SOOO... what we did in class. Last day, we basically just had to do a quiz on percent composition and empirical and molecular formulas. That took up almost the whole block.. because it was SO SO SO HARD! *ahem* Me, as well as other studentts agree that everything should be easy peasy :) sooooo make it easierrr please ^^!
ok..so anyways, we didn't do much after that.. we had to do a flow chart on Lab 4C, and copy out the table. That was basically all we had to do! In the next few classes, we are going to determine the percentage of water in hydrate, the moles of water present in a hydrate, and how to write the empirical formula of the hydrate!!!
This is a gas hydrate.
But anyways today, we did Lab 4C.
In this lab we worked with a bunsen burner... so us, the students could have been in GREAT DANGER!! WAHHH. No... we weren't... but some saftey equipment we needed included a lab apron and safety googles. The purpose of this lab was to determine the percentage of water as well as the number of moles of water present in an unknown hydrate.
First step: Heat the crucible with the bunsen burner and a ring stand to make sure that it is dry enough.
Second step: Let the crucible cool down, then weigh the empty crucible. After, add some of the hydrate into the crucible and weigh that as well. Begin heating the crucible over the bunsen burner for 5 minutes. To make sure the heating is effective, make sure that the tip of the fire is touching the crucible, as it should glow dull red.
Third step: Wait for it to cool down and weigh it again. Now, do a second heating.
This second heating is to confirm the numbers that were found in the first heating, just so we can be more accurate.
Final step: Add drops of water back into the anhydrous compound (the thing that's left in the crucible) and see what you get.
So in last class we learned how to find both Emperical and Molecular formulas. Today we learned how to found the EF of an organic compound, which is some what harder.
Basically, we can find the EF of any organic compund by burning the compund (collecting/weighing product). We then have a burnt product. We then can calculate the moles of each element from the original organic (unburnt) product.
Now let's try an example to see if we can calculate it right.
EX. What is the EF of a compound that when a 15.00 gram sample is burned it produces 25.0 grams of CO2 and 10.0 grams of H20
Step 1: Calculate all moles of CO2 and H2O
25.0 CO2 x 1mol CO2/ 44.0g CO2 = 0.568 mol CO2
10.0 H2O x 1mol H2O/18.0 H2O = 0.556 mol H2O
Step 2: Find moles of C and moles of H in CO2 and H2O
0.568 mol CO2 x 1molC/1 molCO2 = 0.568 mol
0.556 mol H2O x 2molH/1molH20 = 1.112 mol
Step: 3: Divide both moles by smallest molar amount
C = 0.568/0.568 = 1
H = 1.112/0.568 = 2
There for the Emperical Formula is CH2
Step 4: Check answer.
Convert moles back into grams. The result should be 15.0g. If the answer does not match, that means Oxygen is a component. To find the amount of Oxygen use this formula.
Mass of O = Mass of compound - mass of C - mass of H
This concludes my blog. Have a good night everyone.
So today............we're going to be doing Empirical and Molecular Formula.
What is the empirical formula? It is the expression of the ratio of atoms in a formula in its lowest terms. By this definition, we can realize that all ionic compounds are empirical formulas.
So, for example, let's say we have the formula C10H14N2, which is Nicotine (the thing that makes you addicted to cigarettes).
Now, to get the empirical formula, let's reduce all the subscripts to the lowest terms. We'd get C5H7N. Of course, this is a whole new compound, but we can say this is the empirical formula of nicotine.
Now, find the empirical formula for a compound consisting of 63% Mn and 37% O.
First, we assume that we have 100g of this compound, just to make everything easier.
Now, we take the 63 g(now grams of Mn, because 63% of 100 is 63) and convert it into moles.
Since the molar mass of Mn is 54.9, we do
63 g X 1 mole = 1.1 moles Mn (in sig figs)
54.9g
Then, we take the 37 g of O and do the same.
37 g X 1 mole = 2.3 moles O
16.0g
Now, since we have both parts of the compound in moles, we now divide each by the smallest molar amount. If you're going like HUH? just wait for the next step.
So, we can all agree that the number 1.1 is less than 2.3 right?
That will be what we are dividing by.
We take the 1.1 moles of Mn and divide by 1.1, giving us the whole number 1.
Then, we take 2.3 moles of O and divide by 1.1, giving us 2.1, which is around 2.
The final part is to scale the ratios into whole numbers. But good thing for us, the numbers in this example are already whole numbers, so we don't need to multiply again. (In other words, we don't get 1.5, which means we would have to multiply by 2 to get a whole number)
We will get 1 Mn, and 2 O.
Our final answer for this question is...... MnO2.
OK..Now let's do some molecular formula. This is a multiple of the empirical formula and it can show the actual number of atoms that are necessary to combine into a molecule.
For this, we have a special formula. That is, the molar mass of a compound divided by molar mass of the empirical formula, will give us the molecular formula.
Let's say
The simplest formula for vitamin C is C3H4O3. Experimental data indicates that the molar mass of vitamin C is about 180 g/mol. What is the molecular formula of vitamin C?
First, we take the empirical formula, and add up the molar masses of each element to get the total molar mass. Get your periodic table for this - you'll need it.
The molar mass of C is 12.0g, and we have 3, so 3 X 12 = 36.0 Then, we have 4 Hydrogen, with each hydrogen at 1.0g each, so 4 X 1 = 4.0 Finally, we have 3 Oxygen at 16.0 g each, so 3 X 16 = 48.0 36.0 + 4.0 + 48.0 = 88.0 g/mol
Now we have both molar mass of compound, and the molar mass of the empirical formula. The last step for us is to put it into the formula.
So 180 g/mol = 2.045, which is around 2. 88 g/mol
This means that we need to take that original formula and multiply it entirely by 2. 2 X C3H4O3 This gives us C6H8O6, which means the molecular formula of vitamin C is C6H8O6
What is percent composition? Well... it is a compound that is a relative measure of the mass, of the different elements that are in the compound. Anyways...you need to be able to figure out the molar mass of each element, so if you are not sure what the molar mass of elements are, please read the previous blogs!!
This is an example of what a element would consist of, if the percent composition was represented by a pie graph
Now... how to actually calculate the percent composition of an element...
Ex 1/ What is the percentage composition of H20
First... you need to find:
Total Molar Mass of H2O.= 18.0g
Molar Mass H = 2.0g/ mol
Molar Mass O = 16.0/ mol
After that you find the % of each element, by dividing it by the total molar mass.
% of H = 2.0g per mol/ 18.0grams x100 = 11.1%
% of O = 16.0 per mol/ 18.0 grams x 100 = 88.9%
11.1%+88.9%= 100%
* Because there were no numbers in the questions, then you don't have to count the significant figures, so just round the number by one decimal place
Ex 2/ What is the percentage composition of Sc2(SO4)3
Total Molar Mass: 333.3 grams
Molar Mass Sc: 45.0 g/mol
Molar Mass S: 96.3 g/mol
Molar Mass O: 192.0 g/mol
% of Sc = 45.0 grams per mol / 333.3 grams x100 = 13.5%
% of S = 96.3 grams per mol / 333.3 grams x 100 = 28.9%
% of O = 192.0 grams per mol / 333.3 grams x 100 = 57.6%
The total percentage totals up to 100%
Ex 3/ What is the percentage composition of Zn(ClO3)2
Total Molar Mass: 232.4 grams
Molar Mass Zn: 65.4 grams/mol
Molar Mass Cl: 71.0 grams/mol
Molar Mass O: 96.0 grams/mol
% of Zn: 65.4 grams per mol / 232.4grams x100 = 28.1%
% of Cl: 71.0 grams per mol / 232.4 grams x100 = 30.6%
% of O: 96.0 grams per mol / 232.4 grams x 100 = 41.3%
The total adds up to 100% YAY!
Ex 4/ If a compound contains 48.0 grams of C, 6.0 grams of H, 64.0 grams of O, and contains some amount of Be and has a total mass of 127.0 grams. Calculate the % composition.
127.0-48.0-6.0-64.0 = 9.0 grams
% of C: 48.0 g per mol / 127.0 grams x 100 = 37.8%
% of H: 6.0 g per mol / 127.0 grams x 100 = 4.7%
% of O: 64.0 g per mol / 127.0 grams x 100 = 50.4%
% of Be: 9.0 g per mol / 127.0 grams x 100 = 7.1%
*Remember, there are numbers in the question, so sig figs count!
FINALLY......
Ex 5/ If a compound contains 137.3 g of Ba, 28.0 g of N, and contains some amount of O and has a total mass of 261.3 grams. Calculate the percent composition.
261.3 - 137.3 - 28.0 = 96.0 grams
% of Ba: 137.3 g per mol/ 261.3 x 100= 52.5%
% of N: 28.0 g per mol / 261.3 x 100 = 10.7%
% of O: 96.0g per mol / 261.3 x 100 = 36.7%
In the case the percentages only add up to 99.9%. Due to the rounding, there may be some cases where the percentages do not add up to 100%. If it is only 0.1% off, it is not a big deal, but to be safe ask Ms. Chen =]
These video's can maybe explain in more detail on how to calculate percent compositions..
After having all that knowledge inputted into your brain... time to use those skills on some other problems!
Whatup? Today we took the mole conversions quiz. Ms.Chen as well as half our class wasn't present today, probably because of the snow storm that hit Vancouver overnight. Anyways, the substitute teacher corrected the mole conversions review in the begining of the class. He then gave us 10 minutes to study. After studying we had the rest of the class to complete the test. I hope everyone did well. See you guys later.
So now that we understand how to convert from grams to moles, moles to grams, particles to moles and moles to particles, let's try converting from particles to grams and grams to particles.
These are now 2 step conversions. Once again we need our periodic table.
Let's start with particles to mass.
If you have 2.78 x 10^22 Fe atoms, how many grams are there?
First, you want to get rid of the atoms, so let's go from atoms to moles.
2.78 x 10^22 atoms of Fe x 1 mole / 6.022 x 10^23 atoms
Now, we also know how to go from moles to grams, so let's get rid of moles.
2.78 x 10^22 atoms of Fe x 1 mole / 6.022 x 10^23 atoms x 55.8g (molar mass of Fe) / 1 mole
Now our final answer is going to be (2.78 x 10^22)(55.8g) / (6.022 x 10^23) in 3 sig figs.
So, we get 2.58g. This means 2.78 x 10^22 Fe atoms weighs 2.58 g.
Now mass to particles...How many atoms of Fe are in 20.0g of Fe?
So, we get rid of grams by multiplying moles/grams.
20.0g of Fe x 1 mole / 55.8g (molar mass of Fe)
Then, since we want particles, we get rid of moles by multiplying particles/moles.
20.0g of Fe x 1 mole / 55.8g (molar mass of Fe) x 6.022 x 10^23 atoms / 1 mole
Our final answer of 20.0g Fe in particles (3 sig figs) is 2.16 x 10^23 atoms.
Well... I'm going to start off with a funny story... My friend VIVIAN CHENG thought the "moles" were the black dots that grow on your face... then she thought it was the ANIMAL mole..... until i explained to her...
Well...we're still learning about moles, and how to convert moles... as much as I dislike moles, we still gotta learn about MOLES!...
Keep in mind Avogadro's Number is 6.022 x 10^23 --> particles/mole
Molar Mass
You can convert molar masses from either:
1. Particles to moles
or
2. Moles to particles
Converting molar masses is somewhat simple. It is just like doing unit conversions.
For example:
1) Convert 4.5 x 10^24 particles to moles
4.5 x 10^24 particles x 1 moles/6.022 x 10^23 particles = 7.5 moles
What I did, was I divided 4.5 x 10^24 by 6.022 x 10^23. The particles cancel out. And because there is only one mole, the answer is simply 7.5 moles.
*Remember that you have to count to the correct number of significant digits.
2) *(2 atoms) Convert 0.82 moles CO₂ to molecules
0.82 moles x 6.022 x 10^23 particles/ 1 moles = 4.9 x 10^23 molecules CO₂
The moles cancel out, so you simply multiply 0.82 x (6.022 x 10^23) particles . Because its 4.9 x 10^23/ 1, it works out to just be 4.9 x 10^23 molecules CO₂
There are two atoms, so there is one more step involved. Well...it's not anything special... but...here goes!
4.9 x 10^23 molecules CO₂ x 2 atoms O/ 1 molecule CO2 = 9.8 x 10^23 atoms of Oxygen
And again... the molecules cancel out, and you just multiply the numbers.
This video may help explain some questions you may have:
Are you tired of converting moles yet? Well we're almost done, so bear with me!
Now.. we are going to learn how to convert moles to grams, and grams to moles. Get your periodic table for this!!! We need it !! To find the molar mass of an element, look for the atomic mass of the element!
Example:
1) Convert 3.06 moles Fluorine to grams
The molar mass of Fluorine is 19.0g/mol
3.06 moles x 19.0g/1 mole = 58.1 grams of F
Here, the moles cancel out, and you just multiply 3.06 x 58.1 grams of F...and.. BAM, you get the answer!
2) To convert 7.65 grams of Fluoride to moles
The molar mass of Fluorine is 19.0g/mol
So today, we learned about moles. No, not the round thing on the skin....the Chemistry mole.
But first, let's understand a hypothesis of a great Italian scientist named Avogadro. He proposed that equal volumes of different gases at the same temperature and pressure will have the same number of particles.
Now, let's define Atomic Mass. This is the mass of 1 atom of the element in the units amu. (atomic mass units) For example, if you look at the atomic mass of an element on the periodic table, it will give you how much 1 atom of that element weighs..
So, let's take Carbon. Carbon's atomic mass is 12.0 amu. This means that every carbon atom weighs 12 atomic mass units.
Now formula mass. This means you add up all the masses of the atoms in the formula of an ionic compound.
So.... let's say we have NaCl.
Na is sodium, and the atomic mass is 23.0. Cl is chlorine, and the atomic mass is 35.5.
So by simple addition: 23.0 + 35.5 = 58.5 amu. This means that NaCl weighs 58.5 atomic mass units.
Molecular Mass: All the atoms of a formula in a covalent compound are added in amu.
So: Carbon Dioxide is C + O2 12.0 + 16.0X2 CO2 = 44.0 amu
Molar Mass is the atomic, molecular, or formula mass of any pure substance, and the unit is in grams per mole. For example, 1 mole of oxygen is just its atomic mass (16.0 amu, but in a different unit, g/mol.)
So 1 mole of oxygen = 16.0 g/mol 1 mole of carbon = 12.0 g/mol
They both have the same number of particles. Hence, the molar atomic mass of an element is the mass of 1 mole of that element. It's just a unit that helps us and scientists count atoms and molecules, without actually counting every single atom. And what Avogadro found, was that the number of particles in 1 mole of any amount of substance is always 6.022 X 10^23 particles per mole.
Now here's a video if you're still somewhat confused.
Well, next class (November 15, a long way I know. It's because of the Remembrance Day holiday coming up) will be our Chapter 3 Chemistry Test.
This test will focus on Significant Digits, Scientific Notation, Uncertainty, Density, the lab, and some conversions and graphing we learned last time.
We reviewed the unit, and then we went to the computer lab to do more graphing.
Please refer to the previous post for instructions on how you can make your own graph! :)
In this Chemistry class, we went to the computer lab and made graphs that demonstrated the relationship between volume in hot and cold water.
Today, we learned how to make these graphs using OpenOffice; however, most of you have Excel on your computer, so that's the alternative.
First, you want to open up Microsoft Excel.
Next, you want to plug in a table of values into the cells.
Let's begin on A1, and labelling it as X values.
Then, put down some numbers for X. At the end you should get:
X values
1
2
3
4
5
In the column beside (so starting with B1), label it as Y values.
In B2, enter an equation that can relate X and Y.
This equation must start with "=" then A_ (whatever cell number your values start at, in this case A2) then an operation.
So, we could have something in cell B2 like..... =A2+5
Once the equation is set, you should see the number 6. (because 1+5 = 6)
So:
X values Y values
1 6
2
3
4
5
Now, click on the cell with the y value 6. There should be a small box in the bottom right hand corner. Drag that all the way down until it covers the last y value.
You should see all the numbers plugged in by now.
X values Y values
1 6
2 7
3 8
4 9
5 10
Now, highlight all the values, and click on the bar graph/chart option in the toolbar.
This will pop out a menu. In this menu, choose the scatter plot, and click next. Then, customize it to your liking, and fill in the x-axis and y-axis information, as well as a title.
After you click finish, a graph should pop out.
To add a linear trend line, right-click a point on the graph and click Add Trendline. You can even show the equation, which will give you the slope of this line you just graphed. (in y=mx+b form)
Adding a trendline:
The finished graph before you customize (i.e. changing colours, fonts):
Format your graph by changing the colours, fonts, etc.
AND You have now created a nice graph that shows your data clearly and effectively.
Another useful tool that can be used to make graphs or many cool geometric things, or even just having fun on the computer, is the Geometer's Sketchpad, introduced to me by my math teacher. :)
Lab 2E occurred on November 3, 2010. In this class we used basic math formulas to calculate the thickness of aluminum foil pieces. These formulas are. V=L*W*H, D=M/V, or V=M/D. Basically each group weighed each piece of aluminum foil. We were then given the density of each piece, which is 2.70g/cm^3. Our next step was to measure the length and width of the aluminum with our ruler. After that we would find the volume by dividing the mass by the density. Once we got the voulme, we divided it by the length and width to get the height (aka Thickness).
We also used a formula to calculate if there was any errors with our test results.
experimental error = estimated measurement - accepted value x 100%
accepted value
Well that wraps about everything up. This lab was fun and easy. It was fun because we didn't have to wear any goggles. LOL, jokes.
Density is a material that is specified as mass per unit volume, or weight per unit volume. Now.. why is density important? Well, density is applied to many chemicals that are being tested. It can find out the buoyancy of an object or fluid. To calculate density, there is a formula...and here it is..
Density = mass/volume
OR
Volume = mass/density
OR
Mass = Density x Volume
*Once you memorize one of them, you can move them around as you would mathematically to calculate density, volume or mass.
now... make sure you REMEMBER THE FORMULAS OF DENSITY!!!
To represent density:
solids - g/cmᵌ
liquid - g/mL
If there is 1cmᵌ of water = 1 mL
then the density of water = 1.0g/mL
= 1000g/L
Ex. Calculate the density of a liquid which has a volume of 28mL and a mass of 26.4g
density = mass ÷ volume
density = 26.4 ÷ 28 = 0.94 g/mL
Unless the fluids are mixed together, fluids that are less dense float on fluids that are more dense.
The Cup of fluids on the left have different density`s and as you see, the fluid on the top ( the clear fluid) is the least dense, the fluid on the bottom (the pink fluid) is the most dense, the fluid in the middle (the blue fluid) has density that is between the clear and pink fluid. The cup of fluid on the right can be mixed together with other fluid.
Before we move onto the answer to the question, let us define a few terms to better understand "uncertainty".
Precision - this is how reproducible a measurement is compared to other similar measurements. In other words, the more decimal places you have, the more precise your number is.
Accuracy - this is how close your measurement (or average measurement, in some cases) comes to the accepted/real value
Let's say you're shooting arrows at a target. If you shoot 5 arrows, and they all end up close to the bullseye, but all of them are far apart from each other, then the shots were accurate, but not precise.
If you shoot 5 arrows and they are far away from the bullseye, but are very close together, as if you shot all 5 of them at the same time into the same area, then your shots were very precise.
If your 5 shots were both very close to the bullseye, and very close together (meaning you can hit the same area consistently), then your shots were both very accurate and very precise.
Now:
Realize that no measurements are exact. Every measurement is just a best estimate, meaning there is room for error and "uncertainty".
However, when you count a set of objects, it is exact. There are 5 humans. There can't be 4.34 humans.
Now that we know the background, what is uncertainty?
A: Uncertainty is the margin of error, usually stated by giving a range of values that contains the real or true value.
There are 2 types of uncertainty: Absolute and Relative.
In absolute uncertainty, there are 2 methods.
Method 1: First, discard any unreasonable data first. You must have at least 3 reasonable measurements in order to use this method. Then, take the average of the measurements. Then, find the largest difference between the average and either the lowest or highest reasonable measurement.
For example:
Trial 1 - 15.3 g
Trial 2 - 15.5 g
Trial 3 - 15.2 g
Trial 4 - 11.9 g (remove it, because it looks very different from the rest of the data)
Average of the reasonable measurements: 15.3g
Difference between average and lowest number = 15.3g - 15.2 g = 0.1 g
Difference between average and highest number = 15.5g - 15.3g = 0.2 g
Absolute certainty based on average: 15.3 ± 0.2 g
Method 2: By using the uncertainty of each instrument
Measure to the best precision, then estimate to 0.1 of the smallest segment on instrument scale.
For example, a ruler's smallest segment is 1 mm. The uncertainty would be the tenth of 1 mm, which is 0.1 mm. So the data recorded for a measurement with a ruler would be:
_____(measured value to the best precision) ± 0.1 mm
In relative uncertainty, it is the ratio of absolute uncertainty to the estimated measurement.
So for example, if the data was 39.3 ± 0.1 g, then the relative uncertainty would be 0.1 / 39.3
This can be expressed in percentage, or in Significant Figures.
The number of significant figures is the relative uncertainty. Refer to the previous blog post for info on Significant Figures.
Well, they help make the number more precise. Numbers which have more significant digits are more precise. The last number in the measurement are usually uncertain.
There are some rules that are associated with making significant figures as precise as possible.
Rules on how to count the number of significant figures:
1) All number's that are not zero are always SIGNIFCANT
ex. 35322 , this number has 5 significant digits
2) Zeros that are in front of the non-zero numbers are NOT SIGNIFICANT
ex. 0.0065 , this number has 2 significant digits
3) Zero's that come after a decimal point ARE COUNTED
ex. 50.601 , this number has 5 significant digits
4) Zero's that come after numbers, but are NOT after a decimal point, are NOT COUNTED
ex. 340000000 , this number only has 2 significant digits
Both Number's have 2 Significant Figures
Some numbers require rounding, however, there are "exact numbers" that require no rounding at all. For example, you cannot say that there are 2.4 numbers of deers. Or say that there are 5.6 jackets.
So now that you know how to count significant numbers, let's learn how to round!!
Here are some rules for rounding:
1) Always look at the number to the right of the digit you want to round
2) If the digit is greater than 5, round up, and if the digit is less than 5,the number stays the same.
3)If the digit is the number 5, and there are more non-zero digits after the 5, round up
4)If that digit ends in the number 5, round the digit to make it an even number, (0,2,4,6,8)
In addition to rounding rules, there are also math rules.
When you add or subtract, remember to round to the number that has the fewest number of decimal places. You determine this by the number's position.
For example,
11.34 L
+ 13.4 L ___________
24.74 L ----> 24.7 L
For this equation, you round to the nearest tenth, because that number has the fewest number of decimal places.
Multiplying and Dividing is a bit different from adding and subtracting significant digits. When you multiply and divide you round to the fewest number of significant digits.
For example,
12.3 Km
x1.2 Km
_______
14.76Km ----> 14.8Km
For this equation, the number with the fewest number of significant digits is 1.2, which has 2 significant figures. So the new number has to have only two significant figures too.
Last class we had a lab. In this lab we experimented with paper chromatography. This particular type of paper is used to separate mixtures, which is one of the techniques that can be used to separate mixures.
So first, we cut 3 thin strips of chromatography paper with a pointy end (like a pencil). On these strips we draw a line 4 cm from the point and place 3 different food colourings on each strip. Later, we test these strips and dip them in test tubes filled with water. The results should be a separation of the food colouring components.
For example we had the green dye. The dye later broke from a yellow color to a blue color. Of course everyone knows blue + yellow =green.
The lab was a success, and it was interesting to see the colors seperate. Well thats a wrap then.
Study hard for the test next class!!!!
Today we will be teaching you several techniques that are used to separate mixtures.
The result after the separation of the mixtures should be different components with different properties.
To do this, you must first realize how to differentiate between components and properties.
For instance, do you want materials to be separated by high density vs. low density, or do you want something reactive vs something inert?
The first method: Hand Separation (solids and solids)
This is the method to use if you want to separate mechanical mixtures or heterogeneous mixtures by using a magnet or some kind of sifting tool.
Second: Evaporation (solids dissolved in a liquid solution)
In this method, you would boil away the liquid, so only the solid would remain.
Third: Filtration (not dissolved solids and liquids)
First, you would pour the mixture containing solid particles through a porous filter.
The result should be solid particles staying on the filter and the liquid compounds passing through.
Fourth: Crystallization (solid in liquid)
Precipitation is the conversion of a solute to solid by chemical or physical change. First, the solids should be separated by filtration or floatation. Then, you need a supersaturated solution of the desired solid and cool it. The result should be pure crystals.
Fifth: Gravity Separation (solids based on density)
In this method, you would use a machine called a centrifuge. This machine would whirl the test tube around at very high speeds, causing the denser materials to move to the bottom.
Sixth: Solvent Extraction
Here, the component moves into a solvent shaken with the mixture.
In a mechanical mixture (a solid with a solid), use a liquid to dissolve one solid, but not the other. This would result in the desired solid becoming left behind.
In a solution, the solvent is insoluble with the solvent that is already present. This solvent dissolves 1 or more substances, leaving the unwanted substances behind.
Seventh: Distillation (liquids in liquids)
If you heat the mixture, the liquid with the lowest boiling point will vaporize first. Then, the vapour moves into the distillation flask and enters the condenser. As the gas cools, it condenses into a liquid, dropping the distillate as a purified liquid.
Eighth: Chromatography
In this method, you move a mixture over material that will retain some components more than others.
The components are distributed between 2 phases. A mixture dissolves in a mobile phase through a stationary phase.
This method can be used to separate very complex mixtures, such as plastics, drugs, and foods; it will also produce highly accurate analyses. There are 2 types of chromatography:
Paper Chromatography:
The stationary phase is a liquid soaked onto a strip of paper.
The mobile phase is the liquid solvent.
Some components tend to spend more time in the stationary phase than others. On the strip of paper, the components should appear as separate spots after drying or developing.
Thin Layer Chromatography:
The stationary phase is a thin layer of absorbent coating a sheet of plastic or gas.
Some of the components will attract to the absorbent strongly. As a result, the components will appear as spots on the sheet.
First of all, how are acids formed?
An acid is formed when a compound composed of Hydrogen ions and a negativley charged ion are dissloved in water, meaning they are aqueous (aq). Once the ions are dissolved in water, they seperate.
Some Guidelines
For acids we add "hydro" at the begining. Then get rid of the last syllable of the non-metal and replace it with (ic). Then you can proceed and add acid at the end.
HCI(aq) -----> Hydrochloric acid
For complex acids replace the last syllables with new ones. Such as.... ___ate ----> ___ic and ____ite -----> ____ous
You dont' need the "hydro" for these complex acids.
Hi everybody! In today's Chemistry class, we reviewed how to write and name ionic and covalent compounds.
An ionic compound is a compound of a 2 or more particles that are oppositely charged. So for example, a metal ion and a non-metal ion.
If we have a K+ and a N3- then what is the compound?
First we look at the charges. K has a positive 1 charge, and N has a negative 3 charge. How can we make the charges equal 0? +1-3=-2, so what we could do is increase the number of K ions to 3. This would give us a positive 3 charge and a negative 3 charge of N, AND +3-3=0 :) So now we know the ionic compound of K+ and N3- is K3N. If we want to name it, it would be Potassium Nitride. Note that the metal is always first.
Now try to write or name these formulas with the Periodic Table:
1) Copper(I) oxide
2) FeO
3) SnCl4 There are also ions called complex anions. Anion indicates that it is a negative ion, and complex just means that it is a group of atoms that behave as one atom.
For example, if we have
Na2SO4 Treat SO4 as basically any other ion. We wouldn't call this sodium sulphur oxide. Instead, there is a specific name for the ion SO4 and that is sulphate. So this compound is called sodium sulphate.
It gets easier with covalent compounds. Covalent compounds are compounds of a non-metal and another non-metal. They share electrons rather than transferring like ionic compounds.
To name covalent compounds, we use the Greek prefixes:
If we want to name CO2, we don't say monocarbon dioxide. We never say mono for the first non-metal. If it was N2O3, then we would put dinitrogen trioxide, but never monocarbon or mononitrogen.
There are also diatomic molecules. These can be memorized by the acroynm HOFBrINCl. These are a special group of molecules composed of 2 identical atoms.
Here is a video on naming ionic compounds:
And here is the covalent compound:
Here are the answers to the practice questions above:
Cu2
Iron(II) oxide
Tin(IV) chloride
This is a gas hydrate. 
