Question

A compound was found to contain 40% carbon, 6.67% hydrogen and 53.33% by mass. Calculate the Empirical formula.

Answer 1

Answer:

Empirical Formula is CH2O

Explanation:

Refer attachment for explanation!!!

Answer 2

Answer..

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The Empirical formula is CH2O

Explanation:

40.00%

40.00%12

40.00%12(40/12)

40.00%12(40/12)(3.33/3.33)=1

40.00%12(40/12)(3.33/3.33)=1H

40.00%12(40/12)(3.33/3.33)=1H6.67%

40.00%12(40/12)(3.33/3.33)=1H6.67%1

40.00%12(40/12)(3.33/3.33)=1H6.67%1(6.67/1)

40.00%12(40/12)(3.33/3.33)=1H6.67%1(6.67/1)(6.667/3.33)=2

40.00%12(40/12)(3.33/3.33)=1H6.67%1(6.67/1)(6.667/3.33)=2O

40.00%12(40/12)(3.33/3.33)=1H6.67%1(6.67/1)(6.667/3.33)=2O53.33%

40.00%12(40/12)(3.33/3.33)=1H6.67%1(6.67/1)(6.667/3.33)=2O53.33%16

40.00%12(40/12)(3.33/3.33)=1H6.67%1(6.67/1)(6.667/3.33)=2O53.33%16(53.33/16)

40.00%12(40/12)(3.33/3.33)=1H6.67%1(6.67/1)(6.667/3.33)=2O53.33%16(53.33/16)(3.33/3.33)=1

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