Balancing equations computations

You must NEVER change the small subscript numbers within a formula. Changing a subscript alters the chemical identity of the substance (for example, changing water, $H_2O$, to hydrogen peroxide, $H_2O_2$).

Calculating Balance: The Tally Method

The most reliable way to balance an equation is to methodically list and count every atom. This ensures the correct molar ratio (the ratio between the amounts in moles of substances) is achieved.

Worked Example: Balancing Magnesium and Hydrochloric Acid

Reaction: Magnesium + Hydrochloric Acid $\rightarrow$ Magnesium Chloride + Hydrogen gas Unbalanced Equation: $Mg + HCl \rightarrow MgCl_2 + H_2$

Step 1: List and Tally (Initial State) List every element under the reaction arrow and count the atoms on both sides.

• Magnesium (Mg): 1 (Reactants) $\rightarrow$ 1 (Products)
• Hydrogen (H): 1 (Reactants) $\rightarrow$ 2 (Products)
• Chlorine (Cl): 1 (Reactants) $\rightarrow$ 2 (Products) Observation: Mg is balanced, but H and Cl are not.

Step 2: Apply Coefficients To balance the 2 hydrogen and 2 chlorine atoms on the right, add a coefficient of 2 in front of the $HCl$ on the left. New Equation: $Mg + 2HCl \rightarrow MgCl_2 + H_2$

Step 3: Update Tally (Final Check)

• Magnesium (Mg): 1 $\rightarrow$ 1
• Hydrogen (H): 2 $\rightarrow$ 2
• Chlorine (Cl): 2 $\rightarrow$ 2 Observation: All atoms match. The equation is balanced.

Step 4: Express as a Molar Ratio The final relationship can be expressed as a molar ratio of $Mg : HCl : MgCl_2 : H_2$. Final Molar Ratio: $1 : 2 : 1 : 1$

Deducing Stoichiometry from Masses (Higher Tier)

If you are given the masses of reactants and products, you can use arithmetic computations to find the balanced equation. You must calculate the moles of each substance and find the simplest whole-number ratio.

$$\text{moles} = \frac{\text{mass (g)}}{\text{Relative Formula Mass } (M_r)}$$

Worked Example: Deducing an Equation from Mass

Problem: $64\text{ g}$ of methanol ($CH_3OH$) reacts with $96\text{ g}$ of $O_2$ to produce $88\text{ g}$ of $CO_2$ and $72\text{ g}$ of $H_2O$. Deduce the balanced equation.

Step 1: Calculate the Relative Formula Mass ($M_r$) for each substance. The Relative Formula Mass ($M_r$) is the sum of the relative atomic masses of all atoms in the formula.

• $CH_3OH = 32$, $O_2 = 32$, $CO_2 = 44$, $H_2O = 18$

Step 2: Calculate moles using the formula.

• $CH_3OH$: $64 / 32 = 2\text{ moles}$
• $O_2$: $96 / 32 = 3\text{ moles}$
• $CO_2$: $88 / 44 = 2\text{ moles}$
• $H_2O$: $72 / 18 = 4\text{ moles}$

Step 3: Establish the simplest molar ratio.

• Ratio ($CH_3OH : O_2 : CO_2 : H_2O$): $2 : 3 : 2 : 4$

Step 4: Write the balanced equation.

• $2CH_3OH + 3O_2 \rightarrow 2CO_2 + 4H_2O$

Verifying Conservation of Mass Arithmetically

You can prove that a balanced equation obeys the Law of Conservation of Mass by checking that the sum of the $M_r$ of the reactants equals the sum of the $M_r$ of the products.

Using the formation of water: $2H_2 + O_2 \rightarrow 2H_2O$

• Reactants side: $2 \times H_2 = 2 \times (2 \times 1.0) = 4.0$; $1 \times O_2 = (2 \times 16.0) = 32.0$. Total = $36.0$.
• Products side: $2 \times H_2O = 2 \times [(2 \times 1.0) + 16.0] = 36.0$.
• Conclusion: The total $M_r$ of reactants ($36.0$) matches the total $M_r$ of products ($36.0$), proving mathematically that mass is conserved.