Asked by sam
A 2.241 g sample of nickel reacts with oxygen to form 2.852 g of the metal oxide. Calculate the empirical formula of the oxide.
Give the number of moles of each element to support you answer.
Give the number of moles of each element to support you answer.
Answers
Answered by
bobpursley
YOU can figure the moles of oxygen, and Ni.
molesNi= 2.241/atomicweightNi
molesO=(2.852-2.241)/15.999
now, get the mole ratio: MolesNi/molesO
(ie, divide both moles by the lowest number).
The formula will be Ni<sub>n</sub>O<sub>m</sub)
where n/m= mole ratio above. If it is a weird number, such as .33, get it to
.33/1=n/m multiply by 3, so that n/m =1/3 In this case it should be weird.
molesNi= 2.241/atomicweightNi
molesO=(2.852-2.241)/15.999
now, get the mole ratio: MolesNi/molesO
(ie, divide both moles by the lowest number).
The formula will be Ni<sub>n</sub>O<sub>m</sub)
where n/m= mole ratio above. If it is a weird number, such as .33, get it to
.33/1=n/m multiply by 3, so that n/m =1/3 In this case it should be weird.
Answered by
bobpursley
It shouldn't be weird, it should be a simple whole number.
Answered by
sam
so
O=.0382
Ni=.0382 also?
O=.0382
Ni=.0382 also?
Answered by
DrBob222
Doesn't that look like 1:1. It does to me.
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