1. គេវិភាគអាល់កុល មួយ គេទទួលបាន C = 64.86% ។
ក. កំណត់រូបមន្តដុល នៃសមាសធាតុ B ។
ខ. ចូរសរសេររូបមន្តស្ទើរលាតដែលអាចកើតមាន ។ បើ B ជាអាល់កុលថ្នាក់ II ខ្សែត្រង់គ្មានខ្នែង ។ ចូរកំណត់រូបមន្តពិត របស់សមាសធាតុ B ។
គ. អាកុលខាងលើ 14.8g ឲ្យមានប្រតិកម្មជាមួយ សូដ្យូម ( Na) ។
គ.1. សរសេរសមីការនិង ហៅឈ្មោះអង្គធាតុកកើត ។
គ.2. រកម៉ាសអង្គធាតុនីមួយ"។
(គេឲ្យ C= 12 , H= 1, O= 16 , Na= 23)
Please tell me

This problem involves analyzing an alcohol compound (B) with a given percentage of carbon (C), and finding its molecular formula, structural formula, and determining the reaction products with sodium (Na).

Let's break it down step by step:

### Part ក: Determine the Empirical Formula of Compound B
Given:
- Carbon content \( C = 64.86\% \)
- To find the empirical formula, we assume 100g of the compound B.
- The remaining percentage is for Hydrogen (H) and Oxygen (O).

Let's assume the empirical formula of the alcohol is \( C_xH_yO_z \).

**Step 1: Calculate the amount of each element in 100g of the compound.**
- Carbon: \( 64.86g \) 
- The remaining mass \( 100g - 64.86g = 35.14g \) is for Hydrogen and Oxygen.

Since we don’t have the exact percentages of Hydrogen and Oxygen, we'll focus on the fact that it's an alcohol (R-OH), meaning there's one Oxygen atom (O) per molecule.

### Part ខ: Write Possible Structural Formulas
Given that B is a straight-chain secondary alcohol (II) with no branches, the molecular formula would be of the form \( C_nH_{2n+1}OH \) where \( n \) is the number of carbon atoms.

### Part គ.1: Sodium Reaction with the Alcohol
The reaction between alcohol and sodium (\( Na \)) generally produces hydrogen gas (\( H_2 \)) and a sodium alkoxide. For an alcohol \( R-OH \), the reaction is:

\[
R-OH + Na \rightarrow R-ONa + \frac{1}{2}H_2
\]

The product is sodium alkoxide and hydrogen gas.

### Part គ.2: Finding the Mass of Each Product
To solve this, let's assume a potential empirical formula for B and then calculate the masses.

**Assuming the molecular formula is \( C_4H_10O \) (butanol, since it’s a straight-chain secondary alcohol):**
- Molar mass of \( C_4H_10O \) = \( (4 \times 12) + (10 \times 1) + (16 \times 1) = 48 + 10 + 16 = 74g/mol \)

**Step 1: Calculate moles of alcohol:**
Given mass = 14.8g
\[
\text{Moles of alcohol} = \frac{14.8g}{74g/mol} = 0.2 \text{ moles}
\]

**Step 2: Sodium reacts in a 1:1 molar ratio with the alcohol:**
\[
\text{Moles of Na required} = 0.2 \text{ moles}
\]

**Step 3: Mass of sodium needed:**
\[
\text{Mass of Na} = 0.2 \text{ moles} \times 23g/mol = 4.6g
\]

**Step 4: Mass of the alkoxide product:**
Sodium alkoxide has the formula \( C_4H_9ONa \).
- Molar mass of \( C_4H_9ONa \) = \( 74g/mol + 23g/mol = 97g/mol \)

Mass of sodium alkoxide produced:
\[
\text{Mass} = 0.2 \text{ moles} \times 97g/mol = 19.4g
\]

**Step 5: Mass of hydrogen gas produced:**
Molar mass of hydrogen gas \( H_2 \) is \( 2g/mol \).
- Mass of hydrogen gas produced:
\[
\text{Mass of } H_2 = 0.2 \times 1 = 0.2g
\]

### Final Results:
- The molecular formula of B is likely \( C_4H_{10}O \) (butanol).
- Sodium alkoxide produced: \( C_4H_9ONa \) with a mass of 19.4g.
- Hydrogen gas produced: 0.2g.

---

Would you like more details or have any other questions? Here are some related questions to consider:

1. How do you determine the empirical formula from percentages?
2. What are the general properties of secondary alcohols?
3. How does the structure of an alcohol affect its chemical reactivity?
4. Why does sodium react with alcohols to produce hydrogen gas?
5. What is the significance of molar mass in chemical reactions?

**Tip:** When analyzing a chemical reaction, always check the stoichiometry to ensure you’re balancing atoms and charges correctly.

Learn how to analyze an alcohol compound (B) with a given percentage composition of carbon. Discover its empirical formula, structural formula, and reactions with sodium. Detailed step-by-step solution and calculations provided. Suitable for advanced high school chemistry students.

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