A carbon compound contains `12.8%` of carbon, `2.1%` of hydrogen and `85.1%` of bromine. The molecular
weight of the compound is `187.9`. Calculatte the
molecular formula of the compound.
(Atomic weight of H `+1.008,` C `= 12.0` and Br = `79. 9))

Options

A. `CH_(3)Br`
B. `CH_(2)Br_(2)l`
C. `C_(2)H_(4)Br_(2)`
D. `C_(2)H_(3)Br_(3)`

Answer

To find the molecular formula of the compound based on the given percentages of carbon, hydrogen, and bromine, we can follow these steps:

### Step 1: Convert the percentages to grams
Assuming we have 100 grams of the compound, the masses of each element can be directly taken from the percentages:
- Mass of Carbon (C) = 12.8 g
- Mass of Hydrogen (H) = 2.1 g
- Mass of Bromine (Br) = 85.1 g

### Step 2: Convert the masses to moles
Next, we convert the mass of each element to moles using their atomic weights:
- Moles of Carbon = Mass of C / Atomic weight of C = 12.8 g / 12.0 g/mol = 1.067 moles
- Moles of Hydrogen = Mass of H / Atomic weight of H = 2.1 g / 1.008 g/mol = 2.083 moles
- Moles of Bromine = Mass of Br / Atomic weight of Br = 85.1 g / 79.9 g/mol = 1.066 moles

### Step 3: Determine the simplest mole ratio
Now, we need to find the simplest whole number ratio of moles:
- Moles of C = 1.067 / 1.066 ≈ 1
- Moles of H = 2.083 / 1.066 ≈ 2
- Moles of Br = 1.066 / 1.066 = 1

Thus, the ratio of C:H:Br is approximately 1:2:1.

### Step 4: Write the empirical formula
Based on the mole ratio, the empirical formula of the compound is:
\[ \text{Empirical Formula} = \text{C}_1\text{H}_2\text{Br}_1 \text{ or simply } \text{C}_1\text{H}_2\text{Br} \]

### Step 5: Calculate the empirical mass
Now, we calculate the empirical mass:
- Empirical mass = (1 × 12.0) + (2 × 1.008) + (1 × 79.9)
- Empirical mass = 12.0 + 2.016 + 79.9 = 93.916 g/mol (approximately 94 g/mol)

### Step 6: Determine the molecular formula
Given that the molecular weight of the compound is 187.9 g/mol, we can find the ratio of the molecular weight to the empirical mass:
- Ratio = Molecular weight / Empirical mass = 187.9 g/mol / 94 g/mol ≈ 2

### Step 7: Write the molecular formula
To find the molecular formula, we multiply the subscripts in the empirical formula by this ratio:
- Molecular formula = \( \text{C}_{1 \times 2}\text{H}_{2 \times 2}\text{Br}_{1 \times 2} = \text{C}_2\text{H}_4\text{Br}_2 \)

### Conclusion
The molecular formula of the compound is:
\[ \text{C}_2\text{H}_4\text{Br}_2 \]