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

Answer:

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

Answer:

$V_2O$

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:

$\% V=\frac{20.38g}{23.58g}*100\%=86.43\% \\\% O=\frac{23.58g-20.38g}{23.58g}*100\%=13.57\%$

- Then the representative moles:

$n_V=\frac{86.42}{51}=1.695\\n_O=\frac{13.57}{16}=0.8482$

- Finally X and Y:

$X=\frac{1.695}{0.8482}=2\\\\Y=\frac{0.8482}{0.8482}=1$

Thus, the empirical formula is:

$V_2O$

Best regards.