The percentage by mass of Oxygen in MgSO\(_4\).7H\(_2\)O is [ M\(_r\) of MgSO\(_4\).7H\(_2\)O = 246 ]
Given:
M\(_r\) of MgSO\(_4\).7H\(_2\)O = 246
M\(_r\) of Oxygen = 16
Percentage by mass of Oxygen = \(\frac{Total mass of Oxygen}{Total mass of MgSO_4.7H_2O}\times 100\)
% by mass of Oxygen = \(\frac{11moles of oxygen}{MgSO_4.7H_2O}\times 100\) (4moles of oxygen + 7 moles of oxygen from 7H\(_2\)O )
= \(\frac{11 X 16}{246}\times 100\)
= \(\frac{176}{246}\times 100\)
= 71.544%
= 71.5%
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