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Chapter 11: Problem 56

The combustion of \(20.63 \mathrm{mg}\) of compound \(\mathrm{Y}\), which contains only \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O}\), with excess oxygen gave \(57.94 \mathrm{mg}\) of \(\mathrm{CO}_{2}\) and \(11.85 \mathrm{mg}\) of \(\mathrm{H}_{2} \mathrm{O}\). (a) Calculate how many milligrams of \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O}\) were present in the original sample of \(Y\). (b) Derive the empirical formula of Y. (c) Suggest a plausible structure for \(\mathrm{Y}\) if the empirical formula is the same as the molecular formula.

Short Answer

Expert verified
The empirical formula for compound Y is C6H6O and a plausible structure could be turpentine.

Step by step solution

01

Calculate the mass of carbon and hydrogen in the compound Y

Since combustion of carbon produces CO2, and hydrogen produces H2O, the masses of C and H in the compound Y can be calculated as follows: \nMass of C in Y = Mass of C in CO2 = 57.94 * 12/44 = 15.78 mg,\nMass of H in Y = Mass of H in H2O = 11.85 * 2/18 = 1.31 mg.
02

Calculate the mass of oxygen in compound Y

The mass of O in compound Y can be calculated by subtracting the total mass of C and H from the total mass of Y. \nMass of O in Y = Mass of Y - (Mass of C in Y + Mass of H in Y) = 20.63 - (15.78 + 1.31) = 3.54 mg.
03

Calculate the moles of each element in compound Y

The moles of Carbon, Hydrogen, and Oxygen in compound Y can be calculated by dividing their masses by their atomic weights: \nMoles of C = 15.78 / 12 = 1.315, \nMoles of H =1.31 / 1 = 1.31, \nMoles of O = 3.54 / 16 = 0.221.
04

Determine the empirical formula

To determine the empirical formula, divide the number of moles of each element by the smallest number of moles calculated: \nC = 1.315 / 0.221 = 6, \nH = 1.31 / 0.221 = 6, \nO = 0.221 / 0.221 = 1, \nso the empirical formula of compound Y is C6H6O.
05

Suggest a plausible structure of compound Y

Given the number of carbon, hydrogen, and oxygen atoms, a plausible structure for Y could be turpentine if the empirical formula same as molecular formula.

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Most popular questions from this chapter

Sulfuric acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right)\) adds to the double bond of alkenes as \(\mathrm{H}^{+}\) and \(-\mathrm{OSO}_{3} \mathrm{H}\). Predict the products when sulfuric acid reacts with (a) ethylene and (b) propene.

A compound having the molecular formula \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}\) does not react with sodium metal. In the presence of light, the compound reacts with \(\mathrm{Cl}_{2}\) to form three compounds having the formula \(\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OCl}\). Draw a structure for the original compound that is consistent with this information.

The compound \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3}\) is hydrogenated to an alkene using platinum as the catalyst. If the product is the pure cis isomer, what can you deduce about the mechanism?

Alkenes exhibit geometric isomerism because rotation about the \(\mathrm{C}=\mathrm{C}\) bond is restricted. Explain.

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