Question

how many moles of s are present in 3.6 × 10²⁴ formula units of al₂s₃?
path: particle al₂s₃ → mol al₂s₃ → mol s
avogadros number internal ratio
? moles s
hint: which of these internal ratios correctly allows units to cancel?
(1 mol al₂s₃)/(3 mol s) or (3 mol s)/(1 mol al₂s₃)

Explanation:

Step1: Convert formula units to moles of \( \text{Al}_2\text{S}_3 \)

Use Avogadro's number (\( 6.022 \times 10^{23} \) formula units/mol) to convert the given formula units of \( \text{Al}_2\text{S}_3 \) to moles.
The formula is \( \text{Moles of } \text{Al}_2\text{S}_3 = \frac{\text{Formula units of } \text{Al}_2\text{S}_3}{\text{Avogadro's number}} \)
Substituting the values: \( \text{Moles of } \text{Al}_2\text{S}_3 = \frac{3.6 \times 10^{24}}{6.022 \times 10^{23}} \approx 5.98 \, \text{mol (or approximately 6 mol for simplicity, but we'll use the precise value here)}. \)

Step2: Convert moles of \( \text{Al}_2\text{S}_3 \) to moles of \( \text{S} \)

From the formula \( \text{Al}_2\text{S}_3 \), the internal ratio of \( \text{S} \) to \( \text{Al}_2\text{S}_3 \) is \( \frac{3 \, \text{mol S}}{1 \, \text{mol Al}_2\text{S}_3} \) (since each mole of \( \text{Al}_2\text{S}_3 \) contains 3 moles of \( \text{S} \)).
Multiply the moles of \( \text{Al}_2\text{S}_3 \) by this ratio:
\( \text{Moles of S} = \text{Moles of } \text{Al}_2\text{S}_3 \times \frac{3 \, \text{mol S}}{1 \, \text{mol Al}_2\text{S}_3} \)
Substituting the moles of \( \text{Al}_2\text{S}_3 \) from Step 1:
\( \text{Moles of S} = \frac{3.6 \times 10^{24}}{6.022 \times 10^{23}} \times \frac{3 \, \text{mol S}}{1 \, \text{mol Al}_2\text{S}_3} \)

First, calculate \( \frac{3.6 \times 10^{24}}{6.022 \times 10^{23}} \approx 5.98 \, \text{mol (of } \text{Al}_2\text{S}_3\text{)} \).
Then multiply by 3: \( 5.98 \times 3 \approx 17.94 \, \text{mol} \).

(Alternatively, combining the steps for simplicity:
\( \text{Moles of S} = \frac{3.6 \times 10^{24} \times 3}{6.022 \times 10^{23}} \)
Calculate numerator: \( 3.6 \times 10^{24} \times 3 = 1.08 \times 10^{25} \)
Denominator: \( 6.022 \times 10^{23} \)
Then \( \frac{1.08 \times 10^{25}}{6.022 \times 10^{23}} \approx \frac{1080}{6.022} \approx 17.93 \, \text{mol} \), which rounds to approximately 18 mol, or more precisely ~17.9 mol.)

Answer:

Approximately \( \boldsymbol{18} \) moles (or more precisely \( \approx 17.9 \) moles) of \( \text{S} \) are present.

(Note: If we use \( 6.02 \times 10^{23} \) for Avogadro's number for simplicity:
\( \frac{3.6 \times 10^{24}}{6.02 \times 10^{23}} \approx 5.98 \, \text{mol Al}_2\text{S}_3 \), then \( 5.98 \times 3 \approx 17.94 \, \text{mol S} \), which is approximately 18 mol.)