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Understanding Chemistry: Percent Composition & Formulas

Learn how to determine compound composition, calculate empirical formulas, and differentiate between empirical and molecular formulas in chemistry. Explore practice problems and discover the relationship between molar mass and formula units.

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Understanding Chemistry: Percent Composition & Formulas

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  1. Chapter 10 : The Mole Sec. 10.4: Empirical & Molecular Formulas

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

  3. Synthetic chemists are often involved in developing new compounds for industrial, pharmaceutical and home uses.

  4. It is the analytical chemist’s job to identify the elements that a compound contains and determine its chemical formula.

  5. Percent Composition • The first step in this chemist’s job is to determine the percent by mass of the elements in the compound. • Recall that the percent by mass of any element in a compound is found by: Mass of Element x 100 Mass of Compound

  6. The percent by mass of each element in a compound is the percent compositionof a compound. • Percent composition can be determined • from • experimental data • the chemical formula

  7. From Experimental Data A chemist has a 100.0 g sample of XY. The sample contains 55.0 g X and 45.0 g Y. Percent of X = 55.0 g/100.0 g x 100 = 55% Percent of Y = 100 – 55 = 45%

  8. From the Formula Assume you have 1 mol of the compound. 1 mol of H2O = 18.0 g In 1 mol of H2O, there are 2 moles of H atoms and 1 mol O atoms 2 mole H atoms x 1.0 g = 2.0 g H 1 mol H atoms 1 mol O atoms = 16.0 g % H = 2.0g/18.0g = 11 % % O = 100 – 11 = 89%

  9. Practice Problems • Calculate the percent composition of sodium sulfate (Na2SO4). • Which has the larger percent by mass of sulfur, H2SO3 or H2S2O8? • What is the percent composition of a compound that contains 2.644 g of gold and 0.476 g of chlorine?

  10. Empirical Formula • The empirical formula is the simplest formula for a compound. • A molecular formula is the same as the empirical formula or it is a multiple of the empirical formula; it is the actualnumber of atoms of each element in one molecule or formula unit of the compound.

  11. Empirical Formula Benzene Molecular formula C6H6 Empirical Formula CH

  12. Empirical Formula Acetylene Molecular Formula C2H2 Empirical Formula ?

  13. Empirical Formula Glucose Molecular Formula C6H12O6 Empirical Formula ?

  14. Empirical Formula Carbon Dioxide Molecular Formula CO2 Empirical Formula ?

  15. Empirical Formula • The empirical formula may or may not be the same as the molecular formula. • If they are different, the molecular formula will be a multiple of the empirical formula.

  16. To Determine Empirical Formula Molar Mass

  17. Empirical Formula • The percent composition of a compound was found to be 40.05% S and 59.95% O. What is the empirical formula for the compound?

  18. Empirical Formula • Step One: Assume you have 100 g of the compound**; Step Two: use molar mass to determine moles of each element. • 40.05 g S x 1 mol S = 1.248 mol S 32. 1 g • 59.95 g O x 1 mol O = 3.747 mol O 16 g ** If you are given mass data instead of % composition, determine moles directly from the grams.

  19. Empirical Formula • Step 3: Divide both mole values by the smaller of the two to get the mole ratio of the elements. 1.248 mol S = 1 mol S 1.248 3.747 mol O = 3 mol O 1.248

  20. Empirical Formula • Now use the simplest, whole number mole ratio as subscripts in the empirical formula: 1 mol S: 3 mol O SO3

  21. Practice Problems • What is the empirical formula for a compound that contains 10.89% Mg, 31.77% Cl, and 57.34% O? • Determine the empirical formula for a compound that contains 74.19% Na and 25.81% O. • When an oxide of potassium is decomposed, 19.55 g of K and 4.00 g of O are obtained. What is the empirical formula of the compound?

  22. Honors Practice Problem • Determine the empirical formula for a compound that contains 35.98% aluminum and 64.02% sulfur. • The chemical analysis of aspirin indicates that the molecule is 60.00% C, 4.44% H, and 35.56% O. Determine the empirical formulas for aspirin.

  23. Molecular Formula • Compounds with the same empirical formula can have very different properties. • Remember, the empirical formula does not always indicate the actual number of moles in the compound. So, different compounds can have the same empirical formula. • Acetylene (C2H2) and benzene (C6H6) are different compounds with the same empirical formula, CH.

  24. Molecular Formula • In order to distinguish between different compounds with the same empirical formula, a chemist must go one step further and determine the compound’s molecular formula. • The molar mass of the compound is determined through experimentation and compared with the molar mass represented by the empirical formula.

  25. Molecular Formula • Suppose a compound has an empirical formula ofClCH2 and a molar mass of 98.96 g/mol. How can its molecular formula be determined? • Step One: Find the molar mass of the empirical formula. ClCH2 = 35.5 + 12.0 + 2(1.0) = 49.5 g/mol

  26. Molecular Formula • Step 2: Divide the molar mass of the compound by the molar mass of the empirical formula. 98.96 g/mol = 1.999 = 2 49.5 g/mol The molecular formula is this multiple of the empirical formula.

  27. Molecular Formula • Step 3: Multiply the subscripts in the empirical formula by this multiple. ClCH2 becomes Cl2C2H4.This is the molecular formula of the compound.

  28. Practice Problems • The empirical formula of a compound is found to be C2H3O2. It has a molar mass of 118.1 g/mol. Determine the molecular formula for the compound. • The molar mass for a compound having the empirical formula of CH is found to be 78.1 g/mol. What is the molecular formula for the compound?

  29. Honors Practice Problem • A colorless liquid composed of 46.68% nitrogen and 53.32% oxygen has a molar mass of 60.01 g/mol. What is the molecular formula? • A compound was found to contain 49.98 g C and 10.47 g H. The molar mass of the compound is 58.12 g/mol. Determine the molecular formula.

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