Final answer:
By calculating the difference in mass before and after writing with chalk and assuming calcium carbonate is the composition, we deduce the mass of chalk used. Converting this to moles and accounting for three oxygen atoms per formula unit, we estimate there are approximately 3.25 × 10¹⁸ oxygen atoms deposited.
Step-by-step explanation:
To determine how many oxygen atoms have been deposited on the surface of the chalkboard, we first calculate the mass of chalk that was used by subtracting the final mass of the chalk from its initial mass. The initial mass is 22.48 g, and after writing the name, the mass is 22.47982 g. Therefore, the mass of chalk used is:
22.48 g - 22.47982 g = 0.00018 g of chalk.
Assuming the main component of the chalk is calcium carbonate (CaCO3), when it's used on a chalkboard, it primarily leaves behind particles of calcium carbonate. However, to simplify the problem as per the instructions, we will only focus on the oxygen part of the compound. One mole of calcium carbonate has three moles of oxygen because the compound has the formula CaCO3. Using the molar mass of calcium carbonate (100.09 g/mol) and knowing that one mole of any substance has 6.022 × 1023 atoms, we can calculate the number of moles of calcium carbonate and thus the number of moles of oxygen:
0.00018 g ÷ 100.09 g/mol = 1.799 × 10-6 moles of CaCO3}
Since there are three atoms of oxygen in one mole of CaCO3, the number of moles of oxygen is:
1.799 × 10-6 moles × 3 = 5.397 × 10-6 moles of oxygen.
Therefore, the total number of oxygen atoms is:
5.397 × 10-6 moles × 6.022 × 1023 atoms/mole ≈ 3.25 × 1018 oxygen atoms.