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