The compounds in exercise 48 are:
a) P4O6,
b) Ca3 (PO4)2, and
c) Na2 H PO4
So, proceed with the calculus for each compound.
a) Molecular formula: P4O6
Molar mass: 4 * 31 g/mol + 6* 16g/mol = 220 g/mol
Number of moles in 1.00 grams of compound = mass in grams / molar mass =
= 1.00 g / 220 g/mol = 0.004545 mol
0.004545 mol of P4O6 contains 4 * 0.004545 = 0.01818moles of atoms of P.
=> 0.01818 moles * 6.022 * 10^23 atoms / mol = 1.095 * 10^ 22 atoms of P.
Answer: 1.095 * 10^22 atoms of P.
b) Ca3 (PO4)2
molar mass = 3 * 40.1 g/mol + 2 * 31.0 g/mol + 8 * 16 g/mol = 310.3 g/mol
number of moles in 1.00 g of Ca3 (PO4)2 = 1.00 g / 310.3 g/mol = 0.00322 mol
0.00322 mol of compound * 2 mol P / mol of compound = 0.00644 mol P
0.00644 mol P * 6.022 * 10^23 atom / mol = 3.878 * 10 ^ 21 atoms P
Answer: 3.878 * 10^21 atoms P
c) Na2 H PO4
molar mass = 2 * 23.0 g/mol + 1 g/mol + 31.0 g/mol + 4 * 16g/mol = 142.0 g/mol
number of moles = 1.00 g / 142.0 g/mol = 0.0070 moles Na2HPO4
=> 0.0070 moles P
=> 0.0070 * 6.022 * 10^23 = 4.215 * 10^21 atoms of P
Answer: 4.215 * 10^21 atoms P
