Q. \( \mathrm{C_4H_{10}} \) empirical formula.

Q: What is the empirical formula of \mathrm{C_4H_{10}} ?

The empirical formula is \mathrm{C_2H_5} . Divide subscripts by the greatest common divisor \gcd(4,10)=2 : \mathrm{C_4H_{10}} \rightarrow \mathrm{C_2H_5} .

Q: Does the molecular formula \mathrm{C_4H_{10}} equal the empirical formula \mathrm{C_2H_5} ?

No. \mathrm{C_4H_{10}} is the molecular formula and is a multiple of the empirical formula. Here \mathrm{C_4H_{10}} = 2 \times \mathrm{C_2H_5} .

Q: If the empirical formula is \mathrm{C_2H_5} , what molecular formula corresponds to a multiplier n=2?

Multiply subscripts by n: \mathrm{C_2H_5} \times 2 \rightarrow \mathrm{C_4H_{10}} . In general, \mathrm{(C_2H_5)_n} = \mathrm{C_{2n}H_{5n}} .

Q: What is the percent composition of \mathrm{C_4H_{10}} and how does it relate to the empirical formula?

Empirical reduction does not change mass percentages. So \mathrm{C_4H_{10}} and \mathrm{C_2H_5} have identical percent composition because both represent the same ratio of atoms.

Q: How can I quickly check whether a formula is already in empirical form?

Look for a common factor in all subscripts. If \gcd of subscripts is 1, it’s already empirical. For \mathrm{C_4H_{10}} , \gcd=2 , so it is not empirical.

Answer

Let the molecular formula be C4H10. Empirical formula is found by dividing all subscripts by their greatest common divisor.

\(4\) and \(10\) have \(\gcd(4,10)=2\).

\[
\text{Empirical subscripts}=\frac{4}{2},\frac{10}{2} = 2,5
\]

Final result: \( \mathrm{C_2H_5} \)

Detailed Explanation

Step 1: Identify the molecular formula.

The molecular formula given is \( \mathrm{C_4H_{10}} \).

Step 2: List the atom subscripts.

For \( \mathrm{C_4H_{10}} \), the number of carbon atoms is \(4\) and the number of hydrogen atoms is \(10\).

Step 3: Find the greatest common factor (GCF) of the subscripts.

We compare \(4\) and \(10\).

The divisors of \(4\) are \(1, 2, 4\).

The divisors of \(10\) are \(1, 2, 5, 10\).

The common divisors are \(1\) and \(2\).

So the GCF is \(2\).

Step 4: Divide all subscripts by the GCF.

Divide the carbon subscript by \(2\):

\( \frac{4}{2} = 2 \)

Divide the hydrogen subscript by \(2\):

\( \frac{10}{2} = 5 \)

Step 5: Write the empirical formula.

Using the reduced whole-number subscripts, the empirical formula is:

\[ \mathrm{C_2H_5} \]

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General Chemistry FAQs

What is the empirical formula of \( \mathrm{C_4H_{10}} \)?

The empirical formula is \( \mathrm{C_2H_5} \). Divide subscripts by the greatest common divisor \( \gcd(4,10)=2 \): \( \mathrm{C_4H_{10}} \rightarrow \mathrm{C_2H_5} \).

How do I find the \(\gcd\) of subscripts in an empirical formula problem?

Compute the greatest common divisor of all subscripts. Here \( \gcd(4,10)=2 \). Then divide each subscript by \(2\). Always reduce until no common factor greater than 1 remains.

Does the molecular formula \( \mathrm{C_4H_{10}} \) equal the empirical formula \( \mathrm{C_2H_5} \)?

No. \( \mathrm{C_4H_{10}} \) is the molecular formula and is a multiple of the empirical formula. Here \( \mathrm{C_4H_{10}} = 2 \times \mathrm{C_2H_5} \).

If the empirical formula is \( \mathrm{C_2H_5} \), what molecular formula corresponds to a multiplier \(n=2\)?

Multiply subscripts by \(n\): \( \mathrm{C_2H_5} \times 2 \rightarrow \mathrm{C_4H_{10}} \). In general, \( \mathrm{(C_2H_5)_n} = \mathrm{C_{2n}H_{5n}} \).

What is the percent composition of \( \mathrm{C_4H_{10}} \) and how does it relate to the empirical formula?

Empirical reduction does not change mass percentages. So \( \mathrm{C_4H_{10}} \) and \( \mathrm{C_2H_5} \) have identical percent composition because both represent the same ratio of atoms.

How can I quickly check whether a formula is already in empirical form?

Look for a common factor in all subscripts. If \( \gcd \) of subscripts is 1, it’s already empirical. For \( \mathrm{C_4H_{10}} \), \( \gcd=2 \), so it is not empirical.

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