What number of atoms of phosphorus are present in 1.00g of each of the compounds in exercise 48?

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