# Question 2d8b3

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"#

$\text{moles" xx "units"/"Mole" = "units}$

$\text{units sugar" xx (12 "C atoms")/"unit sugar" = "C atoms}$

The first one needs to have the molecular weight $\left(\text{grams"/"mole}\right)$ of the sugar. You get that by adding up the atomic weights of each element.
$C = 12 \times 12 = 144$
$O = 16 \times 11 = 176$
$H = 1 \times 22 = 22$
TOTAL = $342 \frac{g}{m o l}$

The second part needs Avogadro's Number (units/mole),
$6.022 \times {10}^{23}$

Followed by the percentage of the element of interest (C) in the whole.
$\text{moles sugar" xx (12 "C atoms")/"mole sugar}$

EXAMPLE Worked Problem:
How many atoms of oxygen are in 3.9g of methanol ($C {H}_{3} O H$)?

Molecular weight of methanol:
$C = 12 \times 1 = 12$
$O = 16 \times 1 = 16$
$H = 1 \times 4 = 4$
TOTAL = $32 \frac{g}{m o l}$

$3.9 \frac{g}{32 \text{g/mol}} = 0.122$ mol methanol

$0.122 m o l \times 6 , 022 x {10}^{23} = 7.34 \times {10}^{22}$ $C {H}_{3} O H$ molecules (unit)
$\frac{1 \text{C atom}}{u n i t} \times 7.34 \times {10}^{22}$ units $= 7.34 \times {10}^{22}$ C atoms