What Is the Empirical Formula of Crocetin?

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Discussion Overview

The discussion revolves around determining the empirical formula of crocetin based on combustion data. Participants analyze the results of a combustion experiment where crocetin is burned to produce carbon dioxide and water, leading to various interpretations and calculations regarding the empirical formula.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents the problem of calculating the empirical formula of crocetin based on combustion data, noting difficulties in arriving at the correct answer.
  • Another participant suggests that the empirical formula could be C5H6O, but does not provide specific feedback on the original calculations.
  • A third participant references Wikipedia, stating that crocetin's molecular formula is C20H24O6 and suggests dividing by four to find the empirical formula, though this is contested.
  • Concerns are raised about the validity of the empirical formula derived from combustion products, with one participant arguing that the assumption of oxygen content in the products is flawed.
  • Another participant points out a mistake in the oxygen count, clarifying that crocetin contains four oxygen atoms, not six.
  • One participant outlines a method for calculating the mass of carbon and hydrogen from the combustion products to determine the mass of oxygen in the original sample, emphasizing the need for independent problem-solving.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the empirical formula of crocetin. Multiple competing views and calculations are presented, with some participants agreeing on certain formulas while others challenge the assumptions and methods used.

Contextual Notes

There are unresolved issues regarding the assumptions made in the calculations, particularly concerning the mass balance of the combustion products and the original sample. The discussion reflects varying interpretations of the data and the formulas involved.

DavidQT
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Crocetin consists of elements carbon, hydrogen and oxygen. Determine the empirical formula of crocetin, if 1.00g of crocetin forms 2.68g of carbon dioxide and 0.657g of water when it undergoes complete combustion.

I have tried this question several times but I can't get the right answer and I don't know what I am doing wrong. I have asked some classmates and they got the answer to be C5H60.
 
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C5H6O looks OK to me. Hard to comment on what you did wrong not seeing what you did.

Please note this should land in homework forums. Moving.
 
Wikipedia says crocetin is C20H24O6
Divide it all by four and your answer is correct :)
 
ArielRodriguez said:
Wikipedia says crocetin is C20H24O6
Divide it all by four and your answer is correct :)

It either is, or is not. It fits the formula, but we don't know if it fits data they were given (and they will be graded for the latter, not for their ability to google).
 
Still Dont get iit

Ok I have a test tomorrow and I still don't get this question.

Crocetin consists of elements carbon, hydrogen and oxygen. Determine the empirical formula of crocetin, if 1.00g of crocetin forms 2.68g of carbon dioxide and 0.657g of water when it undergoes complete combustion.

This is what i did:
2.68/44.01=0.0609 mols of carbon dioxide. 0.657/18.02=0.0365 mols of water.
0.0609/0.0365=1.67 0.0365/0.0365=1
1.67x3= 5 to get it even for the empirical formula. 1x3=3 Have to multiply by 3 to both.
Empirical formula= (CO2)5 (H20)3=C5H6O13. I am pretty sure the answer is C5H60
 
Dividing by 4 does not work anyways 6 oxygen's divided by 4 does not equal 1 oxygen.
 
Typo, crocetin doesn't contain 6 oxygen atoms but 4.
 
DavidQT said:
2.68/44.01=0.0609 mols of carbon dioxide. 0.657/18.02=0.0365 mols of water.
0.0609/0.0365=1.67 0.0365/0.0365=1
1.67x3= 5 to get it even for the empirical formula. 1x3=3 Have to multiply by 3 to both.
Empirical formula= (CO2)5 (H20)3=C5H6O13. I am pretty sure the answer is C5H60

You are assuming that products of combustion contain the same amount of oxygen as crocetin did. That's obviously impossible - you have 2.68+0.657=3.34 g of products, but there was only 1 g of crocetin. No wonder your oxygen is so off.
 
Then how would I go about fixing my mistake.
 
  • #10
Calculate mass of carbon in 2.68g of CO2, calculate mass of hydrogen in 0.657g of H2O, subtract these from the original sample mass - that's the mass of oxygen in the original sample.

Enough spoon feeding for today, now you are on your own.
 

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