected (theoretical?) value. Did I confuse the actual for the expected?? Do you use 70% at all?


2) In an oxidation reaction, ethanol (C2H5OH) is converted to a ethanoic acid (C2H4O2). In an Experiment, a chemist starts with 149.00g of ethanol and forms 180.00g of ethanoic acid. What is the percentage yield of the reaction? Stoichiometry of the reaction is 1:1 for the mentioned compounds.


% yield = (actual value/expected value) *100

Given: 180.00 g C2H4O2 = actual value g C2H4O2 ?

Expected value of C2H4O2 :
149.00 g C2H5OH (1 mol C2H5OH / 46.068 g C2H5OH) (1 mol C2H4O2 / 1 mol C2H5OH) (60.052 g C2H4O2 / 1 mol C2H4O2) = 194.2291 g C2H4O2 ??

180 g / 194.2291 g C2H4O2 *(100) = 92.674 % ??

Gosh, I hope I didn't confuse my values or molar masses again!

Thank you. I am desperate for help. :cry:
 

Answers and Replies

  • #2
Anyone at all? Please? (Are my calculations correct?)

Thanks.
 
  • #3
Please confirm if I did something correctly.

Thanks for your patience.
 
  • #4
16
0
Your calculations look correct to me.:smile:
 
  • #5
GCT
Science Advisor
Homework Helper
1,728
0
7.81454 mol C2H4O2 (60.052 g C2H4O2 / 1 mol C2H4O2) = 469.2785 g C2H4O2 = actual value ??

70 = (469.2785/expected) * 100
expected = 469.2785/.70 = 670 g C2H4O2 ??

Did I muddle this problem big time? I'm confused with what is the actual value and what is the expected (theoretical?) value. Did I confuse the actual for the expected?? Do you use 70% at all?
what you calculated here is the theoretical yield, now how is your theoretical yield less then you're actual percent yield? Think about it.

2) seems fine
 

Related Threads on Stoichiometry, Part II (and Percentage Yield)

  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
6
Views
21K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
1
Views
7K
Replies
4
Views
871
Replies
1
Views
1K
Replies
0
Views
7K
Top