chem class(22 mac)
TRANSCRIPT
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4S6Welcome To the
Chemistry Class
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The simplest ratio of the number of atoms for each element in a compound.
WHAT IS AN EMPIRICAL FORMULA?
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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
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Formula which shows the actual number of each type of atom in a molecule of a particular compound
WHAT IS MOLECULAR FORMULA?
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Molecular FormulaMolecular Formula Empirical FormulaEmpirical Formula
CC22HH66 CHCH33
CC66HH1212OO66 CHCH22OO
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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
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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
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DRAW A TABLE:
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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
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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
Ratio of moles of atoms
1/1=1
1/1=1
3/1= 3
Simplest ratio 1 1 3
Empirical formula = CaCO3.
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TRY THE EXERCISE ON WORKSHEET
Na2O
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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
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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
Ratio of moles of atoms
0.04/0.04=1
0.0594/0.04=1.5= 3/2
Simplest ratio 2 3
Empirical formula = P2O3.
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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
Ratio of moles of atoms
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.
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WATCH THE VIDEO..
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REMINDER!!
Do ALL the exercises (mole conversion). Revise the mole and chemical formula.
Will have a short quiz any time after holiday.
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AND….
Have a look on the chemistry blog
http://interestingchemistryworld.blogspot.com/
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LASTLY…