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Benzaldehyde

$\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}{\mathbf{C}}_{\mathbf{7}}{\mathbf{H}}_{\mathbf{6}}\mathbf{O}}\mathbf{}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}{\mathbf{C}}_{\mathbf{7}}{\mathbf{H}}_{\mathbf{6}}\mathbf{O}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}{\mathbf{C}}_{\mathbf{7}}{\mathbf{H}}_{\mathbf{6}}\mathbf{O}}}\mathbf{\times}\frac{\mathbf{106}\mathbf{.}\mathbf{118}\mathbf{}\mathbf{g}\mathbf{}{\mathbf{C}}_{\mathbf{7}}{\mathbf{H}}_{\mathbf{6}}\mathbf{O}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}{\mathbf{C}}_{\mathbf{7}}{\mathbf{H}}_{\mathbf{6}}\mathbf{O}}}$** = 106.118 g C _{7}H_{6}O**

$\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}{\mathbf{C}}_{\mathbf{7}}{\mathbf{H}}_{\mathbf{6}}\mathbf{O}}\mathbf{}\mathbf{\times}\frac{\mathbf{7}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{C}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}{\mathbf{C}}_{\mathbf{7}}{\mathbf{H}}_{\mathbf{6}}\mathbf{O}}}\mathbf{\times}\frac{\mathbf{12}\mathbf{.}\mathbf{01}\mathbf{}\mathbf{g}\mathbf{}\mathbf{C}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{C}}}$** = 84.07 g C**

Based on the following structural formulas, calculate the percentage of carbon by mass present in each compound.

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Based on our data, we think this problem is relevant for Professor Dixon's class at UCF.