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Question #2d8b3

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Answer 1 · 2017-11-05T03:06:12

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

Explanation:

$\frac{g r a m s}{\frac{grams}{mole}} = moles$ $(grams)/("grams"/"mole") = "moles"$

$moles \times \frac{units}{Mole} = units$ $"moles" xx "units"/"Mole" = "units"$

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

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

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

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

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EXAMPLE Worked Problem:
How many atoms of oxygen are in 3.9g of methanol ( $C H_{3} O H$ $CH_3OH$ )?

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

$3.9 \frac{g}{32 g/mol} = 0.122$ $3.9g/(32"g/mol") = 0.122$ mol methanol

$0.122 m o l \times 6, 022 x 10^{23} = 7.34 \times 10^{22}$ $0.122 mol xx 6,022 x 10^23 = 7.34 xx 10^22$ $C H_{3} O H$ $CH_3OH$ molecules (unit)
$\frac{1 C atom}{u n i t} \times 7.34 \times 10^{22}$ $(1"C atom")/(unit) xx 7.34 xx 10^22$ units $= 7.34 \times 10^{22}$ $= 7.34 xx 10^22$ C atoms

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Question #2d8b3

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