Question

If you were given a sample of a cotton ball and a glass stirring rod with identical mass (ex: 5.0 g), which sample would contain more oxygen atoms?

Answer 1

The sample which would contain more oxygen atoms is a glass stirring rod.

Step-by-step explanation:

According to the periodic table ,

the glass stirring rod is Silicon dioxide and the cotton ball is cellulose

Here , the molar mass of the glass stirring rod
`SiO_(2)` = 60.08 Grams/mole

Molar mass of the cotton ball
`C_(6) H_(10)O_(5)` = 162.09 gram/mole .

since ,total number of molecules present in oxygen is 32 ,

therefore ,

In glass stirring rod ,

number of oxygen atoms present in 5g =
`(32)/(60.07) * 5`

= 2.66 g of oxygen

`(1)/(16) * 2.66`

= 0.16625 moles

= 0.16625 x
`6.023* 10^(23)`

=
`1.001 *10^(23)` atoms

In cotton ball ,

number of oxygen atoms present in 5g =
`(80)/(162.09) * 5`

= 2.467 g of oxygen

`(1)/(16) * 2.467`

= 0.15418 moles

= 0.15418
`* 6.023 * 10^(23)`

= 0.928
`\\ *10^(23)`

Hence , the glass stirring rod contains more number of oxygen atoms .