Question #2d8b3

1 Answer
SCooke
Nov 5, 2017

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"#

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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

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