Heat Stoichiometry

What do we do when a stoichiometry problem involves heat? Here's an example:
{Note - for this example, I'm using very limited sig figs on purpose to make the math easier to follow.}

28g of nitrogen gas reacts with 7g of hydrogen gas to produce ammonia gas. How much heat is liberated by this reaction?

Approach it just like any other stoichiometry problem following the 4 steps (review here):
1. Write a balanced chemical equation. The only twist here is that since we're dealing with heat, we'll also need the {delta}H for the reaction as well.

N₂(g)  +  3 H₂(g)   -->   2 NH₃(g)

{delta}H_rxn = 1(0 kJ/mol) + 3(0 kJ/mol) + 2(-46 kJ/mol) = -92 kJ/mol

2. Find moles. Again, there's a little twist here... this is a limiting reactant problem. So let's find moles of ammonia from each reactant:

(28g / 28g/mol N₂) (2mols NH₃(g) / 1mol N₂(g)) (17g/mol NH₃) = 34g NH₃(g)

(7g / 2g/mol N₂) (2mols NH₃(g) / 3mol N₂(g)) (17g/mol NH₃) = 40.g NH₃(g)

So we know that N₂(g) is the limiting reactant.

3. Use the mole ratio from the balanced chemical equation to find "moles of interest". We already kind of blew past this in the last step... let's back up a bit now that we know the limiting reactant.

(28g / 28g/mol N₂) (1mol of "reaction" / 1mol N₂(g)) = 1 "mole of reaction"

{What's a "mole of reaction"?}

4. Convert moles of interest to whatever you're looking for:

(28g / 28g/mol N₂) (1mol of "reaction" / 1mol N₂(g)) (92kJ/mole of reaction) = 92kJ liberated

Same process, different details. Learn the process and the details will seem less detailed.

Balanced chemical formulas

An email question:
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I just had some confusion regarding the homework for Chapter 2. I am confused about how to get the ions for the balanced chemical formulas. An example that I got incorrect was: 

The balanced chemical formula for magnesium chlorate contains___magnesium ion(s) and___chlorate ion(s)

I'm not sure how to go about getting the answer for this example. 

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Balancing chemical formulas is a critical skill, but it's also one that many students struggle with until that magic moment when it clicks. Let's see if we can break this question apart a bit...

Many monoatomic ions have charges that can be reasonably predicted based upon where they are on the Periodic Table. Basically, we know from observation that noble gases have very stable numbers of electrons (we'll get deeper into that later in the semester...), so other elements in the Periodic Table tend to lose or gain electrons to get to the same number of electrons as a noble gas. This is especially true when the neutral element has a number of electrons that's pretty close to the number in the noble gas.

Metals:

The alkali metals (lithium, sodium, potassium, etc) almost always lose 1 electron to become +1 in ionic formulas, the alkaline earth metals (beryllium, magnesium, calcium, etc) are almost always +2, aluminum and gallium are almost always +3.

Transition metals and the metals toward the bottom of the B/C/N/O/F columns can have more than one common charge, so those usually have to be specified with a roman numeral; nickel(III) is +3, ruthenium(II) is +2, manganese(V) is +5, etc. The one common exception is zinc, which is (almost) always +2 in ionic compounds.

Non-Metals:

These have similar trends, but now as anions. Halogens (fluorine, chlorine, etc) gain an electron to become -1 charged halides in ionic formulas; chalcogens (oxygen, sulfur, etc) gain 2 electrons to become -2; pnictogens (N, P, As, etc) gain 3 electrons to become -3.

In the context of this specific problem, it's reasonable to assume that the magnesium is Mg⁺².

The other half of this specific problem is a polyatomic ion. There are a number of trends that we could explore in polyatomic ions, but at this point in this course, this is something you should just memorize. Find the list of polyatomic ions in your book, make up some flashcards, and just drill these into your brain. Chlorate is ClO₃^-1.

When writing balanced ionic formulas, the "balance" refers to the balance of charge... nature does not allow random positive or negative charges to be running loose, so for every positive charge in an ionic formula, we must have a corresponding negative charge. Since we have a +2 cation and a -1 anion, we need two anions so the +2 charge of the cation is "balanced" by two negative charges from the anions.

Balance in all things. A yin for every yang.

Sig Fig question - addition and subtraction

I got a question via email:
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Would you help me with the question in the textbook, p.40, Example 1.4 (b)?

1. 4 (b) Subtract 421.23 g from 486 g.

I thought 486's sig fig is 3 and 421.23's is 5. Doesn't the answer have to coincide with the fewest decimal places? The answer was 64.77 so I rounded it up to 64.8, and its sig fig is 3. I don't understand why the correct answer is 65g, its sig fig is 2 then.

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Significant figures are something that takes practice, so let's walk through this one step by step. Before we get too tied up in the math, remember that the whole reason we evaluate significant figures is to help us see where the uncertainty is in a number that we report.

First, let's just "do the math" given in the question... that's the simple part.

486g - 421.23g = 64.77g

Now we get to the more critical part... evaluating the information that we are using to get that answer. For addition and subtraction, we round the result of the addition or subtraction to the least-precise decimal place from the inputs. The "least precise" decimal place tells us where the uncertainty starts.

486g (in this case) really means that the true value is less than 487g but greater than 485g. That's the uncertainty we are trying to keep track of with significant figures. Similarly, 421.23g is really less than 421.24g but greater than 421.22g. When we subtract these values, we round to the "ones" digit because that is the least-precise input.

It's not that unusual to gain or lose sig figs when using the addition & subtraction rules. When adding and subtracting, don't worry as much about counting sig figs, just make sure you're rounding to the correct position in your result. 6.3 and 5.9 each have 2 sig figs, but when you add them together, the result (12.2) has 3 sig figs because we round to the least precise position, in this case the tenths place.

When multiplying and dividing, that's when we count the sig figs. Count the sig figs of the inputs and round the result to the same number of sig figs as the input with the fewest sig figs.

Email question 2013-12-12

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Hello Dr. Bodwin. I've been working on some old exams for studying and on the exam 1a from fall 2011, the last question is asking about the empirical formula, and I was just wondering when solving for each part, where do the last two numbers come from?
For example,
C -> (71.98 g)/(12.011 g/mol) = 5.993 mols -> 4.5 -> 9
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Let me pull up the whole problem from the exam key:

{from:  http://www.drbodwin.com/teaching/exams/c150fe1ak.pdf}

When determining empirical formulas from percent composition data, the first step is to assume that you have 100g of sample. You can assume any amount of sample you like, but 100g simplifies things a little because if I have 100g of a sample and I know that 71.98% of that sample is carbon, then there must be 71.98g of carbon in the sample. Using the percentages given in the problem, we know that the "100g" sample contains 71.98g of carbon, 6.711g of hydrogen, and 21.31g of oxygen. Grams are great, but we want to count the number of different atoms, so we need to convert grams to moles... That's the first step in the calculation that is shown.
Once we have moles, we know the relative amounts of each element present in the sample, and we can write a balanced chemical formula:

C_5.993H_6.658O_1.332

Hmm, that doesn't look quite right... But at this point, we have the correct relationship between the moles of each element, so we can force those relationships to be whole numbers by dividing all of them by the smallest one. Essentially, we're saying "what if 1.332 actually represents 1 oxygen atom?" Dividing them all gives the forumla:

C_4.5H₅O₁

Still not perfect, but it's a lot more "normal" looking than the first formula. Again, since we know that the relationship between moles is correct here, we can multiple all the subscripts by something that gives us a nice, round, whole number ratio. If we double everything, we get:

C₉H₁₀O₂

And now we have a good empirical formula for this elemental analysis.

Other questions? Let me know.

Carbons in a propane sample

Today in class we looked at a problem that some of you didn't quite get to the end of by the time class ended. Here it is. If you haven't already worked it through, give it a good try before you jump ahead to the answer...

How many carbon atoms are in a 37.43L sample of propane gas at 17.52°C and 1.472atm?

This starts out as an Ideal Gas Law problem with a single set of conditions.

PV = nRT

Plugging in the values from the problem:

(1.472atm)(37.43L) = n(0.08206^L.atm/_mol.K)((17.52+273.15)K)

n = 2.3099mols of C₃H₈(g)

{NOTE: I'm in the middle of the problem, so I'm not rounding for significant figures yet, 

but it looks like 4 sig figs would be good at this point…}

Each propane molecule contains 3 carbon atoms, so each mole of propane molecules contains 3 moles of carbon atoms.

(2.3099mols C₃H₈) (3mols C/1mol C₃H₈) = 6.92974mols C

(6.92974mols C) (6.022x10^{23 C atoms}/_{mol C}) = 4.173x10²⁴ C atoms in the sample.

DISCLAIMER: This problem assumes that propane is behaving as an ideal gas under these conditions. There's a pretty good chance that it would not be ideal under these conditions, but for the purposes of this problem let's assume it is.

Lab Reports

The exact requirements for a lab report will vary from field to field, class to class, even instructor to instructor. If you're looking for an example of a "good" lab report for my Gen Chem class, try this one:
http://www.drbodwin.com/teaching/genchemlab/iodinationlabreport12a.pdf
The most common problem I see in lab reports is that students don't always explain the experiment and its results in a way that makes it (somewhat) clear that the concepts behind the experiment are understood. The purpose of a Gen Chem experiment is almost never "We collected a bunch of numerical data, made some observations, and calculated/determined this result". What does that result mean? How is that result related to the concepts we talked about in class? How can that result help inform the exercises and exam questions you'll see in the classroom?
One of the harder things for students to do is get a feel for "reasonable" answers because Gen Chem level students don't have a lot of experience looking at these answers. Activation energy is a great example of this. If you have no feeling for how activation energy relates to the observed rate of a reaction, you might calculate an activation energy of 25 J/mol for some problem. Is that a fast reaction or a slow reaction? If you've only every done on-paper activation energy problems, that might be a hard question to answer. The advantage of doing experiments is that you have personally observed what happened, you've gained experience that will help you make some of these judgement calls. For the iodination of acetone experiment, the reaction is fast enough to easily observe, but it's not so fast that it blows up in your hand. The activation energy for the iodination of acetone is somewhere around 80-100 kJ/mol. If 80-100kJ/mol is the activation energy for a reaction that's "kinda fast, but not super fast", what do you think about that 25 J/mol activation energy reaction? {Pay attention to units.}
Lab experiments are a great way to build your knowledge base. When you're writing a lab report, think about the bigger picture and show the reader that you've recognized the link between classroom exercises and the first-hand experience you've had in the lab.