How to calculate percent yield without an equation

[deleted647690]

Can you calculate percent yield for a reaction without knowing the mole ratio of reactants to products? There is a passage in BR orgo that says, "A student starts with 138 mg of 1,4-dimethoxybenzne (138 g/mol) and treats with 1.32 mL HNO3 (1.5 g/mL, 63 g/mole) dissolved into 2.61 mL H2SO4.....................Product masss is 2,5-dimethoxynitrobenzene (183 g/mole) and is 91.5 mg."

A question asks what the percent yield is, and gives:
A. 25%
B. 50%
C. 75%
D.100%


In their answer, they said the limiting reagent is 1,4-dimethoxybenzene, not nitric acid. And then they go to calculate the percent yield based on that. But how would you know that the limiting reagent is not nitric acid?

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[Mad Jack]

I don't know how to do math without math. Autistic savantism, perhaps?

 

[deleted647690]

I don't know how to do math without math. Autistic savantism, perhaps?

I guess what I'm trying to ask is if there is a way to 'estimate' the limiting reagent based on the given information

 

[Mad Jack]

I guess what I'm trying to ask is if there is a way to 'estimate' the limiting reagent based on the given information

Oh I was just basically saying I have no idea. I'll do the math when I get home and see if I can come up with an easy solution.

 

[Mad Jack]

Let's see... You have half of a thousandth moles of product, 1/1000 mole of reagent A and a serious excess of reagent B if my rough off the top of my head, no calculator math works out. So that makes me lean toward 50% efficiency, as your limited reagent resulted in half as many moles of product. Am I correct?

 

[aldol16]

Weird question, but from the information give, you know that the reaction is most likely 1:1 stoichiometry with respect to reagents because all you're doing is incorporating a nitro group into the substrate and the nitro group comes from nitric acid. So you have 1 mmol of substrate and you end up with 0.5 mmol of product, so you have 50% yield based on substrate.

 

[deleted647690]

Let's see... You have half of a thousandth moles of product, 1/1000 mole of reagent A and a serious excess of reagent B if my rough off the top of my head, no calculator math works out. So that makes me lean toward 50% efficiency, as your limited reagent resulted in half as many moles of product. Am I correct?

Weird question, but from the information give, you know that the reaction is most likely 1:1 stoichiometry with respect to reagents because all you're doing is incorporating a nitro group into the substrate and the nitro group comes from nitric acid. So you have 1 mmol of substrate and you end up with 0.5 mmol of product, so you have 50% yield based on substrate.

Yeah, I guess I sort of understand that the 1,4-dimethoxybenzene would be limiting since that is the substrate upon which you are adding a functional group...

 

[aldol16]

Yeah, I guess I sort of understand that the 1,4-dimethoxybenzene would be limiting since that is the substrate upon which you are adding a functional group...

That's not why it's limiting. It's limiting because the reaction is 1 A + 1 B = 1 C. You have 1 mmol of A and at least an order of magnitude more than that of B.

 

[Mad Jack]

That's not why it's limiting. It's limiting because the reaction is 1 A + 1 B = 1 C. You have 1 mmol of A and at least an order of magnitude more than that of B.

Basically this. You figure out which you have an excess of by comparing the number of moles of each reagent, then use the number of moles of that compared to the number of moles of product to find out your efficiency. You're working with milligrams of a large molecule and well over a gram of a small one, so the large molecule has to be the limiting reagent.

 

[deleted647690]

That's not why it's limiting. It's limiting because the reaction is 1 A + 1 B = 1 C. You have 1 mmol of A and at least an order of magnitude more than that of B.

Alright. But what do you mean that you know it is 1:1 just because you are incorporating a functional group?

 

[aldol16]

Alright. But what do you mean that you know it is 1:1 just because you are incorporating a functional group?

You're taking the aryl compound and putting a nitro on it. The nitro is coming from the nitric acid. If the stoichiometry wasn't 1:1, then what would the additional reagent be doing? So say you think it could be 1:2. What's the extra nitric acid doing and where does it go?

 

[Mad Jack]

Alright. But what do you mean that you know it is 1:1 just because you are incorporating a functional group?

Just going by the nomenclature and the reagents, that is the only logical conclusion. Even if it were 2 or 3 to one, there would still be a massive excess of HNO3.