Simplest (Empirical) and Molecular Formulas

Molecular Formula- shows the actual number of atoms

Example: C6H12O6

Simplest Formula - shows the ratio between atoms

Example: CH2O

Given that a compound contains 12.7% C, 2.1% H and 85.2% Br, calculate its simplest (empirical)

formulaStep 1: Assume you have 100 g of the substance therefore,mass of C = 12.7 g mH = 2.1 g mBr =

85.2gMolarMass=12.01g/mol MH = 1.01 g/mol MBr = 79.90

Step 2:Calculate the number of moles of each using n= m/MnC = nH = nBr = = 1.06 mol = 2.1 mol = 1.07 mol

12.7g 12.01

2.1g 1.01

85.2g79.90

Step 3: Ratio CalculationDivide by the smallest number of moles to

figure out the ratio between the atoms. C= H = Br =

C = 1 H = 1.98 Br = 1.01

Therefore the simplest formula is CH2Br

1.06 mol1.06 mol

2.1 mol1.06 mol

1.07 mol1.06 mol

Calculate the molecular formula for the compound in the previous

example if its molar mass is 190 g/mol

Step 1. Calculate the molar mass for the empirical formula, CH2Br.

M CH2Br = 12.01 + 2(1.01) + 79.90

= 93.93 g/mol

Step 2. Divide the molar mass by the simplest (empirical) formula molar mass.

=

= 2Step 3. Multiply this number by the

empirical formula.2 x CH2Br

Therefore, the molecular formula is C2H4Br2

Molar massSimplest formula molar mass

190 g/mol93.93 g/mol

Example 2: What is the empirical formula of a compound that contains

69.9 g Fe, and 30.1 g O?Step 1: Assume the mass is 100 g.Step 2: Calculate the # of moles of each

element using n=m/MmFe = 69.9g mO = 30.1 gMfe = 55.85 g/mol MO = 16.00

g/molnFe = 1.25 mol nO = 1.88

mol

Step 3: Find the ratio by dividing by the smallest # of moles.

Fe = O =

Fe = 1 x 2 O = 1.5 x 2 = 2 = 3In this case, multiply by a factor (2)to get a

whole number ratio. Step 4: Use the simplest whole number ratio:

Fe2O3

1.25 mol1.25 mol

1.88 mol1.25 mol

Try these:p. 209, 211, 218:#11, 14, 15, 16, 18, 19, 20p. 214# 2, 6p. 230#12, 17, 18aAnswers: page 231