Dec.9--Lab 4C [Formula of a Hydrate]

Equipments:

- pipestem triangle

- iron ring & stand

- Bunsen burner

- dry crucible

  

Procedure:

1.      Put on lab apron and safety goggles.

2.      Set up the pipestem triangle, iron ring, stand and Bunsen burner. The ring should be 5-6 cm above the burner. Place a clean and dry crucible on the triangle, and heat for 3 minutes to make sure it is dry.

3.      Weigh the crucible and record the mass in Table 1.

4.      Place hydrate into it until it is 1/3 full. Record the total mass of crucible and hydrate.

5.      Place the crucible on the pipestem triangle, and begin heating. Gradually increase the heat until the bottom of the crucible is a dull red. Maintain the temperature for 5 minutes.

6.      Turn off the burner and allow the crucible to cool for 5 minutes. Weigh and record the total mass.

7.      Reheat the crucible for 5 minutes to make sure there is no more water left. Cool it down and record the mass. Use the lower mass number between the first and second heat. (The difference should be no more than 0.03 g.)

8.      Add a few drops of water to the contents of the crucible. Note any changes that occur.

Dec 5--Calculating the empirical formula of organic compounds

The empirical formula of an organic compounds can be found by:

-buring the compound (reacting with oxygen)

-collecting and weighing the product

-calculate the mass and then mole

*The mass of the product = the mass of the reaction

Ex.

What is the empirical formula of a compound that when a 10.0g sample is burned produces 20.0g of CO2 and  8.0g of H2O?

CxHy + z O2   TO   x CO2 +  y/2 H2O

1. Calculate the moles of CO2 and H2O

CO2   20.0g x (1mole/44g) =0.455 mol

H2O   8.0g x ( 1 mole/18g)=0.444 mol

2. find the mole of C and mole of H in the H2O and CO2

mol C =0.455 mol x (1mol C / 1mol CO2) = 0.455 mol  

mol H=0.444 mol x( 2mol H / 2mol H2O)=0.888 mol 

3. Divide the smallest molar amount

C  0.455/0.455 = 1

 H  0.888/0.455 = 2

Therefore ,the answer is CH2 .

Dec 3--Empirical + Molecular Formula

Empirical Formula:

gives the lowest term ratio of atoms (or moles) in the formula. (All ionic compounds are empirical formulas.)

Ex. C4H10—molecular formula     C2H5—empirical formula
 

Example#1

Consider that we have 10.87g of Fe and 4.66g of O, what is the empirical formula?

1.Convert g into moles                              
 Fe: 10.87g of Fe* （1 mole/55.8g）=0.195 moles
 O: 4.66g of O * (1 mole/16g) = 0.291 moles

2.Divide both of them by the smallest molar amount
Fe: 0.195/0.195=1       
O: 0.291/0.195 = 1.5

3.The last step is to scale ratios to whole numbers
Fe: 1 * 2 = 2        
O: 1.5* 2 = 3
 

Ans:Fe2O3

Example#2

A compound contains 31.9% k, 28.9%Cl, and 39.2% O. What is the empirical formula?
 
K: 31.9g * (1 mole/ 39.1g) = 0.816 mol
Cl: 28.9g * (1 mole/35.5g) = 0.814 mol
O: 39.2g * (1 mole/16g) = 2.45 mol
Repeat the second and third step.

Molecular Formula: 

is a multiple of the empirical formula and shows the actual number of atoms that combine to form a molecule.
 
n= (molar mass of the compound/molar mass of the empirical formula) 

Example#1

A molecule has an empirical formula of C2H5 and a molar mass of 58g/mol what is the molecular formula?

MM C2H5 = 29g/mol
n = (58g/mol / 29g/mol) = 2
MF = 2(C2H5) = C4H10 = BUTANE

Example#2

The empirical formula of a gas is CH2 what is the molecular formula if the molar mass  is 42g/mol

MM of CH2 = 14g/mol
n= (42g/mol / 14g/mol) = 3
MF = 3(CH2) = C3H6

Example#3

 A compound contains 7.44g C, 1.24g H, and 9.92g O , the molar mass of the compound is 180g what is the molecular formula?

C 7.44g * (1 mole/ 12.0g) = 0.62mole     1
H 1.24g * (1mole/ 1.0g) =1.24 moles      2
O 9.92g * (1 mole/ 16.0g) = 0.62 mole    1
CH2O = 12.0+1.0*2 +16.0=30g/mol
n = (180g/mol / 30g/mol) = 6
MF = 6(CH2O) = C6H12O6