To find the relative formula mass of the compound [tex]\( \text{Ca}_3(\text{PO}_4)_3 \)[/tex], we need to consider the masses and the proportions of each element in the compound.
First, note the relative atomic masses of the elements involved:
- Calcium (Ca): 40
- Phosphorus (P): 31
- Oxygen (O): 16
The compound [tex]\( \text{Ca}_3(\text{PO}_4)_3 \)[/tex] contains:
- 3 atoms of Calcium (Ca)
- 3 units of Phosphate ([tex]\(\text{PO}_4\)[/tex]), each containing:
- 1 atom of Phosphorus (P)
- 4 atoms of Oxygen (O)
Let's calculate the mass contribution of each element step by step.
1. Calcium (Ca):
- There are 3 atoms of Ca.
- Each calcium atom has a relative atomic mass of 40.
[tex]\[
\text{Mass of Ca} = 3 \times 40 = 120
\][/tex]
2. Phosphorus (P):
- There are 3 atoms of P, as there are 3 units of Phosphate ([tex]\(\text{PO}_4\)[/tex]).
- Each phosphorus atom has a relative atomic mass of 31.
[tex]\[
\text{Mass of P} = 3 \times 31 = 93
\][/tex]
3. Oxygen (O):
- Each unit of Phosphate ([tex]\(\text{PO}_4\)[/tex]) contains 4 atoms of O.
- There are 3 units of Phosphate ([tex]\(\text{PO}_4\)[/tex]), so the total number of Oxygen atoms is:
[tex]\[
3 \times 4 = 12
\][/tex]
- Each oxygen atom has a relative atomic mass of 16.
[tex]\[
\text{Mass of O} = 12 \times 16 = 192
\][/tex]
Finally, sum the masses of all the elements to get the relative formula mass:
[tex]\[
\text{Relative formula mass} = \text{Mass of Ca} + \text{Mass of P} + \text{Mass of O}
\][/tex]
[tex]\[
\text{Relative formula mass} = 120 + 93 + 192 = 405
\][/tex]
Thus, the relative formula mass of [tex]\( \text{Ca}_3(\text{PO}_4)_3 \)[/tex] is [tex]\( 405 \)[/tex].
