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.
1)NiO
Give the number of moles of each element to support you answer.
moles of Ni
2) 1.339 mol
moles of O
3)1.34 mol
I know #1 is correct but 2 and 3 are wrong. i need help
Yes,2,3 are wrong.
molesNi=2.241/atomicmassNi
molesO=(2.852-2.241)/15.999
To calculate the number of moles of each element in the metal oxide, we can use the molar masses of nickel (Ni) and oxygen (O).
1) Find the number of moles of nickel (Ni):
Molar mass of Ni = 58.69 g/mol
Mass of Ni in the sample = 2.852 g
Moles of Ni = Mass of Ni / Molar mass of Ni
Moles of Ni = 2.852 g / 58.69 g/mol
Moles of Ni ≈ 0.048 mol (rounded to 3 decimal places)
2) Find the number of moles of oxygen (O):
Molar mass of O = 16.00 g/mol
Mass of O in the sample = mass of the sample - mass of Ni
Mass of O in the sample = (2.852 g) - (2.241 g) = 0.611 g
Moles of O = Mass of O / Molar mass of O
Moles of O = 0.611 g / 16.00 g/mol
Moles of O ≈ 0.038 mol (rounded to 3 decimal places)
Therefore, the empirical formula of the oxide is NiO, and the number of moles of Ni is 0.048 mol, while the number of moles of O is 0.038 mol.
To calculate the empirical formula of a compound, you need to find the ratio of the moles of each element present in the compound.
Let's start by finding the number of moles of nickel (Ni) and oxygen (O) in the given sample.
To find the moles of nickel (Ni), we need to use its molar mass. The molar mass of nickel is 58.6934 g/mol.
Given sample mass of nickel (Ni) = 2.241 g
Molar mass of nickel (Ni) = 58.6934 g/mol
Using the formula:
moles of Ni = mass of Ni / molar mass of Ni
moles of Ni = 2.241 g / 58.6934 g/mol
moles of Ni ≈ 0.0382 mol (rounded to four decimal places)
Therefore, the number of moles of nickel (Ni) is approximately 0.0382 mol.
Now let's find the number of moles of oxygen (O). To do this, we need to use the mass of the oxygen in the metal oxide and subtract it from the total mass of the metal oxide.
Given mass of the metal oxide = 2.852 g
Mass of nickel in the metal oxide = 2.241 g (from earlier calculation)
Mass of oxygen (O) in the metal oxide = mass of the metal oxide - mass of nickel
Mass of oxygen (O) in the metal oxide = 2.852 g - 2.241 g
Mass of oxygen (O) in the metal oxide ≈ 0.611 g
Now, let's find the moles of oxygen (O). The molar mass of oxygen is 15.999 g/mol.
Molar mass of oxygen (O) = 15.999 g/mol
Using the formula:
moles of O = mass of O / molar mass of O
moles of O = 0.611 g / 15.999 g/mol
moles of O ≈ 0.0382 mol (rounded to four decimal places)
Therefore, the number of moles of oxygen (O) is approximately 0.0382 mol.
Now that we have the number of moles of each element, we can determine the empirical formula. The subscripts in the empirical formula represent the mole ratio between different elements in a compound.
The subscripts can be simplified by dividing both the moles of nickel (Ni) and oxygen (O) by the smallest number of moles obtained. In this case, both elements have the same number of moles (0.0382 mol).
So, the empirical formula of the oxide is NiO, as mentioned initially.