Question

what is the percentage composition of Fe and H2O in the compound of FeSO4.7H2O (Fe=56, S=32, O=16, H=1_

Answer 1

FeSO4.7H20

Molecular mass = 56+32+4×16+7×18

= 278g

278g has 56g Fe

100g will have

= 20.14g = 20.14%

278g has 126g H2O

100g will have

= 45.32g = 45.32%

PLZ MARK AS BRAINLIEST

Answer 2

♪♪Question♪♪

What is the percentage composition of Fe and H2O in the compound of FeSO4.7H2O?

[Fe=56u, S=32u, O=16u, H=1u]

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♦♦Answer:♦♦

%Composition of Fe=20.14%

and

%Composition of H₂O=45.32%

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Explanation:

As, the percentage composition of an element in a compound is given by:

%Composition of an element=[(Molar mass of that element)÷(Molar mass of compound)]×100

i.e.,

$\%comp = \binom{M.M. of \: element}{M.M. of \: compound } \times 100$

Here,

Molar Mass of Fe(Iron)=56g/mol.

Molar Mass of H₂O=18g/mol.

Molar Mass of FeSO₄.7H₂O=(56)+(32)+(16×4)+(7)(18)=278g/mol

So,

$\% \: Comp. of Fe= \frac{56}{278} \times 100$

$\% Comp. of Fe= \frac{28}{139} \times 100$

$\% \: Comp. of Fe= 20.143\%$

And,

$\% \: Comp. of H2O = \frac{7(18)}{278} \times 100$

$\% \: Comp. of H2O = \frac{126}{278} \times 100$

$\% \: Comp. of H2O = \frac{63}{139} \times 100$

$\% \: Comp. of H2O = 45.323\%$