Combustion analysis problem? - General Chemistry

Once you have all of the elements, you can calculate the empirical formula. In summary, the conversation is about determining the empirical formula of a compound based on the amounts of carbon dioxide, water, and nitrogen produced when a sample is burned in oxygen. The steps involved include converting the products to moles, simplifying the ratios, and calculating the mass of each element present in the sample. The remaining mass is assumed to be oxygen, which is then used to determine the empirical formula of the compound.
  • #1
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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  • #2
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.
 
  • #3


I would first check if the combustion reaction is balanced. If it is, then the amount of oxygen consumed in the reaction can be calculated based on the stoichiometry of the balanced equation. This information can then be used to determine the empirical formula of the compound.

If the reaction is not balanced, then the products and reactants need to be balanced before proceeding with the calculation. Another thing to consider is the possibility of impurities in the sample that could affect the results of the combustion analysis.

In any case, it is important to double check all calculations and make sure they are consistent with the data provided. If there are still discrepancies, it may be necessary to repeat the experiment or seek the assistance of a chemistry expert.
 

1. What is combustion analysis and why is it important in chemistry?

Combustion analysis is a method used to determine the elemental composition of a substance by burning it in the presence of oxygen. It is important in chemistry because it allows us to identify the types and proportions of elements present in a compound, which can help in determining its properties and potential reactions.

2. How does combustion analysis work?

In combustion analysis, a sample of the substance is heated in the presence of pure oxygen, causing it to combust and release water vapor and carbon dioxide. The amount of these products is measured and used to calculate the amount of carbon and hydrogen in the sample. The remaining residue is then analyzed for other elements, such as nitrogen or sulfur.

3. What is the difference between complete and incomplete combustion?

Complete combustion occurs when a substance is burned in excess oxygen, resulting in the complete conversion of the substance into carbon dioxide and water vapor. Incomplete combustion occurs when there is not enough oxygen present, leading to the formation of carbon monoxide and/or elemental carbon in addition to carbon dioxide and water vapor.

4. What factors can affect the accuracy of combustion analysis results?

The accuracy of combustion analysis results can be affected by factors such as impurities in the sample, incomplete combustion, and errors in measurement. It is important to ensure that the sample is pure and that the combustion is complete to obtain accurate results.

5. How is combustion analysis used in determining the empirical formula of a compound?

Combustion analysis can be used to determine the empirical formula of a compound by providing the mass percentages of the elements present. These percentages can then be converted into moles and used to find the simplest whole number ratio of the elements, which represents the empirical formula of the compound.

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