Transcript Empirical formula

Weds March 5, 2014
• Due: HW 7C, Lab Reports
• Today: Determining Chemical Formulas
• Empirical Formulas
• Molecular Formulas
• Friday: Magnesium Oxide Lab
• Bring your own goggles if you don’t want to wear the ones
in the classroom set!
Necessary skills:
• Multiplication
• Division
• Convert mass to moles
Calculation of Chemical Formulae
• A Molecular formula includes the symbol of elements and the
number of atoms of each element in that molecule
• Empirical formula includes symbols of elements in compounds with
subscripts that show the smallest possible whole numbers that
describe the atomic ratio
Determining Chemical Formula
• Ionic compounds – formula unit IS ALREADY the smallest wholenumber ratio
• Molecular compounds (covalent bonds) – molecule is not always the
smallest whole number ratio!
Methane
• Molecular formula: CH4
• Empirical formula:
• CH4
Ethane
• Molecular formula: C2H6
• Empirical formula:
• CH3
Water
• Molecular formula: H2O
• Empirical formula:
• H2O
Benzene
• Molecular formula: C6H6
• Empirical formula:
• CH
Calculation of Empirical Formula
• 1. Determine Mass Composition
• If given percentages: Use percentage composition to convert to a mass
composition (assume 100 g sample so percent is equal to the mass in g)
• If given mass: Skip this step
• 2. Convert mass to moles for each element
• 3. Find the smallest whole-number ratio by dividing each number of
moles by the smallest number
• 4. These whole numbers are the subscripts in your compound
Example 1 – Step 1
• A compound contains 32.38% sodium, 22.65% sulfur,
and 44.99% oxygen. Find the empirical formula.
• 1. Use % composition to get to mass composition
• 32.38% Na = 32.38 g Na
• 22.65% S = 22.65 g S
• 44.99% O = 44.99 g O
Example 1 – Step 2
• A compound contains 32.38% sodium, 22.65% sulfur, and 44.99%
oxygen. Find the empirical formula.
• 2. Convert mass to moles
• 32.38 g Na / (22.989 g/mol) = 1.408 mol Na
• 22.65 g S / (32.065 g/mol) = 0.706 mol S
smallest
• 44.99 g O / (15.999 g/mol) = 2.812 mol O
Example 1 – Step 3
• A compound contains 32.38% sodium, 22.65% sulfur, and 44.99%
oxygen. Find the empirical formula.
• 3. Find smallest ratio by diving by smallest number of moles
• Na: 1.408 mol / 0.706 = 1.99 = 2
These numbers become the
subscripts in the empirical
• S: 0.706 mol / 0.706 = 1
formula.
• O: 2.812 mol / 0.706 = 3.98 = 4
• 4. Write empirical formula
• Empirical formula = Na2SO4
Example 2 – Step 1
• A compound contains 4.43 g phosphorus and 5.72 g
oxygen. Find the empirical formula.
• 1. Use % composition to get to mass composition (you
are already there!)
• 4.43 g P
• 5.72 g O
Example 2 – Step 2
• A compound contains 4.43 g phosphorus and 5.72 g
oxygen. Find the empirical formula.
• 2. Convert mass to moles
• 4.43 g P / (30.974 g/mol) = 0.143 mol P
smallest
• 5.72 g O / (15.999 g/mol) = 0.358 mol O
Example 2 – Step 3
• A compound contains 4.43 g phosphorus and 5.72 g
oxygen. Find the empirical formula.
• 3. Find smallest ratio by diving by smallest number of
moles
• P: 0.143 mol / 0.143 = 1 x 2 = 2
• O: 0.358 mol O / 0.143 = 2.5 x 2 = 5
• THEY MUST BE WHOLE NUBMERS! YOU CAN’T HAVE
HALF AN ATOM! Multiply by 2 (or 3, if the decimal is
.333 or .666) to get whole numbers
ROUNDING HELPS
• If the number ends in .98, .99, .01, or .02  Round to nearest whole
number
• If the number ends in .50, .51, .52, .49, .48, etc  Round to .5 then
multiply by 2
• If the number ends in .33, .32, .34, etc  Round to .3 then multiply
by 3
• If the number ends in .66, .65, .64, etc  Round to .6 then multiply
by 3
Example 2 – Step 4
• A compound contains 4.43 g phosphorus and 5.72 g
oxygen. Find the empirical formula.
• 4. Write empirical formula
•P=2
These numbers become the
subscripts in the empirical
formula.
•O=5
• Empirical formula: P2O5
Calculation of Molecular Formula
• If you know the empirical formula and the molecular mass of a
compound, you can determine the molecular formula.
• There are four pieces of information. If you know three, you can
solve for the fourth.
• 1. Empirical formula
• 2. Molecular formula
• 3. Empirical mass (is the mass of the empirical formula)
• 4. Molecular mass (is the mass of the molecular formula)
Calculation of Molecular Formula
Molecular
Formula
Methane
CH4
Ethane
C2H6
Water
H2O
Benzene
C6H6
Molecular
Mass
CH4
16.05
Relationship
between
Masses (x)
1
CH3
15.04
2
18.02
H2O
18.02
1
78.12
CH
13.01
6
16.05
30.08
Empirical
Formula
Empirical
Mass
What is x?
The relationship between masses (x) will tell you the relationship
between formulas. Find x by:
x = molecular mass
empirical mass
Steps
• 1. Calculate the empirical mass (like you are used to doing.)
• 2. Solve for x using molecular mass (given in the problem) and
empirical mass.
• 3. Multiply the empirical formula by x
x(empirical formula) = molecular formula
Example 1
• Determine the molecular formula of the compound with an
empirical formula of CH and a molecular mass of 78.110 amu.
• Empirical formula= CH
• Molecular formula= ??
• Empirical mass= mass C + mass H = 12.01 + 1.01 = 13.02 amu
• Molecular mass= 78.110 amu
• x = molecular mass = 78.110
formula mass
13.02
• 6(CH) = C6H6
= 5.999
Example 2
• A sample of a compound with a molar mass of 34.00 g/mol consists of 0.44 g H and 6.93 g O. Find its
molecular formula.
• Empirical formula: Must determine from data given in the problem (See p.6 of notes)
• Molecular formula: ??
• Empirical mass: Can determine from empirical formula
• Molecular mass: 34.00
• H: 0.44 g H x (I mol H) = 0.436/0.433 = 1
(1.01 g H)
• O: 6.93 g O x (I mol O) = 0.433/0.433 = 1
(16.00 g O)
• x = molecular mass = 34.00
= 1.99
empirical mass HO (16.00 + 1.01)
• 2(HO) =H2O2
 HO (empirical formula)