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The koala dines exclusively on eucalyptus leaves. Its digestive system detoxifies the eucalyptus oil, a poison to other animals. The chief constituent in eucalyptus oil is a substance called eucalyptol, which contains \(77.87 \% \mathrm{C}\), \(11.76 \% \mathrm{H}\), and the remainder \(\mathrm{O}\). (a) What is the empirical formula for this substance? (b) A mass spectrum of eucalyptol shows a peak at about 154 amu. What is the molecular formula of the substance?

Short Answer

Expert verified
The empirical formula and the molecular formula of eucalyptol are both \(\mathrm{C}_{10}\mathrm{H}_{18}\mathrm{O}\).

Step by step solution

01

Convert percentages to moles

To convert percentages to moles, we will assume having 100 grams of eucalyptol. So, we will have: \[\mathrm{C}: 77.87 g × \frac{1\, mol }{12.01\, g} = 6.484\,moles\] \[\mathrm{H}: 11.76 g × \frac{1\, mol }{1.008\, g} = 11.67\,moles\] To find the mass of oxygen, we'll use the fact, that the remainder is oxygen: \[\mathrm{mass\, of\, O} = 100\, g - 77.87\,g - 11.76\,g = 10.37\, g\] \[\mathrm{O}: 10.37 g × \frac{1\, mol }{16.00\, g} = 0.648\,moles\]
02

Find the simplest mole ratio

We will divide each mole value by the lowest mole value, which is 0.648 moles of oxygen. \[\mathrm{C}: \frac{6.484\, moles}{0.648\, moles} ≈ 10\] \[\mathrm{H}: \frac{11.67\, moles}{0.648\, moles} ≈ 18\] \[\mathrm{O}: \frac{0.648\, moles}{0.648\, moles} = 1\] So, the simplest mole ratio of C, H, and O is 10:18:1.
03

Write the empirical formula

Using the mole ratios we found in the previous step, the empirical formula for eucalyptol will be: \[\mathrm{C}_{10}\mathrm{H}_{18}\mathrm{O}\] Now, let's find the molecular formula.
04

Calculate the molar mass of the empirical formula

We have to find the molar mass of the empirical formula \(\mathrm{C}_{10}\mathrm{H}_{18}\mathrm{O}\): \[M_{emp} = 10 × 12.01\, g/mol + 18 × 1.008\, g/mol + 1 × 16\, g/mol ≈ 152\, g/mol\]
05

Find the multiple of the empirical formula

To find the molecular formula, we need to determine the whole number multiple (n) of the empirical formula using the given molar mass of eucalyptol (154 amu): \[n = \frac{M_{molecular}}{M_{emp}} = \frac{154\, g/mol}{152\, g/mol} ≈ 1\] Since n is approximately 1, the molecular formula of eucalyptol is the same as the empirical formula.
06

Write the molecular formula

The molecular formula of eucalyptol is: \[\mathrm{C}_{10}\mathrm{H}_{18}\mathrm{O}\] So, the empirical formula and the molecular formula of eucalyptol are both \(\mathrm{C}_{10}\mathrm{H}_{18}\mathrm{O}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molecular Formula
Understanding the concept of a molecular formula is vital in chemistry. It represents the actual number and type of atoms in a molecule, showing the exact composition. Every complex molecule can be broken down into its basic elemental makeup.
In the exercise above, both the empirical and molecular formulas for eucalyptol were found to be \(\mathrm{C}_{10}\mathrm{H}_{18}\mathrm{O}\). But what is the difference between these formulas? The molecular formula illustrates the true number of each atom in a compound, which may be a multiple of the empirical formula.
When analyzing molecules, especially large ones, it is essential to ensure the formula reflects reality accurately. Molecular formulas help in the calculation of molar mass, determining reaction stoichiometry, and understanding molecule behavior. In this case, the molecular formula coincides with the empirical one, meaning the compound's smallest unit is also its entire molecule. However, this isn't always the case, as adjustments may be required when further analyzed.
Percent Composition
Percent composition is the percentage by mass of each element in a compound. It provides valuable information about the makeup of a substance and is an essential first step in determining other chemical properties.
In our example, eucalyptol comprises \(77.87\%\) carbon, \(11.76\%\) hydrogen, and the remainder being oxygen. This tells us about how much of each element is present in the compound.
To find the percent composition:
  • Assume a 100g sample for simplicity.
  • Translate the percentages directly into gram amounts of each element.
  • Convert these masses into moles using the atomic mass from the periodic table.
By understanding the percentual makeup, chemists can trace back to the empirical formula and further understand the structure of complex compounds. Knowing the percent composition is crucial when creating compounds or studying substance properties, making it invaluable for proper empirical determinations.
Mole Ratio
Mole ratio plays a critical role in deriving the empirical formula of a compound from the percent composition. It represents the simplest whole-number ratio of moles of each element within a compound. Mastery of calculating and interpreting mole ratios provides insights into chemical formulas and reactions.
To determine the mole ratio in eucalyptol:
  • Convert each element's percentage to moles.
  • Choose the smallest mole value among the elements to divide all mole values, ensuring a simple whole-number ratio is achieved.
Here, for eucalyptol, calculating this ratio yielded \(10:18:1\) for \(\mathrm{C} : \mathrm{H} : \mathrm{O}\), leading to the empirical formula \(\mathrm{C}_{10}\mathrm{H}_{18}\mathrm{O}\).
Mole ratios are foundational in understanding chemical compositions and reactions, essential for tasks ranging from laboratory experiments to industrial applications. By simplifying complex percentage data into simple ratios, chemists make accurate predictions about the behavior and reactivity of compounds.

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Most popular questions from this chapter

One of the steps in the commercial process for converting ammonia to nitric acid is the conversion of \(\mathrm{NH}_{3}\) to \(\mathrm{NO}\) : $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ In a certain experiment, \(1.50 \mathrm{~g}\) of \(\mathrm{NH}_{3}\) reacts with \(2.75 \mathrm{~g}\) of \(\mathrm{O}_{2}\) (a) Which is the limiting reactant? (b) How many grams of \(\mathrm{NO}\) and of \(\mathrm{H}_{2} \mathrm{O}\) form? (c) How many grams of the excess reactant remain after the limiting reactant is completely consumed? (d) Show that your calculations in parts (b) and (c) are consistent with the law of conservation of mass.

A sample of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), contains \(1.250 \times 10^{21}\) carbon atoms. (a) How many atoms of hydrogen does it contain? (b) How many molecules of glucose does it contain? (c) How many moles of glucose does it contain? (d) What is the mass of this sample in grams?

The molecular formula of allicin, the compound responsible for the characteristic smell of garlic, is \(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{OS}_{2}\). (a) What is the molar mass of allicin? (b) How many moles of allicin are present in \(5.00 \mathrm{mg}\) of this substance? (c) How many molecules of allicin are in \(5.00 \mathrm{mg}\) of this substance? (d) How many \(\mathrm{S}\) atoms are present in 5.00 mg of allicin?

Aluminum hydroxide reacts with sulfuric acid as follows: $$ 2 \mathrm{Al}(\mathrm{OH})_{3}(s)+3 \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+6 \mathrm{H}_{2} \mathrm{O}(l) $$ Which reagent is the limiting reactant when \(0.500 \mathrm{~mol}\) \(\mathrm{Al}(\mathrm{OH})_{3}\) and \(0.500 \mathrm{~mol} \mathrm{H}_{2} \mathrm{SO}_{4}\) are allowed to react? How many moles of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) can form under these conditions? How many moles of the excess reactant remain after the completion of the reaction?

Withoutdoing any detailed calculations (but using a periodic table to give atomic weights), rank the following samples in order of increasing number of atoms: \(3.0 \times 10^{23}\) molecules of \(\mathrm{H}_{2} \mathrm{O}_{2}, 2.0 \mathrm{~mol} \mathrm{CH}_{4}, 32 \mathrm{~g} \mathrm{O}_{2}\)

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