I can't remember combustion analysis

In summary, by finding the moles of each compound in the given combustion reaction and comparing it to the sample mass, it was determined that the empirical formula for the unknown C-H-O compound is C3H6O. However, it is important to note that oxygen is also present in the sample due to its presence in both the compound and the air. This should be taken into consideration when calculating the amount of oxygen in the sample.
  • #1
1MileCrash
1,342
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63.8 mg of a C - H - O compound produces 145.0 mg of CO2 and 59.38 mg of H2O. Find empirical formula.

I THOUGHT that I simply had to find the moles of each compound as a ratio, but in doing that I get an answer that isn't an option.

I calculate moles C as (12/44)(.145) = .0395 grams carbon = .00329 moles carbon
Moles O as .10546 grams + .052 grams = .0097 moles oxygen
Moles H as .00329 moles H

Or CO3H...

What am I doing wrong here? I can't remember this crap.
 
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  • #2
Combustion products contain oxygen both from the compound and from the air.

Check hydrogen moles, you look for H, not for H2.

Check what is the mass of 0.00329 moles C, corrected moles of H and 0.0097 moles of O. Compare with the sample mass. Can you think about a way of finding amount of oxygen in the sample?
 
  • #3
Hows this look?

.145 g CO2 = .003295 Moles C = .0395 g
.05938 g H2O = .00659 Moles H = .0065 g

Grams oxygen = .0638 - .0395 - .0065 = .0178 g = .00106 moles oxygen

= C3H6O
 
  • #4
OK now.
 
  • #5


It seems like you are on the right track with your calculations, but there may be a calculation error or a missing piece of information. To find the empirical formula, you need to determine the ratio of each element present in the compound. In this case, we have 145.0 mg of CO2 and 59.38 mg of H2O.

To find the moles of each element, we need to divide the mass of each compound by its molar mass. The molar mass of CO2 is 44 g/mol and the molar mass of H2O is 18 g/mol.

For carbon: 145.0 mg / 44 g/mol = 0.003295 moles
For hydrogen: 59.38 mg / 18 g/mol = 0.003299 moles
For oxygen: 145.0 mg / 44 g/mol + 59.38 mg / 18 g/mol = 0.0097 moles

Now, we need to find the smallest whole number ratio of these moles. This can be done by dividing each number by the smallest one, which in this case is 0.003295.

For carbon: 0.003295 / 0.003295 = 1
For hydrogen: 0.003299 / 0.003295 = 1.0012 (round to 1)
For oxygen: 0.0097 / 0.003295 = 2.946 (round to 3)

This gives us the empirical formula of CH3O. It is possible that there was a calculation error or a missing piece of information in the original problem, so it is always important to double check your work and make sure all necessary information is provided.
 

What is combustion analysis?

Combustion analysis is a laboratory technique used to determine the elemental composition of a substance. It involves burning a sample in a controlled environment and measuring the amount of carbon, hydrogen, and nitrogen present in the resulting combustion products.

Why is combustion analysis important?

Combustion analysis is important because it allows scientists to determine the chemical composition of a substance, which can provide valuable information about its properties and potential uses. It is commonly used in fields such as chemistry, environmental science, and materials science.

How is combustion analysis performed?

Combustion analysis is typically performed using a specialized instrument called a combustion analyzer. The sample is placed in a combustion chamber and heated to a high temperature in the presence of oxygen. The resulting combustion products are then analyzed using various techniques, such as gas chromatography or mass spectrometry.

What are the limitations of combustion analysis?

While combustion analysis is a useful technique, it does have some limitations. For example, it may not be able to accurately determine the composition of complex mixtures or substances that contain elements that are difficult to combust. Additionally, the accuracy of the results can be affected by factors such as incomplete combustion or sample contamination.

What are some common applications of combustion analysis?

Combustion analysis has a wide range of applications in various fields. For example, it is used to analyze the composition of fuels, such as gasoline and diesel, to ensure they meet quality standards. It is also used in environmental studies to analyze air pollution and in the production of pharmaceuticals and other chemicals.

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