How to Determine the Empirical Formula from Combustion Analysis Data?

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Discussion Overview

The discussion revolves around determining the empirical formula of a compound based on combustion analysis data, specifically involving the combustion of a compound containing carbon, hydrogen, and oxygen. Participants are analyzing the calculations needed to derive the empirical formula from the given mass of products formed during combustion.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant presents a calculation for the mass of carbon produced from CO2 and the mass of hydrogen from H2O, leading to a derived mass of oxygen.
  • Another participant questions the calculation of moles of hydrogen in H2O, indicating a potential error in the initial calculation.
  • A correction is made regarding the number of moles of hydrogen, clarifying that there are 2 moles of hydrogen in each mole of H2O.
  • Further calculations are presented, but a participant notes that rounding down to 1 mole of each element is inappropriate and suggests maintaining decimals and scaling to obtain whole numbers.

Areas of Agreement / Disagreement

There is no consensus on the correct empirical formula, as participants are still working through the calculations and addressing potential errors in the methodology.

Contextual Notes

Participants express uncertainty regarding the proper treatment of mole ratios and rounding in their calculations, indicating that assumptions about the calculations may affect the derived empirical formula.

shap
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Homework Statement


When 2.66 grams of a compound containing only carbon, hydrogen, and oxygen is burned completely, 4.50 grams of CO2 and 1.10 grams of H2O are produced. What is the empirical formula of the compound?
Answer Options:
C5H6O4
C5H6O5
C5H12O5
C4H6O3
C4H8O3
C3H4O3
C4H4O5
C5H8O4
C4H4O3

Homework Equations


Dimensional Analysis


The Attempt at a Solution


4.5gCO2*(1molCO2/44gCO2)*(1molCO2/1molC)*(12gC/1molC)=1.23gC
1.1gH20*(1molH2)/18gH2O*(1molH2O/1molH)*(1gH/1molH)=.122gH
2.66g-1.23g-.122g=1.308gO

1.23gC*(1molC/12gC)=.1023molC/.0819mol=1C
.122gH*(1molH/1gH)=.122molH/.0819mol=1H
1.31gO*(1molO/16gO)=.0819molO/.0819mol=1O
Empirical Formula:CHO<--not an available answer
 
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shap said:
1.1gH20*(1molH2)/18gH2O*(1molH2O/1molH)*(1gH/1molH)=.122gH

how many mol of H in H2O?
 
Sorry that was a typo, in my calculation I used 2 mol of H in H2O.
 
shap said:
1.23gC*(1molC/12gC)=.1023molC/.0819mol=1C
.122gH*(1molH/1gH)=.122molH/.0819mol=1H
1.31gO*(1molO/16gO)=.0819molO/.0819mol=1O
Empirical Formula:CHO<--not an available answer

you can't round down to 1. Keep the decimals and multiply all mole numbers by the same factor until you get whole numbers