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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

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Bob Reply
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The nanotechnology is as new science, to scale nanometric
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Damian
Is there any normative that regulates the use of silver nanoparticles?
Damian Reply
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Renato
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biomolecules are e building blocks of every organics and inorganic materials.
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research.net
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sciencedirect big data base
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Introduction about quantum dots in nanotechnology
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nano basically means 10^(-9). nanometer is a unit to measure length.
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there is no specific books for beginners but there is book called principle of nanotechnology
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fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
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That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
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Mostly, they use nano carbon for electronics and for materials to be strengthened.
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carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
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Graphene has a hexagonal structure
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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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