When `2.86 g` of a mixture of `1-`butene, `C_(4)H_(8)` and butane `C_(4)H_(10)` was burned in excess of oxygen `8.80 g` of `CO_(2)` and `4:14 g` of `H_(2)O` were obtained. What is percentage by mass of butane in the mixture

Explanation

To solve the problem step by step, we will follow the outlined approach:

### Step 1: Write the Combustion Reactions
1. **For 1-butene (C₄H₈)**:
\[ C_4H_8 + 6O_2 \rightarrow 4CO_2 + 4H_2O \]
This means that 1 mole of 1-butene produces 4 moles of CO₂ and 4 moles of H₂O.

2. **For butane (C₄H₁₀)**:
\[ C_4H_{10} + \frac{13}{2}O_2 \rightarrow 4CO_2 + 5H_2O \]
This means that 1 mole of butane produces 4 moles of CO₂ and 5 moles of H₂O.

### Step 2: Define Variables
Let:
- \( x \) = mass of butane (C₄H₁₀) in grams.
- Therefore, the mass of 1-butene (C₄H₈) will be \( 2.86 - x \) grams.

### Step 3: Calculate Molar Masses
- Molar mass of butane (C₄H₁₀):
\[ 4 \times 12 + 10 \times 1 = 48 + 10 = 58 \text{ g/mol} \]
- Molar mass of 1-butene (C₄H₈):
\[ 4 \times 12 + 8 \times 1 = 48 + 8 = 56 \text{ g/mol} \]

### Step 4: Calculate Moles of Each Component
- Moles of butane:
\[ n_{C_4H_{10}} = \frac{x}{58} \]
- Moles of 1-butene:
\[ n_{C_4H_8} = \frac{2.86 - x}{56} \]

### Step 5: Calculate CO₂ Produced
- CO₂ produced from butane:
\[ \text{CO₂ from butane} = 4 \times n_{C_4H_{10}} = 4 \times \frac{x}{58} = \frac{4x}{58} \]
- CO₂ produced from 1-butene:
\[ \text{CO₂ from 1-butene} = 4 \times n_{C_4H_8} = 4 \times \frac{2.86 - x}{56} = \frac{4(2.86 - x)}{56} \]

### Step 6: Total CO₂ Produced
The total CO₂ produced is given as 8.80 g. Therefore, we set up the equation:
\[ \frac{4x}{58} + \frac{4(2.86 - x)}{56} = \frac{8.80}{44} \]

### Step 7: Solve for \( x \)
1. Calculate the right side:
\[ \frac{8.80}{44} = 0.2 \text{ moles of CO₂} \]

2. Combine the left side:
\[ \frac{4x}{58} + \frac{4(2.86 - x)}{56} = 0.2 \]

3. Solve the equation:
\[ \frac{4x}{58} + \frac{11.44 - 4x}{56} = 0.2 \]

4. Clear the fractions by finding a common denominator (which is 58 * 56):
\[ 4x \cdot 56 + (11.44 - 4x) \cdot 58 = 0.2 \cdot (58 \cdot 56) \]

5. Simplify and solve for \( x \):
After solving, we find \( x \approx 1.738 \) g (mass of butane).

### Step 8: Calculate Percentage by Mass of Butane
\[ \text{Percentage by mass of butane} = \left( \frac{x}{2.86} \right) \times 100 = \left( \frac{1.738}{2.86} \right) \times 100 \approx 60.8\% \]

### Final Answer
The percentage by mass of butane in the mixture is approximately **60.8%**.

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