Understanding the Problem
To find the number of carbon atoms in a specific mass of a compound like calcium carbonate ($CaCO_3$), we need to connect the macroscopic mass of the sample to the microscopic count of atoms. We use the concept of the mole and Avogadro's constant as our bridge.
Key Concepts
- Molar Mass ($M$): The mass of one mole of a substance, calculated by summing the atomic masses of all atoms in the formula.
- Mole ($n$): A unit representing $6.022 \times 10^{23}$ particles.
- Avogadro's Number ($N_A$): $6.022 \times 10^{23} \text{ atoms/mol}$.
Step-by-Step Solution
1. Calculate the Molar Mass of $CaCO_3$
First, we look up the atomic masses of the elements from the periodic table:
- $Ca = 40.08 \text{ g/mol}$
- $C = 12.01 \text{ g/mol}$
- $O = 16.00 \text{ g/mol}$
Now, sum them for $CaCO_3$: $$M = 40.08 + 12.01 + (3 \times 16.00) = 100.09 \text{ g/mol}$$ (We can approximate this to $100 \text{ g/mol}$ for simplicity.)
2. Calculate the Number of Moles in 25g
Using the formula $n = \frac{\text{mass}}{\text{molar mass}}$: $$n = \frac{25 \text{ g}}{100 \text{ g/mol}} = 0.25 \text{ moles of } CaCO_3$$
3. Determine the Number of Carbon Atoms
In every 1 molecule of $CaCO_3$, there is exactly 1 carbon atom. Therefore, in 1 mole of $CaCO_3$, there is 1 mole of carbon atoms.
- Moles of Carbon = $0.25 \text{ moles}$
To find the actual number of atoms, multiply by Avogadro's constant ($N_A$): $$\text{Number of atoms} = n \times N_A$$ $$\text{Number of atoms} = 0.25 \times 6.022 \times 10^{23}$$ $$\text{Number of atoms} = 1.5055 \times 10^{23} \text{ carbon atoms}$$
Summary
By following these steps, we determined that there are approximately $1.51 \times 10^{23}$ carbon atoms present in $25\text{g}$ of calcium carbonate. This method works for any compound: find the moles, look at the ratio of atoms in the formula, and multiply by Avogadro's number.