An organic compound contained 78%(by wt.) carbon and remaining % of hydrogen. The right option for the empirical formula of this compound is:

Options

A. `CH_3`
B. `CH_4`
C. `CH`
D. `CH_2`

Answer

To find the empirical formula of the organic compound containing 78% carbon and the remaining percentage as hydrogen, we can follow these steps:

### Step 1: Determine the percentage of hydrogen
Given that the compound contains 78% carbon, the remaining percentage will be hydrogen.

\[ \text{Percentage of hydrogen} = 100\% - 78\% = 22\% \]

### Step 2: Convert percentages to grams
Assuming we have 100 grams of the compound, we can convert the percentages to grams:
- Mass of carbon = 78 grams
- Mass of hydrogen = 22 grams

### Step 3: Convert grams to moles
Next, we convert the mass of each element to moles using their molar masses:
- Molar mass of carbon (C) = 12 g/mol
- Molar mass of hydrogen (H) = 1 g/mol

Calculating the moles:
\[ \text{Moles of carbon} = \frac{78 \text{ g}}{12 \text{ g/mol}} = 6.5 \text{ moles} \]
\[ \text{Moles of hydrogen} = \frac{22 \text{ g}}{1 \text{ g/mol}} = 22 \text{ moles} \]

### Step 4: Determine the simplest mole ratio
Now, we need to find the simplest ratio of moles of carbon to moles of hydrogen:
- The ratio of moles of carbon to moles of hydrogen is:
\[ \text{Ratio} = \frac{6.5}{22} \approx 1:3.38 \]

### Step 5: Simplify the ratio
To simplify the ratio to whole numbers, we can multiply both sides by a factor that will yield whole numbers. In this case, multiplying by 3 gives:
\[ \text{C} : \text{H} = 6.5 \times 3 : 22 \times 3 = 19.5 : 66 \]
This is still not a whole number ratio, so we can try multiplying by 2 to get closer to whole numbers:
\[ \text{C} : \text{H} = 6.5 \times 2 : 22 \times 2 = 13 : 44 \]
This is still not simplified.

### Step 6: Find the closest whole number ratio
To find the closest whole number ratio, we can round the moles of carbon to the nearest whole number:
- Carbon: 6.5 rounds to 7
- Hydrogen: 22 rounds to 22

### Step 7: Write the empirical formula
Thus, the empirical formula can be approximated as:
\[ \text{Empirical formula} = \text{C}_7\text{H}_{22} \]

### Final Answer
The empirical formula of the compound is \( \text{C}_7\text{H}_{22} \).