Question

Magnesium sulphate crystals (MgSO4.7H2O) can be dehydrated by heating . Calculate the percentage by mass of water present in the molecule.

Answer 1

Answer :-

Percentage by mass of water in MgSO₄.7H₂O is 51.21 % .

Explanation :-

We have :-

→ Atomic mass of Magnesium = 24 u

→ Atomic mass of Sulphur = 32 u

→ Atomic mass Oxygen = 16 u

→ Atomic mass of Hydrogen = 1 u

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Molecular mass of Magnesium Sulphate crystals i.e. MgSO₄.7H₂O will be :-

= 24 + 32 + 16 × 4 + 7(1 × 2 + 16)

= 56 + 64 + 7(18)

= 120 + 126

= 246 u

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Now, one molecule of Magnesium Crystal has (7 × 18) = 126 u of water .

So, the mass percentage of water in the molecule will be :-

= (126/246) × 100

= 0.5121 × 100

= 51.21 %

Answer 2

⇝ We Know :-

• Mass of Magnesium (Mg) = 24 u

• Mass of Sulphur (S) = 32 u

• Mass of Oxygen (O) = 16 u

• Mass of Hydrogen (H) = 1 u

⇝ Calculating Molar Mass of MgSO₄.7H₂O :

• Molar Mass = Mass of Mg + Mass of Sulphur + 11 × Mass of Oxygen + 14 × Mass of Hydrogen.

❒ Using Above written Masses :

$\text{ Molar mass} = 24 + 32 + 11 \times 16 + 14 \times 1$

$= 24 + 32 + 176 + 14$

$\large\red{ \bf = 246 \: u}$

⇝ Calculating Molar Mass of 7H₂O :

• Molar Mass = 7 × ( 2× Mass of Hydrogen + Mass of Oxygen)

❒ Using Respective Masses :

$\text{ Molar mass} = 7 \times(2 \times 1 +16)$

$=7\times( 18)$

$\large\red{ \bf = 126 \: u}$

⇝ Calculating Reqrd. percentage :-

$\purple{\sf \small \% \: of \: water = \frac{mass \: of \: water \: in \: molecule}{mass \: of \: molecule \: } \times 100}$

$= \frac{126}{246} \times 100$

$= 0.5121 \times 100$

$\bf= 51.21\%$

Hence,

$\large\underline{\pink{\underline{\frak{\pmb{Required \: \% = 51.21\%}}}}}$