Question

If a compound on analysis was found to contain \(\mathrm{C}=\) \(18.5 \%, \mathrm{H}=1.55 \%, \mathrm{Cl}=55.04 \%\) and \(\mathrm{O}=24.81 \%\) then its empirical formula is: (a) CHClO (b) \(\mathrm{CH}_{2} \mathrm{ClO}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OCl}\) (d) \(\mathrm{ClC}_{2} \mathrm{H}_{5} \mathrm{O}\)

Short answer

The empirical formula is CHClO.

Step-by-step solution

Assume 100g of Compound

Assume that the total mass of the compound is 100 grams. This means the percentage composition can be directly translated to grams. Thus, we have \(18.5\) grams of \(\mathrm{C}\), \(1.55\) grams of \(\mathrm{H}\), \(55.04\) grams of \(\mathrm{Cl}\), and \(24.81\) grams of \(\mathrm{O}\).

Convert Mass to Moles

For each element, convert the mass into moles by dividing by its atomic mass:- Carbon (C): \(\frac{18.5}{12.01} = 1.54\) moles- Hydrogen (H): \(\frac{1.55}{1.008} = 1.54\) moles- Chlorine (Cl): \(\frac{55.04}{35.45} = 1.55\) moles- Oxygen (O): \(\frac{24.81}{16.00} = 1.55\) moles

Divide by the Smallest Number of Moles

Identify the smallest number of moles found in Step 2, which is approximately \(1.54\). Divide all the mole values by \(1.54\):- C: \(\frac{1.54}{1.54} = 1\)- H: \(\frac{1.54}{1.54} = 1\)- Cl: \(\frac{1.55}{1.54} \approx 1\)- O: \(\frac{1.55}{1.54} \approx 1\)

Determine the Empirical Formula

The mole ratios are approximately whole numbers: \(1:1:1:1\) for \(\mathrm{C}\), \(\mathrm{H}\), \(\mathrm{Cl}\), and \(\mathrm{O}\). This gives us the empirical formula \(\mathrm{CHClO}\).

Key concepts

Chemical Analysis

Chemical analysis is a vital method used to understand the composition of substances. By examining the constituent elements, we can deduce valuable information about a compound's structure and formula.

In the analysis of a compound, chemists often start by determining the percentage composition of its elements. These percentages tell us how much of each element makes up the compound, which is crucial for finding the empirical formula.

Here's a handy approach to chemical analysis:

• Assume a sample mass, commonly 100 grams, to make calculations simple.
• Express the percentage composition in terms of mass in grams.
• Use the mass of each element to progress further in understanding the compound.

This method is a cornerstone of deriving empirical formulas, where you figure out the simplest whole-number ratio of atoms in a compound.

Elemental Composition

Elemental composition refers to the breakdown of a compound into its individual elements and the respective quantity of each in proportion to the whole. By knowing the elemental composition, chemists can ascertain empirical formulas that represent these proportions in a simplified manner.

To achieve this, follow these general steps:

• Identify the mass of each element in the compound, often derived from chemical analysis.
• Calculate the number of moles for each element using its atomic weight.
• Express these moles as simple whole-number ratios.

This concept is essential because it allows chemists to understand not just what elements are present, but how they relate to one another within the compound. Whether dealing with straightforward or complex molecules, elemental composition is the first step to understanding chemical relationships.

Mole Calculation

Mole calculation is at the heart of finding an empirical formula. It involves converting the mass of each element in a compound into moles, which are then used to derive the simplest ratio of atoms. This step is crucial since it bridges the gap between mass and the count of atoms, giving a deeper insight into the substance.

Here's how you can perform mole calculations:

• Begin by taking the mass of each element from your analysis, typically in grams.
• Divide this mass by the atomic weight of the element, which you can find on the periodic table.
• The resulting number is the total moles of the element in your sample.

This technique is fundamental because it ensures that the empirical formula accurately reflects the proportions of each element in terms of their atomic structure. Mole calculations are a universal tool, indispensable in chemistry for interpreting and predicting the behavior of materials.