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

Question

4 months ago

11

more oxygen atoms?

1 answer:

Anarel [2.9K] · Answer 1 · 2026-02-24T02:40:41+03:00

Answer:

The sample with more oxygen atoms is a glass stirring rod.

Explanation:

Referring to the periodic table,

the glass stirring rod is composed of Silicon dioxide whereas the cotton ball is made of cellulose.

The molar mass of the glass stirring rod $SiO_{2}$ = 60.08 grams/mole.

The molar mass of the cotton ball $C_{6} H_{10}O_{5}$ = 162.09 grams/mole.

Considering the total number of molecules for oxygen is 32,

therefore,

In the glass stirring rod,

oxygen atoms contained in 5g = $\frac{32}{60.07} \times 5$

= 2.66 g of oxygen

= $\frac{1}{16} \times 2.66$

= 0.16625 moles

= 0.16625 x $6.023\times 10^{23}$

= $1.001 \times10^{23}$ atoms

In the cotton ball,

oxygen atoms contained in 5g = $\frac{80}{162.09} \times 5$

= 2.467 g of oxygen

= $\frac{1}{16} \times 2.467$

= 0.15418 moles

= 0.15418 $\times 6.023 \times 10^{23}$

= 0.928 $\\ \times10^{23}$

Thus, the glass stirring rod has a greater quantity of oxygen atoms.