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To determine the molecular mass and the percentage composition of each element in the compound [tex]$Ca_{10}(PO_4)_6(OH)_2$[/tex], we will use the following steps:
1. Determine the number of each type of atom in one molecule of the compound:
- Calcium (Ca): 10 atoms
- Phosphorus (P): 6 atoms
- Oxygen (O): 26 atoms (24 from [tex]$PO_4$[/tex] groups and 2 from [tex]$OH$[/tex] groups)
- Hydrogen (H): 2 atoms
2. Find the atomic masses of each element:
- Calcium (Ca): 40.08 amu
- Phosphorus (P): 30.97 amu
- Oxygen (O): 16.00 amu
- Hydrogen (H): 1.01 amu
3. Calculate the mass contribution of each element:
[tex]\[ \text{Mass of Ca} = 10 \times 40.08 = 400.8 \text{ amu} \][/tex]
[tex]\[ \text{Mass of P} = 6 \times 30.97 = 185.82 \text{ amu} \][/tex]
[tex]\[ \text{Mass of O} = 26 \times 16.00 = 416.00 \text{ amu} \][/tex]
[tex]\[ \text{Mass of H} = 2 \times 1.01 = 2.02 \text{ amu} \][/tex]
4. Calculate the total formula mass of the compound by summing the masses of all elements:
[tex]\[ \text{Formula mass} = 400.8 + 185.82 + 416.00 + 2.02 = 1004.64 \text{ amu} \][/tex]
5. Determine the percentage composition of each element:
[tex]\[ \text{Percentage of Ca} = \left( \frac{400.8}{1004.64} \right) \times 100 = 39.89 \% \][/tex]
[tex]\[ \text{Percentage of P} = \left( \frac{185.82}{1004.64} \right) \times 100 = 18.50 \% \][/tex]
[tex]\[ \text{Percentage of O} = \left( \frac{416.00}{1004.64} \right) \times 100 = 41.41 \% \][/tex]
[tex]\[ \text{Percentage of H} = \left( \frac{2.02}{1004.64} \right) \times 100 = 0.20 \% \][/tex]
Now, we can fill in the table:
[tex]\[ \begin{array}{|c|c|c|c|c|} \hline \text{Atom} & \text{Number} & \text{Atomic mass (amu)} & \text{Mass (Number $x$ atomic mass)} & \text{Percentage composition (mass/molecular mass $\times 100)$} \\ \hline \text{Ca} & 10 & 40.08 & 400.8 & 39.89\% \\ \hline \text{P} & 6 & 30.97 & 185.82 & 18.50\% \\ \hline \text{O} & 26 & 16.00 & 416.00 & 41.41\% \\ \hline \text{H} & 2 & 1.01 & 2.02 & 0.20\% \\ \hline \text{Formula Mass of $Ca_{10}(PO_4)_6(OH)_2$} & & & & 1004.64 \text{ amu} \\ \hline \end{array} \][/tex]
So, the formula mass of [tex]$Ca_{10}(PO_4)_6(OH)_2$[/tex] is 1004.64 amu and we have completed the table with the relevant information.
1. Determine the number of each type of atom in one molecule of the compound:
- Calcium (Ca): 10 atoms
- Phosphorus (P): 6 atoms
- Oxygen (O): 26 atoms (24 from [tex]$PO_4$[/tex] groups and 2 from [tex]$OH$[/tex] groups)
- Hydrogen (H): 2 atoms
2. Find the atomic masses of each element:
- Calcium (Ca): 40.08 amu
- Phosphorus (P): 30.97 amu
- Oxygen (O): 16.00 amu
- Hydrogen (H): 1.01 amu
3. Calculate the mass contribution of each element:
[tex]\[ \text{Mass of Ca} = 10 \times 40.08 = 400.8 \text{ amu} \][/tex]
[tex]\[ \text{Mass of P} = 6 \times 30.97 = 185.82 \text{ amu} \][/tex]
[tex]\[ \text{Mass of O} = 26 \times 16.00 = 416.00 \text{ amu} \][/tex]
[tex]\[ \text{Mass of H} = 2 \times 1.01 = 2.02 \text{ amu} \][/tex]
4. Calculate the total formula mass of the compound by summing the masses of all elements:
[tex]\[ \text{Formula mass} = 400.8 + 185.82 + 416.00 + 2.02 = 1004.64 \text{ amu} \][/tex]
5. Determine the percentage composition of each element:
[tex]\[ \text{Percentage of Ca} = \left( \frac{400.8}{1004.64} \right) \times 100 = 39.89 \% \][/tex]
[tex]\[ \text{Percentage of P} = \left( \frac{185.82}{1004.64} \right) \times 100 = 18.50 \% \][/tex]
[tex]\[ \text{Percentage of O} = \left( \frac{416.00}{1004.64} \right) \times 100 = 41.41 \% \][/tex]
[tex]\[ \text{Percentage of H} = \left( \frac{2.02}{1004.64} \right) \times 100 = 0.20 \% \][/tex]
Now, we can fill in the table:
[tex]\[ \begin{array}{|c|c|c|c|c|} \hline \text{Atom} & \text{Number} & \text{Atomic mass (amu)} & \text{Mass (Number $x$ atomic mass)} & \text{Percentage composition (mass/molecular mass $\times 100)$} \\ \hline \text{Ca} & 10 & 40.08 & 400.8 & 39.89\% \\ \hline \text{P} & 6 & 30.97 & 185.82 & 18.50\% \\ \hline \text{O} & 26 & 16.00 & 416.00 & 41.41\% \\ \hline \text{H} & 2 & 1.01 & 2.02 & 0.20\% \\ \hline \text{Formula Mass of $Ca_{10}(PO_4)_6(OH)_2$} & & & & 1004.64 \text{ amu} \\ \hline \end{array} \][/tex]
So, the formula mass of [tex]$Ca_{10}(PO_4)_6(OH)_2$[/tex] is 1004.64 amu and we have completed the table with the relevant information.
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