A 400 mg iron capsule contains 100 mg of fe | vedclass

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(D) The molecular formula of ferrous fumarate is $(C_2H_2O_4)Fe$.Atomic mass of $Fe = 56 \ g/mol$.Molar mass of $(C_2H_2O_4)Fe = (2 	imes 12) + (2

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) The molecular formula of ferrous fumarate is $(C_2H_2O_4)Fe$.Atomic mass of $Fe = 56 \ g/mol$.Molar mass of $(C_2H_2O_4)Fe = (2 	imes 12) + (2 	imes 1) + (4 	imes 16) + 56 = 24 + 2 + 64 + 56 = 146 \ g/mol$.Mass of $Fe$ in $100 \ mg$ of ferrous fumarate $= \frac{56}{146} 	imes 100 \ mg \approx 38.36 \ mg$.Percentage of $Fe$ in the $400 \ mg$ capsule $= \frac{38.36 \ mg}{400 \ mg} 	imes 100 \approx 9.59 \%$.Given the options provided,the closest value is $8 \%$.

$A$ $400 \ mg$ iron capsule contains $100 \ mg$ of ferrous fumarate,$(C_2H_2O_4)Fe$. The percentage of iron present in it is approximately $.......... \%$.

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