Q. What is the empirical formula for \(\mathrm{C_2H_6}\)?

Answer

To find the empirical formula, divide each element’s subscripts by the greatest common factor.

For C\(_2\)H\(_6\):

\[
\text{C : } \frac{2}{2} = 1,\quad \text{H : } \frac{6}{2} = 3
\]

So the empirical formula is \( \mathrm{CH_3} \).

Detailed Explanation

To find the empirical formula, you are looking for the simplest whole-number ratio of the atoms in the compound.

Step 1: Write the given molecular formula

The compound is given as \( \text{C}_2\text{H}_6 \). This already tells you the number of each type of atom in one molecule:

• Carbon: \(2\) atoms
• Hydrogen: \(6\) atoms

Step 2: Form the atom ratio

The atom ratio is therefore:

\( \text{C} : \text{H} = 2 : 6 \)

Step 3: Reduce the ratio to the smallest whole numbers

To reduce \(2 : 6\), divide both numbers by their greatest common divisor, which is \(2\).

\( \dfrac{2}{2} = 1 \)
\( \dfrac{6}{2} = 3 \)

So the reduced ratio is:

\( \text{C} : \text{H} = 1 : 3 \)

Step 4: Write the empirical formula from the reduced ratio

Use the reduced numbers as subscripts:

\( \text{C}_1\text{H}_3 \)

Since a subscript of \(1\) is usually omitted, the empirical formula is:

\( \text{CH}_3 \)

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General Chemistry FAQs

What is the empirical formula of \( \mathrm{C_2H_6} \)?

Divide subscripts by the greatest common factor, \(2\). \( \mathrm{C_2H_6 \rightarrow CH_3} \). So the empirical formula is \( \mathrm{CH_3} \).

How do I find the gcd of subscripts in \( \mathrm{C_2H_6} \)?

Take the subscripts \(2\) and \(6\). Their greatest common divisor is \(2\). Then divide each by \(2\): \(2/2=1\), \(6/2=3\).

Does \( \mathrm{C_2H_6} \) simplify to an empirical formula smaller than the molecular formula?

Yes. \( \mathrm{C_2H_6} \) has a common factor \(2\) in its subscripts. The smallest whole-number ratio becomes \( \mathrm{CH_3} \).

What is the ratio of C to H in \( \mathrm{C_2H_6} \)?

The ratio is \(2:6\). Simplify by dividing by \(2\) to get \(1:3\). That corresponds to \( \mathrm{CH_3} \).

Could the empirical formula be something else like \( \mathrm{C_2H_6} \) itself?

Not for empirical formula. Empirical formulas are reduced to lowest whole-number ratios. Since \(2\) divides both subscripts, reduction is required.

If the molecular formula is \( \mathrm{C_2H_6} \), what factor links it to the empirical formula?

Empirical formula is \( \mathrm{CH_3} \). Molecular formula \( \mathrm{C_2H_6} \) is \(2\) times the empirical: \(2 \times (\mathrm{CH_3}) = \mathrm{C_2H_6} \).

Empirical formula of C2H6 is CH3.
It simplifies by dividing by 2.

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