1.547 grams of hydrated MgSO4 is heated in a crucible. After heating, 0.7554 grams of anhydrous MgSO4 remains in the crucible. How many waters of hydration were attached to the MgSO4

7

Explanation:

Let x represent the number of moles of water in the hydrated salt i.e MgSO₄.xH₂O

The following data were obtained from the question:

Mass of MgSO₄.xH₂O = 1.547 g

Mass of anhydrous MgSO₄ = 0.7554 g

Mole of H₂O = x =?

Next, we shall determine the mass of water, H₂O in the hydrated salt, MgSO₄.xH₂O. This can be obtained as follow:

Mass of MgSO₄.xH₂O = 1.547 g

Mass of anhydrous MgSO₄ = 0.7554 g

Mass of H₂O =?

Mass of H₂O = (Mass of MgSO₄.xH₂O) – (Mass of anhydrous MgSO₄)

Mass of H₂O = 1.547 – 0.7554

Mass of H₂O = 0.7916 g

Finally, we shall determine the value of the x as illustrated below:

Mass of MgSO₄.xH₂O = 1.547 g

Molar mass of MgSO₄.xH₂O = 24 + 32 + (16×4) + x[(2×1) + 16]

= 24 + 32 + 64 + x(2 + 16)

= 120 + 18x

Mass of H₂O = 0.7916 g

Molar mass of xH₂O = 18x

Molar Mass of xH₂O/ Molar mass of MgSO₄.xH₂O = mass of xH₂O /Mass of MgSO₄.xH₂O

18x/ 120 + 18x = 0.7916/1.547

Cross multiply

0.7916 (120 + 18x) = 18x × 1.547

94.992 + 14.2488x = 27.846x

Collect like terms

94.992 = 27.846x – 14.2488x

94.992 = 13.5972x

Divide both side by 13.5972

x = 94.992 / 13.5972

x = 7

Thus, the formula for the hydrated salt, MgSO₄.xH₂O is MgSO₄.7H₂O

Number of moles of water, H₂O in the hydrated salt MgSO₄.7H₂O is 7.

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