1. Unit 8 Stoichiometry
Molecular Formulas
Page 65
CH2O
C6H12O6

2. Unit 8 Stoichiometry
Molecular Formulas
Page 65
C2H2
C6H6

3. Relating Molecular and Empirical Formulas
A molecular formula…
is equal or a multiple of its empirical formula
has a molar mass that is the product of the empirical formula mass multiplied by a small integer
molar mass = a small integer
empirical mass
is obtained by multiplying the subscripts in the empirical formula by the same small integer
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4. Diagram of Molecular and Empirical Formulas
A small integer links
a molecular formula and its empirical formula
a molar mass and its empirical formula mass
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5. Some Compounds with Empirical Formula CH2O
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6. Calculating a Molecular Formula from an Empirical Formula
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7. Finding the Molecular Formula
Determine the molecular formula of a compound that
has a molar mass of g and an empirical
formula of CH.
STEP 1 Calculate the empirical formula mass.
Empirical formula mass of CH = g/mol
STEP 2 Divide the molar mass by the empirical formula mass to obtain a small integer.
78.11 g = ~ 6
13.02 g
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8. Finding the Molecular Formula (continued)
STEP 3 Multiply the empirical formula by the small integer to obtain the molecular formula.
Multiply each subscript in C1H1 by 6.
Molecular formula = C1x6 H1x6 = C6H6
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9. Learning Check
A compound has a molar mass of 176.1g and an empirical formula of C3H4O3. What is its molecular formula?
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10. Solution STEP 1 Calculate the empirical formula mass.
C3H4O3 = g/mol
STEP 2 Divide the molar mass by the empirical formula mass to obtain a small integer.
176.1 g (molar mass) = 2
88.06 g (empirical formula mass)
STEP 3 Multiply the empirical formula by the small integer to obtain the molecular formula.
molecular formula = 2 x empirical formula
C3x2H4x2O3x = C6H8O6
Basic Chemistry Copyright © 2011 Pearson Education, Inc.

11. Molecular Formula
A compound contains 24.27% C, 4.07% H, and 71.65% Cl. The molar mass is about 99 g. What are the empirical and molecular formulas?
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12. Solution STEP 1 Calculate the empirical formula mass.
24.27 g C x 1 mol C = mol of C
12.01 g C
4.07 g H x 1 mol H = mol of H
1.008 g H
71.65 g Cl x 1 mol Cl = mol of Cl
35.45 g Cl
Basic Chemistry Copyright © 2011 Pearson Education, Inc.

13. Solution (continued) 2.021 mol C = 1 mol of C 2.021
4.04 mol H = 2 mol of H
2.02 mol Cl = 1 mol of Cl
Empirical formula = C1H2Cl1 = CH2Cl
Empirical formula mass (EM) CH2Cl = g
Basic Chemistry Copyright © 2011 Pearson Education, Inc.

14. Solution (continued)
STEP 2 Divide the molar mass by the empirical formula mass to obtain a small integer.
Molar mass = 99 g = 2
Empirical formula mass g
STEP 3 Multiply the empirical formula by the small integer to obtain the molecular formula. 2 x (CH2Cl)
C1x2H2x2Cl1x2 = C2H4Cl2
Basic Chemistry Copyright © 2011 Pearson Education, Inc.