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Talk:Yield (chemistry)

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Slight Modification

The maximal yield of a chemical reaction would be 100%, a value that is never reached.

There are some reactions that can give 100% yields, if the chemist working them up is especially talented. Very few though. I changed "never" to "rarely"

Propose merge with yield (chemistry). The theoretical yield article does not link from any other article and overlaps with the other yield article. My first attempt at merge was countered. I think I did carefully analyse if there were non-overlapping segments and i did not find any. Somebody else should have a go at it if and when we proceed V8rik 18:32, 21 November 2006 (UTC)[reply]

• I have performed the merge with theoretical yield. It is a is a Yield (Chemistry) and should be part of this page because it can not be understood in the absence of the other yeilds. Please discuss objections here before reverting. M stone 22:52, 15 May 2007 (UTC)[reply]

I find the lack of citations of references disturbing on this page. How can we know what is put here is the true way to work out a reactions yield?

Also if all the definitions of yield are together, it is easier to understand them all with reference of one to the other

• I added the requested reference but do consult the Wikipedia:Scientific citation guidelines and Wikipedia:Assume good faith
• please sign your posts (4 tildes) V8rik 21:43, 18 February 2007 (UTC)[reply]

I found this article to be very helpful with my chem lab proj. and therefor believe i should stay as is unless more detailed info is included and the precent yeild information is cited regaurding a correct experiment ie (40%-50%= acceptable experiment and 90%= expert) i dont think it should merge as yeild chem

I removed the following statement because I thought the ratings were arbitrary. Why were they chosen? Who chose them? While a 97% yield maybe considered acceptable for some purposes however it maybe unacceptable for others.

"Yields above about 90% are called excellent, yields above 80% very good, yields above about 70% are called good, yields below about 50% are called fair, yields below about 40% are called poor."

I think that the statement should be modified or removed to address these concerns. M stone 16:51, 27 May 2007 (UTC)[reply]

• This is silly, the chemical literature especially in organic chemistry is abundant with claims such as excellent, good etc. etc. etc. and there is thankfully at least one source (Vogel) that tries to quantify these statements. Obviously Vogels system is arbitrary but thankfully Vogel is also THE primary source in organic chemistry laboratory practice. See also my comment with edit 21:40, 18 February 2007 V8rik (Talk | contribs) (added reference as requested in talk page, the excellent, very good, good qualification are Vogel's). Deleting valid, relevant and referenced material is vandalism V8rik 17:25, 27 May 2007 (UTC)[reply]

• Personally I think that such an arbitrary rating systems is not relevant, thus deleting it was not vandalism just an edit. I don’t understand why it is important to label a precise number with less precise term? That being said I don’t have a problem with including the statement as long as it is clearly labeled as Vogel’s rating system. I have modified the statement to say just that. I hope that this is an acceptable resolution. M stone 22:08, 27 May 2007 (UTC)[reply]

• HiM stone, thanks for your consideration. V8rik 15:24, 28 May 2007 (UTC)[reply]

Should it be:

${\displaystyle {\mbox{percentage yield}}={\frac {\mbox{actual yield}}{\mbox{theoretical yield}}}\times 100}$

or

${\displaystyle {\mbox{percentage yield}}={\frac {\mbox{actual yield}}{\mbox{theoretical yield}}}\times 100\%}$

Views please.

Ben 09:49, 12 November 2007 (UTC)[reply]

According to "Chemistry the Central Science" it should be:

${\displaystyle {\mbox{percentage yield}}={\frac {\mbox{actual yield}}{\mbox{theoretical yield}}}\times 100}$

M stone 13:11, 12 November 2007 (UTC)[reply]

I would vote for the first one, or for

${\displaystyle {\mbox{yield}}={\frac {\mbox{actual yield}}{\mbox{theoretical yield}}}\times 100\%}$

which is equivalent to

${\displaystyle {\mbox{yield}}={\frac {\mbox{actual yield}}{\mbox{theoretical yield}}}}$

This is based on what I remember reading from some authority (I don't remember if it was NIST, BIPM, or IUPAC), which suggested that the percent sign should be considered exactly equivalent to 1/100. --Itub 13:13, 12 November 2007 (UTC)[reply]

• According to Percentage, percentage is a way of expressing a number as a fraction of 100 (per cent meaning "per hundred") and percentage is not a unit. I would say it is option 1. V8rik 17:18, 12 November 2007 (UTC)[reply]

Maybe

${\displaystyle {\mbox{percentage yield}}={\frac {\mbox{actual yield}}{\mbox{theoretical yield}}}\times 100}$

but

${\displaystyle {\mbox{yield}}={\frac {\mbox{actual yield}}{\mbox{theoretical yield}}}\times 100\%}$

Ben 17:32, 12 November 2007 (UTC)[reply]

I added a small amount of content from the article conversion (chemistry). This is temporary.Oceanflynn (talk) 20:57, 25 June 2020 (UTC)[reply]

@Oceanflynn: Your recent edits are generally useful, and I have only made a few minor corrections. However there is one incomplete sentence which needs to be fixed using the source cited. At the end of the section Definitions, your last sentence says『They wrote that knowing the stoichiometry of a chemical reaction—the numbers and types of atoms in the reactants and products,』What comes after the comma, please? Dirac66 (talk) 16:10, 1 July 2020 (UTC)[reply]

OK, thanks for completing the sentence. Dirac66 (talk) 17:38, 1 July 2020 (UTC)[reply]

Some definitions are placed here more than once.--M97uzivatel (talk) 06:02, 21 September 2023 (UTC)[reply]