/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 92 The koala dines exclusively on e... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 92

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 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 of eucalyptol is \(C_{10}H_{18}O\), and its molecular formula is also \(C_{10}H_{18}O\), as the empirical formula mass (154.26 g/mol) is almost equal to the given peak mass (154 amu).

Step by step solution

01

Calculate moles of each element

To determine the moles of each element, we will first assume that we have 100 grams of eucalyptol. This allows us to interpret the percentages as the mass of each element in grams. For Carbon (\(C\)): 77.87 g For Hydrogen (\(H\)): 11.76 g For Oxygen (\(O\)): (100 - (77.87 + 11.76)) g = 10.37 g Using the molar masses of each element, we can calculate their respective moles. Moles of \(C\) = \(\frac{77.87 \ \mathrm{g}}{12.01 \ \mathrm{g/mol}}\) = 6.485 moles Moles of \(H\) = \(\frac{11.76 \ \mathrm{g}}{1.008 \ \mathrm{g/mol}}\) = 11.667 moles Moles of \(O\) = \(\frac{10.37 \ \mathrm{g}}{16.00 \ \mathrm{g/mol}}\) = 0.648 moles
02

Determine empirical formula

To find the empirical formula, we will divide the moles of each element by the smallest number of moles calculated (in this case, it is moles of \(O\)) to get the smallest whole number ratio. Ratio of \(C: H: O\) = \(\frac{6.485}{0.648} : \frac{11.667}{0.648} : \frac{0.648}{0.648}\) 鈮� 10 : 18 : 1 Thus, the empirical formula for eucalyptol is \(C_{10}H_{18}O\).
03

Determine molecular formula

We are given that the peak in the mass spectrum for eucalyptol is 154 amu. First, let's calculate the mass of the empirical formula: Empirical formula mass = \(10 \times 12.01\ \mathrm{g/mol}\ (C) + 18 \times 1.008\ \mathrm{g/mol}\ (H) + 1 \times 16.00\ \mathrm{g/mol}\ (O) = 154.26\ \mathrm{g/mol}\) The empirical formula mass is almost equal to the given peak mass (154 amu). This indicates that the empirical and molecular formulas are the same, since there is no whole number multiple necessary to match the given mass. Therefore, the molecular formula of eucalyptol is also \(C_{10}H_{18}O\).

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Most popular questions from this chapter

Aluminum sulfide reacts with water to form aluminum hydroxide and hydrogen sulfide. (a) Write the balanced chemical equation for this reaction. (b) How many grams of aluminum hydroxide are obtained from \(14.2 \mathrm{~g}\) of aluminum sulfide?

(a) Determine the chemical formula of the product formed when the metallic element aluminum combines with the nonmetallic element bromine, \(\mathrm{Br}_{2}\). Write the balanced chemical equation for the reaction. (b) What products form when a compound containing \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O}\) is completely combusted in air? Write a balanced chemical equation for the combustion of acetone, \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}(l),\) in air.

Hydrogen cyanide, HCN, is a poisonous gas. The lethal dose is approximately \(300 \mathrm{mg}\) HCN per kilogram of air when inhaled. (a) Calculate the amount of HCN that gives the lethal dose in a small laboratory room measuring \(12 \times 15 \times 8.0 \mathrm{ft}\). The density of air at \(26^{\circ} \mathrm{C}\) is \(0.00118 \mathrm{~g} / \mathrm{cm}^{3}\). (b) If the \(\mathrm{HCN}\) is formed by reaction of \(\mathrm{NaCN}\) with an acid such as \(\mathrm{H}_{2} \mathrm{SO}_{4}\), what mass of \(\mathrm{NaCN}\) gives the lethal dose in the room? \(2 \mathrm{NaCN}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{HCN}(g)\) (c) HCN forms when synthetic fibers containing Orlon \(^{\oplus}\) or Acrilan \(^{\otimes}\) burn. Acrilan \(^{\oplus}\) has an empirical formula of \(\mathrm{CH}_{2} \mathrm{CHCN},\) so \(\mathrm{HCN}\) is \(50.9 \%\) of the formula by mass. \(\mathrm{A}\) rug measures \(12 \times 15 \mathrm{ft}\) and contains 30 oz of Acrilan \(^{\otimes}\) fibers per square yard of carpet. If the rug burns, will a lethal dose of HCN be generated in the room? Assume that the yield of HCN from the fibers is \(20 \%\) and that the carpet is \(50 \%\) consumed.

The compound \(\mathrm{XCl}_{4}\) contains \(75.0 \% \mathrm{Cl}\) by mass. What is the element \(\mathrm{X} ?\)

The fizz produced when an Alka-Seltzer \(^{\circledast}\) tablet is dissolved in water is due to the reaction between sodium bicarbonate \(\left(\mathrm{NaHCO}_{3}\right)\) and citric acid \(\left(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}\right)\) $$ \begin{aligned} 3 \mathrm{NaHCO}_{3}(a q)+\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(a q) \longrightarrow \\ & 3 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Na}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(a q) \end{aligned} $$ In a certain experiment \(1.00 \mathrm{~g}\) of sodium bicarbonate and \(1.00 \mathrm{~g}\) of citric acid are allowed to react. (a) Which is the limiting reactant? (b) How many grams of carbon dioxide form? (c) How many grams of the excess reactant remain after the limiting reactant is completely consumed?
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