Respuesta :
Answer:
The empirical formula is = [tex]NiO[/tex]
Explanation:
Given that:- Mass of nickel = 2.241 g
Mass of the oxide formed = 2.852 g
Mass of the oxygen reacted = Mass of the oxide formed - Mass of nickel = 2.852 g - 2.241 g = 0.611 g
Molar mass of nickel = 58.6934 g/mol
Moles of nickel = [tex]\frac{2.241}{58.6934}\ mol[/tex] = 0.03818 mol
Molar mass of oxygen = 15.999 g/mol
Moles of nickel = [tex]\frac{0.611}{15.999}\ mol[/tex] = 0.03818 mol
Taking the simplest ratio for Ni and O as:
0.03818 : 0.03818 = 1 : 1
The empirical formula is = [tex]NiO[/tex]
The empirical formula for the oxide formed from 2.241-g sample of nickel and 0.611 g of oxygen is NiO.
Nickel + oxygen → Metal oxide
2.241g + x g = 2.852 g
Using the law of conservation of mass,
mass of oxygen = 2.852 g - 2.241 = 0.611 g
Therefore, lets calculate the empirical formula
atomic mass of Ni = 58.6934g
moles of Ni = 2.241 / 58.6934 = 0.03818146503 moles
atomic mass of Oxygen = 16 g
moles of oxygen = 0.611 / 16 = 0.0381875 moles
The simplest ratio of Nickel to Oxygen is as follows:
0.03818146503 : 0.0381875 is 1:1
Therefore, the empirical formula is NiO
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