# Basic Chem: Is This Problem Worded Right?
**URL:** https://boards.straightdope.com/t/basic-chem-is-this-problem-worded-right/284143
**Category:** Factual Questions
**Created:** [January 11, 2005, 2:06am UTC](https://boards.straightdope.com/t/basic-chem-is-this-problem-worded-right/284143 "2005-01-11T02:06:29Z")
**Posts on this page:** 7
**Page:** 1
### Author:  [@Jinx](https://boards.straightdope.com/u/Jinx)
#### Post date: [January 11, 2005, 2:06am UTC](https://boards.straightdope.com/t/basic-chem-is-this-problem-worded-right/284143/1 "2005-01-11T02:06:29Z")
I was tutoring a boy in high school chem who had the following problem… Find the percent yield of copper if 1.87g Al reacts to give 4.65g Cu in the following equation: 2Al + 3CuSO4 → Al2(SO4)3 + 3Cu
Is this problem worded right? First, in the example in the book, they show a compound decompising, like CaCO4 -\> CaCO + 2O2, and they show how to find the percent yield of CaCO.
The copper in IS the copper out, right? There is no percent yield!!! Is it proper to talk about pecent yield of an element??? Perhaps, they meant to find the grams of copper yielded (actual) / grams of copper yielded in the balanced equation?
- Jinx 😕
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### Author:  [@spingears](https://boards.straightdope.com/u/spingears)
#### Post date: [January 11, 2005, 2:58am UTC](https://boards.straightdope.com/t/basic-chem-is-this-problem-worded-right/284143/2 "2005-01-11T02:58:46Z")
> [@Jinx](#):
>
> I was tutoring a boy in high school chem who had the following problem… Find the percent yield of copper if 1.87g Al reacts to give 4.65g Cu in the following equation: 2Al + 3CuSO4 → Al2(SO4)3 + 3Cu
Untill a chemist comes along…
All of the Cu in the CuSO4 is converted to free Cu and the conversion is 100%.
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### Author:  [@Ruken](https://boards.straightdope.com/u/Ruken)
#### Post date: [January 11, 2005, 3:03am UTC](https://boards.straightdope.com/t/basic-chem-is-this-problem-worded-right/284143/3 "2005-01-11T03:03:54Z")
Assuming a complete reaction w/no side pdts and no limiting reagent (equal moles of both starting materials,) one should get ~0.1 mol Cu, not the ~0.07 mol indicated in the problem. What you have here is a ~70% yield. You may want to work through the math on your own. I shouldn’t be doing math in my head when I’m tired.
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### Author:  [@Ruken](https://boards.straightdope.com/u/Ruken)
#### Post date: [January 11, 2005, 3:05am UTC](https://boards.straightdope.com/t/basic-chem-is-this-problem-worded-right/284143/4 "2005-01-11T03:05:21Z")
> [@Ruken](#):
>
> equal moles of both starting materials
Rather, the correct ratio of moles of the starting materials (2:3 Al:CuSo4)
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### Author:  [@gtbiehle](https://boards.straightdope.com/u/gtbiehle)
#### Post date: [January 11, 2005, 3:14am UTC](https://boards.straightdope.com/t/basic-chem-is-this-problem-worded-right/284143/5 "2005-01-11T03:14:53Z")
You start with 0.0693 moles of aluminum. If you perform the reaction correctly, you expect to get 0.1040 moles of copper. If you get fewer than that, as you do in this problem, only 0.0732 moles, then chemists talk about a per cent yield. It does not really matter to them that the copper did not come out of the aluminum atoms. I agree that the language “per cent yield” makes it sound as if the product comes out of the reactant you started with, but that is just not how the language is used. So yeah, the problem is properly phrased. (I was a chemistry major.)
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### Author:  [@Squink](https://boards.straightdope.com/u/Squink)
#### Post date: [January 11, 2005, 3:17am UTC](https://boards.straightdope.com/t/basic-chem-is-this-problem-worded-right/284143/6 "2005-01-11T03:17:32Z")
1.87g Al = 0.069 moles
4.65g Cu = 0.073 moles
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### Author:  [@spingears](https://boards.straightdope.com/u/spingears)
#### Post date: [January 11, 2005, 9:54pm UTC](https://boards.straightdope.com/t/basic-chem-is-this-problem-worded-right/284143/7 "2005-01-11T21:54:59Z")
> [@spingears](#):
>
> Untill a chemist comes along…
> All of the Cu in the CuSO4 is converted to free Cu and the conversion is 100%.
:smack: It’s been a long long time since chemistry.