January 31st - Volume and Enthalpy: Are they Related?

We are well aware that volume at STP (0^oC and 101.3kPa) of any gas is represented by 22.4L / mol. We will utilise this in our conversations. This is not new, but don’t worry, another concept was introduced today. It was a hot topic known as enthalpy.

From previous science studies, we know that reactions that release energy and exothermic. Ones that absorb energy are endothermic.

Let's try a few examples that involve volume and stoichiometry. In the decomposition of 5.0 g of NaCl, what volume of chlorine gas is produced?


Start with the balanced equation: 2NaCl --> 2Na + Cl2 


Now we simply use conversion factors.
 5.0g NaCl (1 mol NaCl / 58.5 g) (1 mol Cl2 / 2 mol NaCl) (22.4L / 1 mol Cl2)
= 0.96L Cl2


Let's try another one. If 3.0L of Br2 is consumed in the reaction 2Li + Br2 -->2LiBr, how much Li was used? 


Simply, start with 3.0L and use appropriate conversion factors.
3.0L Br2 ( 1 mol Br2 / 22.4 L Br2) (2 mol Li / 1 mol Br2) ( 6.9 g Li / 1 mol Li)
= 1.8 g Li

If you require further practice, please refer to our previous examples and watch the video below.

With that now aside, we can have some fun. Enthalpy is described as the heat content of a system under a given pressure. It can also be defined as the energy stored in chemical bonds. There is also change in enthalpy, which is the amount of energy or heat absorbed in a reaction. Heat, on the other hand, is defined as the thermal energy in transit.

In class, we were given the following equation:

2C₈H₁₈ + 25O₂ à 16CO₂ + 18H₂O + 5076kJ

Enthalpy has units of joules. Since the enthalpy is on the right side of the equation, it means that this process is exothermic. Had it been on the left, it would mean that the reaction required heat and would therefore be endothermic. An exothermic reaction has a negative change in enthalpy. An endothermic reaction has a positive change in enthalpy. The pictures below will demonstrate why this is so.

Example

Determine the amount of heat released when 1 mol of octane reacts.

Begin by making a balanced equation:

2C₈H₁₈ + 25O₂ à 16CO₂ + 18H₂O + 5076kJ

1 mol C8H18 (5076 kJ / 2 mol C8H18) = 2538 kJ

Example

Using 2NH₃ + 46.2kJ à N₂ + 3H₂, determine the maximum number of N₂ moles produced in 712kJ of energy are used in the reaction.

712kJ (1 mol N₂ / 46.2kJ) =15.4 mol N₂

More on Enthalpy

The production of methane is shown in the equation below:

C(s) + 2H2(g) --> CH4 + 74kJ

This equation tells is that when we produce the one mole of CH4, we get a change of 74kJ.

The change in enthalpy of a system during a process that occurs under constant pressure, represented as /\H is equal to the head given off or absorbed by the system during that process. When talking about enthalpy, it is important to specify what state we are talking about. For example, the enthalpy of liquid water at 25C is different from gaseous water at 25C.

By convention the enthalpy change for a reaction /\H_rxn is taken to be the enthalpy of the system in going from reactants to products.

H_reaction = Sum (H_products) – Sum (H_reactants)

­

If the enthalpy of the products is less than that of the reactants, /\H will be negative in sign. If the products have a larger enthalpy than the reactants, /\ is positive and this is an endothermic process.

Example

Given the equation CH4 (g) + 2O2(g) à CO2(g) + 2H2O(g) + 802kJ, determine the change in enthalpy.

Since this is an exothermic reaction, we know that the change in enthalpy will be negative. Therefore, /\H = -802kJ per mole of CH4 reacted.As a consequence of the law of conservation of energy, the amount of heat associated with a reaction is directly proportional to the amount of substances involved. Thus, combustion of 2 mol of CH4 will produces 1604 kJ.

Example

How much heat is produced when 4.50g of methane gas are burned in a constant-pressure system?

To solve this, we will need to use stoichiometry.

802 kJ are produced when a mole of CH₄ is burned. A mole of CH₄ has a mass of 16.0g.

Heat produced = (4.50 g CH₄)(1 mol CH₄ / 16.0 CH₄)(802 kJ/mol CH₄)

=226 kJ

It is important to specify the states of the reaction. For example, if the product of methane were liquid water instead of gaseous, /\H would be -890kJ instead of -802kJ.

Hess’ Law of Constant Head Summation

According the this law, if a reaction can be carried out in a series of steps, /\H for the reaction will be equal to the sum of the enthalpy changes for each step (the enthalpy changes are additive).

·         For example, the enthalpy change for the combustion of methane to form carbon dioxide and liquid water can be calculated from /\H for the combustion of methane giving gaseous water:

o   CH₄ (g) + 2O₂(g) à CO₂ (g) + 2H₂O (g)                   /\H = -802 kJ

o   (Add)   2H₂O (g) à 2H₂O (l)                          /\H = -88 kJ

o   CH₄ (g) + 2O₂(g)+ 2H₂Oà CO₂ (g) + 2H₂O (l) + 2H₂O (g)  /\H = -890 kJ

o   Net equation:

o   CH₄ (g) + 2O₂(g) à CO₂ (g) + 2H₂O (l)                    /\H = -890 kJ

To obtain the net equation, the sum of the reactants of the two equations is placed on one side of the arrow, and the sum of the products on the other.

* Since H₂O (g) occurs on both sides of the arrow, it can be cancelled like an algebraic quantity that is on both sides of an equal sign.

The first law of thermodynamics, in the form of Hess’ law, teaches us that we can never expect to obtain more or less energy from a chemical reaction by changing the method of carrying out the reaction.

Heat of Formation

The heat of formation of a compound /\H_f is the enthalpy change involved in forming a mole of that compound from its constituent elements.

A standard enthalpy change /\H_f^o is one that takes place with all reactants and products in their standard states.

By convention, the standard heat of formation of the stable form of any element is zero. (This will be shown in the following example.)

Example:

Find the heat of this reaction:

4NH₃ (g) + 5O₂ (g) à 4NO (g) + 6H₂O (l)

1)    Find heat of formation of the reactants.

Gaseous ammonia has a heat of formation of -45.90 kJ / mol. (Look on Wikipedia, etc)

Since oxygen is in the form it always takes, its heat of formation is zero.

Gaseous NO is +90.29 kJ / mol.

Lastly, the heat of formation of water is -285.83 kJ / mol.

2)    Since heat of formation deals with “per mole” We need to multiply each heat of formation by the number of moles.

NH₃ = -45.90 x 4         NO = 90.29 x 4           H₂O = -285.83 x 6

3)    Solve for /\H_f.

/\H_f = 90.29 x 4 + (-285.83 x 6) – (-45.90 x 4) + 0

      = 361.16 – 1714.98 + 183.6

      = -1170 kJ for this reaction

This is possible because of Hess’ Law, which tells us that no matter which path we take to find the heat of formation, it will be the same.

Fun eh? Let’s do one more.

C₃H₈ (l) + 5O₂ (g) à 3CO₂ (g) + 4H₂O (l)

1)    Find the heat of formation of the reactants.

2)    Since heat of formation deals with “per mole” We need to multiply each heat of formation by the number of moles.

3)    Solve for /\H_f.

Propane = - 104.7 kJ / mol

O₂ = 0 kJ / mol

Carbon dioxide = -393.509 kJ / mol

Water = -285.82 kJ / mol

Propane = - 104.7 kJ / mol                             x1

O₂ = 0 kJ / mol

Carbon dioxide = -393.509 kJ / mol               x3

Water = -285.83 kJ / mol                                x4

/\H_f = /\products - /\reactants

      = (-1180 + -1143) – (-104.7)

      = -2218 kJ

Therefore, for every mole of propane that is combusted, 2218kJ will be produced.

C₃H₈ (l) + 5O₂ (g) à 3CO₂ (g) + 4H₂O (l) + 2218kJ

Posted by Andrew.

January 27th - Lab: Testing the Stoichiometric Method

Today, we did a lab. The problem: does stoichiometry accurately predict the mass of products produced in chemical reactions?

In this experiment, we disolved [2.00g] of Strontium nitrate in 50 mL of water and then reacted it with [3.00g] of Copper (II) sulphate. The product is a precipitate (Strontium sulphate) and Copper (II) nitrate. After mixing the solutions, the precipitate is separated by filtration, dried, and weighed.

Behold Copper (II) sulphate!

The balanced equation for the reaction is:

Sr(NO₃)₂   +   CuSO₄   à   SrSO₄   +   Cu(NO₃)₂  

If 2.00g of Strontium nitrate completely reacts, 1.74g of Strontium sulphate is produced. Here are the calculations:

2.00g   x   1 mol/(87.6g +  2(14.0g) +6(16.0g))   x   1/1   x   (87.6g + 32.0g + 4(16.0g))   =   1.74g

Carrying out the experiment, we first measured 3.00g of Copper (II) sulphate. We then dissolved it in 30mL of water. We did the same with 2.00g of Strontium nitrate. We poured the solutions together (remembering to pour along the stir rod) into a filter and funnel which emptied into an Erlenmeyer flask. It is important to note that we measured the mass of the filter paper before the filtration. After filtration, we poured the filtrate into the sink and placed the filter paper in the drying oven. We then measured the mass when it had dried. Subtracting the weight of the filter paper from the mass of both the precipitate and the filter paper, we found the mass of Strontium (II) sulphate.

Here is a collection of all of our observations:

• Mass of Copper (II) sulphate used: 3.00g
• Mass of Strontium nitrate used: 2.00g
• Mass of filter paper: 2.27g
• Mass of precipitate and filter paper: 4.13g
• Mass of precipitate: 1.86g

The number of moles of precipitate formed was .0101 mol. Here are the calculations:

1.86g  x   mol/183.7g   =   0.0101 mol

Strontium nitrate being formed.

According to our original calculations above, we should have had [g] of precipitate. Therefore, we have a slight margin of error. The percent error is:

100((1.86g - 1.74g)/1.86g)   =   6.45%

Some reasons for our percent error are:

1. The beaker was contaminated. The previous class left Strontium nitrate inside the beaker, which added to our mass of 2.00g. 
2. The filter may not have completely dried. The water that remained in the filter could have contributed to the extra mass.
3. Some estimation was used. Human error is also possible. 

Posted by Michael.

January 25th - Mass to Moles to Moles to Mass

Today, we were able to learn how to go from one mass to another mass. We've already learned all the tools we need to do this, and all we need is this chart:

To correctly move from Mass A to Mass B, we must first go to moles of both substances. To refresh, here is a detailed step-by-step procedure for these type of questions.

1)    Write a balanced chemical formula.

2)    Find the mass of your first substance, and using molar mass, convert it to moles.

3)    Using mole to mole conversions, change mole A to mole B (using coefficients of the balanced equation). Remember, always put what you need over what you have.

4)    Finally, move from moles B to mass B by simply using the molar mass of the new substance.

If you're really good, numbers two through four can be accomplished in a single dimensional analysis equation.

Let’s try an example.

Ex.)  Limestone (calcium carbonate) is heated to produce calcium oxide and carbon dioxide. How much limestone is needed to produce 10kg of calcium oxide?

CaCO₃ à CaO + CO₂

10 000g CaO (1 mol CaO/56.1g CaO) (1 mol CaCO₃ / 1 mol CaO) ( 100g CaCO₃ / 1 mol CaO)
= 1.78 x 10⁴ g or 17.8kg.

That is all you have to do! All it takes is one additional conversion from our previous equations. And since we have all the tools needed, it should be a snap!

For additional practice, the following video is provided:

Posted by Andrew. 

January 23rd - Moles to Mass & Mass to Moles

Today, adding onto our last stochiometric lesson, we learned how to convert from grams to grams in a chemical equation. We must remember that converting from mass A to mass B must always go through moles A and moles B. Confused? This chart should clear it up:

Don’t worry, it’s easy! Converting from moles to mass only requires one additional step. In fact, it’s so simple that we can jump right into a few examples:

Ex. 1) Use the following formula to answer a) and b):

2H₂ + 0₂ à 2HOH

a) How many grams of HOH are produced when 2.50 moles of oxygen are used?

First, we simply convert from moles of oxygen to moles of HOH (we already know how to do this).

2.50 mol   x   (2/2)   =   5.00 mol of HOH

Next, we find the molar mass of oxygen. 2(1.0) + 16.0 = 18.0 g/mol. Then, like we learned in the last chapter, we convert!

5.00 mol   x   18.0 g/mol   =   90.0g of HOH

b) If 3.00 moles of HOH are produced, how many grams of oxygen must be consumed?

Using the same steps, we simply switch around the calculations.

3.00 mol   x   (1/2)   =   1.50 mol of O₂

We then find the molar mass of an oxygen molecule. 2(16.0) = 32.0 g/mol.

1.50 mol   x   32.0 g/mol   =   48.0g of O₂

Wow! That was fun! Now, here's something that'll get you in the mood for example 2:

Ex. 2) Iron (II) phosphate reacts with Tin (IV) nitride to produce Iron (II) nitride and Tin (IV) phosphate. How many moles of Tin (IV) nitride are needed to produce 0.500 mol of Iron (II) nitride? How many grams is this.

First, we must write the chemical formula.

2Fe₃(PO₄)₂ + Sn₃N₄ à 2Fe₃N₂ + Sn₃(PO₄)₄

Next, following what we did above, we convert from moles of Iron (II) nitride to Tin (IV) nitride.

0.500 mol   x   (1/2)   =   0.250 mol of Sn₃N₄

To find how many grams this is, we convert using the molar mass of Sn₃N₄. 3(118.7) + 4(14.0) = 412.1

0.250 mol   x   412.1 g/mol   =   103.0 grams of Sn₃N₄

Great job, everyone!

Here’s a sweet video for more practice:

Posted by Michael.

January 16th - Mole to Mole to More Moles!

What would you do if you needed to know how much table salt you can produce with the handy eight grams of sodium you have? Well, today will we find out. First of all, we need to find some chlorine gas. Be careful, it may be is poisonous.

Now that we have all of our materials, we are set to make some salt. We should now determine the balanced chemical equation for salt, and it is:

2Na _(s) + Cl_{2 (g)} --> 2NaCl

We always were taught that the coefficients are there to abide by Datlon’s atomic theory, that atoms cannot be created nor destroyed. While this is true, a balanced chemical formula contains a more powerful, hidden meaning. Do you want to know? We’ll tell you anyhow, even if you aren’t interested. The coefficients represent the number of moles. For example, if you react two moles of sodium with one mole of chlorine gas you will receive two moles of salt.

So back to our original problem: How much NaCl can be produced with 8.0 grams of sodium? Firstly, we will convert the sodium into moles.

8.0 g Na (1 mol Na/ 23.0g Na)

= 0.35 mol Na

Now we will use the mole to mole conversions to do the next step. Thus far we know that for

2 moles of Na used, 2 moles of NaCl are produced. So we can set up another equation:

0.35 mol Na (2 mol NaCl / 2 mol Na)

= 0.35 mol NaCl

= 20 grams NaCl

There you have it! By using eight grams of sodium and having the chlorine provided, you have successfully made twenty grams of salt. The conversion factor that was used above can be used. However, we learned a quicker method in class. The conversion is:

what you need

what you have

If you don’t believe us, try it out yourself. Here is a video for more information:

Posted by Andrew.

January 16th - An Intro to Stoichiometry

Today, we started our new unit! Of course, that unit is stoichiometry!

What is stoichiometry, you ask? Well, stoichiometry is a branch of chemistry that deals with the quantitative analysis of chemical reactions. It is a generalization of mole conversions to chemical reactions. Simply put, this means that it is an analysis of the ratio of moles found in a chemical reaction.

Just as Mr. Doktor said, understanding the six types of chemical reactions is key to understanding stoichiometry.

The key to stoichiometry.

The six types of reactions are synthesis, decomposition, single replacement, double replacement, neutralization, and combustion.

First, if you need practice balancing equations, now is the time:

Moving on:

• In a synthesis reaction, A + B à AB. It is usually found in the shape of element + element à compound.

Ex. 1) Write the formula equation for the following reaction: Barium oxide (BaO) reacts with water to form barium hydroxide.

BaO + HOH à Ba(OH)₂

We know that barium has a change of 2+ and hydroxide has a charge of 1-. This forms Ba(OH)_2.

Oh look, it’s already balanced. How convenient!

• In a decomposition reaction, AB à A + B. It is the opposite of synthesis. In this course, we can always assume that compounds decompose into elements during decomposition.

Ex. 2) Manganese(II) iodide decomposes to form manganese and iodine.

MnI₂ à Mn + I₂

Manganese(II) has a change of 2+ (remember that it is multivalent), while iodine has a charge of 1-. Remember, though, it is diatomic.

Again, it seems that fate is on our side. It’s already balanced!

• In a single replacement reaction, A + BC à B + AC. The stand-alone element switches with the other cation/anion in the reaction.

Ex. 3) Write the formula equation for the following reaction: Zinc reacts with copper(II) nitrate (Cu(NO₃)₂) to form zinc nitrate and copper.

Zn + Cu(NO₃)₂ à Zn(NO₃)₂ + Cu

As you can see, we simply switch the copper out for the zinc. Remember to check the charge of zinc, as it could change the subscript after the nitrate ion.

Oh, and it’s already balanced. How about that? >_>

• In a double replacement reaction, AB + CD à AD + BC. Like single replacement reaction, the elements switch compounds. However, in this case, there are two switches. There are no stand-alone elements.

Ex. 4) Write a balanced equation for the reaction in which sulphuric acid (H₂SO₄) reacts with potassium hydroxide (KOH) to produce potassium sulphate (K₂SO₄) and water.  

H₂SO₄ + KOH à K₂SO₄ + HOH

This is a peculiar reaction, as one of the products is water. Nonetheless, we can still switch the potassium atom and the hydrogen atoms. We can balance the final equation like so:

H₂SO₄ + 2KOH à K₂SO₄ + 2HOH

• In a neutralization reaction, acid + a base à salt + water. It is a special case of double replacement. Remember, acids have hydrogen on the left side of the chemical equation (water is an exception). Bases have hydroxide on the right side of the chemical equation. Salts are simply ionic compounds.

Ex. 5) Write the formula for a reaction between hydrochloric acid and sodium hydroxide, which react to form sodium chloride and water.

HCL + NaOH à NaCl + HOH

Luckily, this equation is a pinch. There’s nothing to balance! Also, you may notice that through neutralization, we have just performed double replacement. Still, we produced a salt and water from an acid and a base. Superb!

Here's a picture of a salt for no reason!

• In a combustion reaction, a hydrocarbon + oxygen à carbon dioxide + water + energy. A hydrocarbon is a compound containing only hydrogen and carbon. Oxygen is found in its diatomic form in this reaction. Combustion reactions also release energy.

Ex. 6) Write the chemical equation for a reaction between methane (CH₄) and oxygen in a combustion reaction.

CH₄ + 2O₂ à CO₂ + HOH + energy

I’ve gone ahead and balanced everything for you. Why, you ask? Well, because you know the formula of the hydrocarbon, everything else is done for you!

Fun, eh? Need more practice? Here’s an introduction to stoichiometry. It even includes some stuff that will be covered in the next lesson:

Posted by Michael.

January 11th - The Big Test

Today was the day of our test. Next, we begin Unit 4 - Stoichiometry. What is stoichiometry? Well, according to Mr. Doktor:

Stoichiometry builds on the concept of the mole and allows chemists to perform quantitative calculations on the results of chemical reactions. In essence the stoichiometric method allows us to calculate how much of a product will be produced in a given reaction or how much reacts will be needed in specific circumstances.

Here's a great introduction video:

We can't wait!

Posted by Michael.

Jan 9th - Mole, mole, molecular Formulas!

To finish off our unit about moles, we learned about molecular formulas. To figure out the molecular formula of a substance, we need to know two pieces of information. This includes the empirical formula and molar mass of the substance. Once we have these, the molecular formula can be easily determined.

Before we begin to solve questions, let us describe why a molecular formula is important. Previously, we learned that the empirical formula is the lowest ratio of atoms in a compound. Unfortunately, this does not contain much information. For example, let us consider NO. Knowing just the empirical formula, the actual molecular formula can have infinite possibilities, such as NO, N₂O₂, N₄O₄, N₁₆O₁₆­, etc.

This is why molecular formulas are important. Firstly, these formulas tell us how many of each atom is in a compound. In addition, it narrows down the possibilities of different arrangement of the atoms, giving us the shape of the substance.

Empirical	Molecular
(formula)	(formula)
(molar mass)	(molar mass)

This chart will help us accomplish our task. It requires us to fill in the boxes with the appropriate information.

Now that we are equipped with the knowledge about molecular formulas, we can take on some problems. Let’s begin with an easy one.

1) A substance is made up of palladium and hydrogen and has a mass of 43.36g. It contains 42.56g of palladium. The molar mass of the compound is 216.8 g/mol.

First, we begin by determining the empirical formula. Let’s make our chart.

Atom	Mass	Molar Mass	Moles	Mole/ Smallest Mole	Ratio
Pa H	42.56g 0.80g	106.4g/mol 1.0 g/mol	0.40mol 0.80 mol	1 2	1 2

We have determined our empirical formula to be H₂Pa. *The order of writing the elements, at the moment, does not matter. However, sometimes there is obviously a correct way.

We will now form another chart, which will be our aid to determining molecular formulas.

Empirical	Molecular
H2Pa	H4Pa2
108.4g/mol	218.6g/mol

Our empirical formula gives us only half of our needed mass. In order to achieve this, we have to simply multiplying each atom in our compound by two. It’s as simple as that. Therefore, our molecular formula is: H₄Pa₂.

Let’s quickly try another one.

A substance in our bodies is made up of 95.992 grams carbon, 10.985 grams hydrogen, 47.996 grams oxygen, and 14.027 grams nitrogen and has a molar mass of 338g/mol. Determine the molecular formula.

Using one mole of the substance, we receive:

Atom	Mass	Molar Mass	Moles	Mole/ Smallest Mole	Ratio
C H O N	95.992g 10.985g 47.996g 14.027g	12.0g/mol 1.0g/mol 16.0g/mol 14.0g/mol	8 11 3 1	8 11 3 1	8 11 3 1

Our empirical formula is: C₈H₁₁NO₃.

Empirical	Molecular
C8H11NO3	C16H22N2O6
169g/mol	338g/mol

Our molecular formula is: C₁₆H₂₂N₂O₆! That’s it!

Posted by Andrew.

January 5th - Empirical Formulas

After our two week break, we're going to dive right into our next topic: empirical formulas. Empirical formulas are the simplest formulas of a compound. They exist in contrast to molecular formulas. Here are a few examples:

Empirical          Molecular

C₄H₉               C₈H₁₈

NO₂                 N₂O₄

P₂O₅               P₄O₁₀

Another way of saying it is that the empirical formulas are simply the molecular formulas in lowest terms. Empirical formulas only show a ratio of the respective elements in a compound; molecular formulas show the actual number of atoms that bond to form a compound. 

Note: distinguishing between empirical formulas and molecular formulas can be tricky. For example, take the molecular formula N₂O₄. It looks like this:

The name of this compound is dinitrogen tetraoxide. The empirical formula for the same compound is NO₂, which is nitrogen dioxide. It looks like this:

 In this case, you happen to be dealing with completely different compounds.

To determine an empirical formula, we need to know the ratios of each element in a compound. To determine the ratios, we can construct a chart, like the one in the following example:

Ex.) Determine the empirical formula of a compound containing 0.928g of gallium and 0.412g of phosphorus.

As you can see, in the first column is the name of the element/atom. In the second, we have the mass, which is usually given in the question. If not, you can figure it out through simple subtraction. In the third, we have the molar mass. This can be found in the periodic table for each element. Next, we have the moles. Using the conversions we learned earlier in this unit, we can figure 'em out:

0.928g   x   mol/69.7g   =   0.0133 mol of gallium

0.412g   x   mol/31.0g   =   0.0133 mol of phosphorus

Oddly, in this example, we end up with the same number of moles for both compounds.

The fifth column is where we find out number of gallium atoms and phosphorus atoms in the compound. To find this out, we divide the mole for gallium by the smallest mole out of all the calculations we made in column four. We would then do the same for phosphorus. Strangely, in this compound, the ratio would be 1:1 because the moles are equal. 0.0133/0.0133 = 1. In this column, we put 1:1.

The sixth column would contain the lowest term ratio of atoms in the compound (remember: it is the empirical formula!), but in this case we have already obtained the lowest term ratio. We're done! The compound is GaP, or Gallium phosphide. 

Note: in some cases, you may get a ratio that ends up being 1.33 or 1.66. In this case, you would need to multiply it so it becomes a whole number. You would multiply both of those by 3, so your new ratio would be 4 or 5. It's simple!

I hope you enjoyed the lesson. If you want more examples, here are a few great videos that include all sorts of variations of this process:

Posted by Michael.