chemistry 103 lecture 13. outline i. the mole continued…. ii. determining chemical formulas ...

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 = 6.022 x 1023

Avogadro’s number is equal

to 1 mole Makes working with large numbers

easier

Molar Mass from Periodic TableMolar 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

32.07 g 55.85 g 58.44 g 294.20 g 342.30 g

Conversion Factors

1 mole = 6.022 x 1023

(when the question asked for “individual” numbers of something)

1 mole = molar mass

(relationship between mass and number)

The Mole

Molar mass 1 mole 6.022 x 1023

Mass in grams Individual particles

Conversion Factors

1 mole = 6.022 x 1023

1 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 (C9H8O4).

52.1g C9H8O4 (1 mol C9H8O4)

180.16g C9H8O4.

Learning Check

Calculate the number of moles of hydrogen in 52.1 g of aspirin (C9H8O4).

52.1g C9H8O4 x (1 mol C9H8O4) (8moles H)

180.16g C9H8O4 1 mol C9H8O4

Learning Check

Calculate the number of grams of hydrogen in 52.1 g of aspirin (C9H8O4).

Learning Check

Calculate the individual number of hydrogen atoms in 52.1 g of aspirin (C9H8O4).

Learning Check

Allyl sulfide C6H10S is a compound that has the odor of garlic. How many moles of C6H10S are in 225 g?

Learning Check

Allyl sulfide C6H10S is a compound that has the odor of garlic. How many moles of C atoms are in 225g of C6H10S?

Learning Check

Allyl sulfide C6H10S is a compound that has the odor of garlic. How many grams of C are in 225g of C6H10S?

Learning Check

How many H2O molecules are in 24.0 g H2O?

a) 4.52 x 1023

b) 1.44 x 1025

c) 8.02 x 1023

Learning Check

How many H atoms are in 24.0 g H2O?

a) 4.01 x 1023

b) 1.60 x 1024

c) 8.02 x 1023

Learning Check

Determine the mass of 6 molecules of O2 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 CO2.

CO2 = 1 C(12.01g) + 2 O(16.00 g) = 44.01 g/mol

12.01 g C x 100 = 27.29 % C 44.01 g CO2

32.00 g O x 100 = 72.71 % O 44.01 g CO2 100.00 %

What is the percent composition of lactic acid, C3H6O3, a compound that appears in the blood after vigorous activity?

Learning Check

STEP 13C(12.01) + 6H(1.008) + 3O(16.00) = 90.08 g/mol

36.03 g C + 6.048 g H + 48.00 g O

STEP 2%C = 36.03 g C x 100 = 40.00% C

90.08 g

%H = 6.048 g H x 100 = 6.714% H 90.08 g

%O = 48.00 g O x 100 = 53.29% O 90.08 g

Solution

The molecular formula Is the true or actual number of the atoms in a

moleculeThe empirical formula Is the simplest whole number ratio of the atoms

(this is the formula for ionic compounds)

H2O2 HOmolecular 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 C4H8?

a) C2H4 b) CH2 c) CH

2. What is the empirical formula for C8H14?

a) C4H7 b) C6H12 c) C8H14

Learning Check

1. What is the empirical formula for C4H8?

a) C2H4 b) CH2 c) CH

2. What is the empirical formula for C8H14?

a) C4H7 b) C6H12 c) C8H14

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) SN4

3) S4N4

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) SN4

3) S4N4

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%) 3.70

10.0g (78.7%) 12.7g 2.7g (21%) 3.7

23.7g (78.7%) 30.1g 6.4g (21%) 3.7

73.4g (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 NumbersWhen 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 numbermolar 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 CH2O

Molecular Formula from Empirical If the molar mass of the compound with

empirical formula P2O5 is 284g/mol, what is the molecular formula?

A compound has a molar mass of 176.1g and an empirical formula of C3H4O3. What is the molecular formula?

A) C3H4O3

B) C6H8O6

C) C9H12O9

Learning Check