What is the percent composition of iron(II) sulfate hexahydrate?Group of answer choices21.5% Fe; 12.3%S; 24.6% O; 41.6% H21.5% Fe; 12.3%S; 61.5% O; 4.7% H4.2% Fe; 4.2% S; 41.6% O; 50.0% H36.8% Fe; 21.1%S; 42.1% O

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Solution 1

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Solution 2

The percent composition of a compound can be calculated by dividing the mass of each element by the total mass of the compound and then multiplying by 100 to get a percentage.

The formula for iron(II) sulfate hexahydrate is FeSO4.6H2O.

First, we need to calculate the molar mass of the compound.

The molar mass of Fe (iron) is approximately 55.85 g/mol.
The molar mass of S (sulfur) is approximately 32.07 g/mol.
The molar mass of O (oxygen) is approximately 16.00 g/mol, and there are 4 in the sulfate ion and 6*2 in the water molecules, so we multiply this by 16.
The molar mass of H (hydrogen) is approximately 1.01 g/mol, and there are 2*6 in the water molecules, so we multiply this by 12.

Adding these together, we get:

55.85 g/mol (Fe) + 32.07 g/mol (S) + 16.00 g/mol * 16 (O) + 1.01 g/mol * 12 (H) = 278.05 g/mol

Now we can calculate the percent composition:

The percent composition of Fe is (55.85 g/mol / 278.05 g/mol) * 100 = 20.08%
The percent composition of S is (32.07 g/mol / 278.05 g/mol) * 100 = 11.53%
The percent composition of O is (16.00 g/mol * 16 / 278.05 g/mol) * 100 = 91.52%
The percent composition of H is (1.01 g/mol * 12 / 278.05 g/mol) * 100 = 4.36%

However, the sum of these percentages is greater than 100%, which is not possible. This is because we have counted the oxygen in both the sulfate ion and the water molecules, and the hydrogen in the water molecules.

To correct this, we need to subtract the contribution of the water molecules from the total:

The corrected percent composition of O is 91.52% - (16.00 g/mol * 6 / 278.05 g/mol) * 100 = 64.52%
The corrected percent composition of H is 4.36% - (1.01 g/mol * 6 / 278.05 g/mol) * 100 = 2.18%

So, the correct percent composition of iron(II) sulfate hexahydrate is approximately 20.08% Fe, 11.53% S, 64.52% O, and 2.18% H.

None of the provided answer choices match these calculations.

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