EMPIRICAL FORMULA • The empirical formula represents the smallest ratio of atoms present in a compound. • The molecular formula gives the total number of atoms of each element present in one molecule of a compound. The empirical formula is the simplest formula and the molecular formula is the “true” formula.
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EMPIRICAL FORMULA • The empirical formula represents the
smallest ratio of atoms present in a
compound.
• The molecular formula gives the total
number of atoms of each element present
in one molecule of a compound.
The empirical formula is the simplest
formula and the molecular formula is the
“true” formula.
EMPIRICAL FORMULA
Assume 100g sample Calculate mole ratio
Use Atomic Masses
Mass % of
elements
Grams of
each
element
Moles of
each
element
Empirical
Formula
% = mass
mass to moles
divide by smallest
multiply till whole!
Step 1: If given the % composition, assume a 100g
sample then convert % to grams.
Step 2: Use the atomic masses to convert grams to
moles.
Step 3: Divide the moles of each element by the
SMALLEST mole fraction.
Step 4: The results from step 3 should be a whole
number, if not, make it so by multiplying by a
common factor. Helpful Hint: if a # has a fraction:
.50 X by 2 .25 & .75 X by 4 .33 & .66 X by 3
EMPIRICAL FORMULA 1. Calculate the empirical formula from a sample
containing 43.4% Na, 11.3% C, and 45.3% O.
smallest
43.4% 43.4 g Na (1 mole / 23 g/mol) =1.887 moles Na
11.3% 11.3 g C (1 mole / 12 g/mol) = 0.9417 moles C
45.3% 45.3 g O (1 mole / 16 g/mol) = 2.831 moles O
1.887/0.9417 =2.00 Na
2.831/0.9417 = 3.00 O .
9417/.9417 = 1.00 C
Empirical Formula = Na2CO3
EMPIRICAL FORMULA 3. A compound was found to have a composition of 33.0 % Sr, 26.8
% Cl, and 40.2 % water. Calculate the empirical formula of this
hydrate.
smallest
33.0% 33.0 g Sr (1 mole/87.6 g/mol) = 0.3767 moles Sr