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We are asked to find the mass percent of CaCO_{3} in the mixture.

Assume 100 of the mixture.

Mass CO_{3} = 61.1 g

$\mathbf{Moles}\mathbf{}{\mathbf{CO}}_{\mathbf{3}}\mathbf{}\mathbf{=}\frac{\mathbf{61}\mathbf{.}\mathbf{1}\mathbf{}\mathbf{g}}{\mathbf{60}\mathbf{}\mathbf{g}\mathbf{/}\mathbf{mol}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{02}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{CO}}_{\mathbf{3}}\mathbf{}$

Molar mass CaCO_{3} = 100 g/mol

Molar mass (NH_{4})CO_{3} = 96 g/mol

Calculate mass CaCO_{3} in the mixture:

$\frac{\mathbf{mass}\mathbf{}{\mathbf{CaCO}}_{\mathbf{3}}}{\mathbf{100}}\mathbf{+}\frac{\mathbf{100}\mathbf{}\mathbf{-}\mathbf{mass}\mathbf{}{\mathbf{CaCO}}_{\mathbf{3}}}{\mathbf{96}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{02}\mathbf{}\phantom{\rule{0ex}{0ex}}\mathbf{96}(\mathrm{mass}{\mathrm{CaCO}}_{3})\mathbf{+}\mathbf{10}\mathbf{,}\mathbf{000}\mathbf{}\mathbf{-}\mathbf{100}(\mathrm{mass}{\mathrm{CaCO}}_{3})\mathbf{=}\mathbf{}\mathbf{9792}\phantom{\rule{0ex}{0ex}}\mathbf{mass}\mathbf{}{\mathbf{CaCO}}_{\mathbf{3}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{52}\mathbf{}\mathbf{g}$

A mixture of CaCO_{3} and (NH_{4} )_{2} CO_{3} is 61.1 % CO_{3} by mass.

Find the mass percent of CaCO_{3} in the mixture.

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Based on our data, we think this problem is relevant for Professor N/A's class at Ryerson University.