Accepted answer
Answer included
Key idea for atoms to moles
The conversion from atoms to moles uses Avogadro’s number (Avogadro’s constant), \(N_A = 6.022 \times 10^{23}\ \mathrm{mol^{-1}}\), which means \(6.022 \times 10^{23}\) atoms correspond to \(1\ \mathrm{mol}\) of atoms.
Two inverse relationships control the entire topic:
\[ n = \frac{N}{N_A} \qquad\text{and}\qquad N = n \times N_A \]
where \(N\) is the number of atoms (a count) and \(n\) is the amount in moles.
Worked example 1: \(1.50 \times 10^{24}\) atoms of Al to moles
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Identify given and desired quantities.
Given \(N = 1.50 \times 10^{24}\) atoms Al and \(N_A = 6.022 \times 10^{23}\ \mathrm{mol^{-1}}\). Desired \(n\) in moles. - Use the atoms to moles formula. \[ n = \frac{N}{N_A} = \frac{1.50 \times 10^{24}}{6.022 \times 10^{23}}\ \mathrm{mol} \]
- Compute the power-of-ten part and the decimal part. \[ \frac{10^{24}}{10^{23}} = 10^{1} \] \[ n = \left(\frac{1.50}{6.022}\right) \times 10^{1}\ \mathrm{mol} = 0.249 \times 10^{1}\ \mathrm{mol} = 2.49\ \mathrm{mol} \]
- Significant figures. \(1.50 \times 10^{24}\) has 3 significant figures, so the result is reported as \(2.49\ \mathrm{mol}\).
Worked example 2: \(0.250\ \mathrm{mol}\) of Al to atoms
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Identify given and desired quantities.
Given \(n = 0.250\ \mathrm{mol}\). Desired \(N\) in atoms. - Use the moles to atoms formula. \[ N = n \times N_A = 0.250 \times (6.022 \times 10^{23}) \]
- Multiply and report in scientific notation. \[ N = (0.250 \times 6.022)\times 10^{23} = 1.5055 \times 10^{23} \approx 1.51 \times 10^{23}\ \text{atoms} \]
Summary table
| Conversion goal | Relationship | Example input | Example result |
|---|---|---|---|
| Atoms to moles | \(n = \dfrac{N}{N_A}\) | \(N = 1.50 \times 10^{24}\) atoms Al | \(n = 2.49\ \mathrm{mol}\) |
| Moles to atoms | \(N = n \times N_A\) | \(n = 0.250\ \mathrm{mol}\) Al | \(N = 1.51 \times 10^{23}\) atoms |
Visualization: the two-way “count ⇄ amount” map
Final answers
\(1.50 \times 10^{24}\) atoms of Al correspond to \(2.49\ \mathrm{mol}\) of Al, and \(0.250\ \mathrm{mol}\) of Al contains \(1.51 \times 10^{23}\) atoms.
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