Molarity of Solutions

REVIEW
Solution: homogeneous mixture where one substance is dissolved in another substance
      Solute: small quantity
      Solvent: larger quantity

We need to compare the amount of solute dissolved in a certain volume of a solution.
Molarity= number of moles of solute in one liter of a solution

FORMULA
molarity =    mass of solute    ....also known as.... M= mol
                    volume of solution                                              L


This formula can be altered so that any three of these elements can be found as long as you have the other two.

The molar concentration triangle!

Now lets look at a few examples.
Example:  If there is 5 moles of NaCl in 2.5 L of solution, what is the molar
               concentration?
M=  mol        5 mol
         L          2.5 L                        = 2 mol/L NaCl

               

How many moles of Ca(OH)2 is there in 1.30L of a .75 M solution?
         
                 M=mol/L
           = mol=ML
           = mol= (1.30L)(.75M)
                    = 0.975 (SIG. FIGS!)
                    = 0.98 mol Ca(OH)2




                 What volume in mL of 0.1 mol/L NaOH must you take to contain .030 mole of NaOH?
           
                  M= mol/L
                  L= mol/M
                  L= .030mol / 0.1 mol/L = 0.3L
                                                       = 0.3 *     1L     = 300 mL
                                                                   1000 mL

Hopefully now you understand the basics of it.  Its now time to practice!  This game includes some examples of molarity and mole conversions.  Good luck!

Billionaire-molar concentration

Lab 4C: Formula of a Hydrate

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

Supplies
Equipment                                                    ring stand and ring
lab burner                                                     centigram or digital balance
crucible and lid                                              lab apron
crucible tongs                                                safety goggles
pipestem triangle
Chemical Reagents
Approximately 5g of a hydrate
water

Procedure
1. Put on safety goggles
2. Set up pipestem triangle, iron ring, stand and bunsen burner like in 4C-1 in your workbork (Page 46)
3. Heat the crucible for 3 min and let it cool for 3 min
4. Determine the mass of empty crucible and record the mass in Table 1
5. Place hydrate into the crucible. Determine and record the mass of the crucible and hydrate
6. Place the crucible and contents on the pipestem triangle. Increase heat till the bottom is red and maintain this temperature for 5 min.
7. Turn off the burner and let the crucible cooldown for 5 min. Then determine and record the mass of the crucible and its contents.
8. Reheat the crucible for 5 min then let it cool for 5 min and then determine and record the mass. Step 6 and 7 should be within 0.03g, if it's not, reheat it again.
9. After the final mass reading, and crucible is cooled, add a drop of water into the contents of the crucible and note any observation/changes.

Table 1 

Analysis of Results
1.      g H₂0       x 100%   =    % of water  
      g hydrate

2. mass of salt = mass after heating - mass of empty crucible
    moles of salt = mass of salt x 1 mol  
                                                159.6g
3. mass of hydrate = mass of crucible/hydrate - mass of crucible
    mass of H₂0 = mass of hydrate - mass of salt
    moles of H₂0 = mass of H₂0 x 1 mol  
                                                     18g
4. moles of H₂0  = _____          
      moes salt
5. CuSO₄  x  __ H₂0  (copper (II) sulphate pentahydrate)

Empirical Formula of Organic Compounds

Organic Compounds are classified as any substance that contains carbon.
They can be found by:

ʘ Burning (as it reacts with oxygen gas [O2])

ʘ Collecting and weighing the products

                From the mass of the product, the mass of each element in the original Organic Compound can be found.

Keep in mind that the Empirical Formula of a Organic compound is CxHy.

The balanced chemical equation for the burning of a organic compound can be written as:

CxHy + 2O2 => XCO2 + Y/2H2O

The aim is to find the x and the y (which are the simplest ratio of atoms of mole in the elements)

From the equation, we can see that all of the C and H atoms in CxHy we used while making CO2 and H2O. It can therefore be said that the moles of carbon and hydrogen in the product are equal to the moles of Carbon in CO2 and the moles of Hydrogen in H2O.

Example:
A 3.27g sample is burned and produces 10.27 g of CO2 and 4.20g of H20. What is it's empirical formula?
Step 1: Find the moles of each
CO2  =>  C = 12                   = 44     10.27g CO2 x 1mole CO2/44g CO2 = 0.233 moles of CO2
                O = 16 x 2 = 32

H2O  =>  H = 1 x 2 = 2         = 18     4.20g H2O x 1mole H2O/18g H2O = 0.233 moles of H2O
                O = 16
Step 2: Find the moles of Carbon and Hydrogen
0.233 mol CO2 x 1 mol C/1 mol CO2 = 0.233 mol H
0.233 mol H2O x 2 mol H/1 mol H2O = 0.466 mol C
Step 3: Divide by smaller number
C 0.233 / 0.233 = 1
H 0.466 / 0.233 = 2
Therefore Empirical Formula = CH2
Step 4: Check Masses
0.466 mol H  x 1g H/1 mol H    = 0.466 g H
0.233 mol C x 12g C/1 mol C  = 2.80 g H
0.466g H + 2.80g C = 3.27g
If there is a difference between the mass of carbon and hydrogen versus the mass of the compound, there is oxygen in the compound. To find the amount of oxygen,
Mass of O = Mass of compound - Mass of C + H

For further explanations, please watch this video:

http://www.youtube.com/watch?v=9qGo9YTdnMg

Percent Composition

Percent Composition: The percent in mass that a certain element in a compound contains.  In other words, the percentage of the molar mass that is taken up by a certain element.

          % Composition =    mass of element      *100
                                      mass of compound


What percentage of glucose (C6H12O6) is oxygen?

Molar mass of Glucose:
        6  C @ 12g/mol =     72g/mol
       12 H @   1g/mol =     12g/mol
        6  O @ 16g/mol =  + 96g/mol
                                        180g/mol

Calculate Percent Composition

  96g/mol    *100 = 53.3 %
180g/mol  

Now add the step.  How about if there is more then just one molecule?  Pull out you mole map because this is where you will need it!

What is the percent composition of 5.00g of Vitamin A (C20H30O)?

Step 1:  Molar mass:
      20  C  @  12g/mol = 240g/mol
      30  H  @    1g/mol =   30g/mol
        1  O  @  16g/mol = +16g/mol 
                                        286g/mol

Step 2:  Percentage of total molar mass
  240g/mol C     *100= 83.9%              30g/mol H     *100= 10.5%             16g/mol O     *100= 5.6%                  
286g/mol total                                    286g/mol total                                   286g/mol total

Step 3: Check your work
(always be sure to check that the percentages that you came up with add up to 100%)

  83.9%
  10.5%
 + 5.6%
  100.0%

Now one question is done.  Yes, I know it is a long process.  Going through all the steps is very helpful though and reminds you what is going on.  Do them even when it does start feeling tedious!


Now that you have seen some examples, try out this quiz!!!  Make sure you have some scrap paper and a periodic table at hand!

Percent Composition Quiz

Calculating the Empirical & Molecular Formula

What is Empirical formula:

• Empirical formula gives the LOWEST TERM RATIOS of atoms or moles
• All formula of ionic compounds are empirical
• e.g. NaCl, MgO, Al2O3

How to determine the Empirical formula given the mass of each atoms?

2 simple steps: 

1. Convert all the mass of elements (in grams) to moles
2. Find the simplest whole number ratio between these amounts

e.g. Determine the empirical formula of a 100 g compound of phosphorus and oxygen that contains a 43.64% of phosphorus by mass.

Solution:

In 100 g, there are 43.64 g of phosphorus and 56.36 g (100 - 43.64 g) of oxygen.

Amount of phosphorus = 43.64 g / 30.97 g mol-1 = 1.409 mol

Amount of oxygen = 56.36 g / 16.00 g mol-1 = 3.523 mol

The whole number ratio of phosphorus to oxygen bay be found by dividing through by the smaller number:

Ratio of P : O = 1.409 : 3.523 = 1 : 2.5 (dividing by 1.409) = 2 : 5 (multiplying by 2)

Therefore, the empirical formula is: P2O5

What is Molecular formula:

• Molecular formula gives all atoms which makes up a molecule
• Ionic or covalent compound formulas can be molecular
• e.g. C3H8, Na2CO3

IMPORTANT EQUATIONS!!!

Molecular formula (MF) = Empirical formula (EF) x Whole Number (N)

MF mass = EF mass x N

Mass of one mole = EF mass (in grams) x N

Tool kit

Here is an awesome online empirical formula calculator.  You can check if your answers are correct by typing in the mass percentage of element and the name of the element.

http://firstyear.chem.usyd.edu.au/calculators/empirical_formula.shtml

Practice questions

If you want to practice on solving empirical formula problems, check out these:

http://kurtniedenzu.cmswiki.wikispaces.net/file/view/Empirical+Formula+Worksheet+5.pdf

http://www.fremont.k12.ca.us/cms/lib/CA01000848/Centricity/ModuleInstance/6112/Empirical_WS.pdf

Two Step Mole Conversions

Two Step Conversions: Grams --> Formula Units/particles/molecules/atoms
The process is much the same except now you have to go two steps.  If you want to go from grams to formula units/particles/molecules/atoms then you multiply the number by 1 mol/grams and by 6.022 x 10^23/1 mol.

Ex.  There are 250g worth of pennies.  How many molecules of atoms of copper are present?
Atomic mass= 58.9
molar mass= 58.9g/mol

250g x   1 mol    x 6.022 x 10^23 -- 2.6 x 10^24 molecules C
           58.9g/mol         1 mol

Ex.  An Olympic sized pool has 2.5 x 10^9 g of water in it.  How many atoms of oxygen are present in an Olympic sized pool?
water=H2O
molar mass= 2 O @ 16.0
                     1 H @   1.0
                                        =33g/mol

2.5 x 10^9g x 1mol x 6.022 x 10^23 x   2 atoms   -- 9.1 x 10^31 atoms O
                      33.0g         1 mol            1 molecule  

Two Step Conversions: Formula Units/particles/molecules/atoms --> Grams
If you want to go from formula units/particles/molecules/atoms to grams you go through a similar process.  This time you multiple the number of molecules by 1 mol/6.022 x 10^23 and then by grams/1 mol.  If it is atoms and you have to account for multiple atoms in each molecule, you have to account for that and divide as necessary.

Ex. There are 300 atoms in a sheet of aluminum.  What is its mass in grams?
atomic mass= 27.0
molar mass  = 27.0g/mol

300 atoms x         1 mol        x 27.0g -- 1.3 x 10 ^-20g
                     6.022 x 10^23    1 mol

Ex. There are 2500 molecules of table salt (NaCl) in a salt shaker.  How many grams of salt is there present?
molar mass= 1 Na @ 23.0
                     1 Cl  @ 35.5
                                        = 58.5g/mol

2500 molecules x        1 mol        x 58.5g -- 2.4 x 10^-19g
                            6.022 x 10^23   1 mol

Mole Conversions

Particle/Atom/Formula Unit ---> Moles

ex. How many moles of Ag are present in 3.0 x 10¹⁶ atoms of Silver?
3.01 x 10²⁴ Ag atoms x         1 mol Ag               
                                     6.022x10²³ Ag atoms
= 5.0 x 10^-8 mol Ag

Moles ---> Particles/Atoms/Formula Units
ex. How many atoms are present in 2 moles of Potassium?
2 mol K x 6.022 x 10²³ atoms K
                        1 mol K
=1x10²⁴ mol K

Moles ---> Grams
ex. What is the mass in grams of 3.5 moles of Sulfur?
3.5 mol S x 32.1 g S
                   1 mol S
= 110 g S

Grams ---> Moles
ex. How many moles are there in 2.49 grams of Tin?
2.49 g Sn x   1 mol Sn   
                   118.7 g Sn
= 0.0210 mol Sn

The Mole

The mole is a unit, like dozen. It is equal to the number of carbon-12 atoms, in 12 grams of carbon.
1 Mole is also known as Avogadro's number. It creates a simpler way for chemists to count atoms and molecules.
Avogadro's number = 6.022 x 10²³ KNOW THIS NUMBER!!!!
Keep in mind that equal volumes of a different gasses have consistent values.


Relative Mass
Written as a comparison against the mass of another object.
  amu = atomic mass unit
Avogadro's Hypothesis
Explains that equal volumes of different gasses that are currently at the same temperature and pressure will  have the same number of particles.

Li. 7.0                    N. 14.0

=>It therefore states that if they have the same number of particles, the mass ratio is due to the mass of the particles.

Atomic Mass

Mass of the atoms of an element. Generally written in amu's, (or u's, of dalton's)

              Eg: Carbon. 16.0 amu

Formula Mass

Applies only to ionic compounds. Sum of the mass of atoms, written in amu's.

              Eg: NaCl - Na = 23.0  + Cl = 35.5 => 58.5 amu

Molecular Mass

Applies only to covalent compounds. Sum of the mass of atoms, written in amu's.

              Eg: CH4 (Methane) C = 12 + H4 = 1 * 4 = 4 => 16 amu

Molar Mass

Molar mass is the mass of one Mole of each element. (written in grams/mole)

              Eg: 1 Mole of Fe = 55.8 g/mol, 1 mole of Cs = 132.9 g/mol

Can also apply to the atomic/molecular/formula mass of any pure substance.

REMEMBER: ALL HAVE THE SAME NUMBER OF PARTICLES!

Graphing in Excel 2010

A graph represents the relationship between two variables.  The horizontal axis and the vertical axis each show the measurements for the two different variables.

This post will just illustrate the basic steps to making a decent scatter point graph.  

Step 1: Type in your data.  Highlight it with your courser so as all the data you want in your graph is selected. 

Step 2:  Go to the top of the page and look at the menu bar.  In the section that lists the different types of graphs, there is an icon labeled scatter.  Pull your courser over it and you should get a small window showing the options below.

After you select the format that you would like, then you will get a graph. The dots will not necessarily be in a straight line like what is shown below.  The graph will represent the data that you entered.

Step 3: Now that you have a simple graph, you can add a trend line.  If you right click a data point, this menu will come up.  Then you select the trend line option.

Step 4:  Now that you have gotten to this menu, you have quite a few options.  First, you can choose the best tread line for your graph.  If your line is straight, then linear would be best where as if it is curved, you would want to choose polynomial etc.  At the bottom of this first menu, there is also a box.  Beside it says display equation.  If you check this box, you get a slope equation put on your graph.  There is also several options on the side of the menu which you can adjust to make your graph more visually appealing.

Step 5: Your graph now has the absolute basics and maybe even is looking a little bit colorful. Its now time to change your tittle and add axis labels.  To get to the menu that allows you to change the

. Step 6:  Select the label chart button to change the tittle.  The icon for adding axis labels is also located here.

You can also just click on the current tittle and change it to whatever you would like it to be

Step 7:  Click the label axis icon and go down to the horizontal axis.  From here you can add on.  A text box will show up underneath your horizontal axis and you can change the label to what is fitting for the graph you are creating.

Step 8:  Now that you have a horizontal axis label, go back and do the same thing except go to vertical axis.  There are a few different options as to how you can format that (words turned to be written sideways, horizontal beside data, or words written vertically).

Once you have choosen this, you have a basic excel scattered point graph

Lab 2E: Determining the Aluminum Foil Thickness

Objectives
1. to calculate the thickness of a sheet of aluminum foil and express the answer in terms of proper scientific notation and significant figures

Supplies
Equipment
3 rectangular pieces of aluminum foil (minimum 15cm x 15cm)
metric ruler
centigram balance


Procedure
1. measure the length and width of each piece of foil and record the measurements in Table 1
2. use a centigram balance to find the mass of each piece of aluminum foil and record the masses

Table 1

Density

• a physical property of matter that describes the ration of mass to volume.

Density = Mass/Volume

                 possible units: g/mL, g/L, g/cm^3, Kg/L

1cm^3 = 1mL

Float vs. Sink experiment

Density of Liquid > Density of object                  Float

Density of Liquid < Density of object                  Sink

For example:
       Say there is a filled water balloon filled with an unidentifiable liquid.  There is 350mL with a mass of 450g.  What is the density of the liquid in the balloon?

D= M/V
D= 450/350
= 1.28mL/g

WORD PROBLEMS

1.  A block of aluminum occupies a volume of 15.0 mL and weighs 40.5 g. What is its density?

2.  Mercury metal is poured into a graduated cylinder that holds exactly 22.5 mL. The mercury used to fill the cylinder weighs 306.0 g. From this information, calculate the density of mercury.

3.  What is the weight of the ethanol that exactly fills a 200.0 mL container?

 The density of ethanol  is 0.789 g/mL.

4.  A rectangular block of copper metal weighs 1896 g. The dimensions of the block are 8.4 cm by 5.5 cm by 4.6 cm. From this data, what is the density of copper?  (hint:  find the volume of a block first)

5. What volume of silver metal will weigh exactly 2500.0 g. The density of silver is 10.5 g/cm³.

6. Find the mass of 250.0 mL of benzene. The density of benzene is 0.8765 g/mL.

7.  A block of lead has dimensions of 4.50 cm by 5.20 cm by 6.00 cm. The block weighs 1587 g. From this information, calculate the density of lead.

8. 28.5 g of iron shot is added to a graduated cylinder containing 45.50 mL of water. The water level rises to the 49.10 mL mark, From this information, calculate the density of iron.

Answers

1) 2.70g/mL

2) 13.6g/mL

3)  253g

4) 8.922g/cm^3

5) 238 cm^3

6) 219.1 g

7) 11.3g/cm^3

8)7.92g/mL

Measurement and Uncertainty

-no measurement is exact

-it's your best estimate

-if you count an object, it's the exact number. 

     ex. 10 pens stays as 10

Absolute Uncertainty

-uncertainty is expressed in the units of measurement not as a ratio

Method: 1- make at least 3 measurements. Get the average. Absolute uncertainty is the LARGEST difference between the average of lowest and highest of reasonable measurement

Ex.

                    Trial #                                            Mass of an object

                        1                                                         14.4g

                        2                                                         14.2g

                        3                                                         17.5g (remove)

                        4                                                         14.3g  

                        5                                                         14.1g

-Average: (of four reasonable measurements)

    =14.3g

-Difference between the average and low measurement:

    =14.3-14.1 = 0.2g

-Difference between high and average:

    =14.4-14.3 = 0.1g

Mass would be recorded as 14.3+0.2g

Method: 2- always measure to best precision as you can. So estimate to a fraction 0.1 of smallest segment on the instrument scale

ex. beaker: 25mL. Best precision is 2.5 mL

Relative Uncertainty & Significant Figures

Relative Uncertainty = Absolute Uncertainty

                                 estimated measurement

PRACTICE

Find the average, and the absolute uncertainty of these pictures

Significant Figures

Definitions:

• Precision: How easy it is to reproduce a measurement
• Accuracy: How close the measurement is to the actual value
• Significant Figures: Measured or meaningful digits. The higher number of significant figures = more precise number

Accuracy Vs Precision

More PowerPoint presentations from Chloe

Significant Figures

Part I:

• The last number is always uncertain as it can easily be a digit higher or lower
• Significant digits include all certain digits (known digits) plus one uncertain
• Eg: 3.87329 
• Certain/Uncertain
• 6 Sig Figs

Part II:

• Leading Zero's are NEVER counted
• Zero's at end, if they are after a decimal ARE counted. If no decimal, they ARE NOT.
• Eg:
• 0.00000006 = 1 SF
• 12.000 = 5 SF
• 12000 = 2 SF

Part III: Exact Numbers

• Some amounts are defined to an exact point - no rounding needed
• Other's have an infinite number of significant digits
• Eg:
• If you have two light bulbs, you have two whole light bulbs, not 1.899999934583 light bulbs

Part IV: The Rules of Rounding

• Usually rounded the same way as in math
• EXCEPTIONS
• if the digit is equal to 5, and there are more non-zero digits after it, round up.
• If the digit is equal to 5 and there are NO more non-zero digits after it, make the last digit even
• Eg:
• 9.86356 to 3 SF = 9.86
• 4.87625 to 4 SF = 4.8763
• 12.065 to 2 SF = 12.06
• 12.075 to 2 SF = 12.08

Part V: Connections with Math

• When Adding or Subtracting, round to the fewest number of decimals

• When Multiplying or dividing, round to the fewest number of significant digits.
• Cannot have any more than the least number of significant digits
• DO NOT ROUND UNTIL CALCULATION IS COMPLETE!!!

PRACTICE QUIZ

Significant Figures and Rounding Quiz

Writing name and formulas for Covalent and Ionic compounds

Ionic Compounds

For Ionic compounds with metals that have more then one possible charge; Roman Numerals need to be put after the metal.  For example: Lead (IV) Oxalate

Roman Numerals

1 = I

2 = II

3 = III

4 = IV

5 = V

6 = VI

7 = VII

8 = VIII

9 = IX

10 = X

Covalent Compounds



Pracitce Questions

 Name the following:

KHCO3

SiS2

ZnF2

SF3

P2O5

Write the formula for the following:

Tetracarbon decahydride

Chlorine gas

Calcium Sulphide

Aluminum nitrate

Sodium Iodide

Answers:

Potassium bicarbonate

Silicon desulphide

Zinc Fluoride

Sulfur trifluroide

Diphosphorous pentoxide

C4H10

Cl2

CaS

Al(NO3)3

NaI

Lab 3b: Separation of a Mixture by Paper Chromatography

Objectives
1. to assemble and operate a paper chromatography apparatus
2. to study the meanilng and significance of Rf values
3. to test various food colorings and to calculate their Rf  values
4. to compare measured Rf  values with standard Rf  values
5. to separate mixtures of food colorings into their components
6. to identify the components of mixtures by means of their Rf  values

Supplies
Equipment                                                                            Chemical Reagents
per class:                                                                                 set of food coloring
5 glass stirring rods                                                                  (yellow, green, blue, red)
several pairs of scissors                                                           unkown mixture of food colorings
per lab station:
3 large test tubes
     (25 mm x 200 mm)
3 Erlenmeyer flasks (250 mL)
metric ruler
pencil
chromatography paper strips
      (2.5 cm wide x 66 cm long)








Procedure
Part I: Setting Up
1. Get 3 Erlenmeyer flasks and put a test tube in each one.
2. Get 3 chromatography paper and draw a line across each strip 4cm from one end then cut a point from the end.
3. Put water on test tubes, about 2cm deep
Part II: Rf  Values of Individual Food Colorings
1. You get to test one food coloring to test. Get one of your chromatography paper and place a dot of food coloring on the line.
2. Put the strip in the test tube. Don't push the strip down too far, tip should just touch the bottom. Don't let the strip touch the wall of the test tube.
3. Observe.
4. Continue observing for 10 min. Try to identify 2 colors as they move up the paper.
5. After about 20 minute and there's no more color separating, immediately draw a line across the top edge of the water before it evaporates.
6. Measure d1 and d2 then record the values on tables 1 and 2. Then calculate the Rf  value.

Part III: Separation of Mixtures into Their Components

1. Second strip will be spotted with a sample of green coloring and the third will be spotted with an unknown mixture of food coloring.

2. Follow directions from part II from 2-6.

3. Record data on Table 3

4. Clean up.

Table 1 : Results for Lab Station

Table 2 : Class Results 

Table III

Acids

How acids are formed

H + Cl = HCl(g)
ionic "non-acid" hydrogen chlordie
HCl(g) + H2O(l) = H30(aq) + Cl(aq)
hydrochloric acid

Rules for naming simple acids
1. hydrogen becomes hydro
2. the last syllabe becomes "-ic"
3. acid is added at the end
ex.
H2S     =     hydrogen chlorate   =   chloric acid
HCl      =     hydrogen chloride   =   hydrochloric acid

Rules for naming complex acids
1. no more hydrogen (hydro)
2. if second element ends with "-ate" it becomes "-ic"
    if it's "-ite" it becomes "-ous"
3. acid is added at the end
ex.
HClO4     =     hydrogen chlorate     =      chloric acid
HlO          =     hydrogen hypochlorite =   hypochlorous acid
-a way to remember this is:
we ate - ic - y sushi and got appendic - ite - ous

Law of Definite Composition (Proust's Law)
The statement that is in a pure compound the elements are always combined in fixed proportions by weight/mass.

ex. H2O has 2 atoms of H and 1 atom for O for a total of 18g

Law of Multiple Proportion (Dalton's Law)

The same elements can combine in more than one proportion to form different compunds