(a) Nitrogen has relatively stable isotopes (half-life greater than 1 second) of mass numbers 13, 14, 15, 16, and 17. (All except 14N and 15N are radioactive.) Calculate how many protons and neutrons are in each of these isotopes of nitrogen. (b) Write the electronic configurations of the third-row elements shown in the partial periodic table in Figure 1-6.
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Textbook Solutions for Organic Chemistry
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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