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Stoichiometry 25: Calculate the mass of anhydrous Na₂CO₃ in 500 mL of 0.1 M solution
Introduction
To begin with, if you’re preparing a sodium carbonate solution for a chemistry experiment.
You need to prepare 500 mL of a 0.1 M Na₂CO₃ solution.
Here’s the big question:
How many grams of anhydrous sodium carbonate (Na₂CO₃) do you need to weigh out?
That’s where stoichiometry meets solution chemistry.
If you’ve ever wondered how chemists get the exact amount of solute to make a solution of the right molarity, this article will walk you through the process.
Step 1: Understand what molarity means
Chemists define molarity (M) as moles of solute per liter of solution.
So, when someone says “0.1 M Na₂CO₃”, they’re saying there are 0.1 moles of sodium carbonate in every 1 liter of solution.
But we only want 500 mL, which is half of a liter.
So the first step is simple:
Figure out how many moles are in 0.5 L of a 0.1 M solution.
Step 2: Calculate the number of moles
To begin with, use the molarity formula:
Moles = Molarity × Volume (in liters)
So we get:
Moles = 0.1 M × 0.5 L = 0.05 moles
That’s it.
You now know you need 0.05 moles of Na₂CO₃ to make your solution.
Step 3: Convert moles to grams
As a starter, when we go from moles to grams, we need the molar mass of sodium carbonate.
Let’s calculate it using atomic masses:
- Sodium (Na) = 23 g/mol × 2 = 46 g/mol
- Carbon (C) = 12 g/mol
- Oxygen (O) = 16 g/mol × 3 = 48 g/mol
Molar mass of Na₂CO₃ = 46 + 12 + 48 = 106 g/mol
Now use the formula:
Mass = Moles × Molar Mass
So:
Mass = 0.05 × 106 = 5.3 grams
That’s how much anhydrous sodium carbonate you need to weigh out.
Step 4: Why “anhydrous” matters
In chemistry, anhydrous means “without water.”
Some compounds like Na₂CO₃ can absorb water from the air and form hydrated crystals, which have water molecules trapped inside the solid.
These extra water molecules increase the mass but don’t contribute to the chemical reactivity of the sodium carbonate.
So, when a question specifies “anhydrous,” it directs you to ignore any water content and calculate based only on the pure, dry substance.
Step 5: What if the volume changes?
Let’s say someone asks you to make 250 mL of the same solution.
You’d still follow the same process.
First, convert volume to liters:
250 mL = 0.25 L
Then:
Moles = 0.1 M × 0.25 L = 0.025 moles
Mass = 0.025 × 106 = 2.65 grams
So for half the volume, you’d use half the mass.
Step 6: Avoiding common errors
Many students mistakenly use milliliters directly without converting to liters.
Consequently, that leads to wrong mole values.
Another error is using the wrong molar mass.
If you accidentally use the molar mass of hydrated sodium carbonate, you’ll end up weighing too much.
Finally, make sure you’re using the correct formula:
Na₂CO₃ for anhydrous,
Na₂CO₃·10H₂O for decahydrate,
Na₂CO₃·H₂O for monohydrate, and so on.
Each one has a different molar mass.
Step 7: Applications of sodium carbonate
In fact, sodium carbonate isn’t just found in labs.
It’s used in glass manufacturing, detergents, paper production, and even in water softening.
When preparing solutions for these applications, precision matters.
An extra gram or two could lead to incorrect concentrations, poor reactions, or waste.
This is especially important in pharmaceutical and industrial settings where mass-scale production is involved.
Step 8: Why chemists prefer molarity
Molarity allows chemists to work with predictable, reproducible quantities of substances in solutions.
If you know the molarity and volume, you can quickly determine how much of any compound you’re working with.
That’s the whole idea behind standard solutions.
Once a solution is prepared, it can be used in titrations, analytical chemistry, or controlled reactions with consistent outcomes.
Step 9: Let’s check our logic one more time
You were asked to make 500 mL of 0.1 M Na₂CO₃ solution.
You converted that to moles:
0.1 × 0.5 = 0.05 moles
You multiplied by molar mass:
0.05 × 106 = 5.3 grams
So you now know that to prepare the solution, you simply need to dissolve 5.3 g of pure, anhydrous sodium carbonate into 500 mL of water.
That’s chemistry made clear.
Final Wrap-Up
To prepare 500 mL of a 0.1 M solution of anhydrous sodium carbonate, you need to weigh out 5.3 grams of Na₂CO₃.
This is calculated using the molarity formula, molar mass, and careful unit conversions.
The key takeaway here is that solution chemistry isn’t just theory — it’s practical math that supports everything from lab work to industrial production.
Once you master the simple logic behind molarity and mass conversions, you’ll find these kinds of questions much easier to tackle.
Still practicing? Want more clarity?
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