Calculating the Mass of Calcium Phosphate

This problem requires us to calculate the mass in grams of a given number of moles of calcium phosphate ($Ca_3(PO_4)_2$). We are provided with the number of moles and the atomic masses of the elements calcium (Ca), phosphorus (P), and oxygen (O).

Understanding the Concepts

To find the mass from moles, we need two key pieces of information:

• The number of moles of the substance (given as $0.25$ mol).
• The molar mass of the substance ($Ca_3(PO_4)_2$).

The molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of all atoms present in one molecule or formula unit of the compound.

Step 1: Determine the Molar Mass of Calcium Phosphate ($Ca_3(PO_4)_2$)

First, let's identify the number of atoms of each element in the chemical formula $Ca_3(PO_4)_2$.

• Calcium (Ca): There are 3 atoms.
• Phosphorus (P): The subscript outside the parenthesis (2) applies to everything inside. So, there are $2 \times 1 = 2$ atoms of P.
• Oxygen (O): Similarly, there are $2 \times 4 = 8$ atoms of O.

Now, we use the provided atomic masses to calculate the molar mass:

• Atomic mass of Ca = $40.08 \text{ g/mol}$
• Atomic mass of P = $30.97 \text{ g/mol}$
• Atomic mass of O = $16.00 \text{ g/mol}$

The molar mass of $Ca_3(PO_4)_2$ is calculated as follows:

Molar Mass = $(3 \times \text{Atomic mass of Ca}) + (2 \times \text{Atomic mass of P}) + (8 \times \text{Atomic mass of O})$

Molar Mass = $(3 \times 40.08 \text{ g/mol}) + (2 \times 30.97 \text{ g/mol}) + (8 \times 16.00 \text{ g/mol})$

Molar Mass = $(120.24 \text{ g/mol}) + (61.94 \text{ g/mol}) + (128.00 \text{ g/mol})$

Molar Mass = $310.18 \text{ g/mol}$

Here is a summary of the atomic masses used:

Step 2: Calculate the Mass from Moles

We can now use the formula relating mass, moles, and molar mass:

Mass = Moles $\times$ Molar Mass

We are given:

• Moles = $0.25$ mol
• Molar Mass of $Ca_3(PO_4)_2$ = $310.18 \text{ g/mol}$

Substituting these values into the formula:

Mass = $0.25 \text{ mol} \times 310.18 \text{ g/mol}$

Mass = $77.545 \text{ g}$

Conclusion

The calculated mass is $77.545$ grams. Rounding this to two decimal places gives $77.55$ grams, which matches one of the options provided.

Therefore, the mass of $0.25$ moles of calcium phosphate ($Ca_3(PO_4)_2$) is approximately $77.55 \text{ g}$.