What is $\text{Stoichiometry}$? 📏
**$\text{Stoichiometry}$** (stoy-kee-AHM-uh-tree) is the calculation of $\text{reactants}$ and $\text{products}$ in a chemical $\text{reaction}$. It's like using a recipe to predict exactly how much of each ingredient you need and how much final product you'll make. The key is to use the **$\text{Mole}$ $\text{Concept}$** and a **$\text{Balanced}$ $\text{Equation}$**.
The $\text{Mole}$ and $\text{Molar}$ $\text{Mass}$ ⚖️
The $\text{Mole}$ ($\text{mol}$)
The SI unit for the amount of substance. It's a counting unit, like a dozen.
$\text{Avogadro}$'s $\text{Number}$ ($\text{N}_A$)
One $\text{mole}$ of any substance contains $6.02 \times 10^{23}$ particles ($\text{atoms}$, $\text{molecules}$, etc.).
$\text{Molar}$ $\text{Mass}$ ($\text{MM}$)
The mass in $\text{grams}$ of one $\text{mole}$ of a substance. It is numerically equal to the $\text{atomic}$ $\text{mass}$ (or $\text{molecular}$ $\text{mass}$) on the $\text{Periodic}$ $\text{Table}$, measured in $\text{g}/\text{mol}$.
The $\text{Mole}$ $\text{Ratio}$ ➗
The coefficients (the large numbers in front of the formulas) in a **$\text{balanced}$ $\text{chemical}$ $\text{equation}$** represent the $\text{mole}$ $\text{ratio}$ between the $\text{reactants}$ and $\text{products}$. This ratio is the $\text{conversion}$ $\text{factor}$ used in $\text{Stoichiometry}$.
Example $\text{Reaction}$: $\text{The}$ $\text{Haber}$ $\text{Process}$
$$\text{N}_{2} (g) + 3\text{H}_{2} (g) \longrightarrow 2\text{NH}_{3} (g)$$$\text{Mole}$ $\text{Ratio}$: 1 $\text{mol}$ $\text{N}_2$ : 3 $\text{mol}$ $\text{H}_2$ : 2 $\text{mol}$ $\text{NH}_3$
To convert moles of $\text{H}_2$ to moles of $\text{NH}_3$, you would use the conversion factor: $\frac{2 \text{ mol } \text{NH}_3}{3 \text{ mol } \text{H}_2}$
Interactive $\text{Stoichiometry}$ $\text{Practice}$ (Mole-to-Mole) ❓
Using the $\text{Haber}$ $\text{Process}$ equation from above ($\text{N}_{2} + 3\text{H}_{2} \longrightarrow 2\text{NH}_{3}$), calculate the moles of $\text{NH}_3$ produced.
Problem:
If you start with $5.0$ $\text{moles}$ of $\text{N}_2$, how many $\text{moles}$ of $\text{NH}_3$ can be produced?
$$\text{Calculation}: 5.0 \text{ mol } \text{N}_2 \times \frac{\text{Coefficient of } \text{NH}_3}{\text{Coefficient of } \text{N}_2} = ? \text{ mol } \text{NH}_3$$
⚡ $\text{Mole}$ $\text{Concept}$ $\text{Check}$!
Click the correct value for each $\text{Stoichiometry}$ term.
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