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Chapter 3: Problem 39

Cinnamic alcohol is used mainly in perfumery, particularly in soaps and cosmetics. Its molecular for\(\begin{array}{lll}\text { mula is } & \text { C }_{9} \text { H }_{10} \text { O. }\end{array}\) (a) Calculate the percent composition by mass of \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O}\) in cinnamic alcohol. (b) How many molecules of cinnamic alcohol are contained in a sample of mass \(0.469 \mathrm{~g} ?\)

Short Answer

Expert verified
The mass percentage compositions of each element in cinnamic alcohol are: 72.73 \% C, 8.56 \% H, and 18.71 \% O. The 0.469 g sample of cinnamic alcohol contains approximately \(2.4 * 10^{21}\) molecules.

Step by step solution

01

Finding the molar mass of cinnamic alcohol

First, find the molar mass of cinnamic alcohol using the molecular formula \(C_{9}H_{10}O\) and the atomic masses of Carbon (C), Hydrogen (H), and Oxygen (O). The molar mass is calculated as follows: \(9 * MolarMass(C) + 10 * MolarMass(H) + 1 * MolarMass(O)\). The atomic masses of C, H and O are approximately 12.01 g/mol, 1.008 g/mol and 16.00 g/mol respectively. Plugging these values and calculating yields the molar mass as \(118.18\) g/mol.
02

Calculating the percent composition

The percent by mass of each element is calculated by dividing the total mass contributed by the element in the molecule by the molar mass of the molecule, and then multiplying by 100. For Carbon: \((9 * MolarMass(C)/ MolarMass(C9H10O)) * 100\%. For Hydrogen: \((10 * MolarMass(H) / MolarMass(C9H10O)) *100\%. And for Oxygen: \(MolarMass(O) / MolarMass(C9H10O) *100\) \%. Inputting the values for each element and calculating yields the mass percentage of Carbon as 72.73 \%, Hydrogen as 8.56 \% and Oxygen as 18.71 \%.
03

Calculating the number of molecules

To find the number of molecules in the given mass of cinnamic alcohol we first find the number of moles of cinnamic alcohol in 0.469 g using the equation Moles = mass/molar mass, yielding the moles of cinnamic alcohol in 0.469 g as 0.004 moles. We then multiply the moles of cinnamic alcohol by Avogadro's number \(6.022 * 10^{23}\) to find the number of molecules. The calculated result is approximately \(2.4 * 10^{21}\) molecules.
04

Final Conclusions

The percent composition by mass of \(C_{9}H_{10}O\) is approximately 72.73 \% C, 8.56 \% H and 18.71 \% O. The number of molecules of cinnamic alcohol in 0.469 g is approximately \(2.4 * 10^{21}\) molecules.

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