Problem 87
Question
An organic compound has \(\mathrm{C}\) and \(\mathrm{H}\) percentage in the ratio \(6: 1\) and \(\mathrm{C}\) and \(\mathrm{O}\) percentage in the \(3: 4\). The compound is (a) \((\mathrm{COOH})_{2}\) (b) \(\mathrm{CH}_{3} \mathrm{OH}\) (c) \(\mathrm{HCHO}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\)
Step-by-Step Solution
Verified Answer
None of the given options match the ratio conditions exactly.
1Step 1: Understanding the Carbon to Hydrogen Ratio
The problem states that the carbon to hydrogen ratio is \(6:1\). This means for every 6 parts of carbon, there is 1 part of hydrogen. We'll compare this with the molecular formulas of the given compounds.
2Step 2: Understanding the Carbon to Oxygen Ratio
The carbon to oxygen ratio given is \(3:4\). This means for every 3 parts of carbon, there are 4 parts of oxygen. We'll check this ratio against the provided compounds.
3Step 3: Analyze Option (a)
The molecular formula for (a) \((\mathrm{COOH})_2\) is \(\mathrm{C}_2\mathrm{H}_2\mathrm{O}_4\). The carbon to hydrogen ratio is \(2:2 = 1:1\) and the carbon to oxygen ratio is \(2:4 = 1:2\). Neither ratio matches the requirements.
4Step 4: Analyze Option (b)
The molecular formula for (b) \(\mathrm{CH}_3\mathrm{OH}\) is \(\mathrm{C}_1\mathrm{H}_4\mathrm{O}_1\). The carbon to hydrogen ratio is \(1:4\), and the carbon to oxygen ratio is \(1:1\). These do not match the requirements.
5Step 5: Analyze Option (c)
The molecular formula for (c) \(\mathrm{HCHO}\) is \(\mathrm{C}_1\mathrm{H}_2\mathrm{O}_1\). The carbon to hydrogen ratio is \(1:2\), and the carbon to oxygen ratio is \(1:1\). These do not match the requirements.
6Step 6: Analyze Option (d)
The molecular formula for (d) \(\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}\) is \(\mathrm{C}_2\mathrm{H}_6\mathrm{O}_1\). The carbon to hydrogen ratio is \(2:6 = 1:3\), and the carbon to oxygen ratio is \(2:1\). Again, these do not match the required ratios.
Key Concepts
Molecular FormulasCarbon to Hydrogen RatioCarbon to Oxygen Ratio
Molecular Formulas
In organic chemistry, a molecular formula is like a map that guides us through the structure of a compound. It shows us exactly how many atoms of each element are present in a molecule. For example, consider the molecular formula \(\mathrm{C}_2\mathrm{H}_6\mathrm{O}\). This tells us that in each molecule of this compound, there are 2 carbon atoms, 6 hydrogen atoms, and 1 oxygen atom.
The molecular formula is crucial because:
The molecular formula is crucial because:
- It describes the composition of the molecule.
- Helps in determining the ratios of atoms, such as the carbon to hydrogen ratio or carbon to oxygen ratio.
- Allows us to derive empirical formulas, which show the simplest whole-number ratio of elements in a compound.
Carbon to Hydrogen Ratio
The carbon to hydrogen ratio is a key detail that helps to identify and differentiate organic compounds. It shows how often carbon atoms repeat compared to hydrogen atoms in a compound. For instance, a \(6:1\) ratio means there are 6 carbon atoms for every hydrogen atom. This ratio is not only important for identifying compounds but also for understanding their chemical properties.
For example, let's consider various compounds:
For example, let's consider various compounds:
- In \(\mathrm{CH}_3\mathrm{OH}\), the ratio is \(1:4\) which doesn't match \(6:1\), making it chemically different.
- For \(\mathrm{CH}_2\), the ratio of \(1:2\) still doesn't fit \(6:1\), indicating another mismatch.
Carbon to Oxygen Ratio
The carbon to oxygen ratio is an essential parameter in the classification and study of organic compounds. This ratio represents how carbon atoms compare to oxygen atoms in a given compound. A \(3:4\) carbon to oxygen ratio implies that for every 3 carbon atoms, there are 4 oxygen atoms.
Analyzing carbon to oxygen ratios can help:
Analyzing carbon to oxygen ratios can help:
- Identify potentially similar molecular compounds.
- Predict physical aspects like boiling points or solubility.
- Understand possible reactions a compound can undergo, like oxidation reactions which are vital in metabolic pathways.
Other exercises in this chapter
Problem 85
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