Lab 4C

 For Lab 4C, "Determining the Empirical Formula of a Compund", we got an unknown hydrate that filled 1/3 of a crucible. We place the crucible on a pipestem triangle with a bunsen burner underneath it; once the crucible turns dull red with heat, we let it cool for around 5 minutes. After weighing it, we heat it once again for approximately 5 minutes and then determine the mass. We reheat it so that we are sure there is no more water left so the hydrate becomes an anhydrous salt.

Molar Volume of a Gas at STP

    Since gases expand and contract depending on the pressure and temperature, we use STP (Standard Temperature and Pressure to make sure all gases are measured accurately. 1 mole of gas is to take up 22 L, at 1 atmostphere of pressure and a temerature of 0°.
   Converting from molà litres
   Ex: Calculate the volume (L) of 20mole of ammonia at STP
       
  20mol x      22.4L             = 448L
                __________
                  1 mole



 Converting from litresà mol
 Ex: Calculate the number of moles in 5.54 L of CO2 at STP

 5.54L x  1 mole         =  0.247 mol
              _________
               22.4 L

-Victoria

Diluting Solutions to Prepare Workable Solutions

A Quick Recap:

Solvent = the liquid that does the dissolving

Solute = the substance which dissolves

Solution = is prepared by dissolving a solute in a solvent.

-         When a solution is diluted more solvent is added to it.

-         The moles of solute is always constant (the only difference is that there is more solvent in the less concentrated solution)

The total number of solutes in the solution remains the same after dilution, but the volume of the solution becomes greater.

Thus we have the equation:

            Moles of solute before = Moles of solute after

M₁V₁ = M₂V₂

Lets try an example:

Concentrated HNO₃ is 15.4 mole/L.  How would you prepare 2.50 L of 0.375 M HNO₃?

M₁V₁ = M₂V₂

15.4 mole/L x V₁  = 0.375 M x 2.50 L

→ Re organize the equation

 V₁  = 0.375 M x 2.50 L

                 15.4 M

V₁ = 0.0609 L

V₁ = 60.9 mL

*Remember Sig Figs

To calculate how much water we need in order to dilute the solution we take our known volume (V₂) and subtract it from V₁.

       2.50 L – 0.0609 L  = 2.44L

So, you would need to add 2.44 L of water to make the equation true.

Youtube Time:

http://www.youtube.com/watch?v=KM2uzc8msas

 By Candace

Molarity Concentration / "Molarity" of Solutions

Molar Concentration

Here's a quick review of a past chapter:

Homogeneous mixture, when one substance is dissolved in another it is called a solution.

• the smaller quantity (being dissolved) is called a solute
• the larger quantity (dissolving) is called a solvent

Ex: sugar in tea
The sugar would be the solute and the tea is the solvent.

Molar Concentration / Molarity is the number of moles of solute in one litre of a solution. We use the letter "M" to represent molar concentration, which has the units of "moles/L"

Formulae:

         - Molarity= moles of solute (mol)
                           volume of solution (L)

         - or put more simply:  M=mol
                                                 L

         - mol= MxL           and           L=mol
                                                            M

      
Check out this link for some more help :)

http://www.youtube.com/watch?v=h0cdLIfus8c

Try a practice problem :

Find the molar concentration of a solution that has 0.310 moles of LiOH² in 1.200 L of solution.

-Lauren

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