After subtracting the weight of the container, the Vanadium metal inside is
found to weigh 8.34 grams. After reacting with Oxygen the newly formed
compound weighs 21.43 grams. What is the empirical formula?
Find moles V.
8.34/molar mass.
Find moles oxygen.
21.43-8.43 = 13.09 g oxygen
moles = 13.09/16 = ??
Now find the ratio in small whole numbers. The easy way to do that is to divide the smaller number of itself (making it 1.000), then divide the other number by the same small number, found to a whole number and that should be the formula.
To determine the empirical formula, we need to find the relative number of atoms of each element in the compound. We can do this by comparing the masses of the elements present.
Let's start by finding the mass of Oxygen in the compound. We subtract the weight of the Vanadium from the total weight of the compound.
Mass of Oxygen = Total weight of the compound - Weight of Vanadium
Mass of Oxygen = 21.43 grams - 8.34 grams
Mass of Oxygen = 13.09 grams
Now, we can use the masses of Vanadium and Oxygen to find the ratio of their atoms.
To obtain the ratio, we need to convert the masses of each element to moles. We can use the molar mass of each element for this calculation. The molar mass of Vanadium (V) is 50.94 g/mol, and the molar mass of Oxygen (O) is 16.00 g/mol.
Moles of Vanadium = Weight of Vanadium / Molar mass of Vanadium
Moles of Vanadium = 8.34 grams / 50.94 g/mol
Moles of Vanadium ≈ 0.163 moles
Moles of Oxygen = Mass of Oxygen / Molar mass of Oxygen
Moles of Oxygen = 13.09 grams / 16.00 g/mol
Moles of Oxygen ≈ 0.818 moles
Next, we divide the mole values of both elements by the smallest mole value to determine the simplest whole-number ratio.
Vanadium: Oxygen ≈ 0.163 moles / 0.163 moles : 0.818 moles / 0.163 moles
Vanadium: Oxygen ≈ 1 : 5
Since the ratio is 1:5, the empirical formula is V2O5.