Calculate the percentage of metal deficiency | vedclass

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Question

(A) In $Fe_{0.94}O$,the total positive charge must balance the total negative charge to maintain electrical neutrality.Let the number of $Fe^{2+}$ ion

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) In $Fe_{0.94}O$,the total positive charge must balance the total negative charge to maintain electrical neutrality.Let the number of $Fe^{2+}$ ions be $x$ and the number of $Fe^{3+}$ ions be $(0.94 - x)$.The charge on $O^{2-}$ is $-2$.So,$2x + 3(0.94 - x) = 2$.$2x + 2.82 - 3x = 2$.$-x = -0.82$,so $x = 0.82$.The number of $Fe^{3+}$ ions is $0.94 - 0.82 = 0.12$.The total iron ions are $0.94$.The fraction of $Fe^{3+}$ ions is $(0.12 / 0.94) 	imes 100 \approx 12.77\%$.However,the question asks for the metal deficiency percentage.The total iron expected is $1.00$,but we have $0.94$.Deficiency = $1.00 - 0.94 = 0.06$.Percentage deficiency = $(0.06 / 1.00) 	imes 100 = 6\%$.

Calculate the percentage of metal deficiency defect in $Fe_{0.94}O$.

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