Question

An organometallic compound on analysis was found to contain, C = 64.4%, H = 5.5% and Fe = 29.9%. Determine its empirical formula (At mass of Fe = 56 u).

Answer 1

An organometallic compound contains Carbon = 64.4%; Hydrogen = 5.5%, and Iron = 29.9% (Given atomic weight of Fe = 56)

Element Name: C

Percentage (%): 64.4

Atomic mass: 12

Relative Atomic Mass: $\frac { 64.4 }{ 12 } =5.36$

Divide by the least number: $\frac { 5.36 }{ 0.53 } =10.1$

Simplest Ratio: 10

Element Name: H

Percentage (%): 5.5

Atomic mass: 1  

Relative Atomic Mass: $\frac { 5.5 }{ 1 } =5.5$

Divide by the least number: $\frac { 5.5 }{ 0.53 } =10.4$

Simplest Ratio: 10

Element Name: Fe

Percentage (%): 29.9

Atomic mass: 56

Relative Atomic Mass: $\frac { 29.9 }{ 56 } =0.53$

Divide by the least number: $\frac { 5.53 }{ 0.53 } =1$

Simplest Ratio: 1

Finally, the simplest ration we obtain is,

Carbon = 10 numbers; Hydrogen = 10 numbers; and Iron = 1 number.

Therefore, the Empirical formula = ${ C }_{ 10 }{ H }_{ 10 }Fe$

Answer 2

Answer:

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