Compute the empirical and molecular formulas for each of the following elemental | StudySoup

Textbook Solutions for Organic Chemistry

Chapter 1 Problem PROBLEM 1-14

Question

Compute the empirical and molecular formulas for each of the following elemental analyses. In each case, propose at least one structure that fits the molecular formula.

\(\begin{array}{ccccc} \text { C } & \mathbf{H} & \mathbf{N} & \mathbf{C l} & \text { MW } \\ \text { (a) } 40.0 \% & 6.67 \% & 0 & 0 & 90 \\ \text { (b) } 32.0 \% & 6.67 \% & 18.7 \% & 0 & 75 \\ \text { (c) } 25.6 \% & 4.32 \% & 15.0 \% & 37.9 \% & 93 \\ \text { (d) } 38.4 \% & 4.80 \% & 0 & 56.8 \% & 125 \end{array}\)

Solution

Step 1 of 8

The molecular formula gives the exact amount of elements present in a compound, whereas the empirical formula denotes the reduced form of the molecular formula. The empirical formula is multiplied with a particular coefficient to obtain the molecular formula.

Part (a)

The percentage of carbon is 40.0%, the percentage of H is 6.67%, the percentage of N is 0 and the percentage of Cl is 0.

The percentage of oxygen can be found as follows:

\(\begin{aligned} \text { Percentage of oxygen } & =100-(40+6.67) \\ & =100-46.67 \% \\ & =53.3 \% \end{aligned}\)

The moles of carbon can be found as follows:

\(\begin{aligned}\text{ Moles }&=\frac{\text{ Mass }}{\text{ Molar mass }}\\ &=\frac{40\mathrm{\ g}}{12\mathrm{\ g}/\mathrm{mol}}\\ &=3.33\mathrm{\ mol}\end{aligned}\)

The moles of hydrogen can be found as follows:

\(\begin{aligned}\text{ Moles }&=\frac{\text{ Mass }}{\text{ Molar mass }}\\ &=\frac{6.67\mathrm{\ g}}{1\mathrm{\ g}/\mathrm{mol}}\\ &=6.67\mathrm{\ mol}\end{aligned}\)

The moles of oxygen can be found as follows:

\(\begin{aligned}\text{ Moles }&=\frac{\text{ Mass }}{\text{ Molar mass }}\\ &=\frac{53.33\mathrm{\ g}}{16\mathrm{\ g}/\mathrm{mol}}\\ &=3.33\mathrm{\ mol}\end{aligned}\)

Among these, the smallest value of mole is 3.33 mol.

Hence, divide all the mole values with the value 3.33.

The molar ratio of the elements carbon, hydrogen, and oxygen can be obtained as follows:

\(\begin{aligned} \text { Mole ratio } & =\mathrm{C}: \mathrm{H}: \mathrm{O} \\ & =\frac{3.33}{3.33}: \frac{6.67}{3.33}: \frac{3.33}{3.33} \\ & =1: 2: 1 \end{aligned}\)

Hence, the empirical formula of the compound is \(\mathrm{CH}_{2} \mathrm{O}\) and the molar mass is 90.

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full solution

Title Organic Chemistry 9 
Author Leroy G. Wade, Jan W. Simek
ISBN 9780321971371

Compute the empirical and molecular formulas for each of the following elemental

Chapter 1 textbook questions

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