How many formula units of (NH4)3PO4 are present in 12.25 g of K3PO4? Please explain so I can do these in the future! Very confused.
No formula units of ammonium phosphate is present in potassium phosphate? Goodness.
I have the same question in my homework. Anybody know how to do this?
To determine the number of formula units of (NH4)3PO4 in 12.25 g of K3PO4, you need to follow a series of steps:
1. Find the molar mass of K3PO4:
- K (potassium) has a molar mass of 39.10 g/mol.
- P (phosphorus) has a molar mass of 30.97 g/mol.
- O (oxygen) has a molar mass of 16.00 g/mol.
Multiply the molar mass of each element by the number of atoms in the formula, then add them together:
(3 x 39.10 g/mol for K) + (1 x 30.97 g/mol for P) + (4 x 16.00 g/mol for O) = 164.29 g/mol.
2. Determine the number of moles of K3PO4 in 12.25 g:
Divide the mass of K3PO4 by its molar mass:
12.25 g / 164.29 g/mol = 0.0746 mol.
3. Use the stoichiometry of the balanced chemical equation to find the number of moles of (NH4)3PO4:
The balanced chemical equation for the reaction between K3PO4 and (NH4)3PO4 is:
3(NH4)3PO4 + 2K3PO4 → 6NH4PO4 + 3K2PO4
From the equation, you can see that 2 moles of K3PO4 react with 6 moles of (NH4)3PO4.
So, for every 2 moles of K3PO4, there are 6 moles of (NH4)3PO4.
4. Calculate the number of moles of (NH4)3PO4:
Multiply the number of moles of K3PO4 (0.0746 mol) by the mole ratio of (NH4)3PO4 to K3PO4:
0.0746 mol x (6 mol (NH4)3PO4 / 2 mol K3PO4) = 0.224 mol (NH4)3PO4.
5. Convert the number of moles of (NH4)3PO4 to formula units:
Avogadro's number, 6.022 x 10^23 formula units/mol, tells us that there are 6.022 x 10^23 formula units in 1 mole of any substance.
Multiply the number of moles of (NH4)3PO4 (0.224 mol) by Avogadro's number:
0.224 mol x 6.022 x 10^23 formula units/mol = 1.35 x 10^23 formula units of (NH4)3PO4.
Therefore, there are approximately 1.35 x 10^23 formula units of (NH4)3PO4 present in 12.25 g of K3PO4.