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