December 9, 2010 - 4C

Lab to determine :
1. the percentage of water in an unknown hydrate
2. the moles of water present in each mole of this unknown hydrate, when given the molar mass of the anhydrous salt
3.write the empirical formula of the hydrate

We found the mass of hydrate, water given off, and mass of anhydrous salt.

from this we could find the empirical formula of the hydrate

December 1 2010 - Empirical and molecular formula

Empirical formulas give the lowest term ratio of atoms (or moles) in the formula.
*All ionic compounds are empirical formula.
Ex. -C4H10 (Butane) is a molecular formula.
-C2H5 is the empirical formula

To find empirical formula:
1.Get the mass of each element by assuming a certain overall mass for the sample (100 g is a good mass to assume when working with percentages).
2.Convert the mass of each element to moles of each element using the atomic masses.
3.Find the ratio or the moles of each element by dividing the number of moles of each by the smallest number of moles.
4.Use the mole ratio to write the empirical formula.

Example
1: A 14.0 g sample of a compound that contains only carbon and hydrogen is completely burned. The resulting products are 20.49 g of CO2 and 12.59 g of H2O. If the molar mass of the compound is known to be 30 g/mol, then what is the empirical and molecular formula of the compound?

Step 1: Calculate the number of moles of CO2 that are formed. This is the same as the number of moles of C in the original compound.

CO2 molar mass = C + 2O = (12.01) + 2(16.00) = 44.01 g/mol

(20.49 g CO2)(1mol/44.01 g) = 0.4656 mol CO2 therefore 0.4656 mol C in the original compound.

Step 2: Calculate the number of moles of H2O that is formed. Since there are 2 H atoms per H2O molecule, this is 1/2 the amount of H in the original compound.

H2O molar mass = 2H + O = 2(1.01) + 16.00 = 18.02 g/mol

(12.59 g H2O)(1 mol/18.02 g) = 0.6987 mol H2O therefore twice as much H (1.397 mol) in the original compound.

Step 3: Substitute the moles into a trial formula C0.4656H1.397

Step 4: Divide by the smallest trial subscript, so that you can simplfy to a whole number ratio

Therefore, CH3 is the empirical formula.

Step 5: Calculate a trial molar mass for the empirical formula

CH3 molar mass = 1C + 3H = (12.01) + 3(1.01) = 15.04 g/mol.

Step 6: Divide the actual molar mass of the compound by the trial molar mass, to find how many formula units there are in the real compound

(30 g/mol)/(15.04 g/mol) = 1.995 which is almost 2

So there are 2(CH3) units in the actual molecular formula, C2H6


To find molecular formula:
1. Determine the molecular mass in grams.
2. Divide the molecular mass of the compound by the molecular mass of the empirical formula.
3. Round the quotient to the closest integer.
4. Multiply the rounded number by all the subscripts, using the product as the new subscripts.


Example:

A compound has an empirical formula of CH2 and a molecular mass of 42g. Determine its molecular mass.

Step 1: Carbon = 12 + Hydrogen = 2(1.01)
Molecular Mass (g) = 14.02
Step 2: 42 ÷ 14.02 = 2.999
Step 3: 3
Step 4: CH2 • 3 = C3H6

November 28 2010 - Percent Composition

The percent composition of a compound is a relative measure of the mass of each element present in the compound.  


To calculate the percent composition (percentage composition) of a compound
1.Calculate the molecular mass (molecular weight, formula mass, formula weight), MM, of the compound
2.Calculate the total mass of each element present in the formula of the compound
3.Calculate the percent compositon (percentage composition): % by weight (mass) of element
    = (total mass of element present ÷ molecular mass) x 100


Eg.
Calculate the molecular mass (MM):MM = (2 x 22.99) + 32.06 + (4 x 16.00) = 142.04


Calculate the total mass of Na present:2 Na are present in the formula, mass = 2 x 22.99 = 45.98


Calculate the percent by weight of Na in Na₂SO₄:%Na = (mass Na ÷ MM) x 100 = (45.98 ÷ 142.04) 
x 100 = 32.37%
Calculate the total mass of S present in Na₂SO₄:1 S is present in the formula, mass = 32.06


Calculate the percent by weight of S present:%S = (mass S ÷ MM) x 100 = (32.06 ÷ 142.04) x 100 
= 22.57%


Calculate the total mass of O present in Na₂SO₄:4 O are present in the formula, mass = 4 x 16.00 = 64.00


Calculate the percent by weight of O in Na₂SO₄:%O = (mass O ÷ MM) x 100 = (64.00 ÷ 142.04) x 100 = 45.06%


The answers above are probably correct if %Na + %S + %O = 100, that is,     32.37 + 22.57 + 45.06 = 100

November 19, 2010

Mole Conversions


1. 6.022 * 10^23 particles = 1 mole
a) particles to moles
Particles / (6.022*10^23) = moles
Ex. (3.01*10^24 particles) * (1mole / 6.022*10^23 particles) = 5.00 moles with significant figures
b) moles to particles
moles * 6.022*10^23 = particles
Ex. 0.75 moles * (6.022*10^23 molecules) / 1mole = 4.5 * 10^23 molecules with significant figures
c) grams to moles
mass of something * (1mole / atomic mass)
Ex. 3.45g Carbon to moles
3.45g * (1mole / 12.0g [Carbon's atomic mass]) = 0.288 moles Carbon
Ex. 6.2g of MgCl2 to moles
6.2g * (1mole / 95.3g [MgCl2 atomic mass])  = 0.0650 moles MgCl2

November 1 2010 - Density

A materials density is defined as its mass per unit volume.  It is, essentially, a measurement of how tightly matter is compacted together.  

To calculate the density of an object, you take the mass and divide it by the volume.  The equation will look like this : P=M/V  where P is the Density, M is the mass and V is the volume. 

The unit for solids are g/Cm³ and the unit for liquids are g/mL

1 Cm³ of water = 1mL

The density of water is 1.0g/mL or 1000g/L

If the density of an object is greater than the density of a liquid, then it will sink

If the density of an object is less than the density of a liquid, then it will float.

=) DENSITY VIDEO!!!!!!!!!

Oct 26, Significant Digits

After the test that I have failed, today we have learned significant digits.

Significant Digits(Figures)
-The last digit in a measurement is uncertain.
-Trailing zeros after the decimal point are counted
-Leading zeros aren't counted
-Trailing zeros without a decimal point aren't counted

Example : 2.567 = 4 significant digits
               0.0097 = 2 significant digits
               10000 = 1 significant digits
               10000. = 5 significant digits
Exact Numbers
-Same quantities are defined as exactly a certain amount and no rounding is required and have infinite significant digits
 Ex. Number of people, animals, items etc
Rounding Rules
-if the number you are rounding from is greater than 5 you round up
-if the number you are rounding from is smaller than 5 you keep the number in front
-when the number you are rounding to is 5 then:
                 -if the number you are rounding from is 5, round to make the last digit even
                 -if the number you are rounding from isn't 5 then you round up
Example:
     Rounding to the tenth
1.01 = 1.0
1.06 = 1.1
1.00 = 1.0
4.95493 = 5.0
4.85493 = 4.8
1.05 = 1.1

October 19 2010 - Experiment 3B

Today, we did our third experiment yay!  We learned how to separate mixtures through chromatography.

The first thing we did was we had to set up our experiment by getting 3 large test tubes and put them into  Erlenmeyer flasks.  Water was put into each test tube.  We then cut a tip on 3 of the chromatography paper.  We would then have to put a dot of red/yellow/blue on one strip of chromatography paper, we would then dip that into the water in the test tube and wait for the water to come up through the paper.  We did this for 2 other mixtures, green dye and an unknown substance.  We observed and realized that the three substances had separated to a certain degree.

A video to a chromatography experiment

October, 15/10 - Separation of Mixtures and Solution.

The basis of separating mixtures is by separating different components with different properties form each other.
There are nine ways that we will be learning about how separation occurs when separating.


Hand Separation: Evaporation

• Hand separation (solid and solid)
• Mechanical mixtures separated by using magnets or sieves.
• Evaporation
• Boiling the liquid to leave the desired solid.

Filteration

• This method can be used when the solid is not dissolved in the water.
• The mixture passes through the filter and and the solids are stopped by the pores of the filter if the particles are too big to fit through.
• If the pores are bigger than the particles, the particles will pass.

Crystallization

• Precipation in crystallization converts solutes to solid form by chemical or physical change.
• Solids than separate by filteration or floatation. Floatation is when the denser components sink to the bottom.
• Evaporate and then cool and you will get the desired component(s).

Gravity Separation

• Separates particles by density.
• Heavy particles will sink to the bottom. The light particles will rise to the top.

Solvent Extraction

• A component will move into a a solvent and is shaken with mixtures.
• When dealing with mecahnical mixtures ( solid and solid), liquid will be used to dissolve one component.
• When dealing with solutions, the use of liquid will dissolve one or more substance until it leaves the wanted substances all by itself.

Distillation
(liquid and liquid solution)

• Using the boiling method, the liquid with the lowest boiling point will rise and condense first. Then it cools and drops into another container as pure. This may also leave the other substance as pure.

Chromatography

• Sweeping the mixtures, compounds that are heavier will slow down at a point to come to a gradual stop.
• This is a very good method for separating complex mixtures like drugs.
• The separated components can be collected individually.

Sheet CHromatography

• Known as paper chromatography (pc)
• Stationary Phase - components that spent a long time in a place will soak into the paper, leaving spots in a phase more than others.
• The components appear at separate spots ajd spreads when dried (compenents can be collected).

Thin layer Chromatography (TLC)

• Stationary phase - A thin layer of absorbent is coated on a sheet of plastic or glass.
• Some compounds will bond to it stronger than others and will stay on the absorbent. When dried, it can also be collected.  

Naming Acids Oct.13/2010

Today October 13th 2010 wednesday we learned naming acids.

Acids - formed when a compund composed of HYDROGEN ions and a negatively charged ions are dissolved in water which is also called aqueous.
                           - Ions seperate when dissolved in water. Ions with H₂O to form
                           - H⁺ ions with H₂O to form H₃O⁺(H(YDRONIUM ion)

Naming The Acids (Simple Acids)
1. Use "hydro" as the beginning
2. Last syllable of the non-metal is dropeed and replaced with "-ic"
3. Add "acid" at the end

Examples.
HI(aq) - hydroiodic acid
HCl(aq) - hydrochloric acid
H₂S - hydrosulphuric acid

Naming Complex Acids
1. -ate replaces with -ic
    -ite replaces with -ous
2. Acid at the end of the name

Examples.
H₃ClO - chloric acid
HNO₂  - nitrous acid
HCH₃COO - acetic acid

Naming Acid Part.1

Naming Acid Part.2



October 7 2010 - Writing and Naming Ionic and Covalent Compounds

Ionic Compounds
-composed of 2 or more ions (oppositely charged)
-held together by electrostatic forces
-electrons are transfered from metal to non-metals


Writing and naming ionic compounds 


To make an ionic compound, you need to combine a positive and a negative together/ a metal and non-metal. 


Ex.
Na+ (positive) and Cl- (negative) make NaCl


If the charges are different, then you need more ions to balance it.


Ex.
Mg(+2) has a +2 charge, so you need two Cl- to balance it


The result would be MgCl2

Ex.

  NaCl is composed of sodium and chlorine. Na is always 1+, and Cl is always 1-.  Add the charges of the positive ions, and separately add the charges of the negative ions. Each will equal the other.

Look at CaCl2, for example. Ca always has a charge of 2+. Cl again is always 1-. You need two Cl and one Ca to make a neutral formula. Be sure to write the formula for the smallest whole-number ratio for each ion.

When writing the name for the formulas, you just add the prefix -ide to the end of the second element. So NaCl would be sodium chloride. 

The rule is the same for polyatomic ions.  Just put parenthesis around the polyatomic ion to remember that the entire group of atoms is the charge.  For example, nitrate (NO3)-.  the entire polyatomic ion is 1 negative charge.  Or, phosphate, (PO4)-3. This means that the entire polyatomic ion is a 3 negative charge.  When you combine it, use the same rule as if you were combining single ions together. For example: NaNO3 or Na3PO4


When writing out the names of multivalent elements, you use roman numerals after the name.


Ex.
FeO = Iron(II) Oxide


Covalent Compounds

-share electrons
-non-metals with non-metals
-diatomic  molecules= H_2, N_2, O_2, F_2, Cl_2, I_2, Br₂


Writing and naming covalent compounds


To make a covalent compound, usually you need a negative and a negative together/ a non-metal and non-metal.  Writing covalent compounds is the exact same as writing ionic compounds.


When writing the names of the covalent compounds, you must use greek prefixes.  


Ex.
CO2 = Carbon Dioxide


*note* if only one of the element is needed in a covalent compound, the prefix mono is not needed, but it is needed on the second element.


CO = Carbon Monoxide


Ionic and Covalent Bonding video

October 7, 2010

Finding Out About Matter (p.25-34, 36-39)



Everything thats around us in our environment is made of matter. Matter can be broaden into different categories that distinguishes different types of matter from each other. For an example, all substances have a boiling point but at different temperatures. These characteristics can give matter individuality.

Purfying Matter

When different types of matter are put together, they can either form mixtures or solution. So what are mixtures and what are solutions? Mixtures are when two types of matter that mixes together and the mixture is clearly visible such as having sand mixed in water. Solution is when you mix two different types of matter together but you cannot see with the naked eye such as mixing sugar into water. However, the two elements can be separated through the method of distillation. However, some mixtures of matter are more difficult to separate than others and some solutions may be considered as pure substances until an innovative way of distillating them has been discover.

Characteristics of Pure Substances



Ways to determine the charactersitcs of substances include their boiling point, melting point, and freezing point. Through the characteristics, we can tell the difference between on substance and the other. For example, if we boil the solution of salt wate, the temperature gradually rises when the water boils away. On the other hand, if you just boil pure water, you will find that it boils to a point and continue boiling until the water disappears. This means that pure substances have a constant boiling point. Usually, mixtures don't have constant boiling point. However, some mixtures such as water and grain alcohol have a constant boiling point at 78.2 degrees. Melting and freezing points are also charactersitics to be dicussed. Melting point is the point when a specific type of substance melts while freezing point is the point where a specific type of substance freeze due to the low temperature.

Chemical and Physical Changes

There are two types of changes in matter. One is chemical change and the other is physical change. Chemical changes are changes that do not necessarily have to be seen. The change is determine by the release of energy, colour change, and possibly the smell. After a substance undergoes chemical change, it becomes a new substance and usually it cannot change back to its original form. Chemical changes that break down into two or more new substances is called decompostion. Physical change is the change that you can see with the naked eye and no new substance is formed.

Compounds and Elements

Elements are the bases of a specific type of substances. They can combine with another element that is different from themselves to form compounds such as NaCl (salt). A way to separate compounds to form new substances that can be done with electricity is called electrolysis. This method can decompose substances to form new kinds of matter.



Go to: http://www.youtube.com/watch?v=xYFAj50c7xM to learn more.

 

October 5 2010 - Experiment 2B

Today, which is October the 5th.  We did an experiment on the heating and cooling of a pure substance (dodecanoic acid/lauric acid) and watch the changes of it.

We were given a test tube containing dodecanoic acid in liquid state and had a thermometer in it.  We would then have to clamp it into position into a beaker of room temperature water.  We would monitor the cooling process and record its temperature every 30 seconds until the dodecanoic acid reached a temperature near 25 deg. Celsius.  The temperature would be recorded every 30 seconds.  After that, we would have to put the solid dodecanoic acid into a beaker full of 55 deg. Celsius water.  With hot plate under the beaker. The temperature would also be taken every 30 seconds and be recorded.  Then we cleaned up with rinsing our hand with soap and water. 

-A video of demonstrating melting and freezing of a substance.
Some info on dodecanoic acid

September 29, 2010 - Experiment 2C

Today we did our first experiment of the year.  Yay ^_^.  It was about chemical and physical changes.

What we did was we would mix 2 out of the 4 different solutions together to create 6 different combinations.  We did this on a glass square with a 3 by 4 grid which aided in the creation of the 6 combinations.  We then recorded all the data that we had observed and then cleaned up.

	A	B	C	D
A
B
C
D

Table looked like this ^^

a video on an experiment about chemical and physical changes (click here)
                              

September, 28 2010 - Matter

On September, 28 2010 we reviewed on the things that we had learned in the past about matter. We took some notes on what matter is to help use remember the specific definition of matter.  




 Here is a link to check out this video.

http://www.youtube.com/watch?v=-Ypwtjp0FSE




Matter is divided into two categories. One is pure substances and the other is mixtures.

Pure substances 

• have one set of properties
• have only one kind of particle

Pure substances are spit into two parts. Pure substances are either Elements or Compounds.

Elements:

• are the simplest form (It cannot be decompose)
• are made of atoms

Compounds:

• are made of elements
• are chemically combined
• are the smallest particles
• are molecules (either inonic or covalent)

Mixtures

• have more than one set of properties and substances
• are physically combined

Mixtures are split into two categories as Homogeneous and Hetrogeneous.

Homogeneous:

• are uniformed throughout
• appears to have only one compound. Ex. solution.

Hetrogeneous:

• not uniform
• appears to have mroe than one component. Ex. Water and oil. Salade dressing.

         The properties of matter have physical and chemical changes as the environment around the matter changes. Physical changes are changes that you can see with the naked eye while chemical changes may not necessarily be visible. In physical changes, no new substance is formed. The chemical composition does not change. Matters that undergoes physical changes can be reversible. For example, melting point and boiling point. On the other hand during a chemical change, new substances are produced. Unlike physical changes, the changes that occur in chemcial changes are irreversible. For example, you cannot uncook a food that was already cooked.

September, 23 2010 - Unit Conversion

On September, 23 2010 we reviewed and went further on unit conversion.
We did a lot of review sheets and went over them.

Converting units might be confusing and hard to deal with. There is one way how I remember it.
starting from small it goes k<M<G<T<P. KMGTP.
There's the da, deca for decades.

Say you have to convert 20km to dm.

The equation would be 20km*(10, 000dm/1km) .
The km will get canceled and u will be left with dm.
I always put the unit ur converting from which is km on the bottom first  and then think how many dm there is in a km.  1km = 1000m 1dm = 0.1m which is 10dm = 1m then 1km = 10, 000dm and then just times that by 20.
Here's a chart for you to look at.

And here's a video for you to look at as well.

http://www.youtube.com/watch?v=w0nqd_HXHPQ

Sept 21 2010 - Scientific Notation

Scientific Notation is used to express very large or very small numbers using powers of 10.  Scientific notation is very useful and is often favored by scientists, mathematicians and engineers, who work with large numbers.
  

a ×  10^b

b is the exponent
a is the coefficient






<sup>Ex.
Change the following number into 
scientific notation : 642 000


Step one : put a decimal point into the number so that there is one digit left of the decimal point.

                  6.42000


Step two: count the number of spaces needed to get back to the origonal
                  number. Since you have to move the decimal point 5 spaces to
                  the right, the exponent is 5.


                  6.42000
                     ------> 5 spaces 
                                       
                  6.42000 X 10^5

**When moving the decimal to the right to get back to the original number, the exponent will always be 
positive.  When moving the decimal to the left to get back to the original number, the exponent will always be
negative**</sup>