A 400 mg iron capsule contains 100 mg of ferrous fumarate, (CHCOO)2Fe. The percentage of iron present in it, is approximately

Question

A 400 mg iron capsule contains 100 mg of ferrous fumarate, (CHCOO)2Fe. The percentage of iron present in it, is approximately (A) 33% (B) 25% (C) 14% (D) 8%

Tardigrade Admin · Accepted Answer

Molecular wt. of (CHCOO)2Fe = 170 Fe present in 100 mg of (CHCOO)2Fe =(56/170)×100 mg=32.9 mg This is present in 400 mg of the capsule ∴ % of Fe in the capsule =(32.9/400)×100=8.2

Q. A $400 m g$ iron capsule contains $100 m g$ of ferrous fumarate, $(C H COO)_{2} F e$ . The percentage of iron present in it, is approximately

33%

25%

14%

8%

Molecular wt. of $(C H COO)_{2} F e = 170$
Fe present in 100 mg of $(C H COO)_{2} F e$
$= \frac{56}{170} \times 100 m g = 32.9 m g$
This is present in $400 m g$ of the capsule
$∴ %$ of Fe in the capsule $= \frac{32.9}{400} \times 100 = 8.2$

Q. A 400mg iron capsule contains 100mg of ferrous fumarate, (CHCOO)2​Fe. The percentage of iron present in it, is approximately

33%

25%

14%

8%

Molecular wt. of (CHCOO)2​Fe=170 Fe present in 100 mg of (CHCOO)2​Fe =17056​×100mg=32.9mg This is present in 400mg of the capsule ∴% of Fe in the capsule =40032.9​×100=8.2

Q. A $400 m g$ iron capsule contains $100 m g$ of ferrous fumarate, $(C H COO)_{2} F e$ . The percentage of iron present in it, is approximately

Molecular wt. of $(C H COO)_{2} F e = 170$
Fe present in 100 mg of $(C H COO)_{2} F e$
$= \frac{56}{170} \times 100 m g = 32.9 m g$
This is present in $400 m g$ of the capsule
$∴ %$ of Fe in the capsule $= \frac{32.9}{400} \times 100 = 8.2$