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Combustion analysis of a hydrocarbon produced 33.01 g co2 and 27.02 g h2o. calculate the empirical formula of the hydrocarbon.

Respuesta :

kenmyna

The empirical formula of hydrocarbon is CH4

explanation

Hydrocarbon contain both hydrogen(H) and carbon(C)

calculate the mass of  C  and H

mass of C  from CO2= 12/44x33.01=9.003 g

mass of H from H2O =2/18 x27.02 =3.002 g

find the number  of moles of C and H

mole=mass/molar mass

moles  of C=  9.003g/12g/mol=0.7503 moles

moles of H=  3.002g /1g/mol = 3.002 moles

find the mole ratio of each element by dividing each  mole with smallest number of mole  ( 0.7503 moles)

that is C=0.7503/0.7503=1

          H=3.002/0.7503=4

therefore the empirical formula= CH4

The study of chemicals and bonds is called chemistry. There are different types of elements and these are metal and nonmetals.

The correct answer is The empirical formula of hydrocarbon is CH4

What is combustion?

  • Combustion, or burning, is a high-temperature exothermic redox chemical reaction between a fuel and an oxidant, usually, atmospheric oxygen, that produces oxidized, often gaseous products, in a mixture termed as smoke

Hydrocarbon contain both hydrogen(H) and carbon(C)

  • Mass of C from CO2= [tex]\frac{12}{44}*33.01=9.003 g[/tex]
  • Mass of H from H2O =[tex]\frac{2}{18} *27.02 =3.002 g[/tex]

The number  of moles of C and H

Mole=[tex]\frac{mass}{molar mass}[/tex]

Moles  of C=  [tex]\frac{9.003g}{12g/mol}=0.7503 moles[/tex]

Moles of H=  [tex]\frac{3.002g}{1g/mol}[/tex]= 3.002 moles

The mole ratio of each element by dividing each  mole by the smallest number of moles ( 0.7503 moles) which is C=0.7503/0.7503=1

H=[tex]\frac{3.002}{0.7503}=4[/tex]

Therefore the empirical formula is CH4.

For more information about the combustion, refer to the link:-

https://brainly.com/question/9864045

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