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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

Balance the following equations, and indicate whether they are combination, decomposition, or combustion reactions: (a) \(\mathrm{C}_{3} \mathrm{H}_{6}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (b) \(\mathrm{NH}_{4} \mathrm{NO}_{3}(s) \longrightarrow \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{N}_{2}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{NH}_{3}(g)\) (e) \(\mathrm{K}_{2} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{KOH}(a q)\)

Copper is an excellent electrical conductor widely used in making electric circuits. In producing a printed circuit board for the electronics industry, a layer of copper is laminated on a plastic board. A circuit pattern is then printed on the board using a chemically resistant polymer. The board is then exposed to a chemical bath that reacts with the exposed copper, leaving the desired copper circuit, which has been protected by the overlaying polymer. Finally, a solvent removes the polymer. One reaction used to remove the exposed copper from the circuit board is $$ \mathrm{Cu}(s)+\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}(a q)+4 \mathrm{NH}_{3}(a q) \underset{2 \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}(a q)} $$ A plant needs to produce 5000 circuit boards, each with a surface area measuring \(2.0\) in. \(\times 3.0\) in. The boards are covered with a 0.65-mm layer of copper. In subsequent processing, \(85 \%\) of the copper is removed. Copper has a density of \(8.96 \mathrm{~g} / \mathrm{cm}^{3} .\) Calculate the masses of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\) \(\mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) needed to produce the circuit boards, assuming that the reaction used gives a \(97 \%\) yield.

A manufacturer of bicycles has 4815 wheels, 2305 frames, and 2255 handlebars. (a) How many bicycles can be manufactured using these parts? (b) How many parts of each kind are left over? (c) Which part limits the production of bicycles?

(a) What is the difference between adding a subscript 2 to the end of the formula for \(\mathrm{CO}\) to give \(\mathrm{CO}_{2}\) and adding a coefficient in front of the formula to give 2 CO? (b) Is the following chemical equation, as written, consistent with the law of conservation of mass? \(3 \mathrm{Mg}(\mathrm{OH})_{2}(\mathrm{~s})+2 \mathrm{H}_{3} \mathrm{PO}_{4}(a q) \longrightarrow \mathrm{Mg}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+6 \mathrm{H}_{2} \mathrm{O}(l)\) Why or why not?

What is the molecular formula of each of the following compounds? (a) empirical formula \(\mathrm{HCO}_{2}\), molar mass \(=90.0 \mathrm{~g} / \mathrm{mol}\) (b) empirical formula \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\), molar mass \(=88 \mathrm{~g} / \mathrm{mol}\)

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