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Chapter 3: Problem 41

Cinnamic alcohol is used mainly in perfumery, particularly in soaps and cosmetics. Its molecular formula is \(\mathrm{C}_{9} \mathrm{H}_{10} \mathrm{O}\) (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
a) The percent composition by mass of Cinnamic alcohol: C: 80.5%, H: 7.54%, O: 11.9%. b) There are approximately \(2.10 x 10^{21}\) molecules of cinnamic alcohol in a 0.469 g sample.

Step by step solution

01

Calculate the molecular weight of C9H10O

The molecular weight can be calculated by multiplying the number of atoms of each type by their atomic weights and adding up these values. Atomic weights for C (Carbon), H (Hydrogen) and O (Oxygen) are approximately 12.01 g/mol, 1.01 g/mol, and 16.00 g/mol respectively. Molecular weight of C9H10O = (9 * 12.01 g/mol) + (10 * 1.01 g/mol) + (16.00 g/mol) = 134.15 g/mol.
02

Calculate percent composition by mass

Percent composition by mass is calculated as the mass contribution of element divided by total mass of compound multiplied by 100. Thus percent by mass of C = ((9 * 12.01) / 134.15) * 100 = 80.5%, H = ((10 * 1.01) / 134.15) * 100 = 7.54%, and O = (16 / 134.15) * 100 = 11.9%.
03

Calculate number of molecules in 0.469 g sample

To find the number of molecules, moles are first found by dividing mass by molecular weight. Then multiply the moles by Avogadro's number (6.022 x 10^23). The number of molecules = (0.469 g / 134.15 g/mol) * (6.022 x 10^23 molecules/mol) = approximately 2.10 x 10^21 molecules.

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