4. Hemoglobin, a protein in red blood cells, carries

O2, from the lungs to the body’s cells. Iron (as

Fe2+) makes up 0.33 mass % of hemoglobin. If the

molar mass of hemoglobin is 6.8x104 g/mol, how

many Fe2+ ions are in one molecule?

Assume 1 mol hemoglobin:

mass Fe2+ = 0.0033 (6.8x104 g mol-1) = 224.4 g

Looking at one molecule of hemoglobin:

mass Fe2+ in one molecule = 224.4 amu

# of Fe2+ ions = 224.4 𝑎𝑚𝑢

55.85 𝑎𝑚𝑢= 4.02

4 Fe2+ per molecule of hemoglobin

1

Solution Stoichiometry

A solution is a homogenous mixture of a

solvent plus a solute.

9 July 2013 2

Concentration of Solutions

Amount of solute present in a given

quantity of solvent or solution

Expressed as molarity

𝑀 = 𝑚𝑜𝑙𝑎𝑟𝑖𝑡𝑦 =𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒

𝑡𝑜𝑡𝑎𝑙 𝑙𝑖𝑡𝑒𝑟𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

9 July 2013 3

𝑔𝑟𝑎𝑚𝑠 𝑠𝑜𝑙𝑢𝑡𝑒

𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑠𝑜𝑙𝑢𝑡𝑒

𝐷𝑒𝑛𝑠𝑖𝑡𝑦 =𝑚𝑎𝑠𝑠

𝑣𝑜𝑙𝑢𝑚𝑒

Example

Calculate the molarity of a solution that

contains 12.5 g of pure sulfuric acid

(H2SO4) in 1.75 L of solution (Molar mass

H2SO4 = 98.1 g)

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𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 =12.5 𝑔 𝐻2𝑆𝑂4

1 𝑚𝑜𝑙 𝐻2𝑆𝑂4

98.1 𝑔 𝐻2𝑆𝑂4

1.75 𝐿= 0.0728 M 𝐻2𝑆𝑂4

Concentration

An intensive (not extensive) property

◦ Independent of the volume of solution (like

density or temperature)

4 July 2013 5

Preparing Solutions in the

Laboratory

9 July 2013 6

Practice Problem

How many grams of KI is required to

make 500. mL of a 2.80 M KI solution?

Reworded: Suppose I have 500. mL of a

2.80 M KI solution. How many grams of

KI solute does it contain? (“It” meaning

the entire 500. mL volume.)

9 July 2013 7

Practice Problem

How many grams of KI is required to

make 500. mL of a 2.80 M KI solution?

9 July 2013 8

𝑔𝑟𝑎𝑚𝑠 𝐾𝐼

= 500. 𝑚𝐿1 𝐿

1000 𝑚𝐿

2.80 𝑚𝑜𝑙 𝐾𝐼

𝐿 𝑠𝑜𝑙𝑛

166 𝑔 𝐾𝐼

𝑚𝑜𝑙 𝐾𝐼

= 232 𝑔 𝐾𝐼

Practice Problem

Sucrose has a molar mass of 342.29 g/mol.

It is a fine, white, crystalline powder with

a pleasing, sweet taste. What is the

molarity of an aqueous solution made by

placing 75.5 g of 85% pure sugar in 500.00

mL of water?

9 July 2013 9

What does 3.5 M FeCl3 mean?

1. 3.5 mol of dry, 100% pure FeCl3

dissolved in 1.00 L total solution volume

2. [Fe3+] = 3.5 M and [Cl-] = 3 x 3.5 M

9 July 2013 10

Example

In 2 L of 0.20 M Al2(SO4)3, (a) what is the molarity of Al2(SO4)3? (b) How many moles of Al are there? (c) What is the molarity of [Al3+] and [SO4

2-] ions?

a. [Al2(SO4)3] = 0.20 M

b. mol Al = 2 L 0.20 mol

L= 0.4 mol Al

c. Al2 += 2 × 0.20M

and SO42 −

= 3 × 0.20M

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Example

Hydrochloric acid is sold commercially as

a 12.0 M solution. How many moles of

HCl are in 300.0 mL of 12.0 M solution?

𝑚𝑜𝑙 𝐻𝐶𝑙

= 300.0 𝑚𝐿1 𝐿

1000 𝑚𝐿

12 𝑚𝑜𝑙 𝐻𝐶𝑙

1 𝐿 𝑠𝑜𝑙𝑛

= 3.60 𝑚𝑜𝑙 𝐻𝐶𝑙

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Dilution

9 July 2013 13

Dilution

Add solvent to a concentrated solution

From higher molarity lower molarity

The solution volume increases while the

number of moles of solute remains the

same

𝑀𝑑𝑖𝑙 × 𝑉𝑑𝑖𝑙 = 𝑛𝑜 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 = 𝑀𝑐𝑜𝑛𝑐 × 𝑉𝑐𝑜𝑛𝑐

9 July 2013 14

Example

“Isotonic saline” is a 0.15 M aqueous

solution of NaCl. How would you

prepare 0.80 L of isotonic saline from a

6.0 M stock solution?

𝑀𝑑𝑖𝑙 × 𝑉𝑑𝑖𝑙 = 𝑛𝑜 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 = 𝑀𝑐𝑜𝑛𝑐 × 𝑉𝑐𝑜𝑛𝑐

0.15 𝑀 0.80 𝐿 = 6.0 𝑀 𝑉𝑐𝑜𝑛𝑐

𝑉𝑐𝑜𝑛𝑐 = 0.02 𝐿 = 20. 𝑚𝐿

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A 0.020 L portion of the concentrated solution must be

diluted to a final volume of 0.80 L.

Practice Problem

1. Calculate the maximum number of

moles and grams of H2S that can form

when 158 g of aluminum sulfide reacts

with 131 g of water:

Al2S3 + H2O Al(OH)3 + H2S [unbalanced]

What mass of the excess reactant remains?

9 July 2013 16

Practice Problem

2. Sodium borohydride (NaBH4) is used

industrially in many organic syntheses.

One way to prepare it is by reacting

sodium hydride with gaseous diborane

(B2H6). Assuming an 88.5% yield, how

many grams of NaBH4 can be prepared

by reacting 7.98 g of sodium hydride and

8.16 g of diborane?

9 July 2013 17

Practice Problem

3. Calculate each of the following

quantities:

a. Volume in milliliters of 2.26 M potassium

hydroxide that contains 8.42 g of solute

b. Number of Cu2+ ions in 52 L of 2.3 M

copper(II) chloride

c. Molarity of 275 mL of solution

containing 135 mmol of glucose

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Practice Problem

4. Calculate each of the following quantities:

a. Volume of 2.050 M copper(II) nitrate that must be diluted with water to prepare 750.0 mL of a 0.8543 M solution

b. Volume of 1.63 M calcium chloride that must be diluted with water to prepare 350. mL of a 2.86x10-2 M chloride ion solution

c. Final volume of a 0.0700 M solution prepared by diluting 18.0 mL of 0.155 M lithium carbonate with water

9 July 2013 19

Announcements

9 July 2013 20

Quiz # 3 on July 16 Tuesday

◦ Coverage: Chapter 3 – Stoichiometry of

Formulas and Equations

Long Test # 1 on July 18, Escaler Hall

(tentative)

◦ Coverage: Chapters 1-3

Problem Sets

Chapter Practice problems

1 6, 7, 8, 22, 24, 26, 30, 34, 47, 49, 53, 55

2 11, 13, 17, 33, 61, 65, 67, 69, 71, 73, 75

3 6, 10, 12, 14, 16, 18, 20, 23, 26, 28, 30, 32,

33, 38, 40, 41, 43, 45, 49, 51, 65, 66, 69, 71,

73, 75, 83, 85, 89, 93

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Practice on some MC questions:

http://www.mhhe.com/physsci/chemistry/silberberg/student/

olc/quiz.mhtml