An organic compound was found to contain only C, H, and Cl. When a 1.50-g sample of the compound was completely combusted in air, 3.52 g of CO was formed. In a separate experiment the chlorine in a 1.00-g sample of the compound was converted to 1.27 g of AgCl. Determine the empirical formula of the compound.

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vickydes566 vickydes566

21-02-2020
Physics

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An organic compound was found to contain only C, H, and Cl. When a 1.50-g sample of the compound was completely combusted in air, 3.52 g of CO was formed. In a separate experiment the chlorine in a 1.00-g sample of the compound was converted to 1.27 g of AgCl. Determine the empirical formula of the compound.

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Hashirriaz830 Hashirriaz830 · Answer 1 · 2020-02-21T07:47:53+01:00

Answer:

Hence the emperical formula for the compound will be C6H5Cl

Explanation:

solution:

To Calculate the mass of Carbon atom in CO2 and Chlorine in AgCI and Hydrogen atom from the sample

Mass of Carbon in 3.52 g of CO2 = (3.52 Molar Mass of CO2)* Molar Mass of Carbon atom =(3.52/44) *12 = 0.96g

Similarly Mass of Chlorine and Hydrogen can be calculated.

1.27 Mass of Chlorine = (1.27/143.25)*35.45 = 0.314g

Mass of Hydrogen = Mass of the sample - Mass of Carbon and Chlorine

= 1.50-0.96 + 0.314g

= 0.069g

Now Calculating the number of moles of each 0.96

Number of moles of Carbon = 0.96/12 = 0.08 moles

Number of moles of Chlorine = 0.314/35.5=0.013 moles

Number of moles of Hydrogen = 0.069/1.01=0.068 moles

Dividing each of the obtained moles by 0.013 we get the integer numbers i.e 0.08/0.013 = 6 , 0.013/0.013 = 1 , 0.068/0.013 = 5.

Hence the emperical formula for the compound will be C6H5Cl

tallinn tallinn · Answer 2 · 2022-03-31T15:24:21+02:00

When the empirical formula for the compound will be = C6H5Cl

Computation of Mass of Carbon atom

To Compute the mass of Carbon atom in CO2 Chlorine in AgCI and also Hydrogen atom from the sampling

Then the Mass of Carbon in 3.52 g of CO2 is = (3.52 Molar Mass of CO2)* After that Molar Mass of Carbon atom is =(3.52/44) *12 = 0.96g

Besides, When the Mass of Chlorine and Hydrogen can be calculated.

Then 1.27 Mass of Chlorine is = (1.27/143.25)*35.45 = 0.314g

Now Mass of Hydrogen is = Mass of the sample - Mass of Carbon and Chlorine

Then = 1.50-0.96 + 0.314g

= 0.069g

Now we are Calculating the number of moles each 0.96

Then the Number of moles of Carbon is = 0.96/12 = 0.08 moles

After that Number of moles of Chlorine is = 0.314/35.5=0.013 moles

Now the Number of moles of Hydrogen is = 0.069/1.01=0.068 moles

Then Dividing each of the obtained moles by 0.013 we get the integer After that the numbers i.e 0.08/0.013 is = 6 , 0.013/0.013 = 1 , 0.068/0.013 = 5.

Therefore the empirical formula for the compound will be C6H5Cl

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