What is the empirical formula of vanadium oxide , if 2.74 g of the metal oxide contains 1.53 g of metal ?
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Mass of Vanadium Oxide = 2.74 grams.
Mass of Vanadium = 1.53 grams.
% comp. of Vanadium = (Mass/Mass of the oxide) × 100%
= (1.53/2.74) × 100%
= 55.84%
% comp. of Oxygen =(100-55.84)%
= 44.16%
Let the mass of compound will be 100 g
So mass of vanadium = 55.84 g
mass of oxygen = 44.16 g
Molar mass of vanadium = 52 g/mol
Molar mass of oxygen = 16 g/mol
Number of moles of vanadium=55.84/52
=1.07
Number of moles of oxygen=44.16/16
=2.7
Lowest number of moles is 1.07 so dividing both atoms moles by it .
V=1.07/1.07
=1
O=2.7/1.07
=2.5
So multiply both by 2 to get a whole number that is:
V=1×2
=2
O=2.5×2
=5
So the formula is V₂O₅.
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