Question

Calculate the number of each atom in 2.5 gram of caco3 ​

Answer 1

`2.5\; \rm g` of
`\rm CaCO_(3)` would contain:

• Approximately
  `1.5 * 10^(22)` calcium atoms (approximately
  `0.025\; \rm mol`,)
• Approximately
  `1.5 * 10^(22)` carbon atoms (approximately
  `0.025\; \rm mol`,) and
• Approximately
  `4.5 * 10^(22)` oxygen atoms (approximately
  `0.075\; \rm mol`.)

Step-by-step explanation:

Look up the Avogadro constant:
`N_(\rm A) \approx 6.022 * 10^(23)\; \rm mol^(-1)`.

For example, "
`1\; \rm mol` of carbon atoms" would contain
`N_(\rm A)` carbon atoms (approximately
`6.022* 10^(23)`) by definition.

Look up the relative atomic mass of carbon, calcium, and oxygen on a modern periodic table:

• Calcium:
  `40.078`.
• Carbon:
  `12.011`.
• Oxygen:
  `15.999`.

In other words, the mass of
`1\; \rm mol` of calcium atoms would be
`40.078\; \rm g`. The mass of
`1\; \rm mol\!` of carbon atoms would be
`12.011\; \rm g`, and the mass of
`1\; \rm mol \!\!` of oxygen atoms would be
`15.999\; \rm g`.

As the formula
`\rm CaCO_(3)` suggests, every formula unit of this ionic compound includes one calcium atom, one carbon atom, and three oxygen atoms. The formula mass of
`\rm CaCO_(3)\!` would give the mass of every mole of
`\rm CaCO_(3)\!\!` formula units.

Calculate the formula mass of
`\rm CaCO_(3)` from the relative atomic mass data:

`\begin{aligned} & M({\rm CaCO_(3)}) \\ =\; & 40.078\; \rm g \cdot mol^(-1) \\ & + 12.011\; \rm g \cdot mol^(-1) \\ & + 3 * (15.999\; \rm g \cdot mol^(-1)) \\ =\; & 100.086\; \rm g \cdot mol^(-1)\end{aligned}`.

Calculate the number of
`\rm CaCO_(3)` formula units in that
`2.5\; \rm g` of this compound:

`\begin{aligned}& n({\rm CaCO_(3)}) \\ =\; & \frac{m({\rm CaCO_(3)})}{M({\rm CaCO_(3)})} \\ =\; & (2.5\; \rm g)/(100.086\; \rm g \cdot mol^(-1)) \\ \approx\; & 0.025\; \rm mol\end{aligned}`.

In other words,
`2.5\; \rm g` of
`\rm CaCO_(3)` would contain approximately
`0.025\; \rm mol`
`\rm CaCO_(3)\!` formula units.

Again, there are one calcium atom, one carbon atom, and one oxygen atom in every
`\rm CaCO_(3)` formula unit. That approximately
`0.025\; \rm mol`
`\rm CaCO_(3)\!` formula units would thus contain:

• Approximately
  `1 * 0.025\; \rm mol = 0.025\; \rm mol` calcium atoms,
• Approximately
  `1 * 0.025\; \rm mol = 0.025\; \rm mol` carbon atoms, and
• Approximately
  `3 * 0.025\; \rm mol = 0.075\; \rm mol` oxygen atoms.

Make use of the Avogadro constant to convert the numbers.

For example, the number of calcium atoms in that approximately
`0.025\; \rm mol` of calcium atoms would be:

`\begin{aligned} & N({\text{calcium}) \\ = \; & n({\text{calcium}) \cdot N_(\rm A) \\ \approx \; & 0.025\; \rm mol * 6.022* 10^(23) \cdot mol^(-1) \\ \approx \; & 1.5 * 10^(22) \end{aligned}`.

Likewise, the number of carbon atoms in that approximately
`0.025\; \rm mol` of carbon atoms would be:

`\begin{aligned} & N({\text{carbon}) \\ = \; & n({\text{carbon}) \cdot N_(\rm A) \\ \approx \; & 0.025\; \rm mol * 6.022* 10^(23) \cdot mol^(-1) \\ \approx \; & 1.5 * 10^(22) \end{aligned}`.

The number of oxygen atoms in that approximately
`0.075\; \rm mol` of oxygen atoms would be:

`\begin{aligned} & N({\text{oxygen}) \\ = \; & n({\text{oxygen}) \cdot N_(\rm A) \\ \approx \; & 0.075\; \rm mol * 6.022* 10^(23) \cdot mol^(-1) \\ \approx \; & 4.5 * 10^(22) \end{aligned}`.