Empirical & Molecular Formulas

Unit 4: Stoichiometry Chapter 10 – The Mole

2/6/2014

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Learning Target

:  Understand how to determine the percent composition of a compound.

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Learning Outcome:

 Know how to calculate the percent composition of a compound.

Practice

• Suppose you spend 6 hours texting friends each day. What percentage of each day do you spend texting?

6 ÷ 24 = 0.25 hours 0.25 x 100% = 25%

Percentage Composition

• Definition o The mass of each element in a compound compared to the entire mass of the compound and multiplied by 100 percent

2 ways to find % composition

1) Total mass of compound Percentage 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.

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.

Practice

A 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?

2/7/2014

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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.

Empirical Formula

• Definition • A formula that gives the simplest whole-number ratio of the atoms of elements • For example, H 2 O 2 is the molecular formula for hydrogen peroxide.

• HO is the empirical formula for hydrogen peroxide.

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  (% composition)  Ratio of Atoms chemical formula

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.

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 → moles o 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

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 O 3

Practice

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

Practice

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

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

Molecular Formula

• Definition o 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.

Determining Molecular Formula

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

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 moles

o 42.56 g Pd x 1 mole Pd = 0.40 mol Pd o 106.42 g 0.80 g H x 1 mole H = 0.79 mol H 1.01 g

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 H 2

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 PdH 2 , then the empirical molar mass = 106.42g/mol Pd + 2(1.01g/mol H) = 108.44 g/mol

Continued Once Again

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Step 5 Find, Molar Mass ÷ Formula Mass:

216.8 g/mol 108.44 g/mol = 1.995  2

Step 6:

**MOST IMPORTANT STEP

FINALLY!!! Now adjust the molar formula PdH 2 becomes Pd 2 H 4

Practice Problem

Octane (C 8 ), 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?

Last Practice

• β-carotene, a compound found in carrots, can be broken down to form vitamin A. The empirical formula for β-carotene is C 5 H 7 . The molar mass of β-carotene is 536 g/mol. What is the molecular formula for β-carotene?

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 %

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?