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The percentage by mass of Oxygen in MgSO\(_4\).7H\(_2\)O is [ M\(_r\) of MgSO\(_4\).7H\(_2\)O = 246 ]

Chemistry
WAEC 2024

The percentage by mass of Oxygen in MgSO\(_4\).7H\(_2\)O is [ M\(_r\) of MgSO\(_4\).7H\(_2\)O = 246 ]

  • A. 26.0%
  • B. 45.5%
  • C. 71.5%
  • D. 84.0%
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Correct Answer: Option C
Explanation

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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