Subscribe $4.99/month

Un-lock Verified Step-by-Step Experts Answers.

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

Our Website is free to use.
To help us grow, you can support our team with a Small Tip.

All the data tables that you may search for.

Need Help? We got you covered.

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Products

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Need Help? We got you covered.

Chapter 2

Q. 2.8

Relating Number of Atoms, Amount in Moles, and Mass in Grams

In the sample of sulfur weighing 4.07 g pictured in Figure 2-18, (a) how many moles of sulfur are present, and (b) what is the total number of sulfur atoms in the sample?

Analyze
For (a), the conversion pathway is g S → mol S. To carry out this conversion, we multiply 4.07 g S by the conversion factor (1 mol S/32.07 g S). The conversion factor is the molar mass inverted. For (b), the conversion pathway is mol S → atoms S. To carry out this conversion, we multiply the quantity in moles from part (a) by the conversion factor (6.022 × 10^{23} atoms S/1 mol S).

Fig 2.18

Step-by-Step

Verified Solution

(a) For the conversion g S → mol S, using (1/M) as a conversion factor achieves the proper cancellation of units. The result of this calculation should be stored without rounding it off because it is required in part (b).

? mol S = 4.07 g S \times \frac{1  mol  S}{32.07  \sout{g  S}}= 0.127 mol S

(b) The conversion mol S → atoms S is carried out using the Avogadro constant as a conversion factor.

? atoms S = 0.127 mol S \times \frac{6.022  \times  10^{23}  atoms  S}{1  \sout{mol  S}} = 7.64 \times 10^{22}  atoms  S 

Assess
By including units in our calculations, we can check that proper cancellation of units occurs. Also, if our only concern is to calculate the number of sulfur atoms in the sample, the calculations carried out in parts (a) and (b) could be combined into a single calculation, as shown below.

? atoms S = 4.07 g S \times \frac{1  \sout{mol  S}}{32.07  \sout{g  S}} \times \frac{6.022  \times  10^{23}  atoms  S}{1  \sout{mol  S}} = 7.64 \times 10^{22}  atoms  S

Had we rounded 4.07 g S × (1 mol S/32.07 g S) to 0.127 mol S and used the rounded result in part (b), we would have obtained a final answer of 7.65 \times 10^{22} atoms S. With a single line calculation, we do not have to write down an intermediate result and we avoid round-off errors.