SOLUTION: Combustion analysis pogil ap chem - Studypool - Free Printable

Let's go through the Combustion Analysis worksheet step by step and provide a clear, accurate explanation for each question. You've already started answering some of them — we’ll refine and clarify your responses to ensure they are scientifically precise.

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🔍 Overview: Combustion Analysis

Combustion analysis is a method used to determine the elemental composition (especially carbon and hydrogen) of organic compounds by burning them in excess oxygen. The products are carbon dioxide (CO₂) and water (H₂O), which are measured to find the mass of C and H in the original compound.

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## ✔ Question 1: According to Model 1, what reactant is always required for combustion?

> Your answer: *Hydrocarbons & Oxygen is always required for combustion.*

✔ Refined Answer:
The reactant that is always required for combustion is oxygen (O₂).

🔹 While hydrocarbons are common fuels, combustion can occur with other substances too (e.g., sulfur, metals), but oxygen is essential as the oxidizing agent in all combustion reactions shown here.

👉 So, the correct answer is:
Oxygen (O₂) is always required for combustion.

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## ✔ Question 2: Balance the reactions in Model 1 while keeping the hydrocarbon coefficient as "1".

We are given these unbalanced equations. Let’s balance each one with 1 mole of hydrocarbon:

1. CH₄ + O₂ → CO₂ + H₂O

- Carbon: 1 on both sides → OK
- Hydrogen: 4 H on left → need 2 H₂O (since each has 2 H)
- So: CH₄ + O₂ → CO₂ + 2 H₂O
- Now count O: Right side = 2 (from CO₂) + 2 (from 2 H₂O) = 4 O atoms
- Left: O₂ must supply 4 O → so need 2 O₂

✔ Balanced:
CH₄ + 2 O₂ → CO₂ + 2 H₂O

✔️ Already balanced in the image.

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2. C₂H₆ + O₂ → CO₂ + H₂O

- C₂H₆ → 2 C → need 2 CO₂
- 6 H → need 3 H₂O (since 3 × 2 = 6 H)
- So: C₂H₆ + O₂ → 2 CO₂ + 3 H₂O
- O atoms on right: 2×2 + 3×1 = 4 + 3 = 7 O atoms
- So need 7/2 O₂ on left

✔ Balanced:
C₂H₆ + ⁷⁄₂ O₂ → 2 CO₂ + 3 H₂O

(You wrote this correctly.)

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3. C₂H₄ + O₂ → CO₂ + H₂O

- C₂H₄ → 2 C → 2 CO₂
- 4 H → 2 H₂O
- So: C₂H₄ + O₂ → 2 CO₂ + 2 H₂O
- O atoms: Right = 2×2 + 2×1 = 4 + 2 = 6 O
- So need 3 O₂

✔ Balanced:
C₂H₄ + 3 O₂ → 2 CO₂ + 2 H₂O

✔️ Correct in your image.

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4. C₃H₈ + O₂ → CO₂ + H₂O

- C₃H₈ → 3 C → 3 CO₂
- 8 H → 4 H₂O
- So: C₃H₈ + O₂ → 3 CO₂ + 4 H₂O
- O atoms: 3×2 + 4×1 = 6 + 4 = 10 O → need 5 O₂

✔ Balanced:
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

✔️ Correct in your image.

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## ✔ Question 3: How is the coefficient of CO₂ related to the chemical formula of the hydrocarbon?

> Your answer: *The amount of carbon atoms in the hydrocarbon are the coefficients for CO₂.*

✔ Refined Answer:
The coefficient of CO₂ in the balanced combustion reaction is equal to the number of carbon atoms in the hydrocarbon molecule.

🔹 For example:
- CH₄ → 1 C → 1 CO₂
- C₂H₆ → 2 C → 2 CO₂
- C₃H₈ → 3 C → 3 CO₂

So:
➡️ Each carbon atom produces one molecule of CO₂, so the CO₂ coefficient = number of carbon atoms in the hydrocarbon.

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## ✔ Question 4: How is the coefficient of H₂O related to the chemical formula of the hydrocarbon?

> Your answer: *Hydrogen atoms in the hydrocarbon are half the coefficients of water.*

✔ Refined Answer:
The coefficient of H₂O is half the number of hydrogen atoms in the hydrocarbon.

🔹 Because each water molecule (H₂O) contains two hydrogen atoms, and all hydrogen from the fuel ends up in water.

Examples:
- CH₄ → 4 H → 2 H₂O (4 ÷ 2 = 2)
- C₂H₆ → 6 H → 3 H₂O (6 ÷ 2 = 3)
- C₃H₈ → 8 H → 4 H₂O (8 ÷ 2 = 4)

So:
➡️ H₂O coefficient = (Number of H atoms) / 2

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🧠 Summary of Key Concepts:

| Hydrocarbon | C Atoms | H Atoms | CO₂ Produced | H₂O Produced |
|------------|--------|--------|---------------|--------------|
| CH₄ | 1 | 4 | 1 | 2 |
| C₂H₆ | 2 | 6 | 2 | 3 |
| C₂H₄ | 2 | 4 | 2 | 2 |
| C₃H₈ | 3 | 8 | 3 | 4 |

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✔ Final Answers Recap:

1. Oxygen (O₂) is always required for combustion.
2. All equations are correctly balanced:
- CH₄ + 2 O₂ → CO₂ + 2 H₂O
- C₂H₆ + ⁷⁄₂ O₂ → 2 CO₂ + 3 H₂O
- C₂H₄ + 3 O₂ → 2 CO₂ + 2 H₂O
- C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
3. The CO₂ coefficient equals the number of carbon atoms in the hydrocarbon.
4. The H₂O coefficient equals half the number of hydrogen atoms in the hydrocarbon.

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Let me know if you'd like help with the next page or applying this to real combustion analysis problems (like finding empirical formulas from masses)!