What do we mean by "molar quantities"? How are these used in stoichiometry, and chemistry problems?

Sep 6, 2016

You might have to re-ask this question. I am willing to try to answer. One mole of stuff is equivalent to $6.022 \times {10}^{23}$ individual items of that stuff.

Explanation:

The use of the mole specifies the given number. In other circumstances, if we were poultry farmers of bakers, we might use $\text{dozens}$ or $\text{grosses}$, but these numbers are a little too low to describe molecular quantities. The $\text{mole}$ is the $\text{chemists' dozen}$ and specifies the number $6.022 \times {10}^{23}$.

Why should we choose such a number? Well, it turns out that if we have $6.022 \times {10}^{23}$ individual ""^1H atoms, then we have a mass of $1 \cdot g$ precisely. The mole is thus the link between the micro world of atoms and molecules, which we can't see, to the macro world of grams and litres, which we can conveniently measure and utilize in reactions.

From this we develop the idea of equivalent weight, in which a given mass of stuff can be precisely related to an actual number of molecules or atoms. Atomic mass is quoted on the Periodic Table, and the mass in grams is that possessed by a mole, by Avogadro's number of molecules, ${N}_{A} \equiv 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$.

And thus for a simple reaction we can make the following predictions:

$C + {O}_{2} \rightarrow C {O}_{2}$

$12.0 \cdot g$ of carbon represents ${N}_{A}$ carbon (by definition). An equivalent quantity of dioxygen molecules has a mass of $32.0 \cdot g$. Given complete reaction, we get ${N}_{A}$ carbon dioxide molecules, with a mass of $44.0 \cdot g$. We could also say that $C {O}_{2}$ has a mass of $44.0 \cdot g \cdot m o {l}^{-} 1$.

Anyway, you should review your notes and problems. You are certainly free to post queries. Someone here will help you. Good luck.

See here for an example of an A level problem.