Target Compound:
Li₇ La₃ Zr₂ O₁ ₂
This formula tells us the desired molar ratio of each element:
Element
Required Moles
Li
7
La
3
Zr
2
Step 1: Choose the Base Compound — La₂ O₃ (Lanthanum Oxide)
You’ve chosen Lanthanum Oxide (La₂ O₃ ) as your base compound.
Let’s recall:

One mole of La₂ O₃ contains 2 moles of La atoms.
So, to obtain the 3 moles of La required by the LLZO formula:
So, we fix 1.5 moles of La₂ O₃ for the synthesis.
Step 2: Determine Moles of Other Compounds Based on Stoichiometry
1. Zirconium Oxide (ZrO₂ )
 Each mole of ZrO₂ provides 1 mole of Zr.
 We need 2 moles of Zr (from LLZO formula), so:
2 moles of Zr = 2 moles of ZrO2
2. Lithium Carbonate (Li₂ CO₃ )
 Each mole of Li₂ CO₃ provides 2 moles of Li.
 LLZO requires 7 moles of Li, so:
7
= 3.5 moles of Li2CO3
2
Step 3: Molar Masses of Reagents
Step 4: Calculate the Mass of Each Compound
However, Li is volatile at high temperatures and can be lost during calcination and
sintering.
So we add 10–15% excess Li₂ CO₃ to compensate.
Let’s add 15% excess:
Excess Li₂ CO₃ =258.62×1.15 = 297.41 g
Final Reagent Quantities for 1 Mole LLZO:
To synthesize LLZO using 2 grams of La₂ O₃ as the base, we can scale all the other
reagents accordingly using stoichiometric ratios.
Step 1: Calculate Scaling Factor
We used 488.72 g of La₂ O₃ for the full synthesis.
Now we want to scale it down to 2 g.
Step 2: Apply the Scaling Factor to All Other Reagents
Multiply this factor with the original masses of ZrO₂ and Li₂ CO₃ .
2. ZrO₂
246.44 g × 0.004093 = 1.009 g
3. Li₂ CO₃ (with 15% excess)
297.41 g × 0.004093 = 1.218 g
Final Reagent Masses for Lab (Based on 2 g La₂O₃):