• 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!

• Multiplication
• Division
• Convert mass to moles

• 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

• Ionic compounds – formula unit IS ALREADY the smallest wholenumber ratio
• Molecular compounds (covalent bonds) – molecule is not always the smallest whole number ratio!

• Molecular formula: CH
4
• Empirical formula:
• CH
4

• Molecular formula: C
2
H
6
• Empirical formula:
• CH
3

• Molecular formula: H
2
O
• Empirical formula:
• H
2
O

• Molecular formula: C
6
H
6
• Empirical formula:
• CH

• 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

• 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

• 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

• 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
S: 0.706 mol / 0.706 = 1
These numbers become the subscripts in the empirical formula.
O: 2.812 mol / 0.706 = 3.98 = 4
• 4. Write empirical formula
• Empirical formula = Na
2
SO
4

• 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

• 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

• 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

• 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

• A compound contains 4.43 g phosphorus and 5.72 g oxygen. Find the empirical formula.
• 4. Write empirical formula
• P = 2
• O = 5
These numbers become the subscripts in the empirical formula.
• Empirical formula: P
2
O
5

• 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)

Molecular
Formula
Molecular
Mass
Empirical
Formula
Empirical
Mass
1
Relationship between
Masses (x)
Methane CH
4
16.05
CH
4
16.05
Ethane C
2
H
6
CH
3
30.08
15.04
2
Water H
2
O H
2
O
18.02
18.02
1
Benzene C
6
H
6
CH
78.12
13.01
6

The relationship between masses (x) will tell you the relationship between formulas. Find x by: x = molecular mass empirical mass

• 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

• 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 = 5.999
formula mass 13.02
• 6(CH) = C
6
H
6

• 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) =H
2
O
2
 HO (empirical formula)