1. ចំហេះសព្វ 1.6g អង្គធាតុសរីរាង្គ A គេទទួលបាន ទឹក 3.6g និង ឧស្ម័ន កាបូនឌីអុកស៊ីត 22.4L នៅលក្ខខណ្ឌធម្មតា ។ 
 ក. កំណត់រូបមន្តទូទៅ នៃអង្គធាតុសរីរាង្គ A ។ 
 ខ. គណនាសមាសភាពសតភាព នៃ ធាតុបង្កក្នុង អង្គធាតុសរីរាង្គ A ។ 
 គ. កំណត់រូបមន្តងាយនៃអង្គធាតុ សរីរាង្គ A  ។ 
 ឃ. កំណត់រូបមន្តម៉ូលេគុលនៃអង្គធាតុសរីរាង្គ A នេះ បើដឹងថា វាមាន ដង់ស៊ីតេធៀបនឹង ខ្យល់ ស្មើ នឹង 0.55 ។ 
 ( គេឲ្យ C = 12 , H = 1 , Vm = 22.4L/mol. M(ខ្យល់) = 29 ) 
 Please tell me 

To solve this problem step by step, we need to address each part using the given information. Let's go through each section:

### Part A: Determine the General Formula of Organic Compound A

When the organic compound A (CxHy) is combusted, it produces CO₂ and H₂O. The balanced chemical equation for the combustion of an organic compound is typically:

\[
\text{CxHy} + O_2 \rightarrow x \text{CO}_2 + \frac{y}{2} \text{H}_2\text{O}
\]

Given:
- Mass of organic compound A = 1.6 g
- Mass of water (H₂O) produced = 3.6 g
- Volume of CO₂ produced = 22.4 L at standard conditions (Vm = 22.4 L/mol)

**Step 1:** Determine moles of CO₂:

\[
\text{Moles of CO₂} = \frac{\text{Volume of CO₂}}{\text{Molar Volume}} = \frac{22.4 \text{ L}}{22.4 \text{ L/mol}} = 1 \text{ mol}
\]

**Step 2:** Determine moles of H₂O:

Molar mass of H₂O = \(2 \times 1 + 16 = 18 \text{ g/mol}\)

\[
\text{Moles of H₂O} = \frac{\text{Mass of H₂O}}{\text{Molar mass of H₂O}} = \frac{3.6 \text{ g}}{18 \text{ g/mol}} = 0.2 \text{ mol}
\]

**Step 3:** Determine the amount of C and H in the compound.

- Each mole of CO₂ corresponds to 1 mole of carbon, so 1 mole of CO₂ means 1 mole of C.
- Each mole of H₂O corresponds to 2 moles of hydrogen, so 0.2 moles of H₂O means 0.4 moles of H.

Thus, 1 mole of A contains 1 mole of C and 0.4 moles of H. The empirical formula would be:

\[
\text{Empirical formula} = \text{CH}_2
\]

### Part B: Calculate the Mass Percent Composition of the Elements in Organic Compound A

Let's calculate the mass of C and H in the compound.

- Mass of C = \(1 \times 12 \text{ g/mol} = 12 \text{ g}\)
- Mass of H = \(0.4 \times 1 \text{ g/mol} = 0.4 \text{ g}\)
- Total mass = 1.6 g (Given)

Percentage by mass of C:

\[
\%C = \frac{12 \text{ g}}{1.6 \text{ g}} \times 100\% \approx 75\%
\]

Percentage by mass of H:

\[
\%H = \frac{0.4 \text{ g}}{1.6 \text{ g}} \times 100\% \approx 25\%
\]

### Part C: Determine the Simplest (Empirical) Formula of the Organic Compound A

From the earlier steps, we have already determined that the simplest formula is \( \text{CH}_2 \).

### Part D: Determine the Molecular Formula of Organic Compound A

Given that the compound's density relative to air is 0.55, and the molar mass of air is 29 g/mol:

- Molar mass of A = \(0.55 \times 29 \text{ g/mol} = 15.95 \text{ g/mol}\)

Since the molar mass of the empirical formula CH₂ is 14 g/mol:

- The molecular formula is approximately CH₂ (since 15.95 is close to 14).

So, the molecular formula of A is CH₂.

### Conclusion:

- **General formula of A:** CH₂
- **Mass percent composition:** 75% C and 25% H
- **Simplest formula:** CH₂
- **Molecular formula:** CH₂

Let me know if you need more details or if you have any questions!

### Related Questions:
1. How do you calculate the molar mass of a compound?
2. What is the significance of the empirical formula in chemistry?
3. How does the molar volume relate to gas calculations at standard conditions?
4. Why is the density relative to air important in determining the molecular formula?
5. How do you balance a combustion reaction for an unknown compound?
6. What are some common mistakes to avoid when calculating empirical formulas?
7. Can you determine the molecular formula directly from mass percentages?
8. How would you approach a problem where the gas volume is not at standard conditions?

### Tip:
Always check that the sum of the mass percentages adds up to 100% to ensure accuracy in your calculations.

Learn how to determine the empirical and molecular formulas of organic compound A using combustion analysis. This problem involves calculating percent compositions, molar volumes, and empirical formulas based on combustion products. Suitable for advanced high school students studying chemistry.

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