0.45g of an organic compound on ignition gave 0.44 gram of carbon dioxide and 0.09 gram of H2O determine the empirical formula of the compound.
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Mrs m Kundu
Jan 3, 2015
Calculate the empirical formula of given organic compound?
0.30 gm of an organic compound containing C, H & O on combustion gave 0.44 gm CO2 & 0.18 gm H2O. Determine the empirical formulae of compound.
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Swathi Ambati
Ans: Given data:
Mass of organic compound – 0.30g
The compound contains C , H & O elements.
Mass of CO2 gas produced after combustion – 0.44g
Mass of H2O produced after combustion – 0.18g
Calculation of mass of carbon:
C + O2 → CO2
12g 44g
44g of CO2 contains 12g of carbon
0.44g of CO2 contains 0.44 X 12 \44 = 0.12g
Calculation of mass of Hydrogen
2H2 + O2 → 2H2O
4g 36g
36g of water contains 4g of hydrogen
0.18g of water contains 0.184/36 = 0.02g
% of C = Mass of carbon /Mass of compound X 100
= 0.12/0.30 X 100
= 40%
% of H= 0.02/0.30 X 100 = 6.6%
% of oxygen = 100 -(%C +% H)
= 100 – 46.66
= 53.34%
Element % composition % / Atomic weight Simple ratio
Carbon 40% 40/12 = 3.33 3.33/3.33 = 1
Hydrogen 6.66% 6.66/1 = 6.66 6.66\3.33 = 2
Oxygen 53.34% 53.34/16 = 3.33 3.33/3.33 = 1
Number of carbon atoms present in the given compound is 1
Number of hydrogen atoms present in the given compound is 2
Number of oxygen atoms present in the given compound is 1
The empirical formula of given compound is CH2O