How many atoms of oxygen are present in 4875 milligrams of Ca3(PO4)2?

• A. 1.83 × 10^{22} atoms
• B. 9.15 × 10^{21} atoms
• C. 2.44 × 10^{23} atoms
• D. 3.66 × 10^{22} atoms

Answer: (A)

Verified step by step guidance

1. First, calculate the molar mass of calcium phosphate, Ca\_3(PO\_4)\_2. Use the atomic masses: Ca = 40.08 g/mol, P = 30.97 g/mol, and O = 16.00 g/mol. The formula contains 3 calcium atoms, 2 phosphorus atoms, and 8 oxygen atoms (since there are 2 phosphate groups, each with 4 oxygens). Calculate the total molar mass by summing these contributions.
2. Convert the given mass of calcium phosphate from milligrams to grams by dividing by 1000, since molar mass is in grams per mole.
3. Calculate the number of moles of Ca\_3(PO\_4)\_2 by dividing the mass in grams by the molar mass you found in step 1. This gives you the amount of substance in moles.
4. Determine the total number of oxygen atoms by multiplying the number of moles of Ca\_3(PO\_4)\_2 by Avogadro's number (\$6.022 \times 10^{23}$ atoms/mol) and then by the number of oxygen atoms per formula unit (which is 8).
5. The result from step 4 will give you the total number of oxygen atoms present in 4875 milligrams of Ca\_3(PO\_4)\_2.