Question

For the complete oxidation of \(100 \mathrm{~g}\) of cyclohexanol to cyclohexanone, the quantity of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) requireds (assuming \(100 \%\) chemical yield) [Atomic mass: \(\mathrm{Cr}=52, \mathrm{~K}=39]\). (A) \(294 \mathrm{~g}\) (B) \(147 \mathrm{~g}\) (C) \(98 \mathrm{~g}\) (D) \(195 \mathrm{~g}\)

Short answer

(C) \(98 \mathrm{~g}\)

Step-by-step solution

Write the balanced chemical equation

For the oxidation of cyclohexanol (C_6H_11OH) to cyclohexanone (C_6H_10O), the balanced chemical equation is given by: \[C_6H_{11}OH + \dfrac {1}{3}K_2Cr_2O_7 + 2H_2SO_4 \rightarrow C_6H_{10}O + \dfrac {2}{3}Cr_2(SO_4)_3 + K_2SO_4 + 5H_2O\]

Calculate the moles of cyclohexanol

First, we need to compute the number of moles in 100 g of cyclohexanol. The molecular weight of cyclohexanol (C_6H_11OH) is: \[ (6 \times 12.01) + (11 \times 1.008) + 16.00 + 1.008 = 100.16 \mathrm{~g/mol}\] Now we can determine the number of moles in 100 g of cyclohexanol: \[ \dfrac{100 \mathrm{~g}}{100.16 \mathrm {~g/mol}} = 0.998\ \mathrm{moles}\]

Use stoichiometry to determine moles of K2Cr2O7 required

According to the balanced chemical equation, one mole of cyclohexanol reacts with \(\dfrac {1}{3}\) moles of \(K_2Cr_2O_7\). So, the moles of \(K_2Cr_2O_7\) required for 0.998 moles of cyclohexanol are: \[0.998\ \mathrm{moles\ of\ cyclohexanol} \times \dfrac{1}{3}\ \mathrm{moles\ of\ K_2Cr_2O_7} = 0.333\ \mathrm{moles\ of\ K_2Cr_2O_7}\]

Calculate the mass of K2Cr2O7 required

To calculate the mass of \(K_2Cr_2O_7\) required, we need to determine its molecular weight. The molecular weight of \(K_2Cr_2O_7\) is: \[ (2 \times 39) + (2 \times 52) + (7 \times 16) = 294 \mathrm{~g/mol}\] Now, we can find the mass of \(K_2Cr_2O_7\) needed for the complete oxidation of 100 g of cyclohexanol by multiplying the required moles by the molecular weight: \[0.333\ \mathrm{moles}\times 294 \mathrm{~g/mol} = 98\ \mathrm{g}\] So, the answer is: (C) \(98 \mathrm{~g}\) of \(K_2Cr_2O_7\)

Key concepts

Oxidation Reactions

An oxidation reaction in organic chemistry involves the addition of oxygen or the removal of hydrogen from a molecule. This process is crucial as it helps in converting one compound to another. For example, cyclohexanol is oxidized to cyclohexanone in the presence of an oxidizing agent like potassium dichromate \( (K_2Cr_2O_7) \). In this reaction, the oxidizing agent facilitates the removal of hydrogen from the alcohol group in cyclohexanol, transforming it into a ketone group present in cyclohexanone. To understand oxidation fully, remember:

• Loss of electrons leads to oxidation.
• Gain of electrons leads to reduction.

These reactions often occur together in what is called a redox (reduction-oxidation) reaction, maintaining a balance where one substance is oxidized and another is reduced.

Stoichiometry

Stoichiometry is the part of chemistry that focuses on measuring the quantitative relationships between reactants and products in a chemical reaction. In simple terms, it helps us calculate how much of each substance is needed in a reaction to produce a desired product. For example, to oxidize 100 grams of cyclohexanol to cyclohexanone using potassium dichromate, we used stoichiometry to determine that it required \(0.333\) moles of \(K_2Cr_2O_7\). The steps typically involve:

• Using the balanced chemical equation.
• Determining molecular weights.
• Calculating the moles of each substance.

This method is essential in laboratory settings to ensure reactions occur without excess waste and to maximize the yield of desired products.

Balanced Chemical Equation

A balanced chemical equation accurately represents a chemical reaction using appropriate symbols and formulas. It ensures that the number of atoms for each element is the same on both the reactant and product sides of the equation, following the Law of Conservation of Mass. For instance, the reaction of cyclohexanol to cyclohexanone with potassium dichromate is balanced as follows:\[ C_6H_{11}OH + \frac{1}{3}K_2Cr_2O_7 + 2H_2SO_4 \rightarrow C_6H_{10}O + \frac{2}{3}Cr_2(SO_4)_3 + K_2SO_4 + 5H_2O \]In balancing equations:

• Ensure equal numbers of each atom type on both sides.
• Adjust coefficients in front of molecules, not subscripts.
• Balance elements one at a time, starting with the most complicated molecule.

A balanced equation is essential before performing any stoichiometric calculations.

Molecular Weight Calculation

In chemistry, the molecular weight of a compound is the sum of the atomic weights of all atoms in its molecular formula. It plays a critical role in determining the amount of substance required in a chemical reaction. For instance, to compute the molecular weight of cyclohexanol (\(C_6H_{11}OH\)), you sum the atomic weights of all the atoms:\[ (6 \times 12.01) + (11 \times 1.008) + 16.00 + 1.008 = 100.16 \text{ g/mol} \]This calculation is crucial because it allows us to convert between grams and moles, facilitating proper stoichiometric calculations, as seen when determining the moles of \(K_2Cr_2O_7\) needed to oxidize 100 grams of cyclohexanol. Remember:

• Molecular weight is given in grams per mole \( (\text{g/mol}) \).
• Accurate molecular weights are required for precise reactions.
• Knowing this helps predict how much of each reactant is needed to form a product.

Understanding molecular weight calculations is fundamental to accurately preparing reactions in both academic and professional chemistry settings.