Question

If in an organic compound the weights of \(\mathrm{C}, \mathrm{H}, \mathrm{O}\) are taken in \(6: 1: 8\) ratio. The molecular formula of the compound is: (a) \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{4}\) (c) \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3}\) (d) \(\mathrm{CH}_{2} \mathrm{O}_{2}\)

Short answer

The molecular formula of the compound is \(\mathrm{C}_2\mathrm{H}_4\mathrm{O}_2\).

Step-by-step solution

Calculate Moles Ratio

First, we'll find the mole ratio of each element by dividing the given weight ratio by the atomic masses of the elements. The atomic mass of carbon (C) is 12, hydrogen (H) is 1, and oxygen (O) is 16. So, the mole ratio becomes:\(\text{C} = 6/12 = 0.5\)\(\text{H} = 1/1 = 1\)\(\text{O} = 8/16 = 0.5\)

Simplify Moles Ratio

Now, let's simplify the moles ratio obtained by dividing each by the smallest value from the previous step, which is 0.5:\(\text{C ratio}: 0.5/0.5 = 1\)\(\text{H ratio}: 1/0.5 = 2\)\(\text{O ratio}: 0.5/0.5 = 1\)This simplifies the mole ratio to 1:2:1 for C:H:O.

Determine Empirical Formula

The empirical formula can be determined using the mole ratio which is 1:2:1 for C:H:O. Thus, the empirical formula becomes \(\mathrm{CH}_2\mathrm{O}\).

Verify Options with Empirical Formula

Examine each option to see which one matches with the empirical formula or is a multiple of it:(a) \(\mathrm{C}_2\mathrm{H}_4\mathrm{O}_2\) is \( (\mathrm{CH}_2\mathrm{O}) \times 2\)(b) \(\mathrm{C}_2\mathrm{H}_4\mathrm{O}_4\) is \( (\mathrm{CH}_2\mathrm{O}_2) \times 2\)(c) \(\mathrm{C}_3\mathrm{H}_6\mathrm{O}_3\) is \( (\mathrm{CH}_2\mathrm{O}) \times 3\)(d) \(\mathrm{CH}_2\mathrm{O}_2\) is a modification of \(\mathrm{CH}_2\mathrm{O}\)We see option (a) and (c) both are valid extensions of the empirical formula \(\mathrm{CH}_2\mathrm{O}\) according to the given mass ratio, but in terms of empirical formats often times just one instance or minimal molecule same conformance should be spotted hence choice (a) extends correctly on it.

Key concepts

Mole Ratio Calculation

Calculating the mole ratio is a fundamental step in deriving the empirical formula of a compound. It involves converting the mass proportion of the elements to moles because a chemical formula is expressed in moles rather than mass. To achieve this, you divide each element's given weight by its respective atomic mass.
This results in the mole calculation:

• For Carbon (C): Divide its weight by its atomic mass (12). For this exercise, the calculation is \(6/12 = 0.5\).
• For Hydrogen (H): Since its atomic mass is 1, the mole remains the same, \(1/1 = 1\).
• For Oxygen (O): Divide its weight by 16, giving the result \(8/16 = 0.5\).

These mole ratios can then be simplified by dividing each by the smallest mole value to ensure that they are in their simplest whole number form. In our example, dividing by 0.5 gives a final mole ratio of 1:2:1 - indicating the elemental ratio in the compound.

Organic Compounds

Organic compounds are primarily made up of carbon atoms. In chemistry, these compounds are those that contain carbon in combination with other elements like hydrogen, oxygen, nitrogen, etc. Always built on carbon's versatile backbone, organic compounds include vast ranges of molecules from fuels to biological molecules. This flexibility often results in different structures sharing similar elemental compositions.
The compound in our exercise exemplifies this principle by involving elements common in organic chemistry: carbon, hydrogen, and oxygen. By analyzing the weighted ratios given, chemists can ascertain not only the composition but the empirical formula representing the simplest integer ratio of the elements involved. This information helps link the empirical formulas to possible molecular structures of the organic compound.

Atomic Mass

Atomic mass is the average mass of a chemical element's atoms, measured in atomic mass units (amu). It is crucial in converting mass ratios into mole ratios. Each element has a unique atomic mass, which is based on the weighted average of the isotopes found in nature.
For our chemical problem, knowing the atomic masses of carbon, hydrogen, and oxygen allows us to translate weight into a mole-based formula.

• Carbon has an atomic mass of 12 amu.
• Hydrogen is exceptionally light with an atomic mass of 1 amu.
• Oxygen stands at 16 amu.

Understanding atomic mass is vital in balancing chemical reactions and obtaining empirical formulas, as it converts weights measured on scales into practical chemical numbers rooted in mole concepts.

Chemical Problem Solving

Chemical problem solving involves understanding chemical principles and applying them in logical sequences to derive solutions. This exercise demonstrates key aspects of chemical problem solving. 1. **Given Data Analysis:** The initial step involves interpreting the provided ratio weights of elements. 2. **Conversion to Moles:** Utilize atomic masses to convert weights to mole estimates, a crucial step to bridge experimental data to chemical formulas. 3. **Mole Simplification:** This leads to extracting the greatest common factor amongst mole numbers, refining them to simple whole numbers creating the empirical formula. 4. **Validation Against Known Compounds:** Once an empirical formula is derived, it is compared to known substances or potential multiplicands to explore actual molecular formulas.
This systematic approach fosters deeper comprehension and nurtures skills in dealing with empirical to molecular formula transitions, demonstrating the disciplinary rigor of chemistry as both science and an analytical art.