Question 553bf

Sep 30, 2015

That much calcium carbonate contains $\text{24 g}$ of carbon.

Explanation:

The percentage of carbon in limestone, or calcium carbonate, ${\text{CaCO}}_{3}$, is actually independent of how much calcium carbonate you actually have.

That is the case because the percent composition of calcium carbonate is always the same.

You know that each formula unit of calcium carbonate contains

• one atom of calcium
• one atom of carbon
• three atoms of oxygen

This means that one mole of calcium carbonate will contain

• one mole of calcium
• one mole of carbon
• three moles of oxygen

So, if one mole of calcium carbonate contains one mole of carbon, you can use calcium carbonate's molar mass and carbon's molar mass to get the percent composition by mass of carbon in ${\text{CaCO}}_{3}$

(1 xx 12.011color(red)(cancel(color(black)("g/mol"))))/(100.09color(red)(cancel(color(black)("g/mol")))) xx 100 = "12.0%"

What this means is that regardless of how much calcium carbonate you have, $\text{12%}$ of that mass will be carbon. Therefore, the percentage of carbon in ${\text{CaCO}}_{3}$ will always be 12%.

In your case, $\text{200 g}$ of calcium carbonate will contain

200color(red)(cancel(color(black)("g CaCO"""_3))) * "12.0 g C"/(100color(red)(cancel(color(black)("g CaCO"""_3)))) = color(green)("24 g C")#