What is the percentage by mass of oxygen in magnesium sulfate heptahydrate (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 is:

$$24 + 32 + (4 \times 16) = 24 + 32 + 64 = 120\, \text{g/mol}$$

Molar mass of 7H₂O is:

$$7 \times ((2 \times 1) + 16) = 7 \times (2 + 16) = 7 \times 18 = 126\, \text{g/mol}$$

Molar mass of MgSO₄·7H₂O is:

$$120 + 126 = 246\, \text{g/mol}$$

The MgSO₄ part contains 4 oxygen atoms.
The 7H₂O part contains $7 \times 1 = 7$ oxygen atoms.
Total number of oxygen atoms = $4 + 7 = 11$
Total mass of oxygen in one mole of MgSO₄·7H₂O is:

$$11 \times 16\, \text{g/mol} = 176\, \text{g/mol}$$

The percentage by mass of oxygen is given by the formula:

$$\text{Percentage by mass of oxygen} = \frac{\text{Total mass of oxygen}}{\text{Molar mass of MgSO₄·7H₂O}} \times 100\%$$

Substitute the calculated values:

$$\text{Percentage by mass of oxygen} = \frac{176\, \text{g/mol}}{246\, \text{g/mol}} \times 100\%$$ $$\text{Percentage by mass of oxygen} \approx 0.715447 \times 100\%$$ $$\text{Percentage by mass of oxygen} \approx 71.54\%$$

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

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