What is the percentage by mass of oxygen in MgSO₄·7H₂O?

Molar mass of Mg = $24\, \text{g/mol}$
Molar mass of S = $32\, \text{g/mol}$
Molar mass of O = $16\, \text{g/mol}$
Molar mass of H = $1\, \text{g/mol}$

Molar mass of MgSO₄ = $24 + 32 + (4 \times 16) = 24 + 32 + 64 = 120\, \text{g/mol}$
Molar mass of 7H₂O = $7 \times ((2 \times 1) + 16) = 7 \times (2 + 16) = 7 \times 18 = 126\, \text{g/mol}$
Molar mass of MgSO₄·7H₂O = $120 + 126 = 246\, \text{g/mol}$

Number of oxygen atoms in MgSO₄ = 4
Number of oxygen atoms in 7H₂O = $7 \times 1 = 7$
Total number of oxygen atoms = $4 + 7 = 11$
Total mass of oxygen = $11 \times 16\, \text{g/mol} = 176\, \text{g/mol}$

The calculated percentage is approximately $71.5\%$. This matches option c.

The final answer is \text{c. 71.5%}