U3AOS1 Topic 3: Stoichiometry

$n=\frac{N}{N_A}$Currently fixing formatting

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

 

Recall from unit ½, the mole is just a way chemists quantify the amount of particles. One mole contains 6.02*10²³ particles represented by the symbol n

 

$n=\frac{N}{N_A}$, where N represents number of particles, $N_A$ represents Avogadro's constant (6.02*10²³) 

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

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 Lmol-1 and only at 100 kPa and 25 °C.

$n=\frac{V}{V_m}$, where $v$ represents volume (L), $V_m$ represents molar volume constant (24.8 L/mol)

Extra: Where does 24.8 L/mol come from?

In the previous study design, the universal gas law was utilised in calculations

$\mathrm{<span\; style="font-size:\; 10px;">P</span>}\mathrm{<span\; style="font-size:\; 10px;">V</span>}<span\; style="font-size:\; 10px;">=</span>\mathrm{<span\; style="font-size:\; 10px;">n</span>}\mathrm{<span\; style="font-size:\; 10px;">R</span>}\mathrm{<span\; style="font-size:\; 10px;">T</span>}$PV=nRT

P = Pressure (kPa)

V = Volume (L)

n = Mole (mol)

R = Gas constant (8.31 J/mol/K)

T = Temperature (K)

At SLC the conditions are 100kPa and 25°C

We first convert the 25°C into equivalent kelvin values by adding 273 (approx.)

25+273=298

As Vm is volume per mole of gas at SLC, we substitute n=1 after rearranging formula to find V (subject V)

$V = \frac{nRT}{P} = \frac{1*8.31*298}{100} = 24.7638\approx24.8L$

Then what about the units?

$n = \frac{V}{V_m}$ rearrange to have Vm as subject

$V_m = \frac{V}{n}$,  where V has the unit L whilst n has the unit mol. Thus L/mol or Lmol-1 unit.

To find mole ratio:

$\frac{coefficient\;unknown}{coefficient\;known} = \frac{n(unknown)}{n(known)}$

$\frac{coefficientunknown}{coefficientknown} = \frac{n(unknown)}{n(known)}$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 C2H5OH(l) + 3O2(g) →2CO2(g) + 3H2O(l)

Here’s a simple example question:

Calculate the volume of CO2(g) produced at SLC when we know 100 g of C2H5OH(l) reacted

1. Find number of moles of C2H5OH(l) reacted (always find moles first)

$n(C_2H_5OH) = \frac{m}{Mr(C_2H_5OH)} = \frac{100}{12*2+6*16} = 2.17...$

2. Find moles of 2CO2(g) produced by using mole ratio

$\frac{coefficientCO_2}{coefficientC_2H_5OH} = \frac{2}{1}$
$\frac{coefficient\;CO_2}{coefficient\;C_2H_5OH} = \frac{2}{1}$

Therefore we know that for every mole of C2H5OH is used, 2 mole of CO2 produced thereby we expect a smaller mole number produced as 1<2

Hence, $n(CO_2)=n(C_2H_5OH) * molar\;ratio = 2.17...* \frac{2}{1} = 4.3478...$

3. As question states that the reaction takes place at SLC (Standard Lab Conditions) we use molar volume constant

$V(CO_2) = n(CO_2)*V_m = 4.3478... mol *24.7L/mol = 107.826...L$

4. 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*10²L

Author note: I personally like to stick with scientific notation because the solutions come in a predictable format which requires minimal effort to determine sigfigs.

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.