Question

Vanadium (V) and oxygen (O) form a series of compounds with the following compositions: Mass % V 76.10 67.98 61.42 56.02 Mass % O 23.90 32.02 38.58 43.98 Compound Mass % N 1 33.28 2 39.94 Mass % Si 66.72 60.06 10. What are the relative numbers of atoms of oxygen in the compounds for a given mass of vanadium

Answer 1

Answer:

For every given mass of Vanadium, the relative number of oxygen atoms present or the mole ratio of Oxygen to Vanadium is:

A. 1:1

B. 3:2

C. 2:1

D. 5:2

Note: The question is stated more clearly below:

Vanadium (V) and oxygen (O) form a series of compounds with the following compositions: Mass % V 76.10 67.98 61.42 56.02 Mass % O 23.90 32.02 38.58 43.98 Compound Mass % N 1 33.28 2 39.94.

What are the relative numbers of atoms of oxygen in the compounds for a given mass of vanadium?

Explanation:

Number of moles in 100 g mass = % mass / molar mass

Molar mass of Vanadium, V = 51 g/mol

Molar mass of oxygen atom, O = 16 g/mol

1. Percentage mass of V and O is 76.10% and 23.90% respectively.

Number of moles of each atom;

V = 76.10/51.0 = 1.5 moles

O = 23.9/16 = 1.5 moles

Mole ratio of oxygen to vanadium = 1.5/1.5 = 1 : 1

2. Percentage mass of V and O is 67.98% and 32.02% respectively

Number of moles of each atom:

V = 67.98/51 = 1.33

O = 32.02/16 = 2

Mole ratio of oxygen to vanadium = 2/1.33 = 1.5 : 1 = 3 : 2

3. Percentage mass of V and O is 61.42% and 38.58% respectively

Number of moles of each atom:

V = 61.42/51 = 1.2

O = 38.58/16 = 2.4

Mole ratio of oxygen to vanadium = 2.4/1.2 = 2 : 1

4. Percentage mass of V and O is 56.02% and 43.98% respectively

Number of moles of each atom:

V = 56.02/51 = 1.10

O = 43.98/16 = 2.75

Mole ratio of oxygen to vanadium = 2.75/1.10 = 2.5 : 1 = 5 : 2