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Empirical & Molecular Formulas Unit 4: Stoichiometry Chapter 10 – The Mole

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Empirical & Molecular Formulas. Unit 4: Stoichiometry Chapter 10 – The Mole. 2/6/2014. Learning Target : Understand how to determine the percent composition of a compound. Learning Outcome: Know how to calculate the percent composition of a compound. Practice. - PowerPoint PPT Presentation

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Page 1: Empirical & Molecular Formulas

Empirical & Molecular Formulas

Unit 4: StoichiometryChapter 10 – The Mole

Page 2: Empirical & Molecular Formulas

2/6/2014

Learning Target: Understand how to determine the percent

composition of a compound.

Learning Outcome: Know how to calculate the percent composition of a

compound.

Page 3: Empirical & Molecular Formulas

Practice

• Suppose you spend 6 hours texting friends each day.

What percentage of each day do you spend texting?

6 ÷ 24 = 0.25 hours0.25 x 100% = 25%

Page 4: Empirical & Molecular Formulas

Percentage Composition

• Definitiono The mass of each element in a compound

compared to the entire mass of the compound and multiplied by 100 percent

Page 5: Empirical & Molecular Formulas

2 ways to find % composition

1) Total mass of compoundPercentage of sodium in sodium chloride

22.99 g Na __ x 100% = 39.3% Na

58.44 g NaCl

2) Experimentally a compound’s mass is measured and then decomposed into its individual element.

Page 6: Empirical & Molecular Formulas

Practice

• Find the percentage composition of a compound that contains 1.94 g of carbon, 0.48 g of hydrogen, and 2.58 grams of sulfur in a 5.00 g sample of that compound.

Page 7: Empirical & Molecular Formulas

PracticeA sample of unknown compound with a mass of 0.847 g has the following composition: 50.51 percent fluorine and 49.49 percent iron. When this compound is decomposed into its elements, what mass of each element would be recovered?

•Can we find the number of moles of carbon, hydrogen and sulfur?

Page 8: Empirical & Molecular Formulas

2/7/2014

• Learning Target:– Understand how to determine empirical and

molecular formulas of a substance.

• Learning Outcome:– Know how to calculate the empirical and molecular

formula of a substance using percent composition.

Page 9: Empirical & Molecular Formulas

Empirical Formula

• Definition

• A formula that gives the simplest whole-number ratio of the atoms of elements

• For example, H2O2 is the molecular formula for hydrogen peroxide.• HO is the empirical formula for hydrogen peroxide.

Page 10: Empirical & Molecular Formulas

Determining the Empirical Formula

• Now that you know how to calculate percentage composition, you can use that ratio of masses to find the ratio of atoms which is a chemical formula

• Ratio of Masses Ratio of Atoms(% composition) chemical formula

Page 11: Empirical & Molecular Formulas

Steps for Finding Empirical Formula

• Percentage Mass (g)

• Mass (g) # of Moles

• Once you have moles, find the smallest whole number ratio between the moles.

Page 12: Empirical & Molecular Formulas

Example

• Determine the empirical formula of a compound containing 5.75 g Na, 3.5 g N, and 12.0 g O.

• Since you have mass, convert mass → moleso Na: 5.75 g Na x 1 mol Na = 0.25 mol

22.99 g

o N: 3.5 g N x 1 mol N = 0.25 mol 14.00 g

o O: 12.0 g O x 1 mol O = 0.75 mol 16.00 g

Page 13: Empirical & Molecular Formulas

Example Continued

• Now divide each mole value by the smallest number of moles:

• Na: 0.25/0.25 = 1• N: 0.25/0.25 = 1• O: 0.75/0.25 = 3

• Empirical Formula = Na N O3

Page 14: Empirical & Molecular Formulas

Practice

Determine the empirical formula of a compound containing 0.928 g of gallium and 0.412 g of phosphorus.

Page 15: Empirical & Molecular Formulas

Practice

Determine the empirical formula of a compound containing 2.644 g of gold and 0.476 g of chlorine.

Page 16: Empirical & Molecular Formulas

• Determine the empirical formula of a compound containing 1.723 g of carbon, 0.289 g of hydrogen, and 0.459 g of oxygen.

Page 17: Empirical & Molecular Formulas

Molecular Formula

• Definitiono The formula that gives the actual number of atoms

of each element in a molecular compound

• The molecular formula is always a whole-number multiple of the empirical formula.

Page 18: Empirical & Molecular Formulas

Determining Molecular Formula

• Determine molecular formula by comparing the molar mass of an unknown compound with the molar mass of the empirical formula.

Page 19: Empirical & Molecular Formulas

Example

• Find the molecular formula of a compound that contains 42.56 g of palladium and 0.80 g of hydrogen. The molar mass of the compound is 216.8 g/mol.

Step1 Convert to moleso 42.56 g Pd x 1 mole Pd = 0.40 mol Pd

106.42 go 0.80 g H x 1 mole H = 0.79 mol H

1.01 g

Page 20: Empirical & Molecular Formulas

Example Continued

Step 2: Find mole ratio:Pd: 0.40/0.40 = 1 H: 0.79/0.40 = 2

• Step 3: Write empirical formula:Pd H2

Page 21: Empirical & Molecular Formulas

Example Continued

• We know that the molar mass of the compound (given in question) = 216.8 g/mol

• Step 4: Find the empirical molar mass

If empirical formula is PdH2 , then the empirical molar mass = 106.42g/mol Pd +

2(1.01g/mol H) = 108.44 g/mol

Page 22: Empirical & Molecular Formulas

Continued Once Again• Step 5 Find, Molar Mass ÷ Formula Mass:

216.8 g/mol = 1.995 2108.44 g/mol

**MOST IMPORTANT STEP Step 6:

FINALLY!!! Now adjust the molar formulaPdH2 becomes Pd2H4

Page 23: Empirical & Molecular Formulas

Practice Problem

Octane (C8), a compound of hydrogen and carbon, has a molar mass of 114.26 g/mol. If the compound contains 18.17g/mol hydrogen, what is its molecular formula?

Page 24: Empirical & Molecular Formulas

Last Practice

• β-carotene, a compound found in carrots, can be broken down to form vitamin A. The empirical formula for β-carotene is C5H7. The molar mass of β-carotene is 536 g/mol. What is the molecular formula for β-carotene?

Page 25: Empirical & Molecular Formulas

3/12/12 PROBLEM

• A component of protein called serine has an approximate molar mass of 105.11 g/mole. If the percent composition is as follows, what is the molecular formula of serine?

C = 34.95 % H= 6.844 % N= 13.59 % O = 46.56 %

Page 26: Empirical & Molecular Formulas

3/12/2012 Problem

Sample (3.585g) contains 38.72% C, 9.62% g of H, 51.66% O and its molar mass is 62g/mol. What is molecular formula of this substance?