Stoichiometry & Mole Calculation: The "Why" Behind the Formula
Let's build this from the ground up — not as a list of formulas to memorise, but as a logical chain of reasoning.
1. The Core Question: What is a Mole?
A mole is simply a counting unit, like a dozen (12) or a gross (144). But instead of 12, a mole contains 6.022×1023 particles (Avogadro's number, NA).
Why this number?
It was chosen so that 1 mole of any substance has a mass in grams equal to its atomic/molecular mass in amu.
- Example: 1 atom of carbon-12 has mass 12 amu.
- 1 mole of carbon-12 has mass 12 grams.
This is the bridge between the microscopic (atoms/molecules) and the macroscopic (grams we can weigh).
2. The Fundamental Relationship
The key formula is:
Where:
- n = number of moles
- m = mass of substance (in grams)
- M = molar mass (in g/mol)
Why does this work?
Think of it as a conversion factor:
If 1 mole of a substance weighs M grams, then m grams contains Mm moles.
Derivation logic:
- Molar mass M tells you: "1 mol = M g"
- So the conversion factor is M g1 mol
- Multiply mass m by this factor: m×M1=Mm moles
3. Connecting to Number of Particles
n=NAN
Where:
- N = number of particles (atoms, molecules, ions)
- NA=6.022×1023 particles/mol
Why?
- 1 mole = NA particles
- So N particles = NAN moles
Combine both formulas:
Mm=NAN
This single equation ties mass, molar mass, number of particles, and Avogadro's number together.
4. The Gas Volume Connection (for gases at STP)
For gases only:
n=22.4 L/molV
Why 22.4 L?
From the ideal gas law: PV=nRT
At STP (Standard Temperature and Pressure: 0°C, 1 atm):
- P=1 atm
- T=273.15 K
- R=0.0821 L·atm/(mol·K)
For n=1 mole:
V=PnRT=11×0.0821×273.15≈22.4 L
So 1 mole of any ideal gas occupies 22.4 L at STP. This is a consequence of the gas laws, not a definition.
5. The Stoichiometry Chain: From One Substance to Another
In a balanced chemical equation like:
aA+bB→cC+dD
The coefficients tell you the mole ratio:
moles of Bmoles of A=ba
Why this works: …