Use representations [A] through [I] in Figure P 7.6 to answer questions a-f. a. Which represent

Use representations [A] through [I] in Figure P 7.6 to...

Key Concepts

Combustion Reactions

Combustion is a chemical reaction in which a substance reacts with oxygen to form products such as carbon dioxide and water, often releasing energy in the process. This concept is key when analyzing how different compounds yield various quantities of reaction products under combustion conditions.

Percent Element by Mass

Calculating the percent by mass of an element within a compound helps in comparing different compounds quantitatively. By determining which compound has a higher percentage of an element, such as sulfur or oxygen, one can infer differences in chemical and physical properties, which is vital in both analytical chemistry and environmental studies.

Atmospheric Reactions and Acid Formation

Some compounds can react with water vapor in the atmosphere to form acids, such as sulfuric acid. This concept is essential for understanding environmental chemistry and the impact of chemical emissions on atmospheric processes and acid rain formation.

Percent Composition

Percent composition refers to the percentage by mass of each element in a compound. It provides insight into the relative abundance of each constituent element, and compounds with the same percent composition will have the same empirical formula even if their molecular formulas differ.

Empirical Formula

The empirical formula of a compound represents the simplest whole-number ratio of the elements present. It is derived by reducing the subscripts of the elements in the molecular formula, and different compounds can share the same empirical formula if their elements are arranged in the same basic ratio, despite differing in molecular mass or structure.

Molecular Yield in Combustion

The number of molecules produced during a combustion reaction depends on the composition of the original compound. For example, compounds with a higher number of carbon atoms will typically produce more carbon dioxide molecules upon complete combustion, making it important to assess molecular structures to predict product yields.

Transcript

00:01 Alright, so this is a pretty long -winded question, and it's got a bunch of different parts.

00:04 So we're going to have to go over each one individually.

00:08 We're giving this figure with non -compounds.

00:10 And you guys see here, we're going to have carbon.

00:13 They're the black atoms.

00:16 Oxygen in red, hydrogen in white, nitrogen is blue, and sulfur is yellow.

00:25 So i've gone ahead and written out the molecular formula for each of these, just to make it easier.

00:30 And you could look these up or you could go ahead and count the molecules here whatever you're allowed to do and so for the first question it wants to know which one which ones of these have the same empirical formula so so we don't have to go ahead and write out all of these let's think of it intuitively for a second so we have h2o which is kind of standing on its own here it's water and none of these other compounds are going to share an image empirical formula.

01:02 So we can mark that one out.

01:05 We have these two similar compounds, which are n -o -2 and n205.

01:09 Now, neither one of these can be simplified anymore.

01:13 So these are going to be two separate compounds with different empirical formulas.

01:17 It's the same here for sulfur dioxide and sulfur trioxide.

01:21 Those can't be simplified anymore.

01:23 So it's in between these four on the edges here, acetic acid, diethyl ether, arithrose, and butanol.

01:30 So let's go ahead and and write out the molecular formula for these four.

01:38 It consists of.

01:39 So for cetic acid, which is a, we're gonna have two carbon.

01:45 It's written a bit differently here, because this is how you write acids with the c -o -o -h.

01:51 Show that they have this acid group on the end here.

01:55 But we're gonna write it out this way and just count all the compounds, or count all the atoms.

02:01 Count all the atoms and then we have four hydrogens and finally we have two oxygen and so that'll be acetic acid and we'll give the empirical formula for this and i have everything divided by two as it can be simplified and it's c h2o okay for diethyl ether we're gonna have c4 h10 we can't divide this oxygen molecule anymore so this is as simple as it's going to get for erythros or erythros you could see that we have four carbons eight hydrogens and four oxygens everything here can be divided by four so we're going to simplify this we're going to get one c two h's and so you can see here that acetic acid and erythros have the same empirical formula even though they're not the same compound and so this would be one of the answers to our question so we'll go ahead and say a and g or one pair okay so here we have butanol which is c4 h10 and you can see here that's the same as diethyl ether so diethyl ether ambutinol with the same empirical formula all right so the next one it's going to ask which ones have the same percent composition again we're going to have to think about this intuitively first so we don't have to do all the math so as you can see here and s .o3 and s .o .2.

04:26 They're the only ones that could, of sulfur compounds that could possibly have the same percent composition because they're the only ones with sulfur.

04:34 So this one has two oxygen and this one has three oxygen.

04:38 So we know that they can't have the same percent composition since they only have one sulfur each.

04:47 So next for our nitrogen compounds, we have n -o -2 and n2 -205.

04:52 So we can go ahead and write this out or we can think about it but we'll go ahead and write it out so we have n02 and n205 here we'll have seven two plus five we'll have seven molecules and here we'll have three one two three so we'll have one -third nitrogen here and two seventh nitrogen here so one -third of nitrogen two -sevenths of nitrogen these are not the same so we know these aren't going to have the same percent composition.

05:28 So water kind of stands on its own here again.

05:30 So it's down to these four molecules again.

05:33 So i've gone ahead and written out the molecular formula for these compounds, starting with acetic acid and then diethyl ether, erythros, and then butanol.

05:42 It's going to be easiest to visualize if we make a chart of all the molecules or all the atoms within the molecule.

05:51 So first off in the first column, i'm going to do the total number.

05:54 Of atoms, then the number of carbon atoms, the number of hydrogen atoms, and the number of oxygen atoms.

06:02 So here for acetic acid, we add all of these up, and we get eight molecules in all, with two carbon, four hydrogen, and two oxygen.

06:16 Here in the second one of diethyl ether, we're going to add all of these up, and you can see that we get 15 atoms in all with four carbons, 10 hydrogens, and one oxygen.

06:30 Now let's go on to the next one.

06:32 Erythros, if you add out 4 plus 8 plus 4 again, you'll see that we get 16 total atoms with 4 carbons, 8 hydrogens, and another 4 oxygen.

06:47 And then finally for butanol, we add all these up, we get 15, and then that'll have 4 4 carbons, 10 hydrogens, and 1 oxygen.

07:03 Now, if we look at this chart, you can see that diethyl ether and butanol already have the same molecular formula.

07:11 So since this is the case, they're going to have the same percent composition.

07:16 So one of our pairs of answers is going to be c and i.

07:21 So the next part of our question asks, which one of the compounds produces the most carbon dioxide when it undergoes a combustion reaction? now, a combustion reaction is when a molecule interacts with oxygen to produce carbon dioxide and water.

09:17 So i've written out some formulas here, and i've added our molecule on the left side.

09:21 This first one is acetic acid.

09:24 The second one is diethyl ether and butanol, since they share the same number of atoms.

09:29 For each element.

09:31 And our last one here is erythros.

09:35 So what we need to do is balance this equation and determine which ones have the most carbon dioxide.

09:46 So here we're going to have to start balancing.

09:48 Okay, so on this side we have two carbon.

09:52 And on this side, we have one carbon.

09:54 So we're going to put a two in front of the carbon dioxide.

10:00 And we see let's check the left side and see how many hydrogen we have we have four hydrogens over here and we have two hydrogens over here so we're going to need to put two right there to make four and we'll equal out so next let's count the oxygen we'll have two right here two right here which is four and then we get two times two which is four on the side plus another two which is six so we have six on the right side, four on the left side.

10:35 So we need to add two more, and so that will be four, five, six.

10:41 And so we have six on both sides, and this is a balanced equation.

10:45 So the combustion of acetic acid will produce two molecules of carbon dioxide per one molecule of acetic acid.

10:56 So for the next one, the combustion of diethyl ether are butanol.

11:01 We're going to have to balance this equation and see on this side we have four carbons and on this side we have one.

11:09 So we're going to have to put four there to get four carbons on both sides.

11:15 We have 10 hydrogens on the left side, two hydrogens of the right side.

11:19 So we'll put a 5 right here since 5 times 2 equals 10 and we'll equal out the hydrogen.

11:26 So we have this one oxygen plus these two oxygens right here.

11:30 So that's 3 over here.

11:31 But right here we have 4 -02 molecules which is 8 and then we will have 5 more oxygen molecules so we have 5 plus 8 is 13 we have so we need 13 on this side we have one right here so we will need a 6 right here to make 12 and then 13 and we can see that this is a balanced equation we have four carbons right here, four carbons on the right side.

12:08 10 hydrogen the left side, 10 hydrogens on the right side.

12:13 And finally, we have 1 plus 12, which is 13 oxygens on the left side.

12:20 And then on this side, we have 8 plus 5, which is 13 oxygens on the right side.

12:26 So finally, we'll go to butanol...

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