Final answer:
The percent by mass of water in iron(II) sulfate heptahydrate (FeSO₄ * 7H₂O) is approximately 45.35%, calculated by dividing the molar mass of the water in the compound by the total molar mass of the compound and multiplying by 100%.
Step-by-step explanation:
To find the percent by mass of water in iron(II) sulfate heptahydrate (FeSO₄ * 7H₂O), you first need to calculate the molar mass of the compound and the molar mass of the water within it. The formula for iron(II) sulfate heptahydrate is FeSO₄ * 7H₂O, which means each molecule consists of one iron(II) sulfate (FeSO₄) and seven water molecules (7H₂O).
The molar mass of FeSO₄ is approximately 151.91 g/mol (55.85 for Fe, 32.07 for S, and 4*(16.00) for O). The molar mass of 7H₂O is approximately 7*(18.015) which equals 126.105 g/mol. Therefore, the molar mass of the entire compound is 151.91 g/mol + 126.105 g/mol = 278.015 g/mol.
To calculate the percent by mass, divide the mass of water (126.105 g) by the total molar mass of the compound (278.015 g) and multiply by 100:
(126.105 g/278.015 g) * 100% = 45.35%.
So, the percent by mass of water in iron(II) sulfate heptahydrate is approximately 45.35%.