Question

Question #2d8b3

Answer 1

Convert your mass to moles, extract the molar percentage of the element of interest, and then convert those moles to atoms.

Explanation:

`(grams)/("grams"/"mole") = "moles"`

`"moles" xx "units"/"Mole" = "units"`

`"units sugar" xx (12 "C atoms")/"unit sugar" = "C atoms"`

The first one needs to have the molecular weight `("grams"/"mole")` of the sugar. You get that by adding up the atomic weights of each element.
`C = 12 xx 12 = 144`
`O = 16 xx 11 = 176`
`H = 1 xx 22 = 22`
TOTAL = `342 g/(mol)`

The second part needs Avogadro's Number (units/mole),
`6.022 xx 10^23`

Followed by the percentage of the element of interest (C) in the whole.
`"moles sugar" xx (12 "C atoms")/"mole sugar"`

EXAMPLE Worked Problem:
How many atoms of oxygen are in 3.9g of methanol (`CH_3OH`)?

Molecular weight of methanol:
`C = 12 xx 1 = 12`
`O = 16 xx 1 = 16`
`H = 1 xx 4 = 4`
TOTAL = `32 g/(mol)`

`3.9g/(32"g/mol") = 0.122` mol methanol

`0.122 mol xx 6,022 x 10^23 = 7.34 xx 10^22` `CH_3OH` molecules (unit)
`(1"C atom")/(unit) xx 7.34 xx 10^22` units `= 7.34 xx 10^22` C atoms