Cinnamic alcohol is used mainly in perfurmery, particularly in soaps and cosmetics. Its molecular formula is C9H10O.

a. Calculate the percent composition by masss of C, H, and O in cinnamic alcohol.

b. How many molecules of cinnamic alchol, not acid ,are contained in a sample of mass 0.469 grams.

Please explain how to do this problem and provide the correct answers.

I think this has been done once for you by another helper who made a typo and called the alcohol an acid. However, you can do this yourself the following way.

a.
%C = (mass C/molar mass alcohol)*100 = ?
%H = (mass H/molar mass alcohol)*100 = ?
%O = (mass O/molar mass alcohol)*100 = ?

b.
You know that there are 6.022 x 10^23 molecules in a mol (molar mass) of the alcohol. So how many mols are in 0.469 g?

Post your work if you have trouble and PLEASE tell us what you don't understand. That way we can address your question properly.

What is the answer to part b? I got 2.11 x 10^21 molecules of C9H10O. Is this correct?

Cinnamic alcohol is used mainly in perfurmery, particularly in soaps and cosmetics. Its molecular formula is C9H10O.

a. Calculate the percent composition by masss of C, H, and O in cinnamic alcohol.

b. How many molecules of cinnamic alchol, not acid ,are contained in a sample of mass 0.469 grams.

Please explain how to do this problem and provide the correct answers.

General Chemistry - DrBob222, Monday, May 17, 2010 at 5:45pm
I think this has been done once for you by another helper who made a typo and called the alcohol an acid. However, you can do this yourself the following way.
a.
%C = (mass C/molar mass alcohol)*100 = ?
%H = (mass H/molar mass alcohol)*100 = ?
%O = (mass O/molar mass alcohol)*100 = ?

b.
You know that there are 6.022 x 10^23 molecules in a mol (molar mass) of the alcohol. So how many mols are in 0.469 g?

Post your work if you have trouble and PLEASE tell us what you don't understand. That way we can address your question properly.

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a.
I don't understand how to get the molar mass of alcohol.

b. I do not understand how to do the math for any of these questions.

Can you please post the answers and how to do them mathematically so I can understand them.
Thanks for your help.

Thank you I did the first part but I am still confused about how to do b.

Is a mole a molecule then does the # of moles = .469/134.

If so what do I do with the 1 mole 6.022x10 23. What does the proportion look like?

A mole is not a molecule. A mole is a number of particles; specifically, it is 6.022 x 10^23 particles. Yes, the # moles is 0.469/134 = 0.00350 moles.

So if 1 mole (134 grams) contains 6.022 x 10^23 individual molecules, how many molecules will there be in 0.00350 moles?
(If you have trouble thinking in terms of molecules, think of this kind of problem. There are 100 pennies in 1 dollar, how many pennies are in 1/2 dollar).

So do I multiply 6.022 x 10 23. By .00035

Yes, but you have an extra zero that doesn't belong there. Note how the units cancel when you multiply.

0.00350 moles x (6.022 x 10^23 molecules/1 mole) = ?? molecules. The unit we did not want to keep (moles) cancels and the unit we wanted the answer to be in (molecules) is kept.

To calculate the percent composition by mass of C (carbon), H (hydrogen), and O (oxygen) in cinnamic alcohol, we need to determine the relative amounts of each element in the molecule.

a. Percent composition by mass:
1. Start by finding the molar mass of cinnamic alcohol (C9H10O).
- C (carbon) has a molar mass of 12.01 g/mol.
- H (hydrogen) has a molar mass of 1.01 g/mol.
- O (oxygen) has a molar mass of 16.00 g/mol.

Calculate the molar mass of cinnamic alcohol:
(9 * 12.01 g/mol) + (10 * 1.01 g/mol) + (1 * 16.00 g/mol) = 148.25 g/mol

2. Calculate the mass of each element present in 1 mole of cinnamic alcohol.
- Carbon (C): (9 * 12.01 g/mol) = 108.09 g/mol
- Hydrogen (H): (10 * 1.01 g/mol) = 10.10 g/mol
- Oxygen (O): (1 * 16.00 g/mol) = 16.00 g/mol

3. Calculate the percent composition by mass of each element.
- Carbon (C): (mass of C / molar mass of C9H10O) * 100
= (108.09 g/mol / 148.25 g/mol) * 100 ≈ 72.94%

- Hydrogen (H): (mass of H / molar mass of C9H10O) * 100
= (10.10 g/mol / 148.25 g/mol) * 100 ≈ 6.81%

- Oxygen (O): (mass of O / molar mass of C9H10O) * 100
= (16.00 g/mol / 148.25 g/mol) * 100 ≈ 10.80%

Therefore, the percent composition by mass of C, H, and O in cinnamic alcohol are approximately 72.94%, 6.81%, and 10.80% respectively.

b. To calculate the number of molecules of cinnamic alcohol in a sample of mass 0.469 grams, we need to use the concept of molar mass and Avogadro's number.

1. Calculate the number of moles of cinnamic alcohol.
- Moles = Mass (g) / Molar mass (g/mol)
- Moles = 0.469 g / 148.25 g/mol ≈ 0.00316 mol

2. Use Avogadro's number (6.022 x 10^23 molecules/mol) to calculate the number of molecules:
- Number of molecules = Moles of cinnamic alcohol * Avogadro's number
- Number of molecules = 0.00316 mol * (6.022 x 10^23 molecules/mol)
≈ 1.90 x 10^21 molecules

Therefore, there are approximately 1.90 x 10^21 molecules of cinnamic alcohol in a sample of mass 0.469 grams.

You gain proficiency by doing.

The formula is C9H10O.
The molar mass is determined by adding all of the atomic masses together. You can find them on a periodic table.
Here are the approximate numbers but you need to look them up and do it more accurately.
C = 12, then 12 x 9 = 108
H = 1, then 1 x 10 = 10
O = 16, then 16 x 1 = 16
109 + 10 + 16 = approximately 134

%C = (108/134)*100 = about 80.6% ETC.

b.
# moles = grams/molar mass. You know molar mass and you know grams. Calculate moles. Then you know 1 mole contains 6.022 x 10^23 molecules. You can set up a proportion or do it by dimensional analysis.