Stoichiometric Numbers: The Language of Chemical Equations
What Are Stoichiometric Numbers?
Imagine a recipe for making a sandwich. It might say: 2 slices of bread + 1 slice of cheese + 2 slices of turkey = 1 sandwich. The numbers (2, 1, 2, 1) tell you the exact proportion of ingredients you need to make the final product without any leftovers. Stoichiometric numbers work in exactly the same way for chemical reactions.
In a chemical equation, these numbers are written in front of the chemical formulas. For example, in the combustion of methane:
The numbers 1 (in front of $ CH_4 $), 2 (in front of $ O_2 $), 1 (in front of $ CO_2 $), and 2 (in front of $ H_2O $) are the stoichiometric numbers. They are the multipliers for the entire formula that follows them. A coefficient of 1 is typically not written, but it is always implied.
The Law of Conservation of Mass and Balancing Equations
The entire reason we need stoichiometric numbers is because of a fundamental law of nature: the Law of Conservation of Mass. This law states that matter cannot be created or destroyed in a chemical reaction. Atoms are simply rearranged to form new substances.
Let's look at an unbalanced equation for the formation of water:
$ H_2 + O_2 \rightarrow H_2O $
If we count the atoms on each side, we see a problem:
| Element | Reactants (Left Side) | Products (Right Side) | Balanced? |
|---|---|---|---|
| Hydrogen (H) | 2 atoms | 2 atoms | Yes |
| Oxygen (O) | 2 atoms | 1 atom | No |
We have an extra oxygen atom on the left. To fix this, we use stoichiometric numbers. We cannot change the formula $ H_2O $ to $ H_2O_2 $ because that is hydrogen peroxide, a different compound. We can only change the coefficients.
By placing a 2 in front of $ H_2O $ and a 2 in front of $ H_2 $, we achieve balance:
Now the atom count is correct: 4 H and 2 O on both sides. The stoichiometric numbers 2, 1, and 2 make this possible.
A Step-by-Step Guide to Balancing Equations
Balancing chemical equations is like a puzzle. Here is a reliable method using the formation of ammonia ($ NH_3 $) from nitrogen and hydrogen as our example.
Step 1: Write the Unbalanced Equation (Skeleton Equation)
$ N_2 + H_2 \rightarrow NH_3 $
Step 2: Count the Atoms on Each Side
| Element | Reactants | Products |
|---|---|---|
| Nitrogen (N) | 2 atoms | 1 atom |
| Hydrogen (H) | 2 atoms | 3 atoms |
Step 3: Start with the Most Complex Substance
Start with the formula that has the most different atoms, which is often the product. Here, it's $ NH_3 $. To balance Nitrogen, we need 2 $ NH_3 $ molecules on the product side.
$ N_2 + H_2 \rightarrow 2NH_3 $
Now, Nitrogen is balanced (2 atoms on each side), but Hydrogen is not. We have 6 H atoms on the right (2 molecules × 3 H atoms each) and only 2 H atoms on the left.
Step 4: Balance the Remaining Atoms
To get 6 H atoms on the left, we need 3 $ H_2 $ molecules (3 molecules × 2 H atoms each = 6 H atoms).
Step 5: Do a Final Check
Reactants: N=2, H=6. Products: N=2, H=6. The equation is balanced! The stoichiometric numbers are 1, 3, and 2.
From Ratios to Real-World Calculations
Stoichiometric numbers establish the mole ratio[1] between all reactants and products. This is the bridge between the microscopic world of atoms and the macroscopic world we can measure in the lab.
Let's use our balanced ammonia equation: $ N_2 + 3H_2 \rightarrow 2NH_3 $
The coefficients tell us that:
- 1 molecule of $ N_2 $ reacts with 3 molecules of $ H_2 $ to produce 2 molecules of $ NH_3 $.
- More usefully, 1 mole of $ N_2 $ reacts with 3 moles of $ H_2 $ to produce 2 moles of $ NH_3 $.
Example Problem: How many moles of hydrogen gas ($ H_2 $) are required to produce 10 moles of ammonia ($ NH_3 $)?
Solution:
1. Identify the mole ratio from the balanced equation: The ratio of $ H_2 $ to $ NH_3 $ is 3:2.
2. Set up a proportion:
$ \frac{3 \text{ moles } H_2}{2 \text{ moles } NH_3} = \frac{x \text{ moles } H_2}{10 \text{ moles } NH_3} $
3. Solve for x:
$ x = 10 \text{ moles } NH_3 \times \frac{3 \text{ moles } H_2}{2 \text{ moles } NH_3} = 15 \text{ moles } H_2 $
Therefore, 15 moles of hydrogen gas are needed. This type of calculation is fundamental for chemists to determine how much of each reactant to use and how much product to expect.
Common Mistakes and Important Questions
Q: Can I change the subscript in a chemical formula to balance an equation?
A: No, never. Changing a subscript, like turning $ H_2O $ into $ H_2O_2 $, changes the identity of the chemical compound. Water ($ H_2O $) and hydrogen peroxide ($ H_2O_2 $) are completely different substances with different properties. Only the stoichiometric numbers (coefficients) placed in front of the formulas can be changed.
Q: What is the difference between a coefficient and a subscript?
A: A coefficient is a stoichiometric number placed in front of a chemical formula and multiplies the entire formula. For example, in $ 2H_2O $, the coefficient 2 means two molecules of water, totaling 4 hydrogen atoms and 2 oxygen atoms. A subscript is a number written after and below an element symbol and indicates the number of atoms of that element in a single molecule. In $ H_2O $, the subscript 2 means there are two hydrogen atoms in one molecule of water.
Q: Are stoichiometric numbers always whole numbers?
A: In most introductory chemistry, yes, we use whole numbers to keep things simple. However, sometimes fractions are used as an intermediate step during balancing, but the final equation should have the smallest set of whole number coefficients. For example, you might temporarily write $ H_2 + \frac{1}{2}O_2 \rightarrow H_2O $ to balance oxygen, but you would then multiply the entire equation by 2 to get $ 2H_2 + O_2 \rightarrow 2H_2O $.
Conclusion
Footnote
[1] Mole Ratio: The ratio of the amounts in moles of any two compounds involved in a chemical reaction. It is derived directly from the stoichiometric numbers in the balanced chemical equation. For example, in $ 2H_2 + O_2 \rightarrow 2H_2O $, the mole ratio of $ H_2 $ to $ O_2 $ is 2:1.