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Empirical Formula
and Percent
Composition
• For Lawn Fertilizer Amount of each
nutrient put on lawn at different times of
the year is very important.
• High percent nitrogen content in Spring to
stimulate green leaf growth
• High percent potassium content in Fall to
stimulate root growth
• Relative amount of each element in a
compound expressed as
Percent Composition.
Example:
Calculating Percent???
Relative amount of each element in a compound expressed as
Percent Composition.
K2CrO4
Potassium Chromate
40.3 % Potassium
26.8% Chromium
32.9% Oxygen
40.3 + 26.8 + 32.9 = 100.0
How do you calculate this weight percent for each element???
Elemental Percent
% Mass of Element E = Mass of E
x 100
Mass of Compound
Percent Composition in a Formula
Just like finding the % you get on a test…
For CaCO3 first find the mass…but do it like this…
Then divide each individual part by the whole… then times
by 100 to get %! Make sure to write % to the hundredth
place (2 after the decimal).
Ca = 40.078
C = 12.011
O = 3(15.999)
100.086
/ 100.086 (100) = 40.04 % Ca
/ 100.086 (100) = 12.00 % C
/ 100.086 (100) = 47.96 % O
They need to add to
100.00
Elemental % in Potassium Chromate
Report to two after the decimal.
Using atomic masses: K2CrO4
K = 39.098
Cr = 51.996
O = 15.999
2(39.098) + 51.996 + 4(15.999 )= 194.188
Total Molar Mass of K2CrO4
% K = 2(39.098) / 194.188 x100 = 40.26% K
% Cr = 51.996 / 194.188 x100 = 26.78 % Cr
% O =4(15.999 )= 194.188 x100= 32.96% O
Elemental % in Potassium Dichromate
K2Cr2O7
(Potassium Dichromate)
Molar Mass = 2(39.098) + 2(51.996) + 7(15.999) =
294.18 g
2(39.098) g  100 = 26.58% K
294.18 g
2(51.996)g  100 = 35.35 % Cr
294.18 g
7(15.999) g 100 = 38.07 % O
294.18 g
How do the two compounds differ even
though they are composed of the same
three elements?
MOLE SONG
(Used for Empirical Formulas)
Mass over formula weight,
That’s the way you get the mole.
Then divide the largest by the smallest.
Press the little buttons as you go.
Multiply it, Multiply it, Multiply it,
Till it’s whole!
Empirical Formula
• Gives the lowest whole number ratio of
atoms in a chemical compound
• HO would be empirical formula for H2O2
• % Composition can be used to figure out
Empirical Formula of compound.
Empirical Formula
• The empirical formula may be the same as the molecular
formula ( CO2)
• If empirical formula is not the same as molecular
formula it is just a whole number ratio of the molecular
formula
empirical formula  molecular formula
NH2 
N2H4
CH2O  C6H12O6
Find Empirical Formula
• Elemental Analysis: 25.9 % N , 74.1% O
• Assume 100 g sample and calculate moles
• Get empirical formula in lowest whole
number ratio
25.9 % N , 74.1% O
25.9 g N and 74.1 g O, change to moles.
25.9 g N |
1 mole N = 1.85 mole N
| 14.007 g N
74.1 g O | 1mole O = 4.63 mole O
| 15.999 g O
Then divide all the moles by the
smallest mole to get the ratio.
1.85 mole N = 1.0 N
1.85 mole
4.63 mole O = 2.5 O
1.85 mole
If the number is not a whole number multiply all
the numbers till they are all a whole number.
1.0 N(2) = 2 N
2.5 O(2) = 5 O
Then you have the empirical formula ratio:
N2O5
Empirical Formula
(Simplest Formula)
This is basically reverse of what we just did with percent
composition.
Let’s find the formula for saccharin, an artificial sweetener.
It has the composition
45.90% C, 2.75% H, 26.20% O, 17.50% S and 7.65% N.
The easiest way to do this is to assume a 100.00g sample.
So . . .45.90 g C, 2.75 g H, 26.20 g O,
17.50 g S and 7.65 g N.
Then convert each into moles…
3.021
45.90 g C | 1 mole C =
| 12.011 g C
2.75 g H | 1 mole H =
2.728
| 1.0079 g H
26.20 g O | 1 mole O =
1.638
| 15.999 g O
17.50 g S | 1mole S = 0.5458
| 32.065 g S
7.65 g N | 1 mole N =
0.5462
| 14.007 g N
Then divide each of the moles by
the smallest mole (ratio)
/ 0.5462 = 5.5 = 11 C This is the ratio of
/ 0.5462 =
each.
If it isn’t a whole
5.0 = 10 H
number…
you must multiply
all the numbers
3.0 = 6 O
until all are whole
numbers!
1.0 = 2 S
/ 0.5462 =
1.0
/ 0.5462 =
/
0.5462
=
=2N
C11H10O6S2N2
Molecular Formula
• You can find the Molecular Formula of a
Compound if you know the empirical
formula and the molar mass.
• Calculate the empirical formula mass,
divide into molar mass and then multiply
this number by each atom in the empirical
formula…
Given the empirical formula is CH2 and the
molecular formula mass is 112.21 g/mole find
the molecular formula.
CH2 has a formula mass of 14.027 g/mole.
112.21 g/mole = 8
14.027 g/mole
therefore the molecular
formula is
C8H16
C(1 x 8) H(2 x 8)
Your turn to try…
Make sure to ask questions if
you don’t understand…and
go back and look at slides
to help you through your
problems!