If a sample of Mg3(PO4)2, has a mass of 0.200 g, what is the mass of phosphorus in that sample?
0.200 g x (2*atomic mass P/molar mass Mg3(PO4)2) = ?
.0472
.0471
0.471 g
To determine the mass of phosphorus in the sample of Mg3(PO4)2, we need to first find the molar mass of the compound and then calculate the mass of phosphorus using stoichiometry.
1. Find the molar mass of Mg3(PO4)2:
- The molar mass of Magnesium (Mg) is 24.31 g/mol.
- The molar mass of Phosphorus (P) is 30.97 g/mol.
- The molar mass of Oxygen (O) is 16.00 g/mol.
- There are two phosphorus atoms and eight oxygen atoms in Mg3(PO4)2.
- Therefore, the molar mass of Mg3(PO4)2 is:
(3 * Molar mass of Mg) + (2 * Molar mass of P) + (8 * Molar mass of O)
= (3 * 24.31 g/mol) + (2 * 30.97 g/mol) + (8 * 16.00 g/mol)
2. Calculate the molar mass of Mg3(PO4)2 using the given values and perform the calculations to find its value.
3. Once you have calculated the molar mass of Mg3(PO4)2, you can now determine the mass of phosphorus in the sample using stoichiometry.
- The stoichiometric ratio between Mg3(PO4)2 and phosphorus is 2:1, as there are two phosphorus atoms in Mg3(PO4)2.
- Given that the mass of the sample is 0.200 g and the molar mass of Mg3(PO4)2, you can calculate the moles of Mg3(PO4)2.
- Then, using the stoichiometric ratio, determine the moles of phosphorus.
- Finally, calculate the mass of phosphorus in the sample by multiplying the moles of phosphorus by its molar mass.
These calculations should give you the mass of phosphorus in the given sample.