How many atoms of carbon are present in 4.50kg CH3COOH?

Question

How do you find the answer to this question? Also, my teacher is really concerned with the class learning how to write problems like this out in parentheses....how do I do this??

Answer 1

eventually you want to end up with atoms as the units, so you'll have a parenthesis with avogadro's number in the numerator: (6.022*10^23 atoms/mol)

now you see you need to get rid of moles from denominator, so multiply by number of moles contained in 4.50kg of CH3COOH. This is found by dividing mass by molecular weight (or multiplying by its reciprocal):

molecular weight of acetic acid is
12.01*2 + 1.00794*4 + 15.9994 * 2 = 60.05 g / mol

moles = (4.50 kg) (1 mol / 60.05g)
now get rid of kg by converting into grams:
moles = (4.50 kg) (1 mol / 60.05g) ( 1000g/1kg)

However there are 2 moles of carbon in each mole of acetic acid, so there is a factor of 2 ( it's really mol C / mol acetic acid, but you can ignore that)

Put all the factors together:

(4.50 kg) (1 mol / 60.05g) ( 1000g/1kg)(6.022*10^23 atoms/mol) (2) = 9.03e+25 atoms

kg, g, and mol cancel out, and the units are atoms

Notice that you start out with the mass (or whatever is given) in a numerator position, and end up with another set of units in the numerator position. So the whole process is the conversion from one to another and placing these conversion-factor parentheses the right way (rather than upside-down).

Answer 2

Atomic weights: H=1 C=12 O=16 HC2H3O2=60

Let HC2H3O2 be abbreviated HAc. There are two C's in a molecule.

4.50kgHAc x 1000gHAc/1kgHAc x 1molHAc/60gHAc x 2molC/1molHAc x 6.02x10^23atomsC/1molC = 9.03 x 10^24 atoms C