Lab quiz next class

Last class we did the questions on lab 4c. For next class we will have a lab quiz as usual.
Questions in lab 4c are based on percent composition, mole conversion, empirical formula, ratio and mass calculation. Be sure to know how to do those kind of questions! Also, remember to think of 2 non-human errors that could possibly occur and a conclusion for lab 4c. Good Luck on the lab quiz guys!

- Catherine

Calculating the Empirical Formula of Organic Compounds

First we need to know what an organic compound is!

-         An organic compound is any substance containing carbon

Example: Methane is one of the simplest organic compounds

The empirical formula of an organic compound can be found by

1)      Burning the compound (this involves reacting it with 02)

2)      Collecting and weighing the products

3)      From the mass of the products, the moles of each element in the  original organic compound can be calculated.

Lets do an example together!

What is the empirical formula of a compound that burns to produce 16.9 g of CO2 and 3.46 g of H2O?

Step 1:  CxHy  +     z02   à    xCO2   +  y/2  H2O

This is the balanced chemical equation. From the equation we can see that all of the C and H in CxHy went into making xCO2 and y/2  H2O.

Therefore the moles of C and H are equal to the moles of C in CO2 and the moles of H in H2O.

Step 2:  Convert grams to moles

16.9 g of CO2  x   1 mole                    =   0.384 moles of CO2

                          44g of CO2

                           

3.46 g of H2O      x   1 mole                =  0.192 moles of H2O

                               18g of H2O

Now we need to find the moles of C and H

Mole of C      0.384 moles   x     1 mol C                = 0.384 moles of C

                                                1 mol of CO2

Mole of H   0.192 moles   x  2 mol of H                 =  0.384 moles of H

                                             1 mol of H2O   

Step 3: Not needed in this problem.

Answer: CH

Lets try another problem!

A 7.30 gram sample of hydrocarbon is burned to give 23.8 grams of CO2 and 7.30 grams of H2O. What is the empirical formula?

Mol of CO2    23.8 g of CO2  x  1 mole                  =  0.541 moles of CO2

                                                  44g of CO2

Mol of H2O   7.30 g of H2O  x   1 mole              = 0.405 moles of H2O

                                                   18g of H2O

Mol of C     0.541 moles of CO2   x  1 mol of C     =  0.541 moles of C

                                                        1 mol of CO2

Mol of H    0.405 moles of H2O   x   2 mol of H     = 0.811 moles of H

                                                          1 mol H2O

Step 3: Divide by the smallest molar amount (0.541)

Carbon =  0.541       = 1          X         2     = 2

                 0.541

Hydrogen =  0.811   =  1.5      X         2     = 3

                     0.541

Change ratio to a whole number ration by multiplying.

Answer: C2H3

*Check Mass

0.541 mol C  x 12.0g C    =  6.49g C

                        1 mol C

0.811 mol H  x 1.01g H    =  0.819g H

                        1 mol H

Total Mass = 6.49g + 0.819g = 7.31g

No other elements are present in this sample.

Note: If mass numbers do not add up to the original mass, another element is present.

By Candace

Empirical and Molecular Formula

Empirical formula, gives the lowest term ratio of atoms (moles) in the formula

*All ionic compounds are empirical formulas

Example:

C4H5 (butane) --> molecular formula
C2H5 --> empirical formula of butane

Example:

Consider, 10.87g of Fe and 4.66g of O. What is the empirical formula?

1) First convert from g to moles.

10.87 Fe x 1 mole = 0.195moles        4.66 O x 1mole = 0.291moles
                   55.8g                                              16.0

2) Divide both by the smallest molar amount.

Fe 1    <--0.195/ 0.195
O 1.5   <--0.195/0.291

3) Scale ratios to whole numbers.

Fe 1  x 2 = 2
O 1.5 x 2 = 3          Which becomes: Fe²O³

Molecular formula, is a multiple of the empirical formula and it shows the real amount of atoms that combine to form a molecule.

Calculating a multiple:

n=    molar mass of the compound     
     molar mass of the empirical formula

Try some practice questions:

Find the empirical formulas from the given information.

1) 9.37g of Fe and 13.2g of O

2) 8.64g of Li, 11.1g of C, and 2.43g of H

 Check out this link!

http://www.youtube.com/watch?v=gfBcM3uvWfs

-Lauren

Percent composition

Percent composition is a measure of the mass of each element is present in a compound. The percentages of the elements in a compounds always add up to approximately or exactly 100%. (You can use this tip to check your answers.)

How do you calculate the percent composition of an element in a compound?
Step 1: Find out the total molar mass of the compound
Step 2: Find out the molar mass of each element in the formula
Step 3: Percent Composition = mass of element/ mass of compound x 100%
Step 3: Check if the percentages add up to approximately 100%. If not, the answer is incorrect.


Let's do an example by following these steps.
eg. What is the percent composition of each element in Carbon Dioxide?

Carbon Dioxide = CO2  (MM is 44)C's molar mass = 12
O's molar mass = 16 x 2 = 32 (multiplied by 2 because 2 O's present)

C's percent com. = 12/44 x 100% = 27%
O's percent com = 32/44 x 100% = 73%

Check: 27%+ 73% = 100% <----- When they add up to 100%, it's correct.

Watch this online presentation and try the problems to see if you understand!
http://www.wisc-online.com/objects/ViewObject.aspx?ID=GCH7104


- Catherine

Mole Conversions Part 2

Last class, we learned how to convert between particles to mass (grams); and conversions between mass (grams) to particles.

1) Particles→ Grams

• To find the mass of the molecule, another step is added to the equation from last time.   
• Ex: You want: 7.49x 10^21 atoms of Colbalt into grams,            

                          = 7.49x 10^21 atoms x 1 mole       x       58.9g = 0.733 of Cobalt.
                                                               ‒‒‒‒‒‒              ‒‒‒‒‒
                                                  6.022 x 10^23                1 mole
                           * the grams is the atomic mass of Cobalt


2) Grams → Particles    

• From the example above, the placement of 6.022 x 10^23 and the grams are swapped when converting grams→particles.
• Ex: You want the number of particles in 6.24 grams of Phosphorous.

                      = 6.24g x 1 mole x 6.022 x 10^23 = 1.21 x 10^23 atoms of Phosphorous
                                      ‒‒‒‒‒     ‒‒‒‒‒‒‒‒‒‒‒
                                      31.0g           1 mole

     * Please remember that significant figures always applys to the answers!

Here is a Mole Map to understand this concept visually!



For more expamples and practice visit this website!

http://www.fordhamprep.org/gcurran/sho/sho/lessons/lesson92.htm

Victoria

Chapter 4: The Mole

• there is a constant ratio in equal volumes of different gases

Ex:

    Oxygen             : Hydrogen
    Carbon Dioxide: Hydrogen
    Carbon Dioxide: Oxygen

Avogadra's Hypothesis

Different gases will have the same number of particles if they are also at the same temperature and pressure.
This means...
                    if they have the same amount of particles, the mass ratio is due to the mass of the particles.
 The mass of 1 atom of the element in atomic mass units (amu/u/daltons)

Formula MassAll atoms of a formula of an ionic compound (in amu)

Ex:     
     Potassium Flouride
     K             F
     39.1+19.0=
     KF= amu

Molecular MassAll atoms of a formula of a covalent compound (in amu)

Ex:     Carbon     Dioxide
     C               O²
     12.0          16.0x2
     CO²= 44.0 amu

Atomic/molecular/formula mass of any pure substance
        (in grams per mol)

Ex:      1 mole of Oxygen= 16.0 g/mol
        "      "     Carbon= 12.0 g/mol
        "       "    Potassium= 39.1 g/mol

Avogadra's Number

The number of particles in 1 mole of any amount of substances is...

 

         6.022x10^²³ particles

                                    mol

The mole is important and very useful to chemists because it enable them to count atoms and molecules

Take a look at this link for a brief explanation about Avogadra's number:

http://www.youtube.com/watch?v=AsqEkF7hcII 

By Lauren

Mole Conversions

Now that we have learned how to calculate atomic mass, formula mass, and molecular mass it is time to learn about mole conversions!

*REMEMBER* Avagadro’s Number: 6.022 x 10^23 particles/mole

Here is a chart to help you out along the way!

                Molar Mass                Avagadro’s Number

                 ÷                      ÷

    GRAMS    àà      Moles    à à   Atoms/Molecules

                        ßß                           ß ß  

                          X                                  X

Conversions from particles ↔ moles

            Particles → Moles

            (Particles also represent atoms, molecules, formula units etc…)

            Example:

            1) Lets say you want to convert 3.0 x 10^16 atoms of Silver into moles.

           

            3.0 x 10^16 atoms of Ag   x     1 mole

                                                            ‒‒‒‒‒‒‒

                                                    6.022 x 10^23 atoms

                    = 5.0 x 10^-8 moles of Ag

            2) 3.01 x 10^24 particles of carbon → moles

             3.01 x 10^24 particles    x     1 mole

                                                       ‒‒‒‒‒‒‒‒‒‒

                                                6.022 x 10^23 particles

                = 5.00 moles of carbon

*Remember to use significant figures for all calculations.

Moles → Particles/molecules/formula units/atoms

1)      0.75 moles of CO2 → molecules

0.75 moles x  6.022 x 10^23 molecules 

                      ‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒
                                   1 mole

= 4.5 x 10^23 molecules of CO2

2) 0.75 moles of CO2 → atoms of oxygen
     4.5 x 10^23 molecules CO2    x    2 atoms of O
                                                          ‒‒‒‒‒‒‒‒‒‒‒‒
                                                         1 molecule CO2

            = 9.0 x 10^23 atoms of oxygen

Conversions between moles ↔ grams

Moles→ Grams

1)      2.04 moles of carbon → grams
*use molar mass of Carbon = 12.0 g/mol

   2.04 moles x 12.0 grams
                     ‒‒‒‒‒‒‒‒‒‒
                         1 mole

= 24.5 grams of carbon

2)      0.341 moles NO2 → grams

molar mass of NO2 = 46.0 g/mol

0.341 moles x  46.0 g

                       ‒‒‒‒‒‒

                        1 mole

= 15.7 grams of NO2

Youtube time! Here is another example!

http://www.youtube.com/watch?v=NMdN1LtHuDA

Grams → Moles

1)      3.45g of carbon → moles

3.45  x   1 mole

             ‒‒‒‒‒‒‒
                12.0g

= 0.288 moles of carbon

2) 6.2g of MgCl2 → moles

            6.2g   x  1 mole
                        ‒‒‒‒‒‒‒

                        95.3g

= 0.065 moles of MgCl2

Youtube time! Here is another example!

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

By Candace

Lab 2E quiz last class

   Last class we did the quiz on lab 2E. The questions and answers were all based on the lab book. We also made two graphs based on the volume and density of cold water and hot water. It was fun to decorate and make the graphs pretty! We were told to answer the question "why are cold water and hot water different". About the answer to this question, it is for you to think of it and find out the reason! :)

- Catherine

Lab Experiment

Today during class we did Experiment 2E, "Determining Aluminum Foil Thickness".

We had 3 rectangular pieces of aluminum foil then measured each one using significant figures.
Next, we used a centigram balance to determine the mass of the piece of foil and recorded it on our table.

To find out the thickness of each piece of tin foil, we first had to figure out what the volume by using the equation : V=M/D
When we got the volume, we used the quation V=LWH to determine the thickness of te foil. We expressed our answer with scientific notation.


-Victoria

Density

What is Density?

-         Density is a physical property of matter

-         Density defined in a qualitative manner as the measure of the relative "heaviness" of objects with a constant volume.

-    Density may also refer to how closely "packed" or "crowded" the material appears to be – think of a  styrofoam vs. ceramic cup.

 

 The ceramic cup is much denser than the styrofoam cup.

 Density = Mass / Volume

We can use this density triangle to easily isolate a desired variable.

D = M / V

M = DV

V = M / D

Units

For a solid we usually (but not all the time) use  g/cm3

For a liquid we usually use (but not all the time) use g/ml

Water and Density

1 cm3 of water = 1 mL

Density of water = 1.0 g/mL

                       Or = 1000 g/L

The density of many substances is compared to the density of water. Does an object float on water or sink in the water

D object > D Liquid  = sink  

            Example. A rock sinks into water

D object < D Liquid  = float

            Example. A piece of wood will float on water.

Youtube Time!

Check out this video for a cool density experiment:

http://www.youtube.com/watch?v=B3kodeQnQvU

Check out this video for some density problems:

http://www.youtube.com/watch?v=KcDZx1GBzSE

By Candace

Accuracy and Precision

• No measurement is exact
• all estimates are only best estimates, which have some degree of uncertainty
• you can only get an exact number when you're counting objects

Absolute Uncertainty, the uncertainty expressed in the units of measurement, not as a ratio

Method 1

1. Make at least 3 measurements
2. Calculate the average
3. The absolute uncertainty= the largest difference between the average and the lowest/ highest reasonable measurement

Method 2 (determining the uncertainty of an instrument)

1. When making a measurement, always measure to the best precision that is possible
2. You should estimate the measurement to a fraction of 0.1 of the smallest segment on the instuments scale
3. On a ruler, the smallest segment is 1mm. The best precision should be to break this into 10 equal pieces. (0.1mm)

Relative Uncertainty and Sigfigs

Relative uncertainty = absolute uncertainty

• Can be expressed in percent (%)
• Or expressed using sigfigs
• The number of sigfigs indicates the relative uncertainty

Try some practice questions:

Find the absolute uncertainty:

Trial #                          Mass of juice boxes  

1                                  200.35
2                                  200.34
3                                  200.34
4                                  200.38
5                                  200.33

Trial#                          Mass of a water jug

1                                 500.29
2                                 500.30
3                                 500.32
4                                 500.24
5                                 500.30


-Lauren

Significant Figures

What are significant figures?
- Also known as significant digits or simply "sig figs"
- They are certain digits in a measurement
- Associated with scientific notation and rounding

What are they important?
- A measurement can never be precise. We use sig figs and scientific notation to find a general number.

How do I know which are significant digits?
- Non-zero digits are sig figs
- Zeroes after the decimal point are sig figs
- Zeroes between 2 other significant digits are sig figs
- Leading Zeroes are NOT sig figs
- In scientific notation, all digits in front of the exponential term are sig figs

Rounding
- Digit > 5, the digit in front of it rounds up
- Digit < 5, the digit in front of it keeps the same. All numbers after the rounded digit disappear.
- Digit = 5 + non-zero digits behind it, round up.
- Digit = 5 + no numbers behind it, round to make the digit even.

http://www.sciencegeek.net/APchemistry/APtaters/chap02counting.htm
This is a practice test on counting significant digits. Test yourself to see if you know the stuff!

- by Catherine

Chromatography Lab

  During class, we did Experiment 3B, "Separation of a Mixture by Paper Chromatography". As we learned last class, Chromatography is one of the techniques used to separate components in a mixture. In the lab we did, we demonstrated chromatography by applying a sample of food colouring to a pointed stip of paper then inserted in a test tube with 2cm of water inside, and did the same for two other samples. We observed each one as their colours expanded and rose upwards, for around 20 minutes, then calculated the ratio of the distance traveled by the solute to the distance traveled by the solvent.
Here is the equation we used: Rf=  d¹ / d²

Next class is our Ch. 1 and 2 TEST!!!! Make sure to bring your calculator!

-Victoria

Separation Techniques

Separating mixtures

·         In order to separate mixtures you need to devise a process that separates components with different properties.

·         Some of these properties include:

o   High density/ Low density

o   Volatile/ Non volatile

o   Soluble/ Insoluble

o   Magnetic/ Non magnetic

o   Polar/ Non polar

·         The more similar the properties = the more difficult it is to separate them.

Some Basic Techniques

Hand Separation and Evaporation

·         Hand separation (solids and solids)

o   This is the most basic technique.

o   Magnets or sieves may also be used to separate mechanical or heterogeneous mixtures.

o   Ex.  A dry mixture of salt and sand.  *Note – manually picking out the sand does not change the chemical identity of the salt or the sand.

·         Evaporation (solid dissolved in a liquid solution)

o   Boil away the liquid and the solid remains.

                                                                  Example of Evaporation
                 

·         Filtration (solids(not dissolved) and liquids)

o   Selects components by particle size. Liquid and gaseous components will pass through the filter while the solid particles will be retained.

o   Can use a porous filter or filter paper (residue left in filter paper, filtrate goes through filter paper)

Watch this youtube video on filtration!
 http://www.youtube.com/watch?v=Q0s71cjCNWs

·         Crystallization

o   A change happens to a mixture (either physical or chemical)

o   Solids are then separated by filtration or floatation. Remaining solid comes out as pure crystals and they are then filtered from the remaining solvent.

Youtube Time!
http://www.youtube.com/watch?v=Jd9C40Svt5g&feature=related

·         Gravity Separation (solids based on density)

o   A centrifuge whirls a test tube around at high speeds forcing the denser materials to the bottom.

o   Works best for small volumes

                                                                        Centrifuge

·         Solvent Extraction

o   A technique also called liquid extraction, used to separate liquids.

o   Separation of a substance from a mixture by dissolving that substance in a suitable solvent

o   Works best with solvents that only dissolve one component

o   Mechanical mixture: (solid and solid) use liquid to dissolve one solid but not the other so the desired solid is left behind or dissolved.

o   Solution: solvent is insoluble with solvent already present. Solvent dissolves one or more substances and leaves unwanted substances behind.

·         Distillation

o   (liquid in liquid)

o   Based on differences in volatilities in a boiling liquid mixture

o   Physical separation process. Not chemical.

o   The liquid with the lowest boiling temperature boils first- vapour ascents to distillation flask and enters condenser; gas cools and condenses back to liquid dropping the distillate as a purified liquid.

                                                                  Process of Distillation

·         Chromatography

o   Can separate very complex mixtures

o   Accurate and very precise

o   Separated components can be collected individually

o   Flow the mixture over a material that retains some components more than others, so different components flow over the material at different speeds

o   A mobile phase sweeps the sample over a stationary phase

·         Sheet Chromatography

o   Paper Chromatography (PC)

§  Stationary phase= a liquid soaked into a sheet or strip of paper

§  Mobile phase = a liquid solvent

§  Components appear as separate spots spread out on the paper after drying or “developing”

Youtube Time!
  http://www.youtube.com/watch?v=fLc36wxLrVI

o   Thin layer Chromatography (TLC)

§  Stationary phase = a thin layer of absorbent coating a sheet of plastic or glass

§  Some components bond to the absorbent strongly; others weakly

§  As with paper chromatography components appear as spots on the sheet

By Candace