0.06g of a hydrocarbon occupies 32cm\(^3\) at S.T.P. It's formula is

Explanation

To find the formula of the hydrocarbon, we need to determine its molar mass and the moles of the hydrocarbon. Since we know the mass and volume of the hydrocarbon at Standard Temperature and Pressure (S.T.P.), we can use the molar volume of a gas at S.T.P., which is 22.4 dm\(^3\), or 22,400 cm\(^3\).

First, let's find the moles of the hydrocarbon:

\[\text{moles} = \frac{\text{volume at S.T.P.}}{\text{molar volume at S.T.P.}}\]

\[\text{moles} = \frac{32 \text{ cm}^3}{22,400 \text{ cm}^3}\]

\[\text{moles} = 1.43 \times 10^{-3}\]

Now, let's find the molar mass of the hydrocarbon:

\[\text{molar mass} = \frac{\text{mass}}{\text{moles}}\]

\[\text{molar mass} = \frac{0.06 \text{ g}}{1.43 \times 10^{-3}}\]

\[\text{molar mass} \approx 42 \text{ g/mol}\]

We can now determine the empirical formula of the hydrocarbon. We know that it consists of carbon (C) and hydrogen (H) atoms. The atomic mass of carbon is 12 g/mol and the atomic mass of hydrogen is 1 g/mol. Let's use x as the number of carbon atoms and y as the number of hydrogen atoms in the formula:

\[12x + y = 42\]

We'll now consider the possible options and see which fits the equation:

Option A: C\(_3\)H\(_6\) (Correct)

In this formula, x = 3, and y = 6. The equation becomes 12(3) + 6 = 42, which is true. So, Option A is correct.

Option B: C\(_2\)H\(_2\)

In this formula, x = 2, and y = 2. The equation becomes 12(2) + 2 = 26, which is not equal to 42. So, Option B is incorrect.

Option C: C\(_2\)H\(_4\)

In this formula, x = 2, and y = 4. The equation becomes 12(2) + 4 = 28, which is not equal to 42. So, Option C is incorrect.

Option D: C\(_2\)H\(_6\)

In this formula, x = 2, and y = 6. The equation becomes 12(2) + 6 = 30, which is not equal to 42. So, Option D is incorrect.

Therefore, the correct formula for the hydrocarbon is C\(_3\)H\(_6\).