Hukum Dasar Kimia | Lavoisier | Proust | Kimia kelas 10

Hukum Lavoisier (Hukum Kekekalan Massa)
📌 This law states that in a closed system, the mass of substances before the reaction (reactants) equals the mass of substances after the reaction (products).
💧 Example: If 2g of Hydrogen reacts with 16g of Oxygen to form Water, the resulting mass of Water is $2g + 16g = 18g$.
🧪 For the reaction $\text{Mg} + \text{S} \rightarrow \text{MgS}$, if 6g of Magnesium reacts to form 14g of Magnesium Sulfide, the mass of Sulfur reacted must be $14g - 6g = 8g$, assuming all reactants are consumed.
⚠️ The Law of Lavoisier strictly applies only when all reacting substances are completely consumed in the reaction.

Hukum Proust (Hukum Perbandingan Tetap)
📌 This law dictates that the ratio of the masses of elements combining to form a specific compound is constant and definite, regardless of the initial amounts used, provided all react.
💧 For Water ($\text{H}_2\text{O}$), the mass ratio of Hydrogen to Oxygen is consistently 1:8 (derived from $2g \text{ H} : 16g \text{ O}$ or $3g \text{ H} : 24g \text{ O}$).
⚖️ When initial reactant masses do not perfectly match the fixed ratio (e.g., 4g $\text{H}$ and 40g $\text{O}$ forming 36g $\text{H}_2\text{O}$), one reactant must be in excess, and the mass formed is governed by the limiting reactant adhering to the fixed ratio.
C calculation for $\text{CO}_2$: If 24g of Oxygen reacts, the required Carbon mass is $\frac{3}{8} \times 24g = 9g$, forming $33g$ of $\text{CO}_2$ because the ratio $9:24$ simplifies to $3:8$.

Applying Proust's Law with Excess Reactants
📌 To determine the limiting reactant when given initial masses (e.g., 6g $\text{C}$ and 12g $\text{O}$ for $\text{CO}_2$ with a 3:8 ratio), divide each mass by its corresponding ratio number: $\frac{6}{3}$ (for $\text{C}$) vs. $\frac{12}{8}$ (for $\text{O}$).
📉 Since $\frac{12}{8} = 1.5$ is smaller than $\frac{6}{3} = 2$, Oxygen is the limiting reactant and is completely consumed (12g).
✅ The mass of Carbon that reacts is calculated based on the consumed Oxygen: $\frac{3}{8} \times 12g = 4.5g$ of Carbon, resulting in $12g + 4.5g = 16.5g$ of $\text{CO}_2$ formed.

Determining Mass Ratios using Atomic Mass (Ar)
📌 The fixed mass ratio within a compound can be calculated using the Atomic Relative Mass ($\text{Ar}$) of the constituent elements.
For a general compound $\text{A}_m\text{B}_n$, the mass ratio is $m \times \text{Ar}_A : n \times \text{Ar}_B$.
💧 For Water ($\text{H}_2\text{O}$), using $\text{Ar}(\text{H}) = 1$ and $\text{Ar}(\text{O}) = 16$: $\text{Mass ratio} = (2 \times 1) : (1 \times 16) = 2:16$, which simplifies to 1:8.
🧪 For Carbon Dioxide ($\text{CO}_2$), using $\text{Ar}(\text{C}) = 12$ and $\text{Ar}(\text{O}) = 16$: $\text{Mass ratio} = (1 \times 12) : (2 \times 16) = 12:32$, which simplifies to 3:8.

Key Points & Insights
➡️ Master the application of Lavoisier's Law by ensuring total reactant mass equals total product mass in closed systems.
➡️ Use the Proust's Law ratio method to identify the limiting reactant by comparing initial mass divided by its stoichiometric ratio value.
➡️ When calculating composition for a known final mass of product (e.g., 22g $\text{CO}_2$): determine the sum of the ratio parts ($3+8=11$), then use fractions ($\frac{3}{11}$ for $\text{C}$, $\frac{8}{11}$ for $\text{O}$) multiplied by the total product mass.
➡️ Atomic Relative Masses ($\text{Ar}$) provide a reliable method to determine the fixed mass ratios within any chemical compound, such as $1:8$ for $\text{H}:\text{O}$ in $\text{H}_2\text{O}$.

📸 Video summarized with SummaryTube.com on Jan 26, 2026, 03:45 UTC