Limiting Reactant Question M19

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Chem_Mod
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Limiting Reactant Question M19

Postby Chem_Mod » Thu Sep 22, 2016 4:08 pm

For those who had questions about problem M19, I encourage you to read pages F111-F113 in your book, section M3. Page F113 has the step-by-step procedure to solving this problem, and the paragraph on page F111 explains why you calculate the mass of oxygen last in a combustion reaction problem (Hint: it has to do with the limiting reagent in this problem!)

Chem_Mod
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Re: Limiting Reactant Question M19

Postby Chem_Mod » Mon Sep 26, 2016 10:39 am

Very helpful. Thanks.

Rosalva_Mejia_41
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Re: Limiting Reactant Question M19

Postby Rosalva_Mejia_41 » Mon Sep 26, 2016 3:44 pm

Within the process, why do we have thr add all of the mole amounts and subtract it to caffeine's mass?

Jineava_To_3N
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Re: Limiting Reactant Question M19

Postby Jineava_To_3N » Mon Sep 26, 2016 8:51 pm

I believe we can not subtract the total moles of the three compounds/element from the given caffeine mass because moles and masses are different units. Instead, we can subtract the total mass of C, H, and N from the given caffeine mass to find the mass of oxygen. We know to find oxygen because the problem states it is a combustion. Since nothing is lost of gained in a reaction, C, H, N, and O are all part of caffeine's chemical formula.
What we need to do is convert the given mass of CO2 into the mass of C in caffeine and the given mass of H2O into the mass of H in caffeine. (Nitrogen does not need to be converted because it is already given as an element.) Then we can add the mass of C, H, and N.

Given CO2 is 0.682 g we want to calculate the mass of C in caffeine.
In summary, we go from mass of CO2(given)->moles of CO2(step 1)->moles of C(step 2)->mass of C(step 3)
1) To get moles of CO2, we divide 0.682g by the molar mass of CO2. (0.682/44.016=0.0155 moles of CO2)
2) To get the moles of C in 0.0155g of CO2 we multiply by a ratio of 1 mol C/1 mol CO2 to get 0.0155 mole of C
3) We want 0.0155 mole of C in grams so multiply by the molar mass of C which is 12.011 g to get 0.186 g.

We repeat the same steps for finding the mass of H in the given mass of H2O.
1) mass of H2O->moles of H2O. Divide by H2O molar mass. (0.174/18.016=0.009658 moles H2O)
2) moles of H2O->moles of H. Multiply by a ratio of 2 mol h/1 mol H2O to get 0.0193 mol H
3) mol of H-> mass of H. Multiply 0.0193 by molar mass of H which is 1.008 g to get 0.01947 g of H.

Now we know the masses of C(0.186 g), H(0.01947 g), and N(given as 0.110 g). When we add these three masses, we get a total of 0.31547 g. When we subtract this total from the mass of caffeine(0.376 g), the resulting answer gives us the mass of oxygen in caffeine. (0.376-0.31547=0.06053)

Breakdown thus far:
0.186 g of C
0.01947 g of H
0.110 g of N
0.06053 f og O

Then, using the mass percentage composition of C, H, N, and O, we are able to find the empirical and molecular formula.
Hope this helps!


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