F.13

Isaac Eyler 1E
Posts: 12
Joined: Fri Sep 29, 2017 7:04 am

F.13

F.13 says: "In an experiment, 4.14 g of phosphorus combined with chlorine to produce 27.8 g of a white solid compound. (a) What is the empirical formula of the compound? (b) Assuming that the empirical and molecular formulas of the compound are the same, what is its name?"

At first glance, it seems like if I just subtract the 4.14 g from the 27.8 g, that would give me the mass of Chlorine present (which is what the solution manual says to do). From there I could divide by the molar mass of the particular element to find moles, and then eventually find the empirical formula. However, doesn't this idea of finding the mass of Cl by subtraction (27.8 g - 4.41 g = Cl in grams) go against the idea of a limiting reactant? If the one reactant is 4.14 g, then shouldn't the maximum amount of mass possible in the product be the same?

Chem_Mod
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Joined: Thu Aug 04, 2011 1:53 pm
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Re: F.13

To clarify the confusion, I recommend you do everything in moles. Mole is the fundamental unit that we use in chemistry, and by converting the given masses to moles and then writing out the chemical equation of the process described, you can easily see what is happening, which is the limiting reactant, and how much material you are forming! This question assumes that the reactant molecules are interacting in the ideal stoichiometric ratios, and that the total mass is conserved. I hope this helps!

aaron tang 2K
Posts: 49
Joined: Thu Jul 27, 2017 3:01 am

Re: F.13

you are on the right track by subtracting 27.8 grams by 4.14 grams of P. That will give you 23.66 grams of Cl, so all you have to do is divide each of the element by it's mass.
P: 4.14/30.974 = .1336604894
Cl: 23.16/75.45 = .6533145275
Next, you would divide the smallest amount into each of them, so P would be 1 and Cl would be 4.8878657. You can round Cl to 5. 1 and 5 are the ratio for your empirical formula.

a. P1Cl5
b. Phosphorus Pentachloride

Claudia Luong 4K
Posts: 59
Joined: Fri Sep 28, 2018 12:25 am

Re: F.13

We already know that we have 4.14 grams of phosphorus, so we only need to solve for grams of chlorine.
We know we have 27.8 grams of product, so 27.8 g product - 4.14 g P = 23.66 g Cl2

Convert into moles:
4.14g P x 1mol/30.97g = .134 mol P
23.66g Cl2 x 1mol/70.9g = .334 mol Cl2

Now, divide both values by the smallest number of moles. In this case, it is .134
.134/.134 = 1
.334/.134 = 2.49

Multiply both numbers by 2 to get whole numbers, and we get P2Cl5.
Therefore, the name of this formula is diphosphorous pentachloride.

gwynlu1L
Posts: 62
Joined: Fri Sep 28, 2018 12:19 am

Re: F.13

Be careful to use the right molar mass for the calculations. Pay attention to the wording of the question. Although chlorine usually exists as chlorine gas, Cl2, the question doesn't say that it is chlorine gas. Thus, you would use the molar mass of just chlorine, or 35.45 g/mol.

Sang Hyoun Hong 3G
Posts: 31
Joined: Fri Sep 28, 2018 12:27 am

Re: F.13

I feel like you're confusing the idea of conservation of mass with the stoichiometric coefficients of the equation. In order to solve this problem, you are allowed to subtract the 4.14 grams of Phosphorus from the product because mass is conserved. So, if you subtract 4.14 from the given 27.8 grams of product, you will find that there is 23.66 grams of Chlorine.
With the masses of these elements in grams, you can find the number of moles by using the molar masses of the elements from the periodic table.
4.14g P(1 mol P/30.97g)=0.1337 mol P
23.66g Cl(1 mol Cl/35.45g)= 0.6674 mol CL
After that, you divide by the smaller number to find the empirical formula.
0.1337molP/0.1337=1 mol P
0.6674molCl/0.1337= 5 mol Cl

So the empirical formula is PCl5
the name for this is Potassium Pentachloride

Nina Do 4L
Posts: 61
Joined: Fri Sep 28, 2018 12:27 am

Re: F.13

I began the question by turning the given 4.14 g P into moles.

(4.14 g P)/(30.97g P) = .13367 moles P. (I got 30.97 g from its molar mass on the Periodic Table).

Then I went back to the question and subtracted the total mass of what the equation produced, 27.8 g, and subtracted it with the original given 4.14 g P to get me the grams of Cl. I got 23.66 g Cl on the products side. I then used the 23.66 g Cl and converted it to Moles.

(23.66 g Cl)/(35.45 g Cl) = .66741 moles Cl. (Got 35.45 from PT)

I then looked at what I've solved with both P and Cl to see which is less. Phosphorus was less, with only .13367 moles so I divided it by that. P got me 1, and .66741/.13367 got me 5 for Cl.

In conclusion, the formula would be PCl5.

Phosphorus Pentachloride