Question

what is the empirical formula of vanadium 1 oxide given that 20.38 grams of vandium combines with oxygen to form 23.58 grams of the oxide

Answer 1

The empirical formula is V₂O

Empirical formula of a compound is the formula that shows the simplest whole number ratio of the atoms of elements in a given compound. Empirical formula is normally calculated when the mass of each element in a compound is known or the percentage composition by mass of each element in a compound is known.

Step by Step Explanation:

Step 1: Percentage composition of each element

Percentage composition=(mass of an element/ mass of the compound)100%

Mass of Vanadium = 20.38 g

Mass of the compound = 23.58 g

% composition of Vanadium = (20.38 g/23.58 g) 100%

= 86.43 %

Mass of Oxygen = 23.58 g -20.38 g

= 3.2 g

% composition of oxygen = (3.2/g/23.58 g) 100%

= 13.57%

Step 2: Find the number of atoms of each element in the compound

Number of atoms = percentage composition/ atomic mass

Atomic mass of Vanadium = 50.94 g/mol

Number of atoms of V = 86.43 /50.94

= 1.6967

Atomic mass of oxygen = 16 g/mol

Number of atoms of O = 13.57/16

= 0.8481

Step 3: Find the simplest ratio of atoms

Vanadium : Oxygen

1.6967 : 0.8481

= 1.6967/0.8481 : 0.8481/0.8481

= 2: 1

Whole number ratio = 2 : 1

Therefore; the empirical formula is V₂O

Answer 2

`V_2O`

Step-by-step explanation:

Hello,

In this case, for the given vanadium oxide we have the general formula:

`V_XO_Y`

Whereas both X and Y are determined as shown below:

- Find the mass percentage of both vanadium and oxygen:

`\% V=(20.38g)/(23.58g)*100\%=86.43\% \\\% O=(23.58g-20.38g)/(23.58g)*100\%=13.57\%`

- Then the representative moles:

`n_V=(86.42)/(51)=1.695\\n_O=(13.57)/(16)=0.8482`

- Finally X and Y:

`X=(1.695)/(0.8482)=2\\\\Y=(0.8482)/(0.8482)=1`

Thus, the empirical formula is:

`V_2O`

Best regards.