Sample problems WS 10-3 Practice Problems (pg 21-22)
% Composition -----Proportion # 2
#2 A sample of an unknown compound with a mass of
part = %part
0.847 g has the following composition: 50.51% fluorine
whole
100%
and 49.49 % iron. When this compound is decomposed
into its elements, what mass of each element would be
X g = 50.51%
recovered?
.847
100%
Cross multiply and divide BECAUSE THERE IS AN = SIGN
X = .428 g F
_____________________________________________________________________________
Empirical Formula Calculations
HO = empirical formula for hydrogen peroxide ---In actuality it is really H2O2
The empirical formula gives the simplest whole number ratio of elements and the molecular formula gives the ACTUAL number
of atoms of each element in a compound.
Several substances have the same empirical formula but EACH substance is described by only ONE molecular formula.
To Determine Empirical Formula #12
1.723 g C x 1 mole C = .1435 mole
12.011g
#12 Determine the empirical formula of a
compound containing 1.723 g of carbon, .289
g of hydrogen and .459 g of oxygen.
.289 g H x 1 mole H = .2867 mole
1.0079 g
.459 O x 1 mole O = .0287 mole
15.999g
To find Empirical Formula--Now, divide each mole by the smallest # of moles.
.0287 mole O = 1 mole O
.0287
.1435 mole C
.0287
= 5 mole C
.2867 mole H = 9.98 = 10 mole H
.0287
= C5H10O
To Determine Molecular Formula
40% C = 40g C x 1 mol
= 3.33 mol C
12.011 g C
6.67% H = 6.67g H x 1 mol
1.0079 g H
= 6.60 mol H
53.3% O = 53.3 g O x 1 mol
15.999 g O
= 3.33 mol O
Ribose is an important sugar that is found in DNA
and RNA. Ribose has a molar mass of 150 g/mol
and a chemical composition of 40.o% carbon,
6.67% hydrogen, and 53.3 % oxygen. What is the
molecular formula for ribose?
Mole Ratio 1 mole C : 2 mole H
: 1 mole O
(You can also divide by the smallest mole above to find the ratio)
Empirical Formula = CH2O
Empirical Formula Mass = 30.0 g/mole
Molar Mass
Empirical Formula Mass
(12.011 + 1.0079 + 1.0079 + 15.999)
= 150 g/mole (given in problem) = 5
30.0 g/mole
Molecular Formula = 5 times Empirical Formula = 5 x (CH2O) = C5H10O5
Chapter 10 Overview
Topic
10.1 – 10.2
Conversions
10.3
% Composition
10.3
Empirical
FormulaMolecular
Formula
Purpose
To change
from one
label to
another
To find
how much
of an
element is
in a
compound
A)
Empirical
Formula =
“lowest”
multiple of
any
formula—
there can
be only
one
B)
Molecular
Formula =
Any
multiply of
an
Empirical
Formula
How
g --can change to -- moles
g--can change to -- molecules
molecules--can change to -- moles
g--can change to -- formula units (u)
ect.
1 mole of any gas = 22.4L
Example:
Amt given g x (1 mole)_ x (6.02 x 1023 particles)=
1
---g from
1 mole
periodic table mass
Use a ratio:
Notes
DO NOT cross multiply
and divide
1) multiply across the
top
2) multiply across the
bottom
3) divide the top by the
bottom.
Part = %
Whole
100
Example:
CH2O= Empirical Formula
C5H10O5=Molecular Formula
------------------------------------------------------------------40% C = 40g C x 1 mol
= 3.33 mol C
12.011 g C
6.67% H = 6.67g H x 1 mol
1.0079 g H
Steps:

----------------------------1) Change all to moles
= 6.60 mol H
53.3% O = 53.3 g O x 1 mol
= 3.33 mol O
15.999 g O
------------------------------------------------------------------Mole Ratio 1 mole C : 2 mole H
: 1 mole O
(You can also divide by the smallest mole above to find
the ratio)
------------------------------------------------------------------Empirical Formula = CH2O
-----------------------------------------------------------------Empirical Formula Mass = 30.0 g/mole
(12.011 +
1.0079 + 1.0079 + 15.999)
-----------------------------------------------------------------Molar Mass (given in problem) = 150 g/mole
=5
Empirical Formula Mass
30.0 g/mole
-----------------------------------------------------------------Molecular Formula = 5 times Empirical Formula = 5 x
(CH2O) = C5H10O5
--------------------------2) Find the ratio
(divide by the smallest)
----------------------------3) Write the Empirical
Formula
----------------------------4) Find the mass of the
Empirical Formula
---------------------------5) Get a multiplication
factor
---------------------------6) Multiply the
Empirical formula
by the multiplication
factor to get Molecular
Formula