Transcript Empirical Formula

% Composition & Empirical
Formula
V.4 PERCENTAGE COMPOSITION
• Percentage composition is the percentage
(by mass) of the species in a chemical
formula.
How to calculate it…
For MgO
1) Find the total molar mass of your compound.
2) Find the specific molar mass of the each element you have.
3) Divide the specific molar mass / total mass
4) Multiply by 100%
Example #1: What is the percent composition of ?
a) MgO (what % of the mass is magnesium, what % of the mass is oxygen?)
b) FeCl2
c) (NH4)3PO4
Example #2: What is the percent composition of the bold species?
NiSO47H20
Empirical Formula
V.5 EMPIRICAL & MOLECULAR FORMULA
Empirical Formula is the SMALLEST ‘whole number ratio’ of
atoms which represents the molecular make-up of a compound.
CH2, C2H4, C3H6, C4H8, C5H10 --- all contain twice as many H’s as C’s
Therefore, the empirical formula (or simplest ratio) is CH2
In this case, all the formulae are whole-number multiples of
CH2
C2H4 = 2 x CH2
C3H6 = 3 x CH2
C4H8 = 4 x CH2
The empirical formula is the simplest whole number ratio between
atoms in a compound.
It is determined experimentally by measuring the mass of the elements
that combine to form a compound.
Example: A compound was found to be composed of 38.7
g C, 9.68 g H and 51.6 g O. Calculate the empirical
formula.
A compound was found to be composed of 38.7 g C, 9.68 g H and 51.6 g O.
Calculate the empirical formula.
38.7 g C x 1 mole
12.0 g
= 3.225 mol
3.225 mol
= 1
9.68 g H x 1 mole
1.01 g
= 9.584 mol
3.225 mol
= 3
51.6 g O x 1 mole
16.0 g
= 3.225 mol
3.225 mol
= 1
Change grams to moles first !
Divide by the smallest number to get whole numbers!
Write the formula!
CH3O
A compound is found to contain 63.55 % Ag, 8.23 % N and 28.24 % O. Calculate
the empirical formula.
Assume that you have 100 g- that means 63.55 g Ag, 8.23 g N, and 28.24 g O
Change grams to moles using molar mass!
63.55 g Ag x
1 mole
107.9 g
= 0.5890 mol
0.5879 mol
= 1
8.23 g N
x
1 mole
14.0 g
= 0.5879 mol
0.5879 mol
= 1
28.24 g O
x
1 mole
16.0 g
= 1.765 mol
0.5879 mol
= 3
AgNO3
A compound is found to contain 50.07 % Cu, 16.29 % P and 33.64% O. Calculate the
empirical formula.
Double
Rearrange
Change % into grams first, then into moles.
50.07 g Cu x
16.29 g P
33.64 g O
x
1 mole
63.5 g
=
0.7885 mol
0.5255 mol
x 1 mole
31.0 g
=
0.5255 mol
0.5255 mol
1 mole
16.0 g
=
2.103 mol
0.5255 mol
= 1.5 = 3
= 1
= 2
Cu3P2O8
Cu3(PO4)2
= 4
= 8
Only do this for ionic compounds- rearrange it as a polyatomic ion
Not for covalent!
What is the empirical formula of a compound containing 39.0 % Si
and 61.0 % O?
Page 93
– Common Fractions → Decimal Conversions
– Helpful to be able to recognize these! Saves you time!
2.67, 1.33, 5.67, 3.33, etc involve THIRDS
( x 3 to clear fraction)
1.75, 2.25, 3.75, etc involve QUARTERS
( x 4 to clear fraction)
IMPORTANT: Don’t round intermediate values, keep 3 or 4
decimals...
Homework
Percentage composition questions
Page 91
#44 a, d, g, I, k, n
#45 a, b, f
Empirical Formula questions
Page 93
# 46 (odd)