4.75 Part A

[Isabella Sanzi 2E]

Hi guys,
I know this question is kind of throwing it back to the beginning of the year, but I am confused on part A of question 75 in chapter 4:
An organic compound distilled from wood was found to have a mole mass of 32.04 grams/mol and the following composition by mass: 37.5% C, 12.6% H, and 49.9% O.
A: Write a lewis structure of the compound and determine the bond angles about the carbon and oxygen atoms.
So in trying to find the formula for the compound, I first multiplied each percentage (in decimal form) by the molar mass of the element given and found that there were (32.04 x .375) 12 carbons, 4 hydrogens (32.04 x 0.126) , and 16 oxygens (32.04 x 0.499) in the molecular formula. Then to find the empirical formula at this point I would divide by the smallest number (in this case 4) to get the formula: C3HO4, but the solution manual says that the empirical formula is CH4O, but wouldn't that imply that there was four times as many hydrogen in the molecule than carbon or oxygen (which does not make sense because of the given percentages)? How did they get this as the empirical formula? If anyone can enlighten me on this I would very much appreciate it. Thanks!

[Lena Nguyen 2H]

When you multiplied the molar mass of 32.04 g/mol by the percentage of carbon, hydrogen, and oxygen, the numbers you get are the amount in grams of each element. To get the number of moles, you have to divide the grams by the molar masses of each element. Then you should get the correct molecular formula.

For example,
32.04 g * .375 = 12.015 g C
12.015 g C / 12.014 g/mol C = 1 mol C

[Yijia_Yang_3A]

same as the last post. when you multiply the total mass by the percentage of each element, you have to take what you end up with and divide that by its molar mass, which can be found on the periodic table.

[105405231]

Adding to the previous posts, match the number you get after dividing the molar mass by the mass of the element to the atomic weight of the element on the periodic table.

[Nathaly Cruz 2D]

I know this is an old post, but I was struggling and it really helped to look at how others have helped each other in the past. I'm super thankful for chemistry community!

[Mariah Disc 2C]

I am so glad I stumbled upon this old question as I was struggling with this topic as well! Thank you :)

[Rishab_Haldar_3B]

So happy I stumbled on this post, I was struggling with the same question - thank you!