A 5.00 gm sample of a natural gas, consisting of methane `CH_(4)`, and ethylene,`C_(2)H_(4)` was burnt in excess oxygen , yeilding `44/3` gm of `CO_(2)` and `H_(2)O` as products . What mole percent of the sample was ethylene ?

Answer

To solve the problem step-by-step, let's break it down into manageable parts.

### Step 1: Define Variables
Let:
- \( X \) = weight of ethylene (\( C_2H_4 \))
- \( 5 - X \) = weight of methane (\( CH_4 \))

### Step 2: Calculate Moles of CO2 Produced
Given that the total mass of CO2 produced is \( \frac{44}{3} \) grams, we can calculate the number of moles of CO2 using the formula:

\[ \text{Number of moles of } CO_2 = \frac{\text{mass}}{\text{molar mass}} \]

The molar mass of \( CO_2 \) is 44 g/mol. Thus,

\[ \text{Number of moles of } CO_2 = \frac{\frac{44}{3}}{44} = \frac{1}{3} \text{ moles} \]

### Step 3: Write Combustion Reactions
For methane (\( CH_4 \)):
\[ CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O \]
From the balanced equation, 1 mole of \( CH_4 \) produces 1 mole of \( CO_2 \).

For ethylene (\( C_2H_4 \)):
\[ C_2H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2O \]
From the balanced equation, 1 mole of \( C_2H_4 \) produces 2 moles of \( CO_2 \).

### Step 4: Set Up the Equation for CO2 Production
- Moles of \( CH_4 \) produced from \( 5 - X \) grams:
\[ \text{Moles of } CH_4 = \frac{5 - X}{16} \]
- Moles of \( C_2H_4 \) produced from \( X \) grams:
\[ \text{Moles of } C_2H_4 = \frac{X}{28} \]

Now, the total moles of \( CO_2 \) produced can be expressed as:
\[ \text{Total moles of } CO_2 = \text{Moles from } CH_4 + \text{Moles from } C_2H_4 = \frac{5 - X}{16} + 2 \cdot \frac{X}{28} \]

### Step 5: Equate to Moles of CO2
Set the total moles of \( CO_2 \) equal to \( \frac{1}{3} \):
\[ \frac{5 - X}{16} + 2 \cdot \frac{X}{28} = \frac{1}{3} \]

### Step 6: Solve for X
To solve the equation, first find a common denominator and simplify:
1. The common denominator for 16 and 28 is 112.
2. Rewrite the equation:
\[ \frac{7(5 - X)}{112} + \frac{8X}{112} = \frac{1}{3} \]
3. Multiply through by 112 to eliminate the denominator:
\[ 7(5 - X) + 8X = \frac{112}{3} \]
4. Expand and simplify:
\[ 35 - 7X + 8X = \frac{112}{3} \]
5. Combine like terms:
\[ 35 + X = \frac{112}{3} \]
6. Convert 35 to a fraction:
\[ \frac{105}{3} + X = \frac{112}{3} \]
7. Solve for \( X \):
\[ X = \frac{112}{3} - \frac{105}{3} = \frac{7}{3} \approx 2.33 \text{ grams} \]

### Step 7: Calculate the Mole Percent of Ethylene
Now that we have \( X \):
- The weight of ethylene \( C_2H_4 \) is approximately \( 2.33 \) grams.
- The total weight of the sample is \( 5 \) grams.

Mole percent of ethylene is calculated as:
\[ \text{Mole percent of } C_2H_4 = \left( \frac{X}{5} \right) \times 100 = \left( \frac{2.33}{5} \right) \times 100 \approx 46.6\% \]

### Final Answer
Thus, the mole percent of the sample that was ethylene is approximately **46.6%**.