Chapter 7 Chemical Quantities The Mole Atomic Mass and Formula Mass Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

The Mole Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Collection Terms A collection term states a specific number of items. 1 dozen donuts = 12 donuts 1 ream of paper = 500 sheets 1 case = 24 cans Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

A mole (mol) is a collection that contains The same number of particles as there are carbon atoms in g of carbon x atoms of an element (Avogadro’s number). 1 mol C = x C atoms 1 mol Na = x Na atoms 1 mol Au = x Au atoms A Mole of Atoms

A mole Of a covalent compound has Avogadro’s number of _________________. 1 mol CO 2 = x CO 2 molecules 1 mol H 2 O = x H 2 O molecules Of an ionic compound contains Avogadro’s number of ___________ units. 1 mol NaCl = x NaCl formula units 1 mol K 2 SO 4 = x K 2 SO 4 formula units A Mole of A Compound

Samples of One Mole Quantities Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Avogadro’s number x can be written as an equality and two conversion factors. As an equality: 1 mol particles = x particles As conversion Factors: x particles and 1 mol particles 1 mol particles x particles Avogadro’s Number

Using Avogadro’s Number Avogadro’s number Converts moles of a substance to the number of particles. How many Cu atoms are in 0.50 mol Cu? 0.50 mol Cu x x Cu atoms 1 mol Cu = 3.0 x Cu atoms Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Using Avogadro’s Number Avogadro’s number Converts particles of a substance to moles. How many moles of CO 2 are 2.50 x CO 2 molecules? 2.50 x CO 2 x 1 mol CO x CO 2 = 4.15 mol CO 2

The number of atoms in 2.0 mol Al is 2.0 mol Al x x Al atoms = 1 mol Al 1.2 x Al atoms Learning Check

The number of moles of S in 1.8 x atoms S is 1.8 x S atoms x 1 mol S = x S atoms 3.0 mol S atoms Learning Check

Subscripts and Moles The subscripts in a formula state The relationship of atoms in the formula. The moles of each element in 1 mol of compound. Glucose C 6 H 12 O 6 In 1 molecule: 6 atoms C 12 atoms H6 atoms O In 1 mol: 6 mol C 12 mol H 6 mol O

Subscripts State Atoms and Moles 1 mol C 9 H 8 O 4 = 9 mol C 8 mol H 4 mol O Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Factors from Subscripts Subscripts used for conversion factors Relate moles of each element in 1 mol compound. For aspirin C 9 H 8 O 4 can be written as: 9 mol C 8 mol H 4 mol O 1 mol C 9 H 8 O 4 1 mol C 9 H 8 O 4 1 mol C 9 H 8 O 4 and 1 mol C 9 H 8 O 4 1 mol C 9 H 8 O 4 1 mol C 9 H 8 O 4 9 mol C 8 mol H 4 mol O

Learning Check How many moles O are in mol aspirin C 9 H 8 O 4 ? mol C 9 H 8 O 4 x 4 mol O = mol O 1 mol C 9 H 8 O 4 subscript factor

Learning Check How many O atoms are in mol aspirin C 9 H 8 O 4 ? mol C 9 H 8 O 4 x 4 mol O x x O atoms 1 mol C 9 H 8 O 4 1 mol O subscript Avogadro’s factor number = 3.61 x O atoms

Atomic Mass Atomic mass is the Mass of a single atom in atomic mass units (amu). Mass of an atom compared to a 12 C atom. Number below the symbol of an element. (the average atomic mass)

Periodic Table and Atomic Mass Ag has atomic mass = amu C has atomic mass = amu S has atomic mass = amu

Atomic Mass Factors The atomic mass Can be written as an equality. Example: 1 P atom = amu Can be written as two ___________ ________. Example: 1 P atom and amu amu 1 P atom

Uses of Atomic Mass Factors The atomic mass factors are used to convert A specific number of atoms to mass (amu). An amount in amu to ___________ of atoms.

Using Atomic Mass Factors 1. What is the mass in amu of 75 P atoms? 75 P atoms x amu = 2323 amu (2.323 x10 3 amu) 1 P atom 2. How many Cu atoms have a mass of x 10 5 amu? x 10 5 amu x 1 Cu atom = 7081 Cu atoms amu

Learning Check What is the mass in amu of 75 silver atoms? 75 Ag atoms x amu = 8093 amu 1 Ag atom

Learning Check How many gold atoms have a mass of 1.85 x 10 5 amu? 1.85 x 10 5 amu x 1 Au atom = 939 Au atoms amu

Formula Mass The formula mass is The mass in amu of a compound. The _____ of the atomic masses of the elements in a formula. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Calculating Formula Mass To calculate the formula mass of Na 2 SO 4 Multiply the atomic mass of each element by its subscript, then total the masses of the atoms. 2 Na x amu = amu 1 Na 1 S x amu = amu amu 1 S 4 O x amu = amu 1 O

Learning Check Using the periodic table, calculate the formula mass of aluminum sulfide Al 2 S 3. 2 Al x amu = amu 1 Al 3 S x amu = amu 1 S amu

Molar Mass Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Molar Mass The molar mass Is the mass of one mol of an element or compound. Is the atomic mass expressed in grams. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Molar Mass from the Periodic Table Molar mass Is the atomic mass expressed in grams. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Give the molar mass for: A. 1 mol K atoms = B.1 mol Sn atoms = Learning Check

Molar Mass of a Compound The molar mass of a compound is the sum of the molar masses of the elements in the formula. Example: Calculate the molar mass of CaCl 2. ElementNumber of Moles Atomic MassTotal Mass Ca g/mol g Cl g/mol g CaCl g

Molar Mass of K 3 PO 4 Calculate the molar mass of K 3 PO 4. ElementNumber of Moles Atomic MassTotal Mass in K 3 PO 4 K g/mol g P g/mol g O g/mol g K 3 PO g

Some One-Mol Quantities g g g g g Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

A. K 2 O g/mol 2 mol K (39.10 g/mol) + 1 mol O (16.00 g/mol) g g = g B. Al(OH) g/mol 1 mol Al (26.98 g/mol) + 3 mol O (16.00 g/mol) + 3 mol H (1.008 g/mol) g g g = g Learning Check

Calculations Using Molar Mass Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Molar mass conversion factors Are written from molar mass. Relate grams and moles of an element or compound. Example: Write molar mass factors for methane CH 4 used in gas cook tops and gas heaters. Molar mass: 1 mol CH 4 = g Conversion factors: g CH 4 and 1 mol CH 4 1 mol CH g CH 4 Molar Mass Factors

Acetic acid C 2 H 4 O 2 gives the sour taste to vinegar. Write two molar mass factors for acetic acid. Calculate molar mass: = g/mol 1 mol of acetic acid = g acetic acid Molar mass factors 1 mol acetic acid and g acetic acid g acetic acid 1 mol acetic acid Learning Check

Molar mass factors are used to convert between the grams of a substance and the number of moles. Calculations Using Molar Mass Grams Molar mass factor Moles

Aluminum is used to build lightweight bicycle frames. How many grams of Al are 3.00 mol Al? Molar mass equality: 1 mol Al = g Al Setup with molar mass as a factor: 3.00 mol Al x g Al = 80.9 g Al 1 mol Al molar mass factor for Al Moles to Grams

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 C 6 H 10 S? Action Plan: Calculate the molar mass, and convert 225 g to moles. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings Grams Molar mass factor Moles

Calculate the molar mass of C 6 H 10 S. 6 C (6 x 12.01) + 10 H (10 x 1.008) + 1 S (1 x 32.07) = g/mol Set up the calculation using a mole factor. 225 g C 6 H 10 S x 1 mol C 6 H 10 S g C 6 H 10 S molar mass factor(inverted) = 1.97 mol C 6 H 10 S Solution

Grams, Moles, and Particles A molar mass factor and Avogadro’s number convert Grams to particles molar mass Avogadro’s number (g mol particles) Particles to grams Avogadro’s molar mass number (particles mol g)

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 2 O molecules are in 24.0 g H 2 O? 24.0 g H 2 O x 1 mol H 2 O x x H 2 O molecules g H 2 O 1 mol H 2 O = 8.02 x H 2 O molecules

Learning Check If the odor of C 6 H 10 S can be detected from 2 x g in one liter of air, how many molecules of C 6 H 10 S are present? 2 x g x 1 mol x x molecules g 1 mol = 1 x 10 9 molecules C 6 H 10 S Odor Threshhold – lower detectable limit

Percent Composition and Empirical Formulas Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Percent Composition Percent composition Is the percent by mass of each element in a formula. 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 2

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 3 C(12.01) + 6 H(1.008) + 3 O(16.00) = g/mol g C g H g O = g/mol STEP 2 %C = g C x 100= 40.00% C g cpd %H = g H x 100 = 6.714% H g cpd %O = g O x 100 = 53.29% O g cpd Solution C3H6O3C3H6O3

Learning Check The chemical isoamyl acetate C 7 H 14 O 2 contributes to the odor of pears. What is the percent carbon in isoamyl acetate? Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Molar mass C 7 H 14 O 2 = 7 C(12.01) + 14 H(1.008) + 2 O(16.00) = g/mol Total C = 7 C(12.01) = g % C = total g C x 100 total g cpd % C= g C x 100 = % C g cpd Solution C 7 H 14 O 2

The empirical formula Is the simplest whole number ratio of the atoms. Is calculated by dividing the subscripts in the actual (molecular) formula by a whole number to give the lowest ratio. C 5 H 10 O 5  5 = C 1 H 2 O 1 = CH 2 O actual (molecular) empirical formula formula Empirical Formulas

Some Molecular and Empirical Formulas The molecular formula is the same or a multiple of the empirical. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

1. What is the empirical formula for _________? A) C 2 H 4 B) CH 2 C) CH 2. What is the empirical formula for _________? A) C 4 H 7 B) C 6 H 12 C) C 8 H Which is a possible molecular formula for ______? A) C 4 H 4 O 4 B) C 2 H 4 O 4 C) C 3 H 6 O 3 Learning Check

A compound contains 7.31 g Ni and 20.0 g Br. Calculate its empirical (simplest) formula. Lets ask, what does the empirical formula tell us? The ratio of atoms of each element, ratio of moles of each element. Action Plan: Convert mass (g) to moles. Learning Check

Convert 7.31 g Ni and 20.0 g Br to moles g Ni x 1 mol Ni = mol Ni g Ni 20.0 g Br x 1 mol Br = mol Br g Br Divide by smallest: mol Ni = 1 Ni0.250 mol Br = 2 Br Write ratio as subscripts: NiBr 2 Solution

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. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Aspirin is 60.0% C, 4.5 % H and 35.5 % O. Calculate its empirical (simplest) formula. We have percentages, not grams. However, the percentages are really an expression of grams of an element in 100. (exact) grams of compound. Learning Check

STEP 1. Calculate the moles of each element in 100. g of the compound g aspirin contains 60.0% C or 60.0 g C, 4.5% H or 4.5 g H, and 35.5% O or 35.5 g O g C x 1 mol C = 5.00 mol C g C 4.5 g H x 1 mol H = 4.5 mol H g H 35.5 g O x 1mol O = 2.22 mol O g O Solution

Solution (continued) STEP 2. Divide by the smallest number of mol mol C= 2.25 mol C (decimal) mol H = 2.0 mol H mol O = 1.00 mole O 2.22

Solution (continued) 3. Use the lowest whole number ratio as subscripts When the moles are not whole numbers, multiply by a factor to give whole numbers, in this case x 4. C: 2.25 mol C x 4 = 9 mol C H: 2.0 mol Hx 4 = 8 mol H O: 1.00 mol O x 4 = 4 mol O Using these whole numbers as subscripts the simplest formula is C 9 H 8 O 4

Molecular Formulas Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

A molecular formula Is a multiple (or equal) of its empirical formula. Has a molar mass that is the empirical formula 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

Diagram of Molecular and Empirical Formulas A small integer links A molecular formula and its empirical formula. A molar mass and its empirical formula mass. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Determine the molecular formula of compound that has a molar mass of g and an empirical formula of CH. STEP 1. Empirical formula mass of CH = g STEP 2. Divide the molar mass by the empirical mass g = ~ g STEP 3. Multiply each subscript in C 1 H 1 by 6. molecular formula = C 1x 6 H 1 x 6 = C 6 H 6 Finding the Molecular Formula

Some Compounds with Empirical Formula CH 2 O Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings formaldehyde

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? What do we need to do? From the empirical formula, determine the empirical mass. Then determine the whole number integer. Learning Check

A compound has a formula mass of and an empirical formula of C 3 H 4 O 3. What is the molecular formula? C 3 H 4 O 3 = g/EF g (molar mass) = g (empirical mass) Molecular formula = 2 x empirical formula C 3 x 2 H 4 x 2 O 3 x 2 = C 6 H 8 O 6 Solution

A compound contains C 24.27%, H 4.07%, and Cl 71.65%. The molar mass is about 99 g. What are the empirical and molecular formulas? STEP 1. Calculate the empirical formula. Write the mass percents as the grams in a g sample of the compound. C g H 4.07 g Cl g Molecular Formula

Finding the Molecular Formula ( Continued) Calculate the number of moles of each element g C x 1 mol C = mol C g C 4.07 g H x 1 mol H = 4.04 mol H g H g Cl x 1 mol Cl = mol Cl g Cl

Finding the Molecular Formula ( Continued) Divide by the smallest number of moles: mol C =1 mol C mol H =2 mol H mol Cl=1 mol Cl Empirical formula = C 1 H 2 Cl 1 = CH 2 Cl Calculate empirical mass (EM) CH 2 Cl = g

Finding the Molecular Formula ( Continued) STEP 2. Divide molar mass by empirical mass. Molar mass = 99 g = 2 Empirical mass g STEP 3. Multiply the empirical formula by the small integer to determine the molecular formula. 2 x (CH 2 Cl) C 1 x2 H 2 x 2 Cl 1x 2 = C 2 H 4 Cl 2

A compound is 27.4% S, 12.0% N and 60.6 % Cl. If the compound has a molar mass of 351 g, what is the molecular formula? Learning Check

In 100. g, there are 27.4 g S, 12.0 g N, and 60.6 g Cl g S x 1 mol S = mol S g S 12.0 g N x 1 mol N = mol N g N 60.6 g Cl x 1mol Cl = 1.71 mol Cl g Cl Solution

Divide by the smallest number of moles mol S /0.854 = 1.00 mol S mol N/0.854 = 1.00 mol N 1.71 mol Cl/0.854 = 2.00 mol Cl empirical formula = SNCl 2 = g Molar Mass/ Empirical mass 351 g = g molecular formula = (SNCl 2 ) 3 = S 3 N 3 Cl 6 Solution (continued)