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.

Summer 2020 - Gen Chem Information

Thank you for your interest in General Chemistry during Summer 2020. I am excited to be teaching this course online this summer and look forward to meeting a very diverse array of students! A couple notes and resources to help you all get started:

1. This is a 3-credit, lecture-only course. If you need to take a Gen Chem lab course, you will have to pick that up at another time.

2. There is a pre-requisite math requirement to register for CHEM 150. If you try to register and require an over-ride to get past this hurdle, please send me an email with your DragonID (8-digit number, starts with "14######" or "15########") and a description of the math courses you have taken.

3. The course is asynchronous which means that there are no specific meeting times. I will try to keep you on track during the course, but one of the biggest challenges for most online learners is being responsible for your time management. For the summer courses, I would expect msot students to require 25-35 hours per week to do all the assigned readings, practice, homework, and quizzes.

4. We will be using the Open Educational Resource (OER) textbook from OpenStax, Chemistry 2nd edition. (https://openstax.org/details/books/chemistry-2e) This book is freely available to download as an electronic copy

5. If you prefer a physical textbook, you can get a print copy of the text for a very reasonable cost. The easiest place I have found it is on Amazon (Amazon link). Based upon some of the feedback I see in the comments, it looks like there may be some dissatisfaction with the paperback version of the text, so it might be worth the few extra dollars to get the hardcover. A physical textbook is not required for the course; if you are comfortable using electronic textbooks, the freely available download will serve you perfectly well.

6. Gen Chem I will address chapters 1-9 (approximately) and Gen Chem II will tackle chapters 10-17 & 21.

I think that addresses most of the common questions I've been getting. I hope you are all having a safe and productive Spring, and I look forward to "seeing" all of you this summer.