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Empirical and Molecular Formulas

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1 Empirical and Molecular Formulas
Section 11.4 Chemistry

2 Objectives Explain what is meant by the percent composition of a compound. Determine the empirical and molecular formulas for a compound from mass percent and actual mass data.

3 Key Terms Percent composition Empirical formula Molecular formula

4 Percent Composition Percent by mass of each element in a compound.
Mass of Element x 100 Mass of Compound

5 Example Calculate percent composition of H in H2O
Molar mass of water: g/mol Determine mass of H in 1 mol of H2O

6 Mass of H 1.01 x 2 = 2.02 g H in 1 mol water Atomic mass of H
from periodic table Number of H in H2O

7 Percent Composition of H
2.02 g of H x 100 = 11.2% H 18.02 g of H2O

8 Empirical Formula Formula for a compound with the smallest whole-number ratio of elements. Percent composition can be used to find the chemical formula.

9 Empirical Formula When given percent composition, assume:
The total mass of the compound is 100 g The percent composition of the element is equal to the mass in grams of the element.

10 Example A compound has a percent composition of 40.05% S and 59.95% O.
So in 100 g of the compound, g are S and g are O. Find the amount of mol for each element.

11 Empirical Formula 40.05g S x 1 mol S = 1.249 mol S 32.07g S
59.95g O x 1 mol O = mol O 16.00g O

12 Empirical Formula The element with the smallest number of mol gets the subscript 1.

13 Empirical Formula S has a subscript 1
Then divide the mol O by the mol S 3.747 mol O/ mol S = 3 mol O Then write your empirical formula using your subscripts: SO3

14 Practice Problems Pg. 333

15 Molecular Formula Formula that specifies the actual number of atoms of each element in one molecule or formula unit of a substance. n = molecular formula mass empirical formula mass

16 Example Compound is composed of 40.68% carbon, 5.08% hydrogen, and 54.24% oxygen and has a mass of g/mol. Determine the empirical and molecular formulas for succinic acid.

17 Example 40.68 g C x 1 mol C = 3.387 mol C 12.01 g C
5.08 g H x 1 mol H = 5.04 mol H 1.008 g H 54.24 g O x 1 mol O = mol O 16.00 g O

18 Example 3.387 mol C/ = 1 mol C 5.040 mol H/ = 1.49 = 1.5 mol H 3.390 mol O/ = = 1 mol O Ratio of C : H : O = 1 : 1.5 : 1

19 Example Empirical Formula: C2H3O2

20 Example n = molecular formula mass empirical formula mass
To find molecular formula, calculate n. n = molecular formula mass empirical formula mass Molecular mass is in the problem! Calculate molar mass of empirical

21 Example n = = 2 59.04 Multiply the subscripts of the empirical by n to find the molecular formula. Molecular Formula: C4H6O4

22 Practice Problems Pg 335

23 Homework Section 11.4 Problems on page 877

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