Question

Complete combustion of 0.858g of compoundX gives 2.63g of CO2 and 1.28g of H2o

Answer 1

Answer:

Let us assume that the compound is CxHy. Therefore, on combustion we will get the following results

 CxHy  +  O2  ------------> xCO2 + y / 2 H2O

 0.858 g  excess  2.63g  1.27 g

Number of moles of CO2 obtained = 2.63 / 44 = 0.06 moles of CO2

Number of moles of H2O obtained = 1.27 / 18 = 0.071 moles of H2O

Now moles of CO2 obtained = moles of Carbon in original sample

therefore 0.06 moles of carbon are present in the sample

Now 1 mole of carbon = 12g

Therefore 0.06 mole of carbon = 0.06 X 12 = 0.72 g of carbon are present in the sample

Similarly

moles of H2O = 1/2 moles of H in original sample

Therefore moles of H in the original sample = 2 X moles of H2O obtained

 = 2 X 0.0705 = 0.141 moles

Therefore 0.141 moles of Hydrogen are present in the original sample

1 mole of Hydrogen =  1.008 g

Therefore 0.141 moles of Hydrogen = 0.141 X 1.008

 = 0.142 g of Hydrogen are present in the sample

Within the limits of experimental error, we find that 

mass of sample used = mass of carbon + hydrogen present in the sample

Hence no oxygen is present in the compound

Now % of C in the sample = (0.72 X 100) / 0.858 = 83.92 %