Mole Ratio Lab!

EXPERIMENTATION!!!!!!

 In this lab, we determined the number of moles of copper per mole of iron. Iron reacted with Copper II chloride to produce Copper and Iron III chloride; 2Fe + 3CuCl₂ --> 3Cu + 2FeCl₃ Thus, the ratio of moles of copper produced to moles of iron consumed was 3:2 or 1.5:1.

We used the following materials in our lab: an apparatus, beakers (400mL), a wash bottle, a stirring rod, crucible tongs, a centigram balance, a drying oven, safety goggles, lab apron, plastic gloves, sandpaper (or emery cloth), filter paper, a retort stand, a funnel, and the reagents (Copper II chloride, 2 Iron nails (approx. 5cm) and distilled water).

We first recorded the mass of an empty and dry 400mL beaker to the nearest 0.01g. Then, we found the mass of 8g of Copper II chloride crystals in a beaker. Next, we added 50 mL of distilled water to the beaker and swirled the beaker around to make sure that all of the Copper II chloride crystals were dissolved. We saw that the solution was a bright greeny-blue colour.  After finding the mass of two clean and dry nails, and placing them into the Copper II chloride solution, we left the nails undisturbed for 20 min. We noticed that when we used the tongs to carefully pick up the nails up one at a time, Copper had formed on the nails, and the colour of the solution became slightly darker. We rinsed off any remaining Copper from them with a wash bottle before removing them completely from the beaker. We found the mass of the nails after we dried them on a paper towel. We created a funnel with the filter paper and found its mass. Then, palcing it in a funnel, we filtered the solution. When pouring, we used the stirring rod to direct the liquid into the second beaker. Finally, we found the mass of the filter paper and the copper (after placing it in a drying oven to dry), washed our hands, and cleaned up all of the materials. Here's a visual representation of two steps in this experiment (draw by me!!).

Our experiment had 7% error. We think this is because there were still a few specs of copper even after we had filtered the solution. We didn't have much time, so we couldn't filter the solution even more. In addition, we might not have cleaned the nails thoroughly enough. There may also be inaccuracy in measuring the amount of Copper II chloride crystals and the amount of solution and substance in the beakers and graduated cylinders.



 
Next time: Percent Composition!

Multistep Mole Conversions!

Today Mr. Doktor wasn't there for class as he wasn't feeling well so we were forced to look at multistep conversions by ourselves. Not to worry though, we managed!


By now, we have a very handy map that can help to guide us on our journey to master the science of chemistry - particularly conversions involving the mole. It should look something like this:

As a review, to convert from mass to moles or moles to mass, we use molar mass; to convert from moles to volume or volume to moles, we use 22.4L/mol (@ STP); to convert from moles to molecules or molecules to moles, we use Avogadro's number; and to convert from molecules to atoms or atoms to molecules, we use the subscripts. Complete conversions usually involve more than one step. Shall we take up some examples?


EXAMPLES
-11.5g of H₂ gas are placed in a balloon at STP. Determine the volume of the balloon.
11.5g x 1mol/2.0g x 22.5L/1mol = 129L


-A car tire contains 25.0L of Nitrogen gas at STP. How many grams of Nitrogen is this?
25.0L x 1mol/22.4L x 28.0g/1mol = 31.3g


-How many formula units are there in 3.5g of Nickel (II) oxide?
3.5g x 1mol/74.7g x (6.02 x 10²³)molec/1mol = 2.8 x 10²²FU


-A sample of Oxygen gas contains 3.5 x 10²¹ molecules. How many grams of oxygen is this?
3.5 x 10²¹molec x 1mol/(6.02 x 10²³)molec x 32.0g/1mol = 0.19g


-A container holds 35.0L of Methane gas at STP. How many molecules of Methane is this?
35.0L x 1mol/22.4L x (6.02 x 10²³)molec/1mol = 9.41 x 10²³


-3.5 x 10²³ Chlorine atoms are present in a sample of Chlorine gas. How many litres of gas at STP is this?
3.5 x 10²³atoms x 1molec/2atoms x 1mol/(6.02 x 10²³) x 22.4L/1mol = 6.5L


-Determine the mass of 10 molecules of Carbon dioxide.
10molec x 1mol/(6.02 x 10²³)molec/1mol x 2atoms/1molec = 7.3 x 10^-22g


-A sample of NO₂ occupies 7.50L. How many Oxygen atoms are present in this sample?
7.50L x 1mol/22.4L x (6.02 x 10²³)molec/1mol x 2atoms/1molec = 4.03 x 10²³atoms


 
Next time: Mole Ratio Lab!

Converting Between Molecules, Moles and Atoms!

In today's class we went over how to calculate the amount of molecules in a given number of moles and vice versa. We also went over how to calculate the number of atoms of a certain element in a compound, which I found quite interesting!

As you can see from the diagram above, 6.02 x 10²³ is the number of atoms in one mole. When we convert between molecules and atoms, we use the subscripts.

EXAMPLES

-How many atoms are there in 1.5 mol of Iron?

1.5mol x 6.02x10^23 atom/1 mol =9.0 x 10^23

(All you do to solve this is multiply the 1.5 mol by 6.02 x 10^23)

-How many water molecules are there in 0.65mol?

0.65mol x 6.02 x 10^23 molecules/ 1 mol =3.9 x 10^23 molecules

(All you do to solve this is multiply the 0.65 mol by 6.02 x 10^23)

-How many Hydrogen atoms are there? How many Oxygen atoms?

We know, from the formula for water, that there are two Hydrogen atoms and one Oxygen atom in every molecule. So, this is our conversion unit:

3.9 x 10^23 molecules x 2 H/ 1 molecule = 7.8 x 10^23 Hydrogen atoms.

3.9 x 10^23 molecules x 1 O/ 1 molecule= 3.9 x 10^23 Oxygen atoms.

-A cylinder of Helium contains 4.6 x 10^30 atoms. How many moles of Helium is this?

4.6 x 10^30 atoms x 1mol/ 6.02 x 10^23 atoms = 7.6 x 10^6 moles of Helium

-A sample of Zinc Sulphate contains 5.0 x 10^22 atoms of Oxygen. How many moles of Oxygen is this? How many moles of Zinc Sulphate (ZnSO4) are there?

5.0 x 10^22 x 1mol/ 6.02 x 10^23 atoms = 0.083 mol of Oxygen

0.083 mol Oxygen x 1 ZnSO4/4 O= 0.021 mol of ZnSO4

-A glass contains 5.0 mol of vinegar (CH3COOH). How many molecules are there in the glass?

5.0 mol x 6.02 x 10^23 molecules/ 1 mol = 3.0 x 10^24 molecules of vinegar

-How many Carbon atoms are there?

3.0 x 10^24 molecules x 2 C/1 molecule= 6.0 x 10^24 atoms of Carbon

-How many Hydrogen atoms are there?

3.0 x 10^24 molecules x 4 H/1 molecule= 1.2 x 10^25 atoms of Hydrogen

-How many Oxygen atoms are there?

3.0 x 10^24 molecules x 2 O/1 molecule= 6.0 x 10^24 atoms of Oxygen

-How many atoms are there in total?

6.0 x 10^24 atoms of Carbon + 1.2 x 10^25 atoms of Hydrogen + 6.0 x 10^24 atoms of Oxygen = 2.4 x 10^25 atoms in total

Mr. Doktor in slow motion!

Next time: Multistep Moles Conversions!

Mole Volume Lab!

We performed another lab today; our goal was to experimentally determine the molar volume of a gas. We used Butane - a colorless, highly flammable gas; its source is natural gas; its major uses are for lighter fluid and fuel; and its formula is C₄H₁₀. We already know how to determine the molar mass of a chemical compound; the molar mass of Butane is 58.0g/mol [4(12.0) + 10(1.0) = 58.0]. Check out the structure of Butane:




  In this experiment, the pressure/temperature weren't actually measured so we don't know the EXACT molar volume. At STP, however, it should be 22.4L/mol. Our experiment was not performed at STP (0^o and 101.3kPa).

We used Butane, a sink, water, a 100mL graduated cylinder, and a weigh scale for this experiment. We first filled the sink full of warm water. Then we weighed the Butane lighter to the nearest 0.01g. Next, we submerged the 100mL graduated cylinder into the sink, making sure that it filled completely with water, inverting it (making sure there were not any air bubbles). We then placed the lighter at the bottom of the sink, underneath the graduated cylinder. Holding down the buttom of the lighter, we released Butane to fill the graduated cylinder 70mL. We saw that the water was displaced. After that, we removed the lighter and shook out all of the water before placing it in the drying oven. Once the lighter was dry, we weighed it once more. The mass was lighter than at the beginning of the experiment.

The mass of the Butane lighter before the experiment was heavier than the mass of the Butane lighter after the experiment. To determine the mass of the Butane lighter used, we subtracted the final from the initial mass:



21.17g - 21.01g = 0.16g  To determine the number of moles of Butane used, we divided the number above by the molar mass of Butane:



0.16g x 1mol/58.0g = 2.8 x 10^-3mol  Finally, to determine the molar volume of Butane, we divided 70mL by the number above:



70mL/(2.8 x 10^-3)mol = 25000mL/mol or 25L/mol.  Our experiment had the following percent error:



[(measured - accepted) / (accepted)] x 100 [(25-24.15)/(24.15)] x 100 =3.5% error (using the the value we obtaned as a class with the ideal gas law; remember, the conditions are not at STP)  A number of factors affected our results; perhaps there were some air bubbles inside the graduated cylinder, underneath the water. It's also possible the we didn't release enough or released too much Butane into the cylinder. In addition, we may not have shook out enough water. Finally, there could be slight miscalculations in recording the mass of the Butane lighter before and after the experiment.
Next time: Molecules, Moles and Atoms!

Converting Between Volume and Moles!

Last class we learned how to convert between grams and moles. Today, we learned how to convert between volume and moles. The volume, in this case, is that of a gas.

 At a specific pressure and temperature, one mole of any gas occupies the same volume.

This standard is: At 0^oC and 101.3kPa...

...1 mol = 22.4L

From this, we can assume that the volume occupied by one mole of oxygen gas is the same as the volume occupied by one mole of hydrogen gas, at the standard temperature and pressure (STP).
Thus, 22.4L/mol is the molar volume at STP. Memorize this number!



EXAMPLES
-How many litres will 2.5mol of H_2(g) occupy at STP?
2.5mol x 22.4L/1mol = 56L


-A certain gas is found to occupy 11.6L at STP. How many moles of gas are there?
11.6L x 1mol/22.4L = 0.51786mol or 0.518mol(to three significant digits)


-At a temperature and pressure where molar volume is 45.0L/mol...?
11.6L x 1mol/45.0L = 0.2578mol or 0.258mol (to three significant digits)


-At STP a sample of Oxygen gas contains 11.5mol. How many litres of oxygen gas are there?
11.5mol x 22.4L/1mol = 257.6L or 258L (to three significant digits)


-At STP an unknown gas (note that the fact that the gas is "unknown" does not make a difference) is found to occupy 150mL. How many moles of gas must there be?
150mL x 1L/1000mL = 0.150L
0.150L x 1mol/22.4L = 0.006696428mol or 0.00670mol (to three significant digits)


-Solve the above using an alternative method.
150mL x 1mol/22.4L = 6.70mmol or 0.00670mol


-A certain amount of Chlorine gas occupies 1.6L. Find the number of moles present and then determine the mass of Chlorine.
35.5 (refer to Chlorine in the periodic table for this number) + 35.5 (because Chlorine is diatomic) = 71.0
1.6L x 1mol/22.4L = 0.07143mol or 0.071mol
0.071428571mol x 71.0g/1mol = 5.071428571 or 5.1g (to two significant digits)




 Next time: Molar Volume Lab!

Converting Between Grams and Moles!

The mole really isn't that hard; you simply need to take and do the work! With practice, you'll master the concepts. Today, we're converting between moles and mass. When we do so, we use molar mass as the conversion factor. Be sure to cancel the appropriate units! 

EXAMPLES
-How many grams is there in 1.5 mol of O₂ ? -How many moles are there in 115g of Fe₂O₃ ? -A sample of HCL contains 0.54 mol. How many grams of HCL is this?-How many grams are present in a 2.5g sample of ammonium phosphate?
-How many grams are needed to make 2.5x10^-3 mol of Cobaltous Chloride hexahydrate?

­

-A compound is made of phosphorous and chlorine. It is found to contain 0.200 mol and has a mass of 27.5g. Determine the molar mass of the compound. Suggest a possible formula.
This last one's a toughy; it involves trial and error. To find the molar mass, divide 27.5g by 0.200mol. x(31.0) + y(35.5) should equal 137.5g/mol. Then, use trial and error to determine that a possible formula is PCL₃ [one Phosphorous (ie. the x value) and three Chlorine (ie. the y value)].

Next time: converting between moles and volume.


NOTE: DO NOT FORGET ABOUT SIGNIFICANT DIGITS!!

Next time: Converting Between Volume and Moles!

Molar Mass: The Mass of Atoms!

Next time: Converting between Grams and Moles!
Molar mass is the mass in grams of one mole of a substance. It can be determined using the atomic mass on the periodic table. Molar mass is measured in g/mol.
If we refer to our periodic table, we see that the molar mass of Beryllium is 9.0.


  Molar Mass of Compounds
To determine the molar mass of a compound add the mass of all the atoms together. It really is as simple as that!

EXAMPLES:



 For extra practice:



 And for a great way to end class Mr. Doktor showed us what would happen if you heated hydrogen gas. It went something like this:




Next class: Converting between grams and moles!

Moley, moley, moley, moley, moley!

Today, we were introduced to the wonderful world of moles! We talked about Avogadro's Number, which is 6.02 x 10²³ and were told an amazing story about numerous "peas."

COUNTING ATOMS
When Objects Contain too many atoms to count or weigh individually, we give them the title of Macroscopic. Because of this, Amedeo Avogadro proposed that the number of Atoms in 12,000,000 grams of Carbon should be equal to one mole of Carbon. This number is now the basis of all quantitative chemistry.

Almost exclusively a chemistry term, a mole is a multitude of things, like a pair (two of something), a dozen (twelve of something) or a century (one hundred of something). To get a better sense of the size of a mole, one mole of meters would cross the entire galaxy over 3000 times; and one mole of seconds is almost 100,000 times more than the age of the universe.

Another great example Mr.Doktor gave us was "The Pea Analogy."

100	1	One pea
101	10	10 peas - spoonful
102	100	10 spoonfuls of peas - plateful
103	1000	10 platefuls of peas - bag
104	10 000	10 bags of peas
105	100 000	100 bags of peas - fridgeful
106	1 000 000	10 fridgefuls of peas - an average bedroom
107	10 000 000	Average house full from basement to attic with peas
109	1 000 000 000	100 houses full from basement to attic with peas - a small village
1012	1 000 000 000 000	100 000 houses full from basement to attic with peas - a small city
1015	1 000 000 000 000 000	A large city like New York covered 1 m deep with peas
1018	1 000 000 000 000 000 000	A province like Alberta covered 1 m deep with peas
1020	100 000 000 000 000 000 000	The land mass of Canada covered 1 m deep with peas
1021	1 000 000 000 000 000 000 000	The land mass of the world covered 1 m deep with peas
1023	100 000 000 000 000 000 000 000	The entire surface of the earth plus 99 planets just like earth covered 1 m deep with peas
6.02 x 1023	602 000 000 000 000 000 000 000	602 planets like earth covered 1 m deep with peas = 12 pencil leads = one mole of carbon atoms = 1/18 the surface area of the Sun

As you now know, a mole represents a HUGE number of particles (ie. atoms, molecules, or formula units). Throughout this unit, we will abreviate molecules as "molec" and formula units as "FU." A formula unit is the empirical formula of an ionic or covalent network solid compound used as an independent entity for stoichiometric calculations.

EXAMPLES
-A Sample of Carbon contains 2.47 x 10²⁵ atoms. How many moles of Carbon is this?

(2.47 x 10²³) atoms x 1/(6.02x10²³)

= 41.0mol

-12.5 moles of O2 represents how many molecule?

12.5mol x (6.02x10²³) FU/1mol

=7.53x 10^24 molec

-How many formula units are present in 0.57 mol of BaCl₂?
0.57mol x (6.02 x 10²³) FU/1mol
=3.43 x 10²³FU

Extra practice::

Next time: Molar Mass

TEST: ATOMIC THEORY

We had our unit two test today. All went well!

TOPICS COVERED

• Early Atomic Theories
• Evidence for the Bohr Model
• Bohr Models
• Quantum Theory
• Isotopes & Mass Spectrometers
• Period Tables & Trends
• Electron Dot Diagrams
• Classification & Nomeclature

Burn, Bunsen Burn! (Hydrate Lab)

We had a very short class to experiment with hydrates, as it was post-secondary day. It was quite a fascinating experiment.

Recall that a hydrate is a compound that can form lattices which can bond to water molecules. These crystals contain water inside them which can be released by heating. In other words, hydrates are ionic compounds that contain an inorganic salt compound loosely bound to water. The hydrate we used was cobalt(ous) chloride hexahydrate.

PURPOSE
The purpose of this experiment was to determine the anhydrous (with no water molecules attached) mass of the hydrate and compare this with the actual mass of water that should be present.

MATERIALS
For this experiment, we used a Bunsen burner, a test tube, a test tube rack, a test tube clamp, a weight scale, and a hydrate (cobaltous chloride hexahydrate). For safety reasons, we wore lab coats, rolled up our sleeves so that they would not catch on fire, and wore safety goggles. We felt like we were professional chemists.

PROCEDURE
Our goal was to boil away all of the water so that only the inorganic salt compound was left. We followed the following procedure:

1. Find the mass of the empty test tube, using the weight scale (this is so we can determine the mass of the hydrate by itself).
2. Fill the test tube with about one centimetre of the hydrate.
3. Find the mass of the test tube and the hydrate, using the weight scale.
4. Using extreme caution, connect and light the Bunsen burner. The gas flow should be adjusted so that it is about five centimetres tall. Always assume that the Bunsen burner is on!
5. Using the clamps, pick up the test tube 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. Find the mass of the test tube and the inorganic salt compound, ensuring none of the chemicals inside spill.

It is very important that you do not leave the hydrate in the flame for too long, because the anhydrous solution will begin to react with oxygen.

OBSERVATIONS

• We noticed that the initial mass, compared to the final mass, was greater than the final mass. This is because the water evaporated, and less matter was present in the test tube after heating it with the Bunsen burner.
• While heating the substance, a color change was evident; the substance changed from magenta, to a dark purple, to a light purple, and finally, to a light blue.
• We could hear that a gas was being released. This is the water evaporating.

ANALYSIS
After the experiment, we determined how much water was released during heating, and what percent of the hydrate was water.

To determine how much water was released during heating, you must first find the mass of the substance on its own, since you weighed both the test tube and the substance inside. So, subtract the mass of the empty test tube from the mass you obtained before heating. This is the mass of the substance on its own, before heating. To find the mass of the substance on its own, after heating, subtract the mass of the empty test tube from the mass you obtained after heating. Now that you have the mass of the substance before and after, you can find out how many grams of water were released during heating, by subtracting the mass at the end of the experiment from the mass at the beginning of the experiment.

To determine what percent of the hydrate was water, divide the amount of water by the mass of the substance at the beginning of the experiment and multiply by 100. In our case, 45% of the hydrate was water. Because the actual percent of water in this hydrate IS 45%, our experiment had 0% error.

If you do not know how to calculate percent error, please refer to the September Post: Measurement and Chemistry! The formula for calculating percent error is:

CONCLUSION
We now have a much better understanding of the composition of a hydrate. Here is a video on the experiment we carried out:

Next class is our chapter test on atomic theory!

Naming Molecular Compounds!

Chemical Nomenclature Part 2!

We started off going over the 7 diatomic molecules, as well as the 2 polyatomics. Diatomic molecules consist of two atoms of the same element. Don't forget, H₂, N₂, O₂, F₂, Cl₂, Br₂, and I₂! Polyatomic molecules consist of several atoms of the same element. Don't forget, P₄ and S₈!

NAMING MOLECULAR COMPOUNDS
To name a molecular compound, we use the name of the first element, then we write the second element with an ending prefix of "ide". The first atom usually does not have a prefix. Hydrogen does not get a prefix. Also, if only one atom of the element exists in the compound, the first element in the compound does not have a prefix. The second element, however, must carry a prefix to indicate how many atoms of that element exist in the compound.

Here is a very handy list of prefixes:

We should also memorize the following list of IUPAC names and formulas:

• Water --> H₂O
• Hydrogen Peroxide --> H₂O₂
• Ammonia --> NH₃
• Glucose --> C₆H₁₂O₆
• Sucrose --> C₁₂H₂₂O₁₁
• Methane --> CH₄
• Propane --> C₃H₈
• Octane --> C₈H₁₈
• Methanol --> CH₃OH
• Ethanol --> C₂H₅OH

Examples:
NO → Nitrogen Monoxide H₂S → Hydrogen Sulphide
N₂O₄ → Dinitrogen Tetraoxide
CS₂ → Carbon Disulphide
P₄O₁₀ → Tetraphosphorous Decaoxide

NAMING ACIDS AND BASES
Compounds with Hydrogen are acids.Hydrogen appears first in the formula unless it is part of a polyatomic group. The IUPAC System uses the aqueous hydrogen compound. Polyatomic acids use the suffix '-ic', and binary acids use the prefix 'hydro-'.

Examples:
HCL --> Hydrochloric Acid
H₂SO₄-->Sulfuric Acid
CH₂COOH --> Acetic Acid/Ethanol Acid

We must memorize the following:

Hydrochloric Acid -->HCL
Nitric Acid -->HNO₂
Sulphuric Acid --> H₂SO₄
Phosphoric Acid --> H₂PO₄
Acetic Acid --> C₂COOH

Naming bases is really quite simple! We use cations followed by OH (Hydroxide)

Examples:
NaOH --> Sodium Hydroxide
BA(OH)₂ --> Barium Hydroxide

USING A BUNSEN BURNER
Next class, we will be doing a lab involving the use of a Bunsen burner. Safety is crucial! When using the Bunsen burner, check the tube; make sure it is not brittle, because gas can escape.The gas being used is butane, which is lighter than air. We turn on ventilation to clean out the air from excess gas.To use a Bunsen Burner, first strike the match. Then, put the match over the burner, before you turn on the valve to release the gas.

In class today, Mr. Doktor used the bunsen burner to heat up magnesium, which created a bright white light, which is also seen as the white light in fireworks. Here is a video of the chemical reaction:

ALWAYS ASSUME THE FLAME IS ON! Everything burns quickly; we are all subject to immediate physical harm if the bunsen burner is being used improperly. We could be sent to the hospital, so we must treat the bunsen burner with respect. If you are harmed by the bunsen burner in any way, shape, or form, go straight to the safety shower.

Next class is the hydrate lab! Be prepared!