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.

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  

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂

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