Question 16An acid of phosphorus has the following percentage composition; Phosphorus = 38.27%; hydrogen = 2.47%; oxygen = 59.26%. Find the empirical formula of the acid and it's molecular formula, given that it's relative molecular mass is 162.

Question

Accepted Answer

Element% compositionAt. wt.Relative no. of atomsSimplest ratioPhosphorus38.273138.2731\dfrac{38.27 }{31}3138.27​ = 1.2341.2341.234\dfrac{ 1.234}{1.234 }1.2341.234​ = 1Hydrogen2.4712.471\dfrac{2.47 }{1}12.47​ = 2.472.471.234\dfrac{2.47 }{1.234 }1.2342.47​ = 2Oxygen59.261659.2616\dfrac{59.26}{16}1659.26​ = 3.703.701.234\dfrac{3.70}{ 1.234 }1.2343.70​ = 3Simplest ratio of whole numbers = P : H : O = 1 : 2 : 3Hence, empirical formula is H2PO3Empirical formula weight = 31 + 2(1) + 3(16) = 31 + 2 + 48 = 81Molecular weight = 162n=Molecular weightEmpirical formula weight=16281=2\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \[0.5em] = \dfrac{162}{81} = 2n=Empirical formula weightMolecular weight​=81162​=2So, molecular formula = 2(H2PO3) = H4P2O6

Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Phosphorus	38.27	31	$\dfrac{38.27 }{31}$ = 1.234	$\dfrac{ 1.234}{1.234 }$ = 1
Hydrogen	2.47	1	$\dfrac{2.47 }{1}$ = 2.47	$\dfrac{2.47 }{1.234 }$ = 2
Oxygen	59.26	16	$\dfrac{59.26}{16}$ = 3.70	$\dfrac{3.70}{ 1.234 }$ = 3

Mole Concept & Stoichiometry Miscellaneous Exercises — Question 16

Question 16