What Did I Do Wrong in Finding the Percent Yield of Alum?

Click For Summary

Discussion Overview

The discussion revolves around the calculation of percent yield in a chemistry experiment involving the synthesis of alum. Participants explore the correct chemical formulas and reaction equations relevant to the experiment, as well as the implications of these formulas on the calculated yield.

Discussion Character

  • Homework-related
  • Debate/contested
  • Technical explanation

Main Points Raised

  • One participant calculates the percent yield of alum as 217% based on their initial formula, indicating a mistake due to the impossibility of a yield over 100%.
  • Multiple participants point out that the alum formula used (KAl(OH)4) is incorrect, suggesting it should be KAl(SO4)2 * 12H2O instead.
  • Another participant proposes adding a different reaction equation involving KAl(OH)4 and H2SO4 to clarify the overall reaction.
  • Some participants express skepticism about the validity of their calculated yields, with one reporting a yield of 112% and another reporting a yield of 61.1% using a different formula.
  • One participant questions whether it is acceptable to replace the incorrect formula with the correct one in their calculations.
  • A later reply confirms that a yield of 61.1% appears reasonable based on the revised formula.

Areas of Agreement / Disagreement

Participants generally agree that the initial formula used for alum is incorrect, but there is no consensus on the correct formula or the final percent yield, as multiple calculations yield different results.

Contextual Notes

Participants express uncertainty regarding the correct chemical formulas and their impact on yield calculations. There are unresolved issues related to the accuracy of the reaction equations and the assumptions made in the calculations.

Who May Find This Useful

Students and educators in chemistry, particularly those focused on stoichiometry and yield calculations in chemical reactions.

physicsnobrain
Messages
123
Reaction score
0

Homework Statement


Find the percent yield of alum

In an experiment to synthesize alum I started with 0.766g of aluminum and made 8.246g of alum


Homework Equations


2Al + 6H20 + 2KOH ---> 2KAl(OH)4 + 3H2


The Attempt at a Solution



First I find the LR. Which in this experiment is Al.

I find the moles of aluminum to be 0.02838mol, next I multiply by the mole ratio and by the molar mass of alum to find the theoretical yield of alum.

Now to find the percent yield I do actual/theoretical X100%

8.246g/3.806g x 100% = 217%. I know I am wrong because this is impossible.

What did I do wrong?
 
Physics news on Phys.org
Your calculations are OK, but they are based on the incorrect alum formula - it is not KAl(OH)4. Unless you just use a wrong name of the compound.
 
Borek said:
Your calculations are OK, but they are based on the incorrect alum formula - it is not KAl(OH)4. Unless you just use a wrong name of the compound.

Well we also received another formula in this lab:

KAl(OH)4 +8H2O +2H2SO4 ---> KAl(SO4)2 * 12H20

Should I add this to the formula I listed in my first post to get a overall reaction equation?
 
physicsnobrain said:
Well we also received another formula in this lab:

KAl(OH)4 +8H2O +2H2SO4 ---> KAl(SO4)2 * 12H20

Should I add this to the formula I listed in my first post to get a overall reaction equation?

Hmm I still get a yield more than 100% after doing this. Which is physically impossible
 
Borek said:
Your calculations are OK, but they are based on the incorrect alum formula - it is not KAl(OH)4. Unless you just use a wrong name of the compound.


Is it allowed if I write the formula form the first equation except replace KAl(OH) for KAl(SO4)2
 
Borek said:
Your calculations are OK, but they are based on the incorrect alum formula - it is not KAl(OH)4. Unless you just use a wrong name of the compound.

Also, even if I use the formula u refer to I get a percent yield of 112% which is physically impossible in every sense and wrong.
 
physicsnobrain said:
Well we also received another formula in this lab:

KAl(OH)4 +8H2O +2H2SO4 ---> KAl(SO4)2 * 12H20

Should I add this to the formula I listed in my first post to get a overall reaction equation?

The formula is still wrong. Search for it on google.
 
Pranav-Arora said:
The formula is still wrong. Search for it on google.

I get 61.1% yield. is this correct?
 
physicsnobrain said:
I get 61.1% yield. is this correct?

I got this number using the fomula:

2Al +2KOH +4H2SO4 + 22H2O ---> 3H2 + 2KAl(SO4)2 * 12H2O


This was the only formula that helped produce a reasonable percent yield (61.1%). I am highly sceptical that there is an alternative to this.
 
  • #10
That looks OK to me.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 6 ·
Replies
6
Views
7K
Replies
4
Views
3K
Replies
1
Views
11K
  • · Replies 4 ·
Replies
4
Views
5K