Question

How do you do mass to mass stoichiometry problems?

Answer 1

Using mole concept, we can calculate the number of unknown moles. Then we can calculate the unknown masses by applying the unitary method.

Complete step by step answer:
We will take an example to understand the step by step methodology to solve any mass to mass stoichiometry problem.
Consider the decomposition reaction mentioned below. We have to find the quantity of $B$ and $C$ formed (in${\text{g}}$) upon decomposition of $x{\text{ g}}$ of $A\,({\text{s}})$.
$A\,({\text{s}})\xrightarrow{\Delta }B\,({\text{g}}) + C\,({\text{g}})$
Step 1: Write the balanced equation.
Suppose the balanced equation representing the above chemical change is:
$a\,A\,({\text{s}})\xrightarrow{\Delta }b\,B\,({\text{g}}) + c\,C\,({\text{g}})$
In this reaction, $a$ moles of compound $A$ decompose to form $b$ moles of compound $B$ and $c$ moles of compound $C$.
Step 2: Calculate molecular mass of all species
Suppose, after calculation, the molecular mass of $A$, $B$ and $C$ comes to be $p$, $q$ and $r{\text{ g mo}}{{\text{l}}^{ - 1}}$, respectively.
Step 3: Calculate the number of moles of reactants or products from the known mass(es).
We will use the formula, ${\text{moles = }}\dfrac{{{\text{mass}}}}{{{\text{molar mass}}}}$.
Moles of compound $A = \dfrac{x}{p}$
Step 4: Calculate the number of unknown moles
We will calculate the number of moles of $B$ and $C$ formed upon decomposition of $x{\text{ g}}$ of $A$ by applying the unitary method.
$\because$Decomposition of $a$ moles of $A$ forms $= b$ moles of $B$
$\therefore$Decomposition of $\dfrac{x}{p}$ moles of $A$ forms $= \dfrac{b}{a} \times \dfrac{x}{p}$ moles of $B$
$\because$Decomposition of $a$ moles of $A$ forms $= c$ moles of $C$
$\therefore$Decomposition of $\dfrac{x}{p}$ moles of $A$ forms $= \dfrac{c}{a} \times \dfrac{x}{p}$ moles of $C$
Step 5: Calculate the unknown masses
We will use the formula, ${\text{mass}} = {\text{moles}} \times {\text{molar mass}}$
Mass of $B = \dfrac{b}{a} \times \dfrac{x}{p} \times q{\text{ g}}$
Mass of $C = \dfrac{c}{a} \times \dfrac{x}{p} \times r{\text{ g}}$

Note: The result can be verified by using conservation of mass.
Mass of B$$$$ + Mass of $C$,
$m = \left( {\dfrac{b}{a} \times \dfrac{x}{p} \times q} \right){\text{ + }}\left( {\dfrac{c}{a} \times \dfrac{x}{p} \times r} \right){\text{ g}}$
$\Rightarrow m = \dfrac{x}{{a \times p}}\left( {bq + cr} \right){\text{ g}}$
From the balanced chemical equation, using conservation of mass, it is clear that,
$bq + cr = ap$
Substituting this value in the above equation:
$\Rightarrow m = \dfrac{x}{{a \times p}}\left( {ap} \right){\text{ g}}$
$\Rightarrow m = x{\text{ g}}$
$\Rightarrow$ Mass of B$$$$ + Mass of $C =$ Mass of $A$