Complete combustion of 3.20g of a hydrocarbon produced 9.69g of CO2 and 4.96g of H2O. What is the empirical formula for the hydrocarbon?

Question

21-09-2018
Chemistry

contestada

Complete combustion of 3.20g of a hydrocarbon produced 9.69g of CO2 and 4.96g of H2O. What is the empirical formula for the hydrocarbon?

Respuesta :

Otras preguntas

Consider how hope valley, a popular ski resort, could use capital budgeting to decide whether the $ 8 million river park lodge expansion would be a good inves

In which quadrant does the terminal side of a 314° angle lie?

the unequal sides of an isoceles triangle is 3cm more than one of its equal sides . if the perimeter is 18 cm . find the length of the sides

...invoca el nombre de Dios en su poema/canción

What are the major products when glucose is broken down in glycolysis? NAD+ and FAD ATP and NADH ATP, FADH2, and NADH pyruvic acid and ATP

What is the main theme of the excerpt? Read this excerpt from Guy de Maupassant's "Mother Savage" about an old woman who takes in four Prussian soldiers while

Your professor is about to give a lecture on the history of psychoanalysis. which theorist is your professor most likely to focus his lecture on?

What is the length of segment AB? 7 9 18 25

What is 3 to the third power and 12 to the third power divided by 2 minus 388 plus 33 and then take away two again.still give the lowest common multiple althoug

Jean has always had trouble with her weight. she was weaned too early and is now overcompensating. according to psychoanalysts, jean is fixated in the _____ psy

AsiaWoerner AsiaWoerner · Answer 1 · 2018-09-30T15:57:58+02:00

Let empirical formula for hydrocarbon is CxHy

it will undergo combustion as

CxHy + (x + y/4) O2 ---> xCO2 + (y/2 )H2O

Given that mass of CO2 produced = 9.69 g

So moles of CO2 produced = 9.69 / 44 = 0.22 moles

So moles of carbon present = 0.22 moles

mass of H2O produced = 4.96 g

Moles of H2O produced = mass / molar mass = 4.96 / 18 = 0.28 moles

So moles of H present = 2 X 0.28 = 0.56 moles

Let us divided the moles of each with lowest value of moles

Moles of Carbon = 0.22 / 0.22 = 1 moles

moles of H = 0.56 / 0.22 = 2.55

Multiplying with two to get whole number

the ratio of carbon and hydrogen will be : C:H = 2:5

empirical formula : C2H5

ardni313 ardni313 · Answer 2 · 2020-01-25T03:22:37+01:00

The empirical formula for the hydrocarbon : C₂H₅

Further explanation

Complete combustion of Hydrocarbons with Oxygen will be obtained by CO₂ and H₂O compounds.

If O₂ is insufficient there will be incomplete combustion produced by CO and H and O

Hydrocarbon combustion reactions (specifically alkanes)

[tex]\large {\boxed {\bold {C_nH _ (_2_n _ + _ 2_) + \frac {3n + 1} {2} O_2 ----> nCO_2 + (n + 1) H_2O}}}[/tex]

[tex]\rm Mass~A~in~AxBy=\dfrac{x\times atomic~mass~A}{molar~mass~AxBy}\times mass~AxBy[/tex]

[tex]\rm mass~B~inAxBy=\dfrac{y\times atomic~mass~B}{molar~mass~AxBy}\times mass~AxBy[/tex]

Complete combustion of 3.20g of a hydrocarbon produced 9.69g of CO₂ and 4.96g of H₂O

Then the mass of C in CO₂: (molar mass of CO₂ = 44 g / mol, atomic mass C = 12 g / mol)

[tex]\rm \dfrac{1\times 12}{44}\times 9.69=2.64~grams[/tex]

[tex]\rm mol~C=\dfrac{2.64}{12}=0.22[/tex]

mass H in H2O (molar mass H2O = 18 g / mol, atomic mass H = 1 g / mol)

[tex]\rm \dfrac{2\times 1}{18}\times 4.96=0.5511~grams[/tex]

[tex]\rm mol~H=\dfrac{0.5511}{1}=0.5511[/tex]

For example the hydrocarbons: CxHy

x and y are the mole ratio of C and H atoms

the mole ratio of C: H (from CO₂ and H₂O) is the same as the ratio of C and H in hydrocarbon compounds then:

mol C: mol H = 0.22: 0.5511 = 0.4 = 4: 10

So that the compound molecular formula: C₄H₁₀ or can be simplified in the form of the empirical formula C₂H₅

Learn more

the combustion of ethanol

https://brainly.com/question/4414828

coefficients are needed to balance the equation for the complete combustion of methane

https://brainly.com/question/1971314

synthesize 18.0 mol of no

brainly.com/question/3636135

mass of copper required

brainly.com/question/1680090