(grams)/("grams"/"mole") = "moles"gramsgramsmole=moles
"moles" xx "units"/"Mole" = "units"moles×unitsMole=units
"units sugar" xx (12 "C atoms")/"unit sugar" = "C atoms"units sugar×12C atomsunit sugar=C atoms
The first one needs to have the molecular weight ("grams"/"mole")(gramsmole) of the sugar. You get that by adding up the atomic weights of each element.
C = 12 xx 12 = 144C=12×12=144
O = 16 xx 11 = 176O=16×11=176
H = 1 xx 22 = 22H=1×22=22
TOTAL = 342 g/(mol)342gmol
The second part needs Avogadro's Number (units/mole),
6.022 xx 10^236.022×1023
Followed by the percentage of the element of interest (C) in the whole.
"moles sugar" xx (12 "C atoms")/"mole sugar"moles sugar×12C atomsmole sugar
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EXAMPLE Worked Problem:
How many atoms of oxygen are in 3.9g of methanol (CH_3OHCH3OH)?
Molecular weight of methanol:
C = 12 xx 1 = 12C=12×1=12
O = 16 xx 1 = 16O=16×1=16
H = 1 xx 4 = 4H=1×4=4
TOTAL = 32 g/(mol)32gmol
3.9g/(32"g/mol") = 0.1223.9g32g/mol=0.122 mol methanol
0.122 mol xx 6,022 x 10^23 = 7.34 xx 10^220.122mol×6,022x1023=7.34×1022 CH_3OHCH3OH molecules (unit)
(1"C atom")/(unit) xx 7.34 xx 10^221C atomunit×7.34×1022 units = 7.34 xx 10^22=7.34×1022 C atoms
