What is the empirical formula of vanadium oxide, if 2.74 g of the metal oxide contain 1.53 g of metal?
Answers
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.
Answer:
.
Explanation:
2.74 g metal oxide contains 1.53g metal and 1.21 g oxygen
% metal =
2.74
1.53
×100 = 55.8%
% oxygen =44.2%
no. of atoms of V=
51
55.8
=1.09
no. of atoms of O=
16
44.2
=2.76
simplest ratio : for V = 1.09/1.09 =1
for O= 1.09/2.76= 2.5
whole no. ratio : V= 1 X2=2 , O=2.5X 2=5
Empirical formula is V2O5
Sorry I scanned it after writing
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