Answer:
[tex]2.5\; \rm g[/tex] of [tex]\rm CaCO_{3}[/tex] would contain:
- Approximately [tex]1.5 \times 10^{22}[/tex] calcium atoms (approximately [tex]0.025\; \rm mol[/tex],)
- Approximately [tex]1.5 \times 10^{22}[/tex] carbon atoms (approximately [tex]0.025\; \rm mol[/tex],) and
- Approximately [tex]4.5 \times 10^{22}[/tex] oxygen atoms (approximately [tex]0.075\; \rm mol[/tex].)
Explanation:
Look up the Avogadro constant: [tex]N_{\rm A} \approx 6.022 \times 10^{23}\; \rm mol^{-1}[/tex].
For example, "[tex]1\; \rm mol[/tex] of carbon atoms" would contain [tex]N_{\rm A}[/tex] carbon atoms (approximately [tex]6.022\times 10^{23}[/tex]) by definition.
Look up the relative atomic mass of carbon, calcium, and oxygen on a modern periodic table:
- Calcium: [tex]40.078[/tex].
- Carbon: [tex]12.011[/tex].
- Oxygen: [tex]15.999[/tex].
In other words, the mass of [tex]1\; \rm mol[/tex] of calcium atoms would be [tex]40.078\; \rm g[/tex]. The mass of [tex]1\; \rm mol\![/tex] of carbon atoms would be [tex]12.011\; \rm g[/tex], and the mass of [tex]1\; \rm mol \!\![/tex] of oxygen atoms would be [tex]15.999\; \rm g[/tex].
As the formula [tex]\rm CaCO_{3}[/tex] suggests, every formula unit of this ionic compound includes one calcium atom, one carbon atom, and three oxygen atoms. The formula mass of [tex]\rm CaCO_{3}\![/tex] would give the mass of every mole of [tex]\rm CaCO_{3}\!\![/tex] formula units.
Calculate the formula mass of [tex]\rm CaCO_{3}[/tex] from the relative atomic mass data:
[tex]\begin{aligned} & M({\rm CaCO_{3}}) \\ =\; & 40.078\; \rm g \cdot mol^{-1} \\ & + 12.011\; \rm g \cdot mol^{-1} \\ & + 3 \times (15.999\; \rm g \cdot mol^{-1}) \\ =\; & 100.086\; \rm g \cdot mol^{-1}\end{aligned}[/tex].
Calculate the number of [tex]\rm CaCO_{3}[/tex] formula units in that [tex]2.5\; \rm g[/tex] of this compound:
[tex]\begin{aligned}& n({\rm CaCO_{3}}) \\ =\; & \frac{m({\rm CaCO_{3}})}{M({\rm CaCO_{3}})} \\ =\; & \frac{2.5\; \rm g}{100.086\; \rm g \cdot mol^{-1}} \\ \approx\; & 0.025\; \rm mol\end{aligned}[/tex].
In other words, [tex]2.5\; \rm g[/tex] of [tex]\rm CaCO_{3}[/tex] would contain approximately [tex]0.025\; \rm mol[/tex] [tex]\rm CaCO_{3}\![/tex] formula units.
Again, there are one calcium atom, one carbon atom, and one oxygen atom in every [tex]\rm CaCO_{3}[/tex] formula unit. That approximately [tex]0.025\; \rm mol[/tex] [tex]\rm CaCO_{3}\![/tex] formula units would thus contain:
- Approximately [tex]1 \times 0.025\; \rm mol = 0.025\; \rm mol[/tex] calcium atoms,
- Approximately [tex]1 \times 0.025\; \rm mol = 0.025\; \rm mol[/tex] carbon atoms, and
- Approximately [tex]3 \times 0.025\; \rm mol = 0.075\; \rm mol[/tex] oxygen atoms.
Make use of the Avogadro constant to convert the numbers.
For example, the number of calcium atoms in that approximately [tex]0.025\; \rm mol[/tex] of calcium atoms would be:
[tex]\begin{aligned} & N({\text{calcium}) \\ = \; & n({\text{calcium}) \cdot N_{\rm A} \\ \approx \; & 0.025\; \rm mol \times 6.022\times 10^{23} \cdot mol^{-1} \\ \approx \; & 1.5 \times 10^{22} \end{aligned}[/tex].
Likewise, the number of carbon atoms in that approximately [tex]0.025\; \rm mol[/tex] of carbon atoms would be:
[tex]\begin{aligned} & N({\text{carbon}) \\ = \; & n({\text{carbon}) \cdot N_{\rm A} \\ \approx \; & 0.025\; \rm mol \times 6.022\times 10^{23} \cdot mol^{-1} \\ \approx \; & 1.5 \times 10^{22} \end{aligned}[/tex].
The number of oxygen atoms in that approximately [tex]0.075\; \rm mol[/tex] of oxygen atoms would be:
[tex]\begin{aligned} & N({\text{oxygen}) \\ = \; & n({\text{oxygen}) \cdot N_{\rm A} \\ \approx \; & 0.075\; \rm mol \times 6.022\times 10^{23} \cdot mol^{-1} \\ \approx \; & 4.5 \times 10^{22} \end{aligned}[/tex].
