C3H7OH is propanol. Suppose you have 85.9 g sample of propanol. How many carbon atoms are in the sample?

The molecular formula of propanol is #C_3 H_8 O#.
First, we should calculate its molecular mass. I prefer to work with rounded numbers because that's how I was taught and frankly it makes life much easier. You should have memorised that the molar masses of carbon, hydrogen and oxygen are, respectively, #12g/(mol)#, #1g/(mol)# and #16g/(mol)#.
So let's add all that up:

#M = (3*12)+(8*1)+(1*16)#
#M = 60g/(mol)#

Okay, but that doesn't tell us much by itself. Elements make up substances in different proportions, because they have different masses, so how much of propanol is actually carbon, in terms of percentage? Remember, #36g# of that is carbon:

#{:(60g -> 100%), (36g -> x%) :}} implies x = 60%#

So now we know that any given sample of propanol is sixty percent carbon. Now we can use the information that we were given: #85.9g#. If that's the total mass of propanol, sixty percent of that must be carbon:

#{: (85.9g -> 100%), (ycolor(white)(x)g -> 60%) :}} implies y = 51.54g#

Now, let us return to the concept of a mole. A mole is a quantity of #6*10^23# of something. You might be used to using a more accurate value, but for now let us use that simplified number. Molar mass is the mass of a mole of something; so #6*10^23# atoms of carbon have a total mass of #12g#. With that in mind, we can now work out how many atoms of carbon have a total mass of #51.54g#:

#{:(6*10^23 "atoms" -> 12g), (zcolor(white)(x) "atoms" -> 51.54g) :}} implies z = 2.58*10^24 "atoms"#

My answer might be different from somebody else's because, as I said, I prefer to work with simpler values. I hope, however, that the logic is clear! Cheers.

Propanol: #"C"_3"H"_7"OH"##=##"C"_3"H"_8"O"#

The molar masses of carbon, hydrogen, and oxygen are:
#"C"="12.0107 g/mol"#
#"H"="1.00794 g/mol"#
#"O"="15.999 g/mol"#

#"Molar mass of propanol"=(3xx12.0107)+(8xx1.00794)+(1xx15.999)="60.095 g/mol"#.

Identify the Molar Masses of Propanol and Carbon

Molar mass of propanol: #"60.095 g/mol"#
Molar mass of carbon: #(3xx12.0107)="36.032 g/mol"#

Determine the Percentage of Carbon

#"Percentage of C in propanol"=("molar mass of C")/("molar mass of propanol")xx100=(36.032"g/mol C")/(60.095"g/mol propanol")xx100="59.958% C"#

Determine the Mass of Carbon in 85.9 g of Propanol

In order to determine the mass of carbon present in 85.9 g propanol, we need to multiply the mass of propanol times the percentage of carbon as a decimal (e.g. #(25%)/(100)=0.25#).

#"85.9 g propanol"xx(0.59958 "C")="51.504 g C"#

There are #"51.504 g C"# in #"85.9 g propanol"#

Determining the Number of Carbon Atoms

Use the mass of carbon in the compound and carbon's molar mass to convert to moles of carbon. Then multiply the moles of carbon times #6.022xx10^23 "atoms/mol"#.

#51.504cancel"g C"xx(1cancel"mol C")/(12.0107cancel"g C")xx(6.022xx10^23"atoms C")/(1cancel"mol C")=2.58xx10^24"atoms C"# rounded to three significant figures.