Experience Chemistry Lesson 5.3.2: % Comp & Empirical Formulas

Lesson 5.3.2: Percent Composition & Empirical Formulas

Multiple Choice

What is the molar mass of NaOH?

• 40 g/mol

• 38 g/mol

• 24 g/mol

• 57 g/mol

Multiple Choice

What is the molar mass of UF₆?

• 101 g/mol

• 238 g/mol

• 257 g/mol

• 352 g/mol

Lesson Objectives

• Compare an Empirical Formula to a Molecular Formula

• Identify an empirical formula using ratios

• Use Percent Composition to find an Empirical formula

Using Percent Composition as a Conversion Factor

• You can use percent composition to calculate the number of grams of any element in a specific mass of a compound

Experience Chemistry | Lesson 5.3.2

Using Percent Composition as a Conversion Factor #1

• Example: 11.11 % of water is hydrogen. How much hydrogen is in 20 grams of water?

Experience Chemistry | Lesson 5.3.2

Percent Composition as a Conversion Factor #2

• Calculate the mass of carbon and the mass of hydrogen in 82.0 grams of propane (C₃H₈).

  • Step 1: Calculate your mass of propane

Experience Chemistry | Lesson 5.3.2

Percent Composition as a Conversion Factor #2

• Calculate the mass of carbon and the mass of hydrogen in 82.0 grams of propane (C₃H₈).

  • Step 1: Calculate your mass of propane

    • C₃H₈ =44 g/mol

  • Step 2: Identify the percent composition of both Carbon and Hydrogen in Propane

Experience Chemistry | Lesson 5.3.2

Percent Composition as a Conversion Factor #2

• Calculate the mass of carbon and the mass of hydrogen in 82.0 grams of propane (C₃H₈).

  • Step 1: Calculate your mass of propane

    • C₃H₈ =44 g/mol

  • Step 2: Identify the percent composition of both Carbon and Hydrogen in Propane

    • C= 81.82 %, H= 18.18%

  • Step 3: Use the percent Composition to convert 82 grams of propane into grams of carbon and grams of hydrogen

Experience Chemistry | Lesson 5.3.2

Multiple Choice

What will your conversion factor for Carbon be?

• $$\frac{81.82\ grams\ C}{100\ grams\ C_3H_8}$$

• $$\frac{18.18\ grams\ C}{100\ grams\ C_3H_8}$$

• $$\frac{36\ grams\ C}{44\ grams\ C_3H_8}$$

• $$\frac{8\ grams\ C}{44\ grams\ C_3H_8}$$

Multiple Choice

What will your conversion factor for Hydrogen be?

• $$\frac{81.82\ grams\ H}{100\ grams\ C_3H_8}$$

• $$\frac{18.18\ grams\ H}{100\ grams\ C_3H_8}$$

• $$\frac{36\ grams\ H}{44\ grams\ C_3H_8}$$

• $$\frac{8\ grams\ H}{44\ grams\ C_3H_8}$$

Percent Composition as a Conversion Factor #2

• Calculate the mass of carbon and the mass of hydrogen in 82.0 grams of propane (C₃H₈).

  • Step 1: Calculate your mass of propane

    • C₃H₈ =44 g/mol

  • Step 2: Identify the percent composition of both Carbon and Hydrogen in Propane

    • C= 81.82 %, H= 18.18%

  • Step 3: Use the percent Composition to convert 82 grams of propane into grams of carbon and grams of hydrogen

Experience Chemistry | Lesson 5.3.2

Empirical Formulas

• Sometimes two compounds can have the same elements, but different amounts of each element

  • Example: H₂O = Water, H₂O₂=Hydrogen Peroxide

• Empirical Formula: Gives the lowest whole-number ratio of the atoms/moles of elements in a compound.

  • May or may not be the same as the molecular formula ( the total number of atoms in a molecule)

  • Can be useful in a lab setting for identification, but doesn't indicate the properties of substances

  • For Hydrogen Peroxide: the empirical formula is 1 : 1

Experience Chemistry | Lesson 5.3.2

Empirical Formulas Sample 1

• Butane has the chemical formula of C₄H₁₀ What is the ratio of Carbon to hydrogen

  • There are 4 carbon and 10

    • 4:10

      • This ratio can be simplified by dividing all numbers by a common factor

Experience Chemistry | Lesson 5.3.2

Dropdown

Butane has the chemical formula of C₄H_10. The Ratio of Carbon to Hydrogen is C

Select answer

: ​ H

Select answer

Dropdown

The empirical formula of water (H₂O) is H​

Select answer

: O​

Select answer

.

Dropdown

Acetylene (C₂H₂) is a flammable gas used in welders' torches. It has an empirical formula of ​ C

Select answer

: H

Select answer

Dropdown

Phenolphthalein has the chemical formula C₂₀H₁₄O₄. Its Empirical formula would be ​ C

Select answer

: H

Select answer

:​ O

Select answer

Multiple Choice

What is the empirical formula of a substance with the molecular formula X₂₀Y₁₅?

• X₁₀Y₁₅

• X₅Y₃

• X₄Y₃

• X₂₀Y₁₅

Using Empirical Formulas Sample Problem 1

• ​If given a percent composition but no formula, you can calculate an empirical formula

  • Example: you have a sample that is 78.1% B and 21.9% H

    • ​Start by assuming you have 100.0 g of your compound

Experience Chemistry | Lesson 5.3.3

Calculating an Empirical Formula

Experience Chemistry | Lesson 5.3.3

Calculating an Empirical Formula

Experience Chemistry | Lesson 5.3.3

Practice Problem 2:

• ​Quantitative analysis shows that a compound contains 32.38% sodium, 22.65% sulfur, and 44.99% Oxygen. Find the empirical formula

  • ​First step: go from percentage composition to mass composition

    • ​32.38% Na = 32.38 g Na

    • ​22.65% S = 22.65 g S

    • ​44.99% O = 44.99 g O

  • ​Second step: convert mass to moles

Experience Chemistry | Lesson 5.3.3

Practice Problem 2

• ​Quantitative analysis shows that a compound contains 32.38% sodium, 22.65% sulfur, and 44.99% Oxygen. Find the empirical formula

  • ​First step: go from percentage composition to mass composition

    • ​32.38% Na = 32.38 g Na

    • ​22.65% S = 22.65 g S

    • ​44.99% O = 44.99 g O

  • ​Second step: convert mass to moles

Experience Chemistry | Lesson 5.3.3

Practice Problem 2:

Experience Chemistry | Lesson 5.3.3