We are asked to calculate for the **mass percent (mass %) ****of MgCO _{3}**

To do that we’re going to use the **mass percent formula **shown below:

$\overline{){\mathbf{mass}}{\mathbf{\%}}{\mathbf{}}{{\mathbf{MgCO}}}_{{\mathbf{3}}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{MgCO}}_{\mathbf{3}}}{\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{lime}\mathbf{}\mathbf{stone}}{\mathbf{}}{\mathbf{x}}{\mathbf{}}{\mathbf{100}}{\mathbf{\%}}}$

A sample of dolomite limestone containing only CaCO_{3} and MgCO_{3} was analyzed. When a .3028 gram sample of this limestone was decomposed by heating, .082 L of CO_{2} at 754 mmHg and 22 degrees Celsius were evolved. The initial sample contained .05112 grams of magnesium. What percent of the limestone by mass is MgCO_{3}?

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