Excercise F13

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Jasmin Argueta 1K
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Joined: Fri Sep 28, 2018 12:16 am

Excercise F13

Postby Jasmin Argueta 1K » Tue Oct 02, 2018 11:45 pm

I was figuring out problem F13, which regards solving for the empirical formula. The way I solved the problem was the same way Professor Lavelle did in class, figuring out the percent compositions of Phosphorus(14.89%) and Chlorine(85.1%). After I found the moles of phosphorus(.48 moles) and chlorine (2.4 moles). then divided both by .48 to get PCL5. The answer is correct but in the solutions manual there is a much faster way of doing it and was wondering if anyone could explain that process which involves diving (4.14/30.97) and (27.8/35.45). Where did the denominators come from?

ian_haliburton_1f
Posts: 61
Joined: Fri Sep 28, 2018 12:24 am

Re: Excercise F13

Postby ian_haliburton_1f » Tue Oct 02, 2018 11:57 pm

For this question, as you are given the total mass of the compound, you do not need to concern yourself with mass percentage composition. Rather, you can divide the actual available masses of the elements by their molar masses to find how many moles of each there are. This is the first step that the solutions manual shows. At this point, divide each mole value by the smallest mole value (here Phosphorus at 0.134 mol) to determine the ratios of elements for the empirical formula.

gillianozawa4I
Posts: 31
Joined: Fri Sep 28, 2018 12:27 am

Re: Excercise F13

Postby gillianozawa4I » Wed Oct 03, 2018 12:07 am

From my understanding, the solutions manual combined many of the steps we learned in lecture, but it still solved it in a similar way. First, they found moles of both P and Cl by dividing 4.14 g P (given) by its molar mass (30.97 g/mol) and by dividing 23.66 g Cl (total mass - mass of P) by its molar mass (35.45). We then have .134 mol of P and .667 mol of Cl. We then divide both numbers by the smallest value (.134 mol) to get whole numbers. 0.134/0.134=1, and 0.667/0.134 = 4.98, which is close enough to 5. The formula is therefore PCl5.


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