Mass of cobalt(II) phosphate

The Correct Answer and Explanation is:

To calculate the mass of cobalt(II) phosphate, we need to know the formula of the compound and the molar masses of its components.

The formula for cobalt(II) phosphate is Co3(PO4)2. This indicates that there are three cobalt ions (Co2+) and two phosphate ions (PO4 3-) in each formula unit.

Step 1: Calculate the molar mass of cobalt(II) phosphate

  1. Cobalt (Co) has a molar mass of approximately 58.93 g/mol.
  2. Phosphate (PO4) consists of one phosphorus atom (P) and four oxygen atoms (O). The molar masses of phosphorus and oxygen are:
    • Phosphorus (P) = 30.97 g/mol
    • Oxygen (O) = 16.00 g/mol
    The molar mass of one phosphate ion (PO4) is calculated as: Molar mass of PO4=30.97+(4×16.00)=30.97+64.00=94.97 g/mol\text{Molar mass of PO4} = 30.97 + (4 \times 16.00) = 30.97 + 64.00 = 94.97 \, \text{g/mol}Molar mass of PO4=30.97+(4×16.00)=30.97+64.00=94.97g/mol

Step 2: Add up the molar masses of all components in Co3(PO4)2

Now, we calculate the total molar mass of cobalt(II) phosphate:

  • Three cobalt atoms: 3×58.93=176.79 g/mol3 \times 58.93 = 176.79 \, \text{g/mol}3×58.93=176.79g/mol
  • Two phosphate ions: 2×94.97=189.94 g/mol2 \times 94.97 = 189.94 \, \text{g/mol}2×94.97=189.94g/mol

So, the total molar mass of cobalt(II) phosphate is:176.79+189.94=366.73 g/mol176.79 + 189.94 = 366.73 \, \text{g/mol}176.79+189.94=366.73g/mol

Step 3: Calculate the mass for a given amount of moles

If you are given a specific number of moles of cobalt(II) phosphate, you can calculate its mass using the formula:Mass (g)=Moles×Molar mass (g/mol)\text{Mass (g)} = \text{Moles} \times \text{Molar mass (g/mol)}Mass (g)=Moles×Molar mass (g/mol)

For example, if you have 2 moles of Co3(PO4)2, the mass would be:Mass=2×366.73=733.46 g\text{Mass} = 2 \times 366.73 = 733.46 \, \text{g}Mass=2×366.73=733.46g

In summary, the molar mass of cobalt(II) phosphate is 366.73 g/mol. You can find the mass of any given amount of cobalt(II) phosphate by multiplying the number of moles by this molar mass.

By admin

Leave a Reply