Answer: N3 H12 P O3
Explanation:
From the question :
N = 31.57% H = 9.10% P = 23.27%
O= 36.06%
Divide each of the element by their respective relative atomic masses.
N = 31.57 / 14 = 2.26
H = 9.10/ 1 = 9.10
P = 23.27 / 31= 0.750
O =36.06 / 16 = 2.25
Divide each answer by the lowest of them all, we then have:
N = 2.26/ 0.750 = Approx = 3
H = 9.10 / 0.750 = Approx = 12
P = 0.750/ 0.750= 1
O = 2.25 / 0.750 = Approx = 3
The empiral formula is
N3 H12 P O3
Answer:
The empirical formula for the compound is (NH₄)₃PO₃
Explanation:
Here we have
Nitrogen Molar mass = 14.01 g/mol, Percentage = 31.57%
Hydrogen Molar mass = 1.00784 g/mol, Percentage = 9.10%
Phosphorus Molar mass = 30.973762 g/mol, Percentage = 23.27%
Oxygen Molar mass = 15.999 g/mol, Percentage = 36.06%
Therefore the number of moles of each element per 100 g of the ionic compound is as follows;
Number of moles, n = Mass/(Molar Mass)
Mass of nitrogen, N in 100 g of ionic compound = 31.57 g n = 2.253 moles
Mass of Hydrogen, H in 100 g of ionic compound = 9.10 g, n = 9.029 moles
Mass of Phosphorus, P in 100 g of ionic compound = 23.27 g, n = 0.7513 moles
Mass of Oxygen, P in 100 g of ionic compound = 36.06 g, n = 2.2539 moles
Dividing by the smallest mole ratio, of 0.7513 (that of phosphorus) we have;
N = 3 moles
H = 12.0184 ≈ 12 moles
P = 1 moles
O = 3 moles
Therefore, we have the empirical formula as follows;
N₃H₁₂PO₃ or (NH₄)₃PO₃.
