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Topic: Calculating % yield  (Read 17674 times)

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pinkster101

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Calculating % yield
« on: March 04, 2006, 06:04:47 PM »
Hi, I am getting really conused and any help would be much appreciated.

Last week in the lab, we prepared trans-bis(glycinato)copper(II) monohydrate. The equation for the reaction was:

[Cu(CH3COO)2 ]H20 + 2NH2CH2COOH ---> [Cu(NH2CH2COO)2]H20 + 2CH3COOH

On the product side, the molecualr weight of the copper acetate monohydrate is 199.644, and the molecular weight of the glycine is 75.07.

When we prepared it we used 2.06g copper acetata monohydrate(10mmol) and 3.067g glycine(40mmol). Which I then calculated as being 0.0103mol copper acetate monohydrate and 0.0409mol glycine

The weight of the product we made was 1.52g

I am just having problems figuring out the mass of the product I was supposed to get. Do I start with a mole ratio?

Thanks for the help.
« Last Edit: March 04, 2006, 07:05:09 PM by pinkster101 »

Offline xiankai

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Re:Calculating % yield
« Reply #1 on: March 05, 2006, 06:36:39 AM »
yes, do so and see how many moles u're supposed to get, then convert it to mass.
one learns best by teaching

pinkster101

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Re:Calculating % yield
« Reply #2 on: March 05, 2006, 04:13:12 PM »
Okay thanks, but there are 2 different products on the product side, do I just combine them or is one of them irrelevant?

Offline xiankai

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Re:Calculating % yield
« Reply #3 on: March 07, 2006, 05:25:57 AM »
its just one of them, the [Cu(NH2CH2COO)2]H20 or to quote you, "mass of the product I was supposed to get"

the acetic acid is a by-product.
« Last Edit: March 07, 2006, 05:26:52 AM by xiankai »
one learns best by teaching

pinkster101

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Re:Calculating % yield
« Reply #4 on: March 12, 2006, 04:34:59 PM »
ok thanks! ;D

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