Question

The ratio of number of oxygen atoms (O) in 16.0g oxygen (O), 28.0 g carbon monoxide (CO) and 32.0g oxygen (O₂) is :

(Atomic mass : C =12, O =16 and Avogadro’s constant NA = 6.022 x 1023 mol-1)
a) 3 : 1 : 1
b) 1 : 1 : 2
c) 3 : 1 : 2
d) 1 : 1 : 1​

Answer 1

The ratio of the number of oxygen atoms in 16.0g O, 28.0g CO, and 32.0g O₂ is 1:1:2, as there is 1 mole of oxygen in 16.0g O and 28.0g CO, and 2 moles of oxygen in 32.0g O₂.

Step-by-step explanation:

To answer the question regarding the ratio of the number of oxygen atoms in 16.0g of oxygen (O), 28.0g of carbon monoxide (CO), and 32.0g of oxygen (O₂), we must first calculate the moles of oxygen atoms in each case.

For elemental oxygen (O), 16.0g corresponds to 1 mole since its atomic mass is 16.0 g/mol. Therefore, it has 1 mole of oxygen atoms.
For carbon monoxide (CO), with a molar mass of 28.0 g/mol (12.0 g/mol for carbon and 16.0 g/mol for oxygen), 28.0g of CO contains 1 mole of CO molecules, which means it also contains 1 mole of oxygen atoms because there is one oxygen atom per molecule of CO.

For diatomic oxygen (O₂), the molar mass is 32.0 g/mol. Therefore, 32.0g of O₂ corresponds to 1 mole of O₂ molecules, which contains 2 moles of oxygen atoms.

So, the ratio of the number of oxygen atoms in 16.0g O, 28.0g CO, and 32.0g O₂ is 1:1:2.