Determining an Emperical Formula using Combustion Analysis data

In summary, the problem is that the book does not use the same divisor for the empirical formula as the problem was stated.
  • #1
Crookedsky
4
0
the problem I am working with is:
Naphthalene is a carbon-hydrogen compound that finds use as mothballs. A sample of naphthalene is subjected to combustion analysis, producing 1.100g of CO2 and .1802g of H2O. Based on these data, calculate the empirical formula of naphthalene.

So when i set up the demensional analysis and get the moles of C and H i get for C .02499/
.02499 and for H i get .02000/.02499. For H the book multiplies the fraction for H by 5 because .02000/.02499 = .800320128 which is not close to a whole number.

My question is: How do they determine what the fraction is. Which is in this case is 4/5ths.

The Emperical Formula for napthalene is C5H4
 
Chemistry news on Phys.org
  • #2
0.8=8/10=4/5.

It is not always that obvious, but here it is pretty simple.
 
  • #3
Ya but how do i know when to multiply is by some fraction? Like .99999 is pretty obvious but what about .4154 or .5900 etc etc. What is the rule of thumb?
 
  • #4
Crookedsky said:
Ya but how do i know when to multiply is by some fraction? Like .99999 is pretty obvious but what about .4154 or .5900 etc etc. What is the rule of thumb?

At a simple level, just multiply by 2, 3, 4, 5 etc until you get reasonably close to whole numbers
 
  • #5
Ok thanks Sjb,
Now i have a different problem that i have been working:
Ethyl Alcohol, the alcohol present in alcoholic beverages, is a carbon-hydrogen-oxygen compound. Combustion analysis of a 1.000g sample of ethyl alcohol produces 1.913g of CO2 and 1.174g of H2O. Based on these data, calculate the empirical formula of ethyl alcohol.

So i set the dimensional analysis and solve for grams of Carbon(.5220) and grams of Hydrogen(.1316). Then i add C and H up and get .6536g mass. Next i solve for O by sutracting .6536 from 1.000 and come up with a .3464g of O. Then I convert H, C, and O to moles and come up with .04346 mol of C, .130 mol of H, and .02165 mol of O.

At this point either the book is wrong or i am wrong. I used the moles of C (.04346) as my divisor to calculate the ratio. I came up with an empirical formula of CH3O5 but the book uses the moles of O (.02165) as the divisor and comes up with a empirical formula of C2H6O.

Can someone check my calculations because i have done this problem twice. Thanks
 
  • #6
What is .04346 / .02165 ?
 
  • #7
Numbers of moles you calculated so far are OK. Answer sjb's question.
 

1. How do I collect data for combustion analysis?

To collect data for combustion analysis, you will need to first weigh your sample and record its mass. Then, place the sample in a combustion apparatus and burn it completely. The combustion will produce carbon dioxide and water vapor, which can be collected and measured using appropriate equipment.

2. What is the purpose of combustion analysis in determining an empirical formula?

The purpose of combustion analysis is to determine the relative amounts of carbon, hydrogen, and other elements present in a compound. This information is then used to calculate the empirical formula of the compound.

3. Can combustion analysis be used for all types of compounds?

No, combustion analysis is only suitable for organic compounds, which contain carbon and hydrogen. Inorganic compounds do not produce carbon dioxide and water vapor when burned, so this method cannot be used to determine their empirical formula.

4. How do I calculate the empirical formula using combustion analysis data?

To calculate the empirical formula, you will need to first determine the moles of carbon and hydrogen present in the compound based on the mass of carbon dioxide and water vapor produced during combustion. Then, divide each mole value by the smallest mole value to get the ratio of elements. This ratio represents the empirical formula of the compound.

5. Are there any limitations to using combustion analysis for determining an empirical formula?

Yes, there are several limitations to consider when using combustion analysis. The method assumes that all carbon in the compound is converted to carbon dioxide and all hydrogen is converted to water vapor, which may not always be the case. Additionally, the presence of other elements such as nitrogen and sulfur can complicate the analysis and may require additional steps. It is important to carefully follow the procedure and use accurate equipment to minimize errors in the data collected.

Similar threads

Automotive Loss of gaseous volume because of fuel combustion
    • Mechanical Engineering
Replies
2
Views
205
Butane Combustion Analysis Demo for a High School Chemistry Class
    • Chemistry
Replies
5
Views
2K
Chemistry Find the formula of a hydrocarbon using combustion data
    • Biology and Chemistry Homework Help
Replies
14
Views
2K
Chemistry Combustion Analysis- Empirical and molecular formula
    • Biology and Chemistry Homework Help
Replies
8
Views
3K
How can I calculate heat of combustion and enthalpy using Hess's Law?
    • Chemistry
Replies
4
Views
2K
Specific Heat at High Temperatures
    • Chemistry
Replies
2
Views
2K
I can't remember combustion analysis
    • Chemistry
Replies
4
Views
10K
Combustion calculations and the required amount of air
    • Chemistry
Replies
2
Views
10K
Why is this calculation of the enthelpy of combustion wrong?
    • Introductory Physics Homework Help
Replies
4
Views
2K
Combustion of Coal in a boiler
    • Mechanical Engineering
Replies
1
Views
2K

Hot Threads

Recent Insights

Back
Top