Chemistry, asked by Anonymous, 10 months ago

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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Answered by krishnasinghks8299
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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

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