Chem class(22 mac)

4S6Welcome To the

Chemistry Class

The simplest ratio of the number of atoms for each element in a compound.

WHAT IS AN EMPIRICAL FORMULA?

EXAMPLE:

The empirical formula for a glucose molecule (C6H12O6) is CH2O. All the subscripts are divisible by six.

C6 H12 O6

6 6 6

C H2 O

Formula which shows the actual number of each type of atom in a molecule of a particular compound

WHAT IS MOLECULAR FORMULA?

Molecular FormulaMolecular Formula Empirical FormulaEmpirical Formula

CC22HH66 CHCH33

CC66HH1212OO66 CHCH22OO

5 STEPS TO GO…. Step 1: Write the element symbol Step 2: Show the mass,or percentage, of

each element Step 3. Divide each by the R,A.M. of the

element Step 4. Divide each answer by the smallest Step 5: Get the simplest ratio

CALCULATION OF EMPIRICAL FORMULA

Example

Analysis of an oxide of copper shows that it contains 3.175 g of copper and 0.4 g of oxygen. What is the empirical formula of this compound?

Step 1. Write the symbols for the elements Cu O

Step 2. Show the mass,or percentage, of each element. 3.175 0.4

Step 3. Divide each by the R,A.M. of the element

= 0.05 0.025

3.175

63.5

0.4

16

Step 4. Divide each answer by the smallest

= 2 1

0 .05

0.025

0 .05

0.025

Step 5. Write the formula Cu2O

DRAW A TABLE:

Calculations for you to try.1. Calculate the empirical formula for the compound which

contains 2.40g of carbon and 0.60g of hydrogen.

Empirical formula = CH3.

Element C H

Mass(g) 2.40 0.60

Number of moles 2.40/12= 0.2

0.60/1= 0.60

Ratio of moles of atoms

0.2/0.2=1

0.60/0.2=3

Simplest ratio 1 3

Analysis of a compound showed that it contained 40% calcium,

12% carbon and 48% oxygen by mass. Calculate the empirical formula of the compound.

Element Ca C O

Percentage(%) 40 12 48

Number of moles

40/40= 1

12/12= 1

48/16=3


1/1=1

1/1=1

3/1= 3

Simplest ratio 1 1 3

Empirical formula = CaCO3.

TRY THE EXERCISE ON WORKSHEET

Na2O

NOTE: IF FINAL RATIO

1.7 : 1 => 2:11.4 : 1 => 1:10.5 : 1 => ½ : 1 => 1:21.5 : 1 => 3/2:1 => 3:2

1.22 G OF PHOSPHORUS(P=31) COMBINE WITH 0.95 G OF OXYGEN. WHAT IS THE EMPIRICAL FORMULA FOR PHOSPHORUS OXIDE? Element P O

Mass(g) 1.22 0.95

Number of moles

1.22/31= 0.04

0.95/16= 0.0594


0.04/0.04=1

0.0594/0.04=1.5= 3/2

Simplest ratio 2 3

Empirical formula = P2O3.

A COMPOUND CONSISTS OF 29.08% OF SODIUM, 40.65% OF SULPHUR AND 30.27% OF OXYGEN. FIND THE EMPIRICAL FORMULA OF THE COMPOUND. [RAM: NA,23 ; S,32; O,16]

Element Na S O

Percentage(%)

29.08 40.65 30.27

Number of moles

29.08/23= 1.264

40.65/32= 1.270

30.27/16= 1.892


1.264/1.264=1

1.270/1.264=1

1.892/1.264=1.5=3/2

Simplest ratio

2 2 3

Empirical formula = Na2S2 O3.

WATCH THE VIDEO..

REMINDER!!

Do ALL the exercises (mole conversion). Revise the mole and chemical formula.

Will have a short quiz any time after holiday.

AND….

Have a look on the chemistry blog

http://interestingchemistryworld.blogspot.com/

LASTLY…

Chem class(22 mac)

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