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3 years ago

Aluminum costs about 85 cents per pound (453.592 g). How much is one atom of aluminum?

Chemistry

1 answer:

kakasveta [241]3 years ago

5 0

One aluminum atom will cost 0.84 × 10⁻²³ cents.

Explanation:

Aluminium costs 85 cents per 453.592 grams.

number of moles = mass / atomic weight

number of moles of aluminium = 453.592 / 27 = 16.8 moles

To calculate the number of aluminum atoms we use the Avogadro's number:

if 1 mole of aluminum contains 6.022 × 10²³ aluminum atoms

then 16.8 moles of aluminum contains X aluminum atoms

X = (16.8 × 6.022 × 10²³) / 1 = 101.17 × 10²³ aluminum atoms

Now, taking in account the aluminium price, we devise the following reasoning:

if 101.17 × 10²³ aluminum atoms costs 85 cents

then 1 aluminum atom costs Y cents

Y = (1 × 85) / (101.17 × 10²³) = 0.84 × 10⁻²³ cents

Learn more about:

Avogadro's number

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$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$ .

Assuming complete decomposition of both samples,

$m(\text{Hg}) = m(\text{residure})$
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First compound:

$m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g$

$n = m/M$ ; $0.6498 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.048 \; g / 16 \; g \cdot mol^{-1}= 0.003 \; mol$
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Oxygen and mercury atoms seemingly exist in the first compound at a $1:1$ ratio; thus the empirical formula for this compound would be $\text{Hg} \text{O}$ where the subscript "1" is omitted.

Similarly, for the second compound

$m(\text{O}) = m(\text{loss}) = 0.016 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012 \; g$

$n = m/M$ ; $0.4172 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.016 \; g / 16 \; g \cdot mol^{-1}= 0.001 \; mol$
$n(\text{Hg atoms}) = 0.4012 \; g / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol$

$n(\text{Hg}) : n(\text{O}) \approx 2:1$ and therefore the empirical formula

$\text{Hg}_{2} \text{O}$ .

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