November 30th - Another Day, Another Lab

Today, we did the molar volume lab. It was a complicated experiment requiring much thought and concentration. With this in mind, we were ready to do the lab. No pressure, right?

Our objective is to find out the ratio of the number of moles of iron used up in our reaction to the number of moles of copper produced in the reaction between iron (in the nails) and Copper (II) chloride. The iron is present in the nails and the Copper (II) chloride is present in the form of a powder. Here's what Copper (II) chloride looks like:

The materials used in this lab include two 400mL beakers, a spray bottle, a stir rod, tongs, lab aprons, goggles, a weigh scale, a drying oven, filter paper, a scoopula, a funnel, a graduated cylinder, and a measuring container.

As mentioned before, this lab was fairly complex. Here's what we had to carry out:

1. Take out the scale and gather all your materials. Get a 400 mL beaker and, using a graduated cylinder, measure out about 50 mL of water and put it into the beaker.
2. Place your measuring container on the scale, then reset you scale to zero. Add a sample of Copper (II) chloride to the measuring container using the scoopula. About 7-8 grams is needed.
3. Dump your sample into the beaker and stir gently with a stir rod.
4. Take two nails and measure them on the scale after resetting it to zero. After recording this number, place them into the beaker.
5. Wait approximately 15 minutes. Then, using tongs, remove both nails. Using a spray bottle or scoopula, scrape the newly formed copper into the beaker until your nail appears clean.
6. Weigh the nails again after drying. Subtract the final mass of the nails from the initial mass to find out how much iron was used up in the reaction.
7. Prepare your retort stand/lab clamp, funnel, and another beaker. After placing the funnel in the center of the clamp with the fresh beaker directly below, grab a filter and weigh it. Then, after folding it appropriately, place it in the funnel.
8. Pour your solution of Copper (II) chloride into the funnel and allow it time to filter. Be sure to use your stir rode to decant when pouring. After the substance has filtered, you will be left with copper in your filter. Place the filter in the drying oven to dry.
9. After drying, weigh the filter again. Subtract the initial weight of the filter from the final weight. This is the mass of copper in the filter.
10. Clean up all your materials and safely dispose of all chemicals.

As in every lab, we had observations. They include seeing the copper form on the nail. It had a rust-like appearance. We noticed that it did not hold very strongly to the nail. With just a spray of water, most of it came off. We also saw the copper (II) chloride dissolve in the water and form a bright turquoise colour. Another observation was the solution becoming a darker shade after it had reacted with the nail.

We carried out experiment within 8% accuracy. We made many qualitative measurements in the experiment. For example, the mass of our filter paper after filtration was 2.36g.

We concluded our lab by stating possible reasons of error. One possible reason for the error is pouring too much water above the filter. This caused the copper to leak through the sides. Not too much copper would have passed through though, as we exceeded the filter limit by a small amount. Another reason for our error could include not scraping all the copper off. Other mistakes that could have possibly accounted for the error took place when using the tongs. The tongs could have scraped off and lost some of our copper while we were removing it from the nails.

A similar experiment can be seen here:

Posted by Andrew.

November 28th - Multistep Mole Conversions

Today, we took everything we know about moles and put it all together. We learned about multistep mole conversions, which are mole conversions that include more than one instance of dimensional analysis. Here are the conversions that we have learned so far:

That's pretty much it! Because we already know how to convert between mass, volume, molecules, atoms, and moles, we can dive right into the examples.

Ex. 1) If you have 13g of Cl₂ in a containter at STP, what is the volume of gas that you have?

Knowing that we have 13g of chlorine, and also knowing that moles are the central conversion factor, it is obvious that we must convert mass to moles to volume. The mass of 2 chlorine atoms in the diatomic molecule is (2)35.5 g/mol, or 71 g/mol.

13g   x   1 mol   x   22.4 L   = 4.1 L of chlorine gas

         71 g        1 mol                              

Ex. 2) A container holds 10.0 L of oxygen gas at STP. How many grams of oxygen is this?

Again, using our central conversion factor (the mole), we can convert this unit using molar volume and mass. The molar mass is (2)16.0 g/mol because oxygen is diatomic and the molar volume is 22.4 L/mol in all cases at STP. Let's solve:

10.0 L   x   1 mol   x   (2)16.0g   =   14.3 g of oxygen gas

22.4 L         1 mol                        

Ex. 3) A container holds 50.5 L of carbon dioxide. How many molecules of CO₂ is this? 

Now we have several things to take into consideration. First, we must account for the number of atoms and the different atoms in carbon dioxide. We must also realize that the question is asking for the number of molecules, not the number of atoms or the mass. Observe:

 50.5 L   x   1 mol   x   6.02 x 10²³ molec   =   1.36 x 10²⁴ molecules of carbon dioxide

    22.4 L             1 mol                                                               

It's just that easy! 

Now, with all of this, you can solve for any unit related to moles using dimensional analysis.

Here's a quick video for some more review:

Posted by Michael.

November 24th - Atoms, Molecules, Moles, and More.

Today’s class was a continuation of our learning of moles. We learned how to calculate the number of moes in a given number of molecules, as well as the other way around. We also learned how to calculate the number of atoms of a certain element in a compound. For example, how many oxygen atoms are there in 1.0 moles of water? Well, let’s find out.

Let’s figure out how we will accomplish the task.

First off, we know that there are 6.02 x 10²³ formula units (atoms, molecules, etc) in one mole of a substance. This is commonly known as Avogadro’s number. However, using water, which has the equation HOH, we cannot find the number of atoms simply using this number. This is because we will be finding the number of HOH units, not oxygen atoms. There must be a way to accomplish this? Well, let’s find out.

Each water compound is composed of two hydrogen atoms and one oxygen atom. Knowing there is one oxygen per water, we can form the conversion factor:

1 O atom

1 HOH molecule

Utilizing this conversion factor, we will get the answer of: 6.0 x 10²³.

Too easy, right? Let’s try some more.

A glass contains 5.0 mole of vinegar (CH₃COOH). How many atoms of each element are there in the glass?

5.0 mol x (6.02x10²³/1 mol) x (appropriate conversion factor) 

There are two C atoms, two O atoms, and 4 H atoms.

Simply, multiply the number up top by this. The answers are as follows:  

• 6.0 x 10²⁴ C atoms 
• 6.0 x 10²⁴ O atoms
• 1.2 x 10²⁵ H atoms

Not that hard, eh? Here’s a video to wrap up:

Posted by Andrew.

November 22nd - Another Lab! Hooray!

Today we completed an all new lab! This lab was all about molar volume, which is the volume of one mole of a substance. The gas we used was butane, an alkane with the formula C₄H₁₀. Behold:

 

Calculating the molar mass of butane is quite simple. Like we learned in previous lessons, we add up the total molar masses of all the atoms in the compound, like so:

4(12.0) + 10(1.0) = 58.0 g/mol

Starting the lab, we gathered our materials: a butane lighter, water, a graduated cylinder, and a scale. We measured the original mass of the lighter, then displaced the water in the submerged graduated cylinder with butane from the lighter. Due to its gaseous state, it floated to the surface. We measured until it got to about 53 mL of gas, then dried off the lighter and weighed it again. Using that information, we found out the total mass of the butane lighter after the experiment. It ended up being a change 0.14 g. Knowing this, we could calculate the number of moles of butane we used:

0.14 g     x     1 mol     =    0.0024 mol

53.0 g/mol         

Using this information, we calculated the molar volume of butane. However, since the units for molar volume are in L/mol, we must convert the 53 mL of butane we measured into litres and then divide it by our number of moles:

0.053 L   =   22 L

.0024 mol               

 We can then calculate our percent error for the experiment:

[(measured   –   accepted)   /   accepted]   x   100%

[(22L   –   22.4L)   /   22.4L]   x   100%

=   1.8%

There were a few factors, however, that affected our result. For example, we may not have adequately dried off the lighter before measuring for the second time. Another factor is the difference in temperature from STP, which is 0 degrees Celsius at 101.3 kPa.

Wasn't that a fun lab? Here's a video demonstrating the flammability of butane (just because):

 

KA-BOOM!

Posted by Michael.

November 18th - More Molar

We are expanding our chart! Today, we added volume to it. In addition, we will be learning how to convert from moles to volume, vice versa, and even from volume to mass.

The volumes we are working with pertain to gases. At a specific pressure and temperature, one mole of any gas occupies the same volume. Therefore, the volume occupied by one mole of oxygen gas is equivalent to the volume occupied by one mole of hydrogen. However, the temperature and pressure of these two have to be the same.

Molar volume – to volume occupied by one mole of a substance

When temperature increases, the volume will increase. This is one of the laws of kinetics. It tells us that a particle vibrates and moves faster. This results in more space between particles and a greater volume.

The volume we will be measuring is performed at STP (standard temperature and pressure). This refers to 0^oC and 101.3 kPa. At STP, one mole equals 22.4L of a gas. We can conclude that:

22.4L is the molar volume at STP

Let’s try a few examples.

How many litres will there be in 5 mol of H₂ gas?

5.0 mol x (22.4L/1 mol) = 112L = 1.1 x 10²L

How many litres will there be in 5 mol of O₂ gas?

5.0 mol x (22.4L/1 mol) = 112L = 1.1 x 10²L

It`s exactly the same! This supports the fact that at a specific pressure and temperature, one mole of any gas occupies the same volume.

And remember to use ‘em significant digits!

Let’s try converting more volume to mass.

A box is 6.0m long, 4.0 meters wide, and 3.0 high. If it is filled with oxygen gas, determine the mass of O₂.

The steps to solve this are as follows:

1.    6.0 x 4.0 x 3.0 = 72m³

2.    72m³ x (1000L/m³) = 72000L

3.    72000L x (1 mol/22.4L) = 3000 mol

4.    3000 mol x (32.0g / 1 mol) = 96,000 g O₂ = 9.6 x 10³ g.

That’s about all we did, but let’s briefly explore the ideal gas law. This is the equation of state of a hypothetical ideal gas. It is given by the equation: PV = nRT, where P is pressure, V is volume, n is number of moles, R is the constant (8.3144621), and T is the temperature (K).

SATP is similar to STP, however, the temperature is 25^oC instead of 0^o. Also, we perform this at 100 kPa rather than 1 atm (101.3 kPa). It stands for Standard Ambient Temperature and Pressure.

At SATP, the volume occupied by one mole is 24.8L/mol. 

Posted by Andrew.

November 16th - Mole Conversions

Today we learned all about mole conversions! Mole conversions are the process in which dimensional analysis is used to convert between moles and grams. 

To convert between moles and mass in grams, we use molar mass as the conversion factor. Molar mass can be found on the periodic table (it is the same as atomic mass). In a compound, it is the sum of the masses of all atoms. For example, if you were only using oxygen, the molar mass would be 16.0 g/mol (the molar mass of oxygen). If you were using carbon dioxide, the molar mass would be 12.0 g/mol + (2)16.0 g/mol for a total of 44.0 g/mol. 

Remember to cancel out the appropriate units and remember significant digits!

Here are a few examples:

Ex. 1) How many grams are there in 1.2 mol of chlorine gas?

First, start out by writing what you know. We know that there's 1.2 mol and that there are 2 oxygen atoms in one compound. We can also look up the atom weight, which is 16.0 g/mol. Then, remembering that the unit we are converting to always goes on top, we can calculate:

1.2 mol   x   (2)35.5g   =   85g of chlorine gas

1 mol                  

Ex. 2) How many moles are there in 100g of Fe₃O₄? 

Knowing that we have 100g, that the molar mass of iron is 55.5 g/mol and that the molar mass of oxygen is 16.0 g/mol, we can find the number of moles:

100g                      x                      1 mol                     =                     0.433 mol of Fe₃O₄

(3)55.5 g + (4)16.0 g                   

Here's a video providing another example of such conversions:

Ex. 3) How many grams are needed to make 3.0×10^-3mol ofMgCO₃ ·5 HOH?

We know that we have 3.0 × 10^-3 mol of the substance, that the molar mass of magnesium is 24.3 g/mol, of carbon is 12.0 g/mol, of oxygen is 16.0 g/mol, and of hydrogen is 1.0 g/mol. We add the molar mass of the hydrate to that of the compound in this equation. Using this knowledge, we can solve:

 3.0 × 10^-3 mol   x   24.3 g/mol + 12.0 g + (8)16.0 g + (10)1.0 g   =   0.52g of MgCO₃ ·5 HOH

 1 mol               

 Ex. 4) A compound, made of phosphorus and chlorine, contains 0.200 mol and has a mass of 27.5g. Determine the molar mass and suggest a possible formula.

We know that molar mass is stated in the units of g/mol. We know both the number of mols and the mass, so we can find the molar mass:

27.5g /0.200 mol = 137.5 g/mol

To find the formula that corresponds to this molar mass, we can create a chart:

With this, we know that the compound contains 1 phosphorus atom and 3 chlorine atoms. Therefore, the compound is PCl₃.

That was fun, wasn't it? Here's another video (if you need more practice!):

Posted by Michael.

November 14th - How Many Moles Am I?

Yes! We will find out my mass in moles in this blog. But moles have different masses, known as molar masses. A molar mass is defined as the mass in grams of one mole of a substance.

It can be determined from the atomic mass on the periodic table. It is measured in grams/mol.

Example! Let's use the element oxygen. Referring to the periodic table: 16.0 amu. What does amu stand for? Well, it's the abbreviation for atomic mass units. 16.0 amu is the mass of one element of oxygen. amu is based on 1/12th of the mass of an atom of carbon-12. Therefore, the mass of carbon-12 is exactly 12 amu.

Another important thing to remember is diatomic molecules! Oxygen, fluorine, and chlorine are examples. When alone these elements exist in pairs, so be careful and remember an oxygen molecule actually refers to two oxygen atoms.

How much of a mass is that? Well, since one oxygen is 16.0 amu, then two will have a mass of 32.0 amu.

The amu corresponds directly to the molar mass. To convert, we simply take the amu and change it to grams per mole. It's as easy as that! In chemistry, we prefer this because it is extremely hard to work with one or two atoms. This is where the mole, which expresses a very large quality, helps us in conversions.

Let us use the element tungsten. This element, since it contains its relative abundance of isotopes, has a molar mass of: 183.8 grams per mole.

The conversion from grams to mole is explained in the post above.

I have a mass equivalent to that of 337.32 moles of W. If you are curious to find out how much this is, we suggest you refer to the upcoming post. This will teach you about conversions between grams and moles.

Posted by Andrew.

November 9th - Avogadro's Number

Today was the day we started our second unit: The Mole. We started the class by getting our tests and labs back. We then dove right into the notes for our fresh topic.

As we are well aware from previous studies, atoms and molecules are extremely small. Macroscopic (visible to the naked eye) objects contain way too many atoms/molecules for us to obtain a practical number for the weight or count of the sample. Thankfully, an Italian chemist by the name Amedeo Avogadro thought of a way to represent a large number of atoms with one convenient unit. Behold Avogrado:

Avogadro proposed that the number of atoms in 12.00000g of carbon-12 is equal to a constant. This constant would be known as a mole. Therefore, in 12.00000g of carbon-12, there is 1 mol. This constant can also be applied to other samples to gain a measure of its relative number of atoms!

This value, which is equal to 6.02 x 10²³ atoms, is now used as the basis for all quantitative chemistry!

 

To demonstrate how big this number really is, we were introduced to the pea analogy. 

According to the analogy, if we had one million peas, we could fill an average bedroom. That's a large number of peas, but it's practically nothing compared to a mole of peas. In fact, one million is only 0.000000000000000166113% of a mole. 

Let's jump up a few exponents. If we had one sextillion of peas, we could cover the entire surface of the earth with peas measuring 1 metre deep. Still, that's only 0.0016% of a mole.

Now, let's say that we cover 602 planets, all the same size as Earth, with peas measuring 1 metre deep. That, my friend, is one mole. And that is the same number of atoms that exist in 12 grams of carbon-12.  

To convert between moles and atoms, we can use simply use dimensional analysis. Here's an example:

If we have a sample of iron containing 5.05 x 10²⁴ atoms, how many moles of iron do we have?

We can also use dimensional analysis to convert between moles and ionic or covalent compounds. Here's an example:

If we have a sample containing 7.0 moles of bromine (Br₂), how many molecules do we have?

 

If we are converting to find how many atoms make up x number of moles, we would answer using the element symbol (ex.          moles of Fe).

If we are converting to find how many molecules (of a covalent compound) make up x number of moles, we would answer using the chemical formula (ex.          moles of HOH).

If we are converting to find how many formula units (of an ionic compound) make up x number of moles, we would also answer using the chemical formula (ex.          moles of NaCl).

Here's a nice video explaining the mole in detail:
 

 

Posted by Michael.

November 7th - The Test

That's right! Today, we had our test on atomic theory. Next class, we start a brand-new unit; The Mole. Here's a quick, interesting introduction:

Posted by Michael.

November 3rd - Up, Up, and Away!

As the title says, that is what happened today. We performed a lab in which we had to separate the water from the substance. The water had to be removed and evaporated up, up, and away.

This was an interesting lab because we used a Bunsen burner. Even though it was fun, we always had to think about safety. So, whenever we saw a Bunsen burner, we assumed that it was on.

In our lab, we followed this procedure:

1.    Find the mass of the empty test tube.

2.    Fill a test tube with about one centimetre of the hydrate.

3.    Carefully place the test tube on the scale and record the mass of the hydrate and test tube.

4.    Using extreme caution, connect and light your Bunsen burner. Adjust the gas flow until the flame is about five centimetres tall.

5.    Pick up the test tube with the clamps and carefully hold it in above the Bunsen burner.

6.    Gently heat the test tube by moving the test tube in and out of the flame for about five minutes or until all the water has boiled away.

7.    Carefully re-weigh the test tube ensuring none of the chemicals inside spill.

While following these directions, we observed what was going on. As the hydrate was being heated and the water molecules were evaporating, the solution was changing colour to a light-blue one. After we figured that all the water was boiled away, we measured the mass again. It was important that we did not leave the hydrate in the flame for too long. This is because the anhydrous solution would begin to react with oxygen, changing its composition.

The hydrate changed colour, as shown in the pictures below. The red is how the hydrate looked prior to being separated from water molecules, and the blue was after.

During the experiment, we recorded our mass before heating and mass after heating. We determined how much water was released, and then what percentage of the hydrate was water. Our percent was about 49%, which was four percent off from the actual value of 45%. This means that we still had a little water left to boil out. Finally, we concluded the lab experiment by stating our percent error.

After this lab, we can know understand why ice may burn. Here is the video: 

"By trapping the propane within the ice, the propane is released as the ice burns. Burning the ice is only something cool you can do with the clathrate. One of the main benefits of trapping gas in clathrates is the fact that the solid takes up less space than the gas itself. So, you can ship more gas in a small container."

When we were not performing the lab, we were studying for our upcoming test.

Posted by Andrew.

November 1st - Naming Molecular Compounds

This class, we continued our study of chemical nomenclature. Specifically, we learned all about the naming of molecular compounds. A molecular compound is a compound that is formed by non-metal pure substances.

As a quick review, let's go over the 7 diatomic molecules. They are:

Remembering that diatomic molecules are molecules that consist of two atoms of the same element, we can compare them to polyatomic molecules. Polyatomic molecules are molecules consisting of several atoms of the same element. The two that we study are P₄ and S₈. 

To name molecular compounds, we can start by finding the name of the first element in the compound. As an example, let's use NO. Here, the first element is nitrogen. We then find out the name of the second element and add the suffix '-ide'. In that case, the name of the compound would be Nitrogen oxide. It is important to note that we must always capitalize the first letter of the first element in a compound. However, we aren't done yet. We must also add the proper prefixes. Generally, the first element in the compound does not have a prefix if only one atom of the element exists in the compound. In this case, we leave it as it is. However, the last element must carry a prefix to indicate how many atoms of that element exist in the compound. In the formula NO, there is only one atom of oxygen. In that case, it would get the label 'monoxide', which we get by adding 'mono-' and 'oxide'. With that, we get the final name of the compound: Nitrogen monoxide! Here's a video that explains the process of naming molecules quite well:

It is also important to note that hydrogen generally doesn't carry a prefix.

Here are a few of the molecular compounds that we will study this year:

IUPAC Name/Formula

water/H₂O

Hydrogen peroxide/H₂O₂

Ammonia/NH₃

Glucose/C₆H₁₂O₆

Sucrose/C₁₂H₂₂O₁₁

Methane/CH₄

Propane/C₃H₈

Octane/C₈H₁₈

Methanol/CH₃OH

Ethanol/C₂H₅OH

Another topic we covered was the naming of acids and bases. In most cases, compounds containing hydrogen are acids. For example, the compound HNO₃ is known as nitric acid. It is also important to note that hydrogen is always present on the left of the acid's formula unless it is part of a polyatomic group (such as acetic acid; CH₃COOH).

According to the classical system, the suffix '-ic' and the prefix 'hydro-' indicate the components of a specific acid. To differentiate between the two types of acids, we follow these rules: 

• Polyatomic acids, such as H₃PO₄, use the suffix '-ic'. This compound is made up of hydrogen (an element) and phosphate (a polyatomic molecule) to form phosphoric acid. It is not binary (made of only two elements), so it does not use the prefix 'hydro-'.
• Binary acids, such as HCl, use the prefix 'hydro-'. This compound is made up of only two elements, making it binary. We can name this hydrochloric acid. 

To name hydrogen-containing compounds using the IUPAC system, we simply add 'aqueous' before the chemical name. HCl_(aq) would be named 'Aqueous Hydrogen chloride'.

Bases are formed with a cation and hydroxide (OH), which is usually found on the right side of the formula. For example, Al(OH)₃ would be called aluminum hydroxide. Naming bases is quite simple!

 

Here are a few common acids and bases:

Acid Name/Formula:

Hydrochloric acid/HCl

Nitric acid/HNO₃

Sulfuric acid/H₂SO₄

Phosphoric acid/H₃PO₄

Acetic acid/CH₃COOH

Base Name/Formula:

Ammonia/NH₃

_{Posted  by Michael.}