Question 19(c)10.47 g of a compound contains 6.25 g of metal A and rest non-metal B. Calculate the empirical formula of the compound (At. wt of A = 207, B = 35.5)

Question

Accepted Answer

Element% compositionAt. wt.Relative no. of atomsSimplest ratioMetal A6.252076.25207\dfrac{6.25 }{207}2076.25​ = 0.030.030.03\dfrac{0.03}{0.03}0.030.03​ = 1Non-metal B4.2235.54.2235.5\dfrac{4.22}{35.5}35.54.22​ = 0.110.110.03\dfrac{0.11}{0.03}0.030.11​ = 3.96 = 4Simplest ratio of whole numbers = A : B = 1 : 4Hence, empirical formula is AB4

Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Metal A	6.25	207	$\dfrac{6.25 }{207}$ = 0.03	$\dfrac{0.03}{0.03}$ = 1
Non-metal B	4.22	35.5	$\dfrac{4.22}{35.5}$ = 0.11	$\dfrac{0.11}{0.03}$ = 3.96 = 4

Mole Concept and Stoichiometry — Question 21

Question 19(c)