Determine the molecular formula of an oxide of iron, in which the mass per cent
of iron and oxygen are 69.9 and 30.1, respectively.
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Determine the molecular formula of an oxide of iron, in which the mass per cent
of iron and oxygen are 69.9 and 30.1, respectively.
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Answer :
Given -
To Find -
Solution -
No. of moles = Mass/Molar mass
➸ n = m/M
No. of moles of Iron :
➸ n = 69.9/55.8
➸ n = 1.25 mol
No. of moles of Oxygen :
➸ n = 30.1/16
➸ n = 1.88 mol
Ratio of Fe to O :- 1.25 : 1.88
➸ 1.25/1.25 : 1.88/1.25
➸ 1 : 1.5
➸ 2 : 3
Hence, the empirical formula : Fe2O3.
Empirical formula Mass : (55.8 × 2) + (16 × 3)
➸ 111.6 + 48
➸ 159.6 g
We know that, n = Molecular mass/Empirical Formula mass.
➸ n = 159.6/159.6
➸ n = 1 mol
Hence, Molecular mass = 1*Fe2O3 = Fe2O3.
Verified answer
Answer:
% of iron by mass = 69.9 % [Given]
% of iron by mass = 69.9 % [Given]% of oxygen by mass = 30.1 % [Given]
% of iron by mass = 69.9 % [Given]% of oxygen by mass = 30.1 % [Given]Atomic mass of iron = 55.85 amu
% of iron by mass = 69.9 % [Given]% of oxygen by mass = 30.1 % [Given]Atomic mass of iron = 55.85 amuAtomic mass of oxygen = 16.00 amu
% of iron by mass = 69.9 % [Given]% of oxygen by mass = 30.1 % [Given]Atomic mass of iron = 55.85 amuAtomic mass of oxygen = 16.00 amuRelative moles of iron in iron oxide = %mass of iron by mass/Atomic mass of iron = 69.9/55.85 = 1.25
% of iron by mass = 69.9 % [Given]% of oxygen by mass = 30.1 % [Given]Atomic mass of iron = 55.85 amuAtomic mass of oxygen = 16.00 amuRelative moles of iron in iron oxide = %mass of iron by mass/Atomic mass of iron = 69.9/55.85 = 1.25Relative moles of oxygen in iron oxide = %mass of oxygen by mass/Atomic mass of oxygen = 30.01/16=1.88
% of iron by mass = 69.9 % [Given]% of oxygen by mass = 30.1 % [Given]Atomic mass of iron = 55.85 amuAtomic mass of oxygen = 16.00 amuRelative moles of iron in iron oxide = %mass of iron by mass/Atomic mass of iron = 69.9/55.85 = 1.25Relative moles of oxygen in iron oxide = %mass of oxygen by mass/Atomic mass of oxygen = 30.01/16=1.88Simplest molar ratio = 1.25/1.25 : 1.88/1.25
% of iron by mass = 69.9 % [Given]% of oxygen by mass = 30.1 % [Given]Atomic mass of iron = 55.85 amuAtomic mass of oxygen = 16.00 amuRelative moles of iron in iron oxide = %mass of iron by mass/Atomic mass of iron = 69.9/55.85 = 1.25Relative moles of oxygen in iron oxide = %mass of oxygen by mass/Atomic mass of oxygen = 30.01/16=1.88Simplest molar ratio = 1.25/1.25 : 1.88/1.25⇒ 1 : 1.5 = 2 : 3
% of iron by mass = 69.9 % [Given]% of oxygen by mass = 30.1 % [Given]Atomic mass of iron = 55.85 amuAtomic mass of oxygen = 16.00 amuRelative moles of iron in iron oxide = %mass of iron by mass/Atomic mass of iron = 69.9/55.85 = 1.25Relative moles of oxygen in iron oxide = %mass of oxygen by mass/Atomic mass of oxygen = 30.01/16=1.88Simplest molar ratio = 1.25/1.25 : 1.88/1.25⇒ 1 : 1.5 = 2 : 3∴ The empirical formula of the iron oxide is Fe2O3.
% of iron by mass = 69.9 % [Given]% of oxygen by mass = 30.1 % [Given]Atomic mass of iron = 55.85 amuAtomic mass of oxygen = 16.00 amuRelative moles of iron in iron oxide = %mass of iron by mass/Atomic mass of iron = 69.9/55.85 = 1.25Relative moles of oxygen in iron oxide = %mass of oxygen by mass/Atomic mass of oxygen = 30.01/16=1.88Simplest molar ratio = 1.25/1.25 : 1.88/1.25⇒ 1 : 1.5 = 2 : 3∴ The empirical formula of the iron oxide is Fe2O3.Mass of Fe2O3 = (2×55.85) + (3×16.00) = 159.7 g mol-1
% of iron by mass = 69.9 % [Given]% of oxygen by mass = 30.1 % [Given]Atomic mass of iron = 55.85 amuAtomic mass of oxygen = 16.00 amuRelative moles of iron in iron oxide = %mass of iron by mass/Atomic mass of iron = 69.9/55.85 = 1.25Relative moles of oxygen in iron oxide = %mass of oxygen by mass/Atomic mass of oxygen = 30.01/16=1.88Simplest molar ratio = 1.25/1.25 : 1.88/1.25⇒ 1 : 1.5 = 2 : 3∴ The empirical formula of the iron oxide is Fe2O3.Mass of Fe2O3 = (2×55.85) + (3×16.00) = 159.7 g mol-1n = Molar mass/Empirical formula mass = 159.7/159.6 = 1(approx)
% of iron by mass = 69.9 % [Given]% of oxygen by mass = 30.1 % [Given]Atomic mass of iron = 55.85 amuAtomic mass of oxygen = 16.00 amuRelative moles of iron in iron oxide = %mass of iron by mass/Atomic mass of iron = 69.9/55.85 = 1.25Relative moles of oxygen in iron oxide = %mass of oxygen by mass/Atomic mass of oxygen = 30.01/16=1.88Simplest molar ratio = 1.25/1.25 : 1.88/1.25⇒ 1 : 1.5 = 2 : 3∴ The empirical formula of the iron oxide is Fe2O3.Mass of Fe2O3 = (2×55.85) + (3×16.00) = 159.7 g mol-1n = Molar mass/Empirical formula mass = 159.7/159.6 = 1(approx)Thus, Molecular formula is same as Empirical Formula i.e. Fe2O3.
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