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Generally, the following reaction takes place in combustion analysis:

After burning 1.333 g of a hydrocarbon in a combustion analysis apparatus, 1.410 g of H 2 O and 4.305 g of CO 2 were produced. Separately, the molar mass of this hydrocarbon was found to be 204.35 g/mol. Calculate the empirical and molecular formulas of this hydrocarbon.

Step 1 : Using the molar masses of water and carbon dioxide, determine the moles of hydrogen and carbon that were produced.

Step 2 : Divide the larger molar amount by the smaller molar amount. In some cases, the ratio is not made up of two integers. Convert the numerator of the ratio to an improper fraction and rewrite the ratio in whole numbers as shown.

Therefore, the empirical formula is C 5 H 8 .

Step 3 : To get the molecular formula, divide the experimental molar mass of the unknown hydrocarbon by the empirical formula weight.

Therefore, the molecular formula is (C 5 H 8 ) 3 or C 15 H 24 .

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After burning 1.082 g of a hydrocarbon in a combustion analysis apparatus, 1.583 g of H 2 O and 3.315 g of CO 2 were produced. Separately, the molar mass of this hydrocarbon was found to be 258.52 g/mol. Calculate the empirical and molecular formulas of this hydrocarbon.

The empirical formula is C 3 H 7 , and the molecular formula is (C 3 H 7 ) 6 or C 18 H 42 .

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Compounds containing carbon, hydrogen, and oxygen

Combustion analysis can also be utilized to determine the empiric and molecular formulas of compounds containing carbon, hydrogen, and oxygen. However, as the reaction is performed in an environment of excess oxygen, the amount of oxygen in the sample can be determined from the sample mass, rather than the combustion data ( [link] , [link] ).

A 2.0714 g sample containing carbon, hydrogen, and oxygen was burned in a combustion analysis apparatus; 1.928 g of H 2 O and 4.709 g of CO 2 were produced. Separately, the molar mass of the sample was found to be 116.16 g/mol. Determine the empirical formula, molecular formula, and identity of the sample.

Step 1 : Using the molar masses of water and carbon dioxide, determine the moles of hydrogen and carbon that were produced.

Step 2 : Using the molar amounts of carbon and hydrogen, calculate the masses of each in the original sample.

Step 3 : Subtract the masses of carbon and hydrogen from the sample mass. Now that the mass of oxygen is known, use this to calculate the molar amount of oxygen in the sample.

Step 4 : Divide each molar amount by the smallest molar amount in order to determine the ratio between the three elements.

Therefore, the empirical formula is C 3 H 6 O.

Step 5 : To get the molecular formula, divide the experimental molar mass of the unknown hydrocarbon by the empirical formula weight.

Therefore, the molecular formula is (C 3 H 6 O) 2 or C 6 H 12 O 2 . Possible compound with this molecular formula are shown in ( [link] ).

Structure of possible compounds with the molecular formula C 6 H 12 O 2 : (a) butylacetate, (b) sec -butyl acetate, (c) tert -butyl acetate, (d) ethyl butyrate, (e) haxanoic acid, (f) isobutyl acetate, (g) methyl pentanoate, and (h) propyl proponoate.
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A 4.846 g sample containing carbon, hydrogen, and oxygen was burned in a combustion analysis apparatus; 4.843 g of H 2 O and 11.83 g of CO 2 were produced. Separately, the molar mass of the sample was found to be 144.22 g/mol. Determine the empirical formula, molecular formula, and identity of the sample.

The empirical formula is C 4 H 8 O, and the molecular formula is (C 4 H 8 O) 2 or C 8 H 16 O 2 . Possible compounds with this molecular formula are shown in ( [link] ).

Structure of possible compounds with the molecular formula C 8 H 16 O 2 : (a) octanoic acid (caprylic acid), (b) hexyl acetate, (c) pentyl proponate, (d) 2-ethyl hexanoic acid, (e) valproic acid (VPA), (f) cyclohexanedimethanol (CHDM), and (g) 2,2,4,4-tetramethyl-1,3-cyclobutandiol (CBDO).
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Binary compounds

By using combustion analysis, the chemical formula of a binary compound containing oxygen can also be determined. This is particularly helpful in the case of combustion of a metal which can result in potential oxides of multiple oxidation states.

A sample of iron weighing 1.7480 g is combusted in the presence of excess oxygen. A metal oxide (Fe x O y ) is formed with a mass of 2.4982 g. Determine the chemical formula of the oxide product and the oxidation state of Fe.

Step 1 : Subtract the mass of Fe from the mass of the oxide to determine the mass of oxygen in the product.

Step 2 : Using the molar masses of Fe and O, calculate the molar amounts of each element.

Step 3 : Divide the larger molar amount by the smaller molar amount. In some cases, the ratio is not made up of two integers. Convert the numerator of the ratio to an improper fraction and rewrite the ratio in whole numbers as shown.

Therefore, the chemical formula of the oxide is Fe 2 O 3 , and Fe has a 3+ oxidation state.

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A sample of copper weighing 7.295 g is combusted in the presence of excess oxygen. A metal oxide (Cu x O y ) is formed with a mass of 8.2131 g. Determine the chemical formula of the oxide product and the oxidation state of Cu.

The chemical formula is Cu 2 O, and Cu has a 1+ oxidation state.

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Bibliography

  • J. A. Dumas, Ann. Chem. Pharm. , 1841, 38 , 141.
  • H. Goldwhite, J. Chem. Edu. , 1978, 55 , 366.
  • A. Lavoisier, Traité Élémentaire de Chimie , 1789, 2 , 493.
  • J. Von Liebig, Annalen der Physik und Chemie , 1831, 21 , 1.
  • A. Linan and F. A. Williams, Fundamental Aspects of Combustion , Oxford University Press, New York (1993).
  • J. M. McBride, "Combustion Analysis," Chemistry 125 , Yale University,<http://www.chem.yale.edu/~chem125/125/history99/4RadicalsTypes/Analysis/Liebiganal.html>.
  • W. Prout, Philos. T. R. Soc. Lond. , 1827, 117 , 355.
  • D. Shriver and P. Atkins, Inorganic Chemistry , 5 th Ed., W. H. Freeman and Co., New York (2009).
  • W. Vining et. al., General Chemistry , 1 st Ed., Cengage, Brooks/Cole Cengage Learning , University of Massachusetts Amherst (2014).
  • J. Warnatz, U. Maas, and R. W. Dibble, Combustion: Physical and Chemical Fundamentals, Modeling and Simulation, Experiments, Pollutant Formation , 3 rd Ed., Springer, Berlin (2001).

Questions & Answers

how do you translate this in Algebraic Expressions
linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
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Cied
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
AMJAD
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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Source:  OpenStax, Physical methods in chemistry and nano science. OpenStax CNX. May 05, 2015 Download for free at http://legacy.cnx.org/content/col10699/1.21
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