Stoichiometry

Habari Mwanafunzi! Welcome to Stoichiometry: The Recipe of Chemistry!

Ever tried to bake mandazis or cook the perfect pot of ugali? You know you can't just throw things together randomly. You need the right amount of flour, water, sugar, and oil. If you have a whole packet of flour but only a little bit of sugar, you can only make a certain number of mandazis, right? The sugar limits you!

Believe it or not, chemistry works in the exact same way. A chemical reaction is like a recipe, and stoichiometry (pronounced stoy-kee-OM-eh-tree) is the skill of reading that recipe to figure out "how much" of everything you need and "how much" of the final product you will get. It’s the mathematics behind chemical reactions!

Image Suggestion: A vibrant, colourful digital art image. On the left, a Kenyan student is happily measuring flour to make chapati in a kitchen, with a recipe book open. On the right, the same student is in a modern school science lab, wearing a lab coat and safety goggles, carefully measuring chemicals into a beaker. A glowing, balanced chemical equation floats between the two scenes, connecting them.

The Heart of the Recipe: The Mole Ratio

The secret to all stoichiometry is the balanced chemical equation. It’s our master recipe! The numbers in front of each chemical formula, called coefficients, tell us the exact ratio of ingredients, not in grams or litres, but in moles.

Think of making chai. Your family recipe might be:

2 cups Water + 1 cup Milk + 2 spoons Tea Leaves → ~3 cups of Chai

This is a ratio! For every 2 cups of water, you need exactly 1 cup of milk.

In chemistry, we use the Haber process to make ammonia (NH₃), a key ingredient in fertilizers used by farmers all over Kenya. The balanced equation is:

  N₂(g) + 3H₂(g) → 2NH₃(g)

This tells us the mole ratio:

• 1 mole of Nitrogen gas reacts with
• 3 moles of Hydrogen gas to produce
• 2 moles of Ammonia gas.

This ratio is our "conversion factor" that lets us move from one chemical to another in a calculation. We call it the Mole Bridge!

  ASCII Art: The Mole Bridge

  Known Substance                      Unknown Substance
  =================                      ===================

  [ Mass of A (g) ]                      [ Mass of B (g) ]
         ↑ ↓ (Molar Mass)                       ↑ ↓ (Molar Mass)
  [  Moles of A   ] —-—(Mole Ratio)-—→ [  Moles of B   ]
                      ^^^^^^^^^^^^
                      THE BRIDGE!
                      (from balanced equation)

Let's Do the Math: Grams to Grams Calculation

Okay, let's put this into practice. This is the most common type of problem you'll see in your KCSE exams!

Problem: The Bamburi Cement factory near Mombasa uses a process that involves heating calcium carbonate (CaCO₃). If you heat 150 g of CaCO₃, how many grams of calcium oxide (CaO), or quicklime, are produced?

Here is the 4-step method that will never fail you!

1. Step 1: Write and Balance the Equation.
   The reaction is: CaCO₃(s) → CaO(s) + CO₂(g). Let's check if it's balanced. (1 Ca, 1 C, 3 O on both sides). Yes, it's balanced! The mole ratio between CaCO₃ and CaO is 1:1.
2. Step 2: Convert Known Mass to Moles (Grams A → Moles A).
   We know the mass of CaCO₃ is 150 g. We need its molar mass.
   Molar Mass of CaCO₃ = Ca (40.1) + C (12.0) + 3*O (16.0) = 100.1 g/mol.

   
     Moles of CaCO₃ = Mass / Molar Mass
                    = 150 g / 100.1 g/mol
                    = 1.4985 mol
       

3. Step 3: Use the Mole Bridge (Moles A → Moles B).
   From our balanced equation, the ratio of CaCO₃ to CaO is 1:1. So, if we use 1.4985 mol of CaCO₃, we will produce 1.4985 mol of CaO.

   
     Moles of CaO = 1.4985 mol CaCO₃ * (1 mol CaO / 1 mol CaCO₃)
                  = 1.4985 mol CaO
       

4. Step 4: Convert Moles of Unknown to Mass (Moles B → Grams B).
   Now we find the mass of 1.4985 mol of CaO. We need its molar mass.
   Molar Mass of CaO = Ca (40.1) + O (16.0) = 56.1 g/mol.

   
     Mass of CaO = Moles * Molar Mass
                 = 1.4985 mol * 56.1 g/mol
                 = 84.07 g
       

Answer: Heating 150 g of calcium carbonate will produce approximately 84.1 g of calcium oxide. See? Just like a recipe!

The Limiting Reactant: When You Run Out of an Ingredient!

Imagine you are at home making githeri for your family. The recipe requires maize and beans. You look in the store and find a whole 5kg bag of maize (plenty!), but only one small tin of beans. What happens? You can only make enough githeri until the beans run out. The beans are your limiting reactant; they limit how much food you can make. The maize is the excess reactant; you will have plenty of it left over.

In chemistry, the limiting reactant (or limiting reagent) is the reactant that gets completely used up first in a reaction. Once it's gone, the reaction stops, no matter how much of the other reactants you have left.

Image Suggestion: A close-up shot of two hands preparing to make githeri. One hand is pouring a huge pile of maize into a large sufuria (pot). The other hand is holding a very small, almost empty can of beans, showing the contrast between the excess and limiting ingredients. The mood is one of mild frustration about not having enough beans.

How to find the limiting reactant: The easiest way is to calculate how much product can be made from each reactant. The one that produces the least amount of product is the limiting reactant!

Percentage Yield: Theory vs. Reality

So, you do a calculation like the one above and find you should get 84.1 g of CaO. This is your theoretical yield – the maximum amount you can possibly make, according to your math.

But when you go to the lab, things can happen. Maybe some of the powder spills, or the reaction doesn't complete perfectly. You weigh your final product and find you only made 81.2 g. This is your actual yield – what you actually got in real life.

The percentage yield tells you how successful your experiment was!

  Formula:

  Percentage Yield = (Actual Yield / Theoretical Yield) * 100%

For our example:

  Percentage Yield = (81.2 g / 84.1 g) * 100%
                   = 0.9655 * 100%
                   = 96.6%

A 96.6% yield is very good for a school lab! It shows your technique was excellent.

You've Got This, Future Scientist!

Stoichiometry might seem like a lot of steps, but it's just a logical process, like following a recipe. Remember these key ideas:

• The balanced equation is your recipe.
• The mole ratio is the bridge to get from what you know to what you want to find.
• The limiting reactant is the ingredient that runs out first.
• Percentage yield compares your real-life result to your perfect on-paper calculation.

Don't be afraid of the math. Practice with different problems from your textbook. The more recipes you follow, the better chemist you will become. Kazi nzuri (Good work)!