Chemistry 103 Lecture 13

Outline I. The MOLE continued…. II. Determining Chemical Formulas  Percent Composition (review)  Empirical/Molecular Formulas

Atomic masses in Grams 107.9g of Ag. How many Ag atoms? 12.01g of C. How many atoms of C? 32.07g of S. How many S atoms?

Avogadro’s Number Solution = x Avogadro’s number is equal to 1 mole Makes working with large numbers easier

Molar Mass from Periodic Table Molar mass Is the atomic mass expressed in grams

Molar Mass The molar mass Is the mass of one mole of an element or compound Is the atomic mass expressed in grams Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

Some One-Mole Quantities g g g g g

Conversion Factors 1 mole = x (when the question asked for “individual” numbers of something) 1 mole = molar mass (relationship between mass and number)

The Mole Molar mass 1 mole x Mass in grams Individual particles

Conversion Factors 1 mole = x mole = molar mass Mole ratio = (“n” mole part)/ (1 mole total)

Subscripts State Atoms and Moles 1mole aspirin 9 mol C 8 mol H 4 mol O

Learning Check Calculate the number of moles of aspirin in 52.1 g of aspirin (C 9 H 8 O 4 ). 52.1g C 9 H 8 O 4 (1 mol C 9 H 8 O 4 ) g C 9 H 8 O 4.

Learning Check Calculate the number of moles of hydrogen in 52.1 g of aspirin (C 9 H 8 O 4 ). 52.1g C 9 H 8 O 4 x (1 mol C 9 H 8 O 4 ) (8moles H) g C 9 H 8 O 4 1 mol C 9 H 8 O 4

Learning Check Calculate the number of grams of hydrogen in 52.1 g of aspirin (C 9 H 8 O 4 ).

Learning Check Calculate the individual number of hydrogen atoms in 52.1 g of aspirin (C 9 H 8 O 4 ).

Learning Check Allyl sulfide C 6 H 10 S is a compound that has the odor of garlic. How many moles of C 6 H 10 S are in 225 g?

Learning Check Allyl sulfide C 6 H 10 S is a compound that has the odor of garlic. How many moles of C atoms are in 225g of C 6 H 10 S?

Learning Check Allyl sulfide C 6 H 10 S is a compound that has the odor of garlic. How many grams of C are in 225g of C 6 H 10 S?

Learning Check How many H 2 O molecules are in 24.0 g H 2 O? a) 4.52 x b) 1.44 x c) 8.02 x 10 23

Learning Check How many H atoms are in 24.0 g H 2 O? a) 4.01 x b) 1.60 x c) 8.02 x 10 23

Learning Check Determine the mass of 6 molecules of O 2 in gram units.

Percent Composition Percent composition Is the percent by mass of each element in a formula. Mass element in compound x 100% Mass of compound

Percent Composition Example: Calculate the percent composition of CO 2. CO 2 = 1 C(12.01g) + 2 O(16.00 g) = g/mol g C x 100 = % C g CO g O x 100 = % O g CO %

What is the percent composition of lactic acid, C 3 H 6 O 3, a compound that appears in the blood after vigorous activity? Learning Check

STEP 1 3C(12.01) + 6H(1.008) + 3O(16.00) = g/mol g C g H g O STEP 2 %C = g C x 100= 40.00% C g %H = g H x 100 = 6.714% H g %O = g O x 100 = 53.29% O g Solution

The molecular formula Is the true or actual number of the atoms in a molecule The empirical formula Is the simplest whole number ratio of the atoms (this is the formula for ionic compounds) H 2 O 2 HO molecular formula empirical formula Types of Formulas

Some Molecular and Empirical Formulas The molecular formula is the same or a multiple of the empirical.

1. What is the empirical formula for C 4 H 8 ? a) C 2 H 4 b) CH 2 c) CH 2. What is the empirical formula for C 8 H 14 ? a) C 4 H 7 b) C 6 H 12 c) C 8 H 14 Learning Check

1. What is the empirical formula for C 4 H 8 ? a) C 2 H 4 b) CH 2 c) CH 2. What is the empirical formula for C 8 H 14 ? a) C 4 H 7 b) C 6 H 12 c) C 8 H 14 Learning Check

A compound has an empirical formula SN. If there are four atoms of N in one molecule, what is the molecular formula? 1) SN 2) SN 4 3) S 4 N 4 Learning Check

A compound has an empirical formula SN. If there are four atoms of N in one molecule, what is the molecular formula? 1) SN 2) SN 4 3) S 4 N 4 Learning Check

Empirical Formula Using Experimental data to find the empirical formula of a compound.

Law of Definite Proportions Experimental studies (decomposition reactions) led to the conclusion that the percentage of each element present in a given compound does not vary - Law of Definite Proportions (Dalton)

Experiment - Law of Definite Proportions Measure Mass of a compound (an oxide of tin) Experimentally “decompose” into oxygen and tin. Measure mass of each.

Law of Definite Proportions 4 different experiments with a compound that contains both Sn and Oxygen only. Mass Sn(g) Mass tin oxide(g) Mass O(g) Mass Sn/Mass O 5.00g(78.7%) 6.35g 1.35g (21.3%) g(78.7%) 12.7g 2.7g (21%) g(78.7%) 30.1g 6.4g (21%) g(78.8%) 93.2g 19.8g (21.2%)3.71

Compound of Sn and O Row 1: mass Sn = 5.00 g mass O = 1.35g Row 3: mass Sn = 23.7g mass O = 6.4g

Empirical Formula Problems 1). Convert percentages given to grams 2). Convert grams to moles 3). Divide by the smallest number in order to ascertain the whole number ratio of atoms of different elements in the compound 4). Clear any obvious fractions in step 3.

A compound contains 7.31 g Ni and 20.0 g Br. Calculate its empirical (simplest) formula. Learning Check

A particular compound is 60.0% C, 4.5% H and 35.5% O. Calculate its empirical (simplest) formula. Learning Check

Percentages given Assume 100 g sample (percentages easily translate into gram quantities) Convert grams into moles Divide by the smallest number of moles present. If you do not have whole numbers at this point, clear the fraction by multiplying every number by the whole number that accomplishes this end.

Percentages given Assume 100 g sample (percentages easily translate into gram quantities) Convert grams into moles Divide by the smallest number of moles present. If you do not have whole numbers at this point, clear the fraction by multiplying every number by the whole number that accomplishes this end.

Percentages given Assume 100 g sample (percentages easily translate into gram quantities) Convert grams into moles Divide by the smallest number of moles present. If you do not have whole numbers at this point, clear the fraction by multiplying every number by the whole number that accomplishes this end.

Percentages given Assume 100 g sample (percentages easily translate into gram quantities) Convert grams into moles Divide by the smallest number of moles present. If you do not have whole numbers at this point, clear the fraction by multiplying every number by the whole number that accomplishes this end.

Converting Decimals to Whole Numbers When the number of moles for an element is a decimal, all the moles are multiplied by a small integer to obtain whole number.

Empirical Formula for Aspirin

Learning Check A pure phosphorus/oxygen compound is 43.7% “P” and the remainder “O”. What is the empirical formula of this compound?

A molecular formula Is a multiple (or equal) of its empirical formula Has a molar mass that is the empirical mass multiplied by a whole number molar mass = a whole number empirical mass Is obtained by multiplying the empirical formula by a whole number Relating Molecular and Empirical Formulas

Some Compounds with Empirical Formula CH 2 O

Molecular Formula from Empirical If the molar mass of the compound with empirical formula P 2 O 5 is 284g/mol, what is the molecular formula?

A compound has a molar mass of 176.1g and an empirical formula of C 3 H 4 O 3. What is the molecular formula? A) C 3 H 4 O 3 B) C 6 H 8 O 6 C) C 9 H 12 O 9 Learning Check