# How are stoichiometry and molarity related?

Jul 16, 2018

Well, stoichiometry requires EQUIVALENCE with respect to mass and charge....

#### Explanation:

And for an older treatment of the principles involved, see this old answer. And the fundamental principle of stoichiometry is $\text{garbage in equals garbage out}$. Every chemical reaction must be balanced with respect to mass and charge. A $10 \cdot g$ mass of reactants from all sources yields at most a $10 \cdot g$ mass of products.

$\text{Molarity}$ is a concentration term, i.e. $\text{molarity"="moles of solute"/"volume of solution}$...and as such it has the units of $m o l \cdot {L}^{-} 1$.. And so if we have TWO of the three quantities, say $\text{molarity}$ and $\text{volume}$, we can get the third..$\text{moles of solute}$...

$\text{molarity"xx"volume"="moles}$....and this is certainly consistent dimensionally. What do I mean by this?

And a practical example? Well suppose I gots a $100 \cdot m L$ volume of $H C l \left(a q\right)$, that is $1 \cdot m o l \cdot {L}^{-} 1$ with respect to $H C l$. What mass of sodium hydroxide is required for equivalence?

We write out the stoichiometric equation as a preliminary:

$H C l \left(a q\right) + N a O H \left(a q\right) \rightarrow N a C l \left(a q\right) + {H}_{2} O \left(l\right)$

${n}_{\text{HCl}} = 0.100 \cdot L \times 1 \cdot m o l \cdot {L}^{-} 1 = 0.100 \cdot m o l$

And for equivalence we require equimolar sodium hydroxide...

$0.100 \cdot m o l \times {\underbrace{40.0 \cdot g \cdot m o {l}^{-} 1}}_{\text{molar mass of NaOH}} = 4.00 \cdot g$...i.e. a $4 \cdot g$ mass of hydroxide is required for equivalence.