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Solution


Q.

On complete combustion 1.0 g of an organic compound (X) gave 1.46 g of CO2 and 0.567 g of H2O. The empirical formula mass of compound (X) is ______ g.  

(Given molar mass in g mol-1: C : 12, H : 1, O : 16)                                   [2025]
1 30  
2 45  
3 60  
4 15  

Ans.

(1)

Moles of C in organic compound = moles of CO2=Mass of CO2Molar mass of CO2=1.4644=0.033 mol

Mass of C in organic compound = moles of C in organic compound × molar mass of C = 0.033×12g=0.396g

Moles of H in organic compound = 2×moles of H2O

=2×Mass of H2OMolar mass of H2O=2×0.56718=0.063 mol

Mass of H in organic compound = moles of H in organic compound×molar mass of H =0.063×1 g=0.063 g

Mass of O in organic compound = Mass of organic compound - (mass of C + mass of H) = 1 - (0.396 + 0.063) = 0.541 g

Moles of O in 1 g of organic compound =Mass of O in 1 g organic compoundMolar mass of O =0.54116mol=0.0338mol

Ratio of moles of C, H and O = 0.033 : 0.063 : 0.0338 = 1 : 2 : 1

Thus Empirical formula is CH2O and empirical formula mass is 30.