Question

\(60 \mathrm{~g}\) of a compound on analysis produced \(24 \mathrm{~g}\) carbon, \(4 \mathrm{~g}\) hydrogen and \(32 \mathrm{~g}\) oxygen. The empirical formula of the compound is: (a) \(\mathrm{CH}_{4} \mathrm{O}\) (b) \(\mathrm{CH}_{2} \mathrm{O}_{2}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}\) (d) \(\mathrm{CH}_{2} \mathrm{O}\)

Short answer

The empirical formula is \( \text{CH}_2\text{O} \).

Step-by-step solution

Determine the moles of each element

Calculate the moles of carbon, hydrogen, and oxygen in the compound using their respective atomic masses (Carbon: 12 g/mol, Hydrogen: 1 g/mol, Oxygen: 16 g/mol).1. **Carbon:** \( \frac{24 \text{ g}}{12 \text{ g/mol}} = 2 \text{ mol} \) 2. **Hydrogen:** \( \frac{4 \text{ g}}{1 \text{ g/mol}} = 4 \text{ mol} \)3. **Oxygen:** \( \frac{32 \text{ g}}{16 \text{ g/mol}} = 2 \text{ mol} \)

Find the mole ratio

Divide the moles of each element by the smallest number of moles from Step 1 (which is 2 moles).1. **Carbon:** \( \frac{2}{2} = 1 \)2. **Hydrogen:** \( \frac{4}{2} = 2 \)3. **Oxygen:** \( \frac{2}{2} = 1 \)

Write the empirical formula

Based on the simplest mole ratio from Step 2, the empirical formula is \( \text{C}_1\text{H}_2\text{O}_1 \), which can be written as \( \text{CH}_2\text{O} \).

Key concepts

Chemical Analysis

Chemical analysis is a crucial process to determine the composition of matter. Whether in a lab or a classroom exercise, understanding which elements make up a compound is fundamental. When a compound undergoes chemical analysis, it is broken down into its basic parts, such as carbon, hydrogen, and oxygen, as seen in the example of the compound weighing 60 grams. The goal is to identify the exact weight of each element in the mixture.
By knowing the masses (0 4  2 f) of these elements, you can start to deduce more about their arrangement within the compound. This step provides clear insight into how much of each element is present before proceeding to calculation of moles.

Moles Calculation

Moles are a standardized way in chemistry to express the amount of a chemical substance. Instead of dealing with large numbers of individual atoms or molecules, chemists use the concept of moles, which allows easier calculation and communication. The process of moles calculation starts with determining the number of moles for each element based on its given mass in the compound. This involves dividing the mass by the element's atomic mass.
For example:

• Carbon: \[ \frac{24 \text{ g}}{12 \text{ g/mol}} = 2\text{ mol} \]
• Hydrogen: \[ \frac{4 \text{ g}}{1 \text{ g/mol}} = 4 \text{ mol} \]
• Oxygen: \[ \frac{32 \text{ g}}{16 \text{ g/mol}} = 2 \text{ mol} \]

This step is integral to progressing from basic measurements towards understanding the empirical formula.

Atomic Mass

Atomic mass is a simple yet vital concept in chemistry used to identify and differentiate elements. Every element on the periodic table is identified by its atomic mass, which corresponds to the total number of protons and neutrons in the nucleus.
For our concerned elements:

• Carbon: 12 g/mol
• Hydrogen: 1 g/mol
• Oxygen: 16 g/mol

These values are universally accepted and essential in calculating moles from chemical analysis data. Knowing the atomic mass allows one to convert grams to moles simply and accurately, aiding significantly in determining the empirical formula of compounds.

Elemental Composition

Elemental composition refers to the proportion of elements within a compound. This is expressed as a ratio or percentage of each element's mass relative to the total mass of the compound. Finding the elemental composition is critical for establishing the empirical formula.
In the exercise example, the compound split comprises:

• 40% Carbon (24 g out of 60 g)
• 6.67% Hydrogen (4 g out of 60 g)
• 53.33% Oxygen (32 g out of 60 g)

These compositions help chemists understand not just the quantity present, but how elements are grouped within the compound. Empirical formulas give an overview of this elemental composition by providing a simple whole-number ratio reflecting the amount of each element present, simplifying complex molecular formulas into more recognizable molecular structures.