Your Federal Quarterly Tax Payments are due April 15th

# Empirical _amp; Molecular Formulas by dffhrtcv3

VIEWS: 0 PAGES: 5

• pg 1
Empirical & Molecular Formulas
Empirical Formula: The lowest whole number ratio
of the different atoms in a compound; usually shown by
laboratory analysis.
Chemical Formulas are usually “predicted” on the basis of valence electrons of the
elements in the compound. However, the only certain way to know a compounds
formula is to determine its composition in the laboratory.
**Knowing the percentage composition of a compound is the first step in determining
the chemical formula for that compound.
Example:
In laboratory analysis it is determined that a compound is 87.5% nitrogen and
12.5% hydrogen.
How to solve this problem:
Step 1. Divide the percent composition of each element by that element’s
atomic mass.
For Nitrogen:    87.5 = 6.25                For Hydrogen:     12.5 = 12.4
14.01                                        1.01

Step 2: Express this number as the smallest whole number ratio. To do
this, divide both numbers by the smallest one.

6.25   = 1                     12.4   = 1.98 = 2 (nearest whole number)
6.25                           6.25

The whole number ratio is 1:2
This ratio expresses the ratio of atoms in the compound. The empirical
formula for this compound, then, is NH2.
Let’s try another:
At a crime scene a dark substance is found on the skin of a victim. Forensic
experts take a sample of the substance and determine through analysis that is
63.5 % silver, 8.2% nitrogen, and 28.2% oxygen. What is the empirical formula of
this compound?
Step 1: Divide the percent composition of each element by that element’s atomic mass.
Ag: 63.5 = .589            N:    8.2 = 0.585             O:   28.2 = 1.76
107.9                       14.01                         16.00
Step 2: Express this number as the smallest whole number ratio. To do this,
divide both numbers by the smallest one.

0.589 = 1.006 = 1                   0.585 = 1               1.76     = 3.01 = 3
0.585      (nearest whole number)   0.585                   0.585    (nearest whole number)

The whole number ratio is 1:1:3

The whole number ratio is given in the order that the elements are listed. The
empirical formula is AgNO3.

Empirical Formulas may or may not be the “actual” or real formula of your
compound.
**If it is an ionic compound, the empirical formula is usually the true formula

**If it is a molecular compound, you need to go on to find the Molecular Formula.
Molecular Formulas:         The molecular formula shows
the true number of atoms of each kind of element. It is
always a whole number multiple of the empirical formula.
In order to determine the molecular formula, you need to know the
molar mass (usually determined through experimentation methods).
For Example,      hydrazine is a compound that contains nitrogen and
hydrogen. Its empirical formula is NH2, but its molar mass is 32.0 g/mol.
What is the true formula of hydrazine?

How to solve this problem:

Step 1: Determine the formula mass of the empirical formula
1 N + 2 H = 14.01 + (2 x1.01) = 16.02g/ mol
Step 2: Divide the Molar mass by the empirical formula mass (efm).
Round to the nearest whole number. This will give you the ‘multiple’ of the empirical
formula
32.0 g/mol ÷ 16.02 g/mol = 1.99 = 2 (whole number)
Step 3: Multiply the subscripts in the empirical formula by your
number calculated in step 2.

[ NH2 ] x 2 = N2H4               The molecular formula for
hydrazine is N2H4

Let’s try another:
Benzene is a molecular compound with a molar mass 78.1 g/mol. The empirical
formula is CH. What is the molecular formula of benzene?
Step 1: Determine the formula mass of the empirical formula
1 C & 1 H = 12.01 + 1.01 = 13.02 g/mol

Step 2: Divide the Molar mass by the empirical formula mass (efm).

78.1 g/mol ÷ 13.02 g/mol = 5.99 = 6 (nearest whole number)
Step 3: Multiply the subscripts in the empirical formula by your
number calculated in step 2.
The molecular formula for
[ CH ] x 6 = C6H6                 benzene is C6H6

To top