U3AOS1 Topic 3: Stoichiometry
Stoichiometry involves the calculations based off mole ratios to quantify the amount of reactants reacted or products produced, based off the law of conservation of mass.
The Concept of the Mole
Recall from unit 1/2, the mole is just a way chemists quantify the amount of particles. One mole contains $6.02 \times 10^{23}$ particles, represented by the symbol $n$.
\[ n= \frac{N}{N_A} \]
Where $N$ represents the number of particles, and $N_A$ represents Avogadro's constant $(6.02 \times 10^{23})$.
Calculating Mass in Stoichiometry
The mass of the reactants used or products produced is represented in grams and found using the molar mass of the element/compound:
\[ n = \frac{m}{M} \]
Where $m$ represents mass ($g$) of substance, $M$ represents molar mass ($g/mol$).
| Study design dot point: | calculations related to the application of stoichiometry to reactions involving the combustion of fuels, including mass-mass, mass-volume and volume-volume stoichiometry, to determine heat energy released, reactant and product amounts and net volume or mass of major greenhouse gases (CO2, CH4 and H2O), limited to standard laboratory conditions (SLC) at 25 °C and 100 kPa |
Volume of Gases in Reactions
The volume of gas used/produced in a chemical equation can be quantified by the molar volume constant which only applies at SLC. This value is $24.8 , \text{L/mol}$ and only at $100 , \text{kPa}$ and $25 , \text{°C}$.
\[ n = \frac{V}{V_m}\]
Where $V$ represents volume ($L$), $V_m$ represents molar volume constant $(24.8 , \text{L/mol})$.
Extra Insight: Derivation of $24.8 , \text{L/mol}$
In the previous study design, the universal gas law was utilized in calculations:
\[ PV=nRT \]
Where:
- $P$ = Pressure (kPa)
- $V$ = Volume (L)
- $n$ = Mole (mol)
- $R$ = Gas constant $8.31 , \text{J/mol/K}$
- $T$ = Temperature (K)
At SLC the conditions are $100 , \text{kPa}$ and $25 , \text{°C}$. We first convert the $25 , \text{°C}$ into equivalent kelvin values by adding $273$ (approx.):
$25+273=298$
As $V_m$ is the volume per mole of gas at SLC, we substitute $n = 1$ after rearranging the formula to find $V$ (subject $V$):
\[ V= \frac{P}{nRT} = \frac{100 \times 1 \times 8.31 \times 298}{100} = 24.7638 \approx 24.8 , \text{L} \]
Units in Stoichiometry
\[ Vm = \frac{V}{n} \]
Where $V$ has the unit $L$ while $n$ has the unit $mol$, thus the unit $L/mol$ or $L , \text{mol}^{-1}$.
To Find Mole Ratio
\[ \frac{ \text{coefficient unknown} }{ \text{coefficient known} } = \frac{ n \text{ (unknown) } } { n \text{ (known) } } \]
You don’t need to follow this formula every time, just think of it intuitively
If I need $5$ of $x$ to produce $4$ of $y$, the number of moles for $y$ must be smaller than $x$ and vice versa.
Eg. For $\ce{C2H5OH(l) + 3O2(g) →2CO2(g) + 3H2O(l)}$
Here’s a simple example question:
Calculate the volume of $\ce{CO2(g)}$ produced at SLC when we know $100 g$ of $\ce{C2H5OH(l)}$ reacted
- Find number of moles of $\ce{C2H5OH(l)}$ reacted (always find moles first)
\[ \ce{n(C2H5OH)=\frac{m}{Mr(C2H5OH)} = \frac{200}{12 \times 2 + 6 + 16} =2.17..} \]
Note: $\ce{Mr(C2H5OH)}$ is the molar mass of every element in the compound
- Find moles of $\ce{2CO2(g)}$ produced by using mole ratio
\[ \frac{\text{coefficient $\ce{CO2}$ }}{\text{coefficient $\ce{C2H5OH}$ }} = \frac{2}{1} \]
Therefore we know that for every mole of $\ce{C2H5OH}$ is used, $2$ mole of $\ce{CO2}$ produced thereby we expect a smaller mole number produced as $1 < 2$
Hence,
\[ \ce{n(CO2)=n(C2H5OH)} \times \text{(molar ratio)} = 2.17 \times \frac{2}{1} = 4.3478 \]
- As question states that the reaction takes place at SLC (Standard Lab Conditions) we use molar volume constant
\[ \ce{V(CO2) = n(CO2) \times Vm=4.3478 mol \times 24.7L/mol = 107.826 L} \]
- Check significant figures (because it matters in VCE chemistry)
As we see in the question the only quantity is $100$ which is $3$ significant figures, therefore final answer must be at $3$ significant figures
Ans: $108L$ or $1.08 \times 10^{2}L $
Author note: I personally like to stick with scientific notation because the solutions come in a predictable format which requires minimal effort to determine sig figs.
| Study design dot point: | determination of limiting reactants or reagents in chemical reactions |
What is a limited or excess reagent?
A limited reagent means that the reagent is stoichiometrically limiting the reaction process.
Whilst excess reagent means that the reagent enables the reaction to proceed due to having excess amounts.
Analogy:
A good way to think of this is that it requires two pieces of bread and a slice of ham for a sandwich
$2$ bread + $1$ ham $= 1$ sandwich
However you have four pieces of bread and one slice of ham
You can only cook up one sandwich despite having the required amount of bread, thereby having two extra slices of bread (excess reagent)
In this case; the slice of ham is considered the limiting reagent as it is stopping the cooking process from proceeding.
Example 1