Law of Definite Proportion: Do the numbers indicating moles count?

I need to find the ratio of a substance as it converts from a bicarbonate to a carbonate:

ex:

$\ce{Fe(HCO3)2 -> FeCO3 + CO2 + H2O}$

$\ce{Fe(HCO3)2} / \ce{FeCO3}$ = $3/2$ *using molar masses

But what if the substances aren't the same number of moles?

ex:

$\ce{2Fe(HCO3)2 -> FeCO3 + CO2 + H2O}$

Would the ratio be:

$\ce{2Fe(HCO3)2 / FeCO3} = x$

or still:

$\ce{Fe(HCO3)2} / \ce{FeCO3} = 3/2$

How do moles factor into this equation and why?

• Balance your reactions before you get the ratio. – Jerry Sep 12 '13 at 10:01

The stoichiometric coefficients are important. They make sure you have a valid relationship between reactants and products that obeys the Law of Conservation of Mass.

As Jerry hints at in his comment, there is another fundamental law that you need to consider: the Law of Conservation of Mass. Matter cannot be created or destroyed (outside of nuclear reactions and relativist physics). In chemistry this means you need the same number of atoms of each element on both sides of the equation. You cannot create additional atoms from nothing, nor can extra atoms of something vanish.

Consider your first equation: $$\ce{Fe(HCO3)2 -> FeCO3 + CO2 + H2O}$$

This equation is balanced - it obeys the law of conservation of mass. Each side of the equation has one $\ce{Fe}$, two $\ce{H}$, two $\ce{C}$, and six $\ce{O}$ atoms. This is a valid chemical equation representing a real reaction relationship.

Your second equation is not balanced.

$$\ce{2Fe(HCO3)2 -> FeCO3 + CO2 + H2O}$$

There are now twice as many atoms of each element on the reactant side of the equation as on the product side. What happens to this extra atoms? They cannot vanish! This is not a valid chemical relationship. Balancing this equation generates a scalar multiple of your first equation, which will yield the same ratio of iron (ii) bicarbonate to iron (ii) carbonate.

$$\ce{2Fe(HCO3)2 -> 2FeCO3 + 2CO2 + 2H2O}$$

The Law of Definite Proportions tells you that chemical compounds always contain the same elemental compositions by mass. I am not sure how that is relevant to the ratio of product to reactant in your stoichiometry problem. In other words, it tells you that iron (ii) carbonate is always $\ce{FeCO3}$ and never $\ce{Fe3(CO3)4}$ or $\ce{Fe2(CO3)3}$ or $\ce{FeCO4}$ or anything else. Those other formulas may refer to real compounds (the first two are real), but they are not iron (ii) carbonate.