Combustion analysis problem? - General Chemistry

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SUMMARY

The discussion focuses on determining the empirical formula of a compound after combustion analysis. A sample weighing 0.91 grams produces 2.073 grams of carbon dioxide, 0.304 grams of water, and 0.0944 grams of nitrogen. The moles calculated are 0.0471 for CO2, 0.0169 for H2O, and 0.00337 for N2, leading to an initial formula of C7H5N. The challenge lies in accounting for the oxygen content, which is not directly measured in the combustion products.

PREREQUISITES
  • Understanding of stoichiometry and mole conversions
  • Familiarity with combustion reactions and empirical formulas
  • Knowledge of mass-to-mole conversions for CO2 and H2O
  • Basic principles of chemical analysis and composition
NEXT STEPS
  • Calculate the mass of carbon, hydrogen, and nitrogen from the combustion products
  • Determine the mass of oxygen in the original sample using mass balance
  • Learn about empirical formula calculations and molecular formula derivation
  • Study combustion analysis techniques in organic chemistry
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Chemistry students, educators, and professionals involved in chemical analysis or empirical formula determination will benefit from this discussion.

joe823
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Homework Statement


When a sample that is .91 grams is burned in oxygen, it produces 2.073 grams of carbon dioxide, .304 grams of water, and .0944 grams of nitrogen. What is the empirical formula of the compound?

The Attempt at a Solution


I converted all of the products to moles:
CO2: .0471 moles
H2O: .0169 moles
N2: .00337 moles.

This gave me C.0471H.0338N.00674.

Simplifying the ratios, I got C7H5N.

My problem is that I don't have any idea of what to do with the oxygen. I don't even know if/how much of it it is present in the sample, because the sample is burned.

Thanks in advance for any help.
 
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From the amounts of carbon dioxide, water, and nitrogen produced, you should be able to figure out the mass of carbon, hydrogen and nitrogen in the original sample. The remaining mass is presumably due to oxygen.
 

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