Question

An organic compound has \(\mathrm{C}\) and \(\mathrm{H}\) percentage in the ratio \(6: 1\) and \(C\) and \(O\) percentage in the \(3: 4\). The compound is: (a) \((\mathrm{COOH})_{2}\) (b) \(\mathrm{CH}_{3} \mathrm{OH}\) (c) HCHO (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\)

Short answer

The compound is HCHO (option c).

Step-by-step solution

Understand Ratios

We are given two ratios: the ratio of carbon (C) to hydrogen (H) is 6:1 and the ratio of carbon (C) to oxygen (O) is 3:4.

Calculate the Relative Percentage of C

Assign a value such as '6x' to the percentage of carbon since it is common in both ratios. This implies the percentage of hydrogen is 'x' and the percentage of oxygen is "\(\frac{8}{3}\times 6x\)" from the C:O ratio.

Find the Unknowns 'x' and Verify Total

Since the sum of percentages of C, H, and O should be 100%, set the equation \(6x + x + \frac{8}{3} \times 6x = 100\). Simplify this to find the value of 'x' and verify if the total adds to 100%.

Solve for 'x'

We have the equation \(6x + x + 16x = 100\). Simplifying gives \(23x = 100\). Thus, \(x = \frac{100}{23}\).

Calculate Element Percentages

Substitute \(x = \frac{100}{23}\) back into each component: \(\%C = 6x = \frac{600}{23}\), \(%H = x = \frac{100}{23}\), and \(%O = 16x = \frac{1600}{23}\).

Find Element Mass Comparison to Molecular Formulas

Compare these ratios to each given formula. For choice (c) HCHO: Carbon is 1/2, hydrogen is 1/4, and oxygen is 1/4. This matches the calculated ratios, meaning choice (c) fits.

Key concepts

Molecular Formulas

In the realm of organic chemistry, understanding molecular formulas is essential. A molecular formula denotes the types and numbers of atoms in a molecule. For instance, the molecular formula for water is \(\mathrm{H}_2\mathrm{O}\), which informs us that each molecule of water consists of two hydrogen atoms and one oxygen atom.
One important aspect is that different compounds can share the same molecular formula but have different structures or properties, known as isomers. For example, ethanol and dimethyl ether both have the formula \(\mathrm{C}_2\mathrm{H}_6\mathrm{O}\) but vary significantly in structure and function.
When working through exercises that require you to identify compounds based on formulas, always compare elemental ratios and quantities carefully to determine the correct compound.

Elemental Analysis

Elemental analysis involves determining the percentage composition of each element within a compound. In the exercise, we learned that the ratio of carbon to hydrogen is 6:1, and carbon to oxygen is 3:4. This analytical data provides insights into the compound's overall makeup.
An elemental analysis can help verify the purity of a compound or confirm its identity by comparing empirical data against known standards. It reveals the basic components, expressed as a percentage, of a compound, allowing chemists to deduce molecular formulas.
Such analysis is crucial when synthesizing new compounds or verifying organic substances' purity. Mastering this helps ensure accurate formula identification and understanding chemical behavior.

Chemical Ratios

Understanding chemical ratios is foundational in chemistry. Ratios express the relative amounts of components within a compound or reaction. In this exercise, we focused on the carbon to hydrogen (C:H) and carbon to oxygen (C:O) ratios.
Ratios are pivotal for calculating quantities in reactions, ensuring components are mixed in proper proportions to achieve desired results. In our example, the ratios provided serve as a clue to derive the molecular formula of the compound. By setting carbon's percentage as a baseline, we used these ratios to ensure consistency across the entire compound.
Grasping how to manipulate and apply ratios in chemical contexts helps sharpen problem-solving skills and deepens understanding of molecular compositions.

Stoichiometry

Stoichiometry is the tool chemists use to calculate relationships in chemical reactions. It helps us relate reactants to products through balanced equations and proportionate amounts.
In this exercise, stoichiometry played a role even without a chemical reaction given. By applying stoichiometric principles, we managed to distribute the total percentage (100%) among the elements in the compound based on the given ratios. Solving the equation \(6x + x + 16x = 100\) showcased the use of stoichiometry in calculations of molecular formulas.
By integrating stoichiometry into problem-solving, it becomes possible to predict the amounts of products formed or required reactants, thereby providing critical insight into experimental designs and processes.