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

Answer:

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

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 = $\frac{32}{60.07} \times 5$

                                                                 = 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 cotton ball ,

number of oxygen atoms present 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 $\\ \times10^{23}$

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