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.
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 NinOm</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.
It shouldn't be weird, it should be a simple whole number.
so
O=.0382
Ni=.0382 also?
Doesn't that look like 1:1. It does to me.
To determine the empirical formula of the oxide, we first need to calculate the number of moles of each element present in the compound.
Let's start by calculating the number of moles of nickel (Ni) and oxygen (O) individually.
1. Calculate the number of moles of nickel (Ni):
- Given mass of nickel = 2.241 g
- Molar mass of nickel (Ni) = 58.693 g/mol (from the periodic table)
- Number of moles of nickel (Ni) = mass of nickel (Ni) / molar mass of nickel (Ni)
= 2.241 g / 58.693 g/mol
= 0.03814 mol
2. Calculate the number of moles of oxygen (O):
- Given mass of oxygen = 2.852 g (from the metal oxide)
- Molar mass of oxygen (O) = 16.00 g/mol (from the periodic table)
- Number of moles of oxygen (O) = mass of oxygen (O) / molar mass of oxygen (O)
= 2.852 g / 16.00 g/mol
= 0.17825 mol
Now, we have the number of moles for each element present in the compound.
Next, we need to determine the empirical formula by finding the lowest whole number ratio between the moles of nickel and moles of oxygen.
Divide the moles of each element by the smallest number of moles:
- Divide 0.03814 mol of Ni and 0.03814 mol of O by the smallest number of moles, which is 0.03814 mol.
- Round the resulting values to the nearest whole number.
After dividing, we find that both Ni and O have a mole ratio of 1. Hence, the empirical formula is NiO.
To summarize, the empirical formula of the oxide is NiO, with 1 mole of nickel (Ni) and 1 mole of oxygen (O) in the compound.