https://openstax.org/books/chemistry-2e/pages/3-introduction

Name: ___________________________________

CHEM 1806: College Chemistry-I (Teach the Doc)

Atomic Mass and Molar Mass – Ch-3: TEACH THE DOC

section 3.1 in OpenStax

Total: 20 points

Watch video tutorial on Moles-to-grams Conversions

Molar Mass (Section 3.1) – 4 points

For example, the molar mass of sodium can be obtained from the periodic table: 22.99 g/mol.

1. What is the molar mass of Ca(NO3)2?

(ans. 164.10 g/mole)

Now solve the following problems using dimensional analysis:

2. Calculate the number of moles of iron in a 15.0 g sample of iron.

(ans: 0.269 mol)

3. Calculate the number of atoms in 0.25 mol of silicon.

(ans: 1.5 x 1023 atoms)

4. Calculate the mass in grams needed to have 2.65 x 1022 atoms of calcium.

(ans: 1.76 g Ca)

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CHEM 1806: College Chemistry-I (Teach the Doc)

Chemical Formulas and the Mole Concept – TEACH THE DOC

sections 3.2 in OpenStax

Chemical Formulas (section 3.2)

watch the video tutorial on Chemical Formulas

Now solve the following problems using dimensional analysis: 6 Points

1. Calculate the grams of carbon in 2.76 moles of C2H6O.

(ans: 66.3 g)

2. Calculate the grams of 2.5 moles of Aluminum Sulfate.

Molarity – TEACH THE DOC

section 3.3 in OpenStax

Solution Concentration

Molarity in Calculations – 5 Points

When you are given molarity (M), use it as a conversion factor.

Practice Problem: Calculate the number of moles of sodium hydroxide in 50.0 mL of 2.50 M NaOH.

(ans: 0.125 mol NaOH)

Dilution Calculations – 5 Points

Practice Problem: Calculate the volume of 12.0 M HCl needed to make 500.0 mL of 4.50 M HCl.

(ans: 188 mL)

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CHEM 1806: College Chemistry-I (Teach the Doc)

Thermodynamics and Calorimetry – Ch-5: TEACH THE DOC

Sections 5.1 & 5.2 in OpenStax

TOTAL: 25 points

Enthalpy

Section 5.3 in OpenStax

Enthalpy of Formation – 15 Points

Watch video tutorial on calculations with Enthalpy of Formation

1. Read Examples 5.11 and 5.12 for the conceptual knowledge of

determine which of the following reactions have the same

a) 2 CO(g) + O2(g)

and

and

, then

.

→ 2 CO2(g)

b) 3/2 H2(g) + 1/2 N2(g) → NH3(g)

c) H2S (g) +

2 O2(g) → H2SO4(l)

(ans: b)

2. Calculate

for each of the following reactions (use the

(a)

2 C4H10 (g) +

(b)

4 NH3(g)

13 O2(g) →

+ 5 O2(g)

→

8 CO2(g) +

4 NO(g) +

data in Appendix G):

10 H2O(g)

6 H2O(l)

ans. -5315 kJ

ans. -1166 kJ

“Thermodynamics and Calorimetry Study Guide” by Montgomery College is licensed under CC BY 4.0

CHEM 1806: College Chemistry-I (Teach the Doc)

Hess’s Law – Ch-5: TEACH THE DOC

5.3 in OpenStax

Hess’s Law – 10 Points

Watch the video tutorial on Hess’s Law

Practice:

if

2 CO(g) + O2 (g) → 2 CO2 (g)

H = -566 kJ

then 6 CO(g) + 3 O2 (g) → 6 CO2 (g)

H = _________

(ans. -1698 kJ)

Practice:

if

3 H2(g) + N2(g) → 2 NH3(g)

4 H2(g) + 2 NO2(g) → N2(g) + 4 H2O(g)

H = – 69 kJ

H = – 255 kJ

What is H for the following reaction?

7 H2(g) + 2 NO2(g)

→ 2 NH3(g)

+

4 H2O(g)

H = _________.

(ans: – 324 kJ)

“Thermodynamics and Calorimetry Study Guide” by Montgomery College is licensed under CC BY 4.0

Name: ___________________________________CHEM 1806: College Chemistry-I (Teach the Doc)

Ch-9: GASES – TEACH THE DOC

TOTAL: 55 Points

Pressure (section 9.1) – 5 Points

1. Complete the following conversion table:

kPa

torr

atm

680.0 mm Hg

Gas Laws (section 9.2) – 10 Points

Watch the video tutorials on Boyle’s Law and Charles’ Law.

2. Suppose a gas has an initial pressure of 2.0 atm. What happens to the pressure if the

volume is doubled (assuming the temperature and moles of gas remain unchanged)?

3. Now suppose a gas occupies 1.5 liters at a certain temperature and pressure. What

happens to the volume if the number of moles of gas is tripled?

All the above gas laws can be combined into one overall equation, called the Combined Gas

Law Equation.

P1 V1 P2 V2

=

n1 T1 n2 T2

Any variables which are held constant may be cancelled out of this equation.

The Ideal Gas Law – Ch 9: TEACH THE DOC

Sections 9.2 & 9.3 OpenStax

“Molecular Orbital Theory Study Guide” by Montgomery College is licensed under CC BY 4.0

Name: ___________________________________CHEM 1806: College Chemistry-I (Teach the Doc)

Ideal Gas Law (section 9.2) – 10 Points

Watch the video tutorial on the Ideal Gas Law

4. An ideal gas sample occupies 1.50 L at 25.0oC and 741 mm Hg. How many moles are

there in the sample? (ans: 0.0598 mol)

5. What volume would the gas sample occupy at STP? (ans: 1.34 L)

Density of a Gas (section 9.3) – 10 Points

Read the derivation in Examples 9.11.

6. What is the equation that relates the density of a gas to its molar mass?

7. What is the molar mass of the gas if its density at STP is 0.901 g/L? (ans: 20.18 g/mol)

“Molecular Orbital Theory Study Guide” by Montgomery College is licensed under CC BY 4.0

Name: ___________________________________CHEM 1806: College Chemistry-I (Teach the Doc)

Partial Pressure – Ch 9: TEACH THE DOC

Section 9.3 in OpenStax

Partial Pressures – 20 Points

Watch the video tutorial about Partial Pressure

8. Which component (H2, O2 or CO2) exerts the highest partial pressure in a gaseous

mixture made of 0.50 mol H2, 0.49 mol O2, and 0.48 mol CO2? (ans: H2)

9. Which of the following mixtures exerts the highest total pressure in 5.0-L container?

Explain.

(a) 0.5 mol H2 + 0.6 mol N2 + 0.4 mol Cl2

(b) 0.7 mol H2 + 0.6 mol O2 + 0.1 mol N2

(c) 0.6 mol H2 + 0.3 mol Ar + 0.7 mol He

(ans: c)

10. A 25.0-L cylinder contains 3.00 g of hydrogen gas and 14.0 g of nitrogen gas at 25.0oC.

Explain.

a. Calculate the mole fraction of each gas: (ans: XH2 = 0.75 and XN2 =0.25)

b. Calculate the partial pressure of each gas: (ans: PH2= 1.47 atm and PN2= 0.489 atm)

“Molecular Orbital Theory Study Guide” by Montgomery College is licensed under CC BY 4.0

CHEMISTRY

Chapter 3

Composition of Substances

and Solutions

Dr. Praveen Vadapally, Barton Community College – Adapted from Patricia Takahara, Montgomery College – “OpenStax Chemistry

Enhanced Lecture Slides” – CC BY 4.0 BY 4.0. Takahara, Patricia. “OpenStax Chemistry Enhanced Lecture Slides (Chapters 1 to 11)”.

OER Commons. Institute for the Study of Knowledge Management in Education, 06 Apr. 2018. Web. 07 Jan. 2020. Slides by Montgomery

College licensed under CC BY 4.0.

CHAPTER 3: SUBSTANCES AND SOLUTIONS

3.1

Formula Mass and the Mole Concept

3.2

Determining Empirical and Molecular Formula

3.3

Molarity

3.4

Other Units for Solution Concentration

2

3.1 FORMULA MASS AND THE MOLE CONCEPT

Formula Mass

Covalent Compounds

• Calculate the sum of the average atomic masses of the atoms in

the formula.

• The formula mass for a covalent substance is referred to as the

molecular mass.

• The molecular formula for chloroform is CHCl3. The molecular

mass is the sum of one carbon atom, one hydrogen atom, and

three chlorine atoms.

3

3.1 FORMULA MASS AND THE MOLE CONCEPT

Aspirin

• An aspirin molecule is a covalent compound and has the

formula C9H8O4.

• The molecular mass is the sum of 9 carbon atoms, 8 hydrogen

atoms and 4 oxygen atoms.

• The molecular mass for aspirin is 180.15 amu.

4

3.1 FORMULA MASS AND THE MOLE CONCEPT

Formula Mass

Ionic Compounds

• are composed of discrete cations and anions combined in

ratios that yield an electrically neutral bulk substance.

• do not exist as discrete molecules.

• have a formula mass but not a molecular mass.

The atomic mass of an ion are approximately the same as the atomic mass of the neutral atom.

5

3.1 FORMULA MASS AND THE MOLE CONCEPT

The Mole

• is a counting/amount unit similar to a pair or a dozen.

• is a specific measure of the number of atoms or molecules in a

sample of matter.

• is defined by a sample of pure 12C weighing exactly 12 g.

• 1 mole contains 6.022 x 1023 of anything.

• 6.022 x 1023 is known as Avogadro’s Number (NA).

A mole is a fixed number of particles in a

chemical unit. One mole always contains

the same number of particles, no matter

what the substance.

6

3.1 FORMULA MASS AND THE MOLE CONCEPT

The Mole

• Each sample contains one mole of atoms.

• Why are the masses of the samples different?

7

3.1 FORMULA MASS AND THE MOLE CONCEPT

The Mole

• Molar mass is the mass (grams) of one mole of a substance.

• The units of molar mass are grams per mole (g/mol).

• The molar mass of a substance is numerically equivalent to its

atomic, molecular, or formula mass in amu.

•

•

Aspirin (C9H8O4) has a molecular mass of 180.15 amu and a

molar mass of 180.15 g/mol.

If one mole of aspirin is placed on a balance in the lab, the

balance would read 180.15 g.

One mole of aspirin.

8

3.1 FORMULA MASS AND THE MOLE CONCEPT

The Mole

• Each sample contains 6.022 x 1023 formula units of

a compound or element.

• Calculate the mass of each sample.

octanol

C8H17OH

pure sulfur

S8

mercury(II) iodide

HgI2

methanol

CH3OH

9

3.1 FORMULA MASS AND THE MOLE CONCEPT

The Mole

• The number of water molecules in the droplet of water shown

below is approximately 100 billion times greater than the

population of earth.

• There are roughly 7,442,000,000 people on earth.

• How many moles of water are

contained in the droplet

described above?

• How many grams of water are

in the droplet?

Be sure to express your answers

in scientific notation and to the

correct number of sig figs!

10

3.1 FORMULA MASS AND THE MOLE CONCEPT

Calculations: Formula Mass, Moles, and Grams

• The relationships between formula mass, moles, grams, and

number of molecules or atoms are used to calculate quantities

that describe the composition and amount of chemical

substances.

• Avogadro’s Number (NA)

• 1 mole = 6.022 x 1023 atoms or molecules

• Chemical Formula

• Use the periodic table to calculate formula mass.

• Formula mass is the number of grams per mole (g/mol).

11

3.1 FORMULA MASS AND THE MOLE CONCEPT

Example 3.3 – Grams to Moles

According to nutritional guidelines from the US Department of

Agriculture, the estimated average requirement for dietary

potassium is 4.7 g. What is the estimated average requirement of

potassium in moles?

Know:

4.7 g of potassium (K)

K is 39.10 amu on the periodic table.

g

K: 39.10 mol

Want:

moles of potassium (K)

Convert grams à moles.

1 𝑚𝑜𝑙

4.7 𝑔 𝐾 ×

= 0.12 𝑚𝑜𝑙 𝐾

39.10 𝑔

Do not want grams – put grams in the denominator.

Want moles – put in the numerator.

Grams will cancel, and moles will remain.

4.7 g has two significant

figures, so the answer must

have two significant figures.12

3.1 FORMULA MASS AND THE MOLE CONCEPT

Example 3.4 – Moles to Grams

One liter of air contains 9.2 x 10-4 mol argon. What is the mass of

argon in one liter of air?

Know:

1 L of argon (Ar) = 9.2 x 10-4 mol

Ar is 39.95 amu on the periodic table.

g

Ar: 39.95 mol

4.2 ×

10!”

Want:

grams of Ar

Convert moles à grams.

39.95 𝑔

𝑚𝑜𝑙 𝐴𝑟 ×

= 0.037 𝑔 𝐾

1 𝑚𝑜𝑙

Do not want moles – put moles in the denominator.

Want moles – put in the numerator.

Moles will cancel, and grams will remain.

9.2 x 10-4 mol has two significant

figures, so the answer must

have two significant figures.

13

3.1 FORMULA MASS AND THE MOLE CONCEPT

Example 3.5 – Grams to Moles to Atoms

Copper is commonly used to fabricate electrical wire. How many

copper atoms are in 5.00 g of copper wire?

Know:

5.00 g Cu wire

g

Cu: 63.55 mol

Want:

atoms in 5.00 g Cu

Convert grams à moles à atoms.

When calculating the number of atoms

or molecules, use Avogadro’s Number.

NA = 6.022 x 1023 atoms

mol

1 𝑚𝑜𝑙

6.022 × 10#$ 𝑎𝑡𝑜𝑚𝑠

5.00 𝑔 𝐶𝑢 ×

= 0.07868 𝑚𝑜𝑙 𝐶𝑢 ×

63.55 𝑔

1 𝑚𝑜𝑙

= 4.74 × 10## 𝑎𝑡𝑜𝑚𝑠 𝑜𝑓 𝐶𝑢

14

3.1 FORMULA MASS AND THE MOLE CONCEPT

Example 3.6 – Grams to Moles

Our bodies synthesize proteins from small molecules called

amino acids. One such amino acid is glycine (C2H5O2N). How

many moles are in 28.35 g of glycine?

Know:

28.35 g glycine

g

C2H5O2N: 75.07 mol

Want:

moles in 28.35 g glycine

Convert grams à moles.

Start by finding the molar mass.

1 𝑚𝑜𝑙

28.35 𝑔 𝑔𝑙𝑦𝑐𝑖𝑛𝑒 ×

= 0.3776 𝑚𝑜𝑙 𝑔𝑙𝑦𝑐𝑖𝑛𝑒

75.07 𝑔

15

3.1 FORMULA MASS AND THE MOLE CONCEPT

Example 3.7 – Moles to Grams

Vitamin C, or ascorbic acid, is a compound with the molecular

formula C6H8O6. The recommended daily dietary allowance of

vitamin C for children aged 4-8 years is 1.42 x 10-4 mol. What is

the daily allowance in grams?

Know:

1.42 x 10-4 mol vitamin C

g

C6H8O6: 176.124 mol

vitamin C

Want:

grams of vitamin C

Convert moles à grams.

1.42 ×

10!”

176.124 𝑔

𝑚𝑜𝑙 𝑣𝑖𝑡𝑎𝑚𝑖𝑛 𝐶 ×

1 𝑚𝑜𝑙

= 0.0250 𝑔 𝑣𝑖𝑡𝑎𝑚𝑖𝑛 𝐶

16

3.1 FORMULA MASS AND THE MOLE CONCEPT

Example 3.8 – Grams to Moles to Molecules

A packet of artificial sweetener contains 40.0 mg of saccharin

(C7H5NO3S).

a. What is the molar mass of saccharin?

b. How many moles of saccharin are present

in the sample?

c. How many molecules of saccharin are

present in the sample?

d. How many moles of carbon atoms are

present in the sample?

e. How many carbon atoms are present in

the sample?

17

3.2 EMPIRICAL AND MOLECULAR FORMULAS

Percent Composition

• Chemical formulas are used to represent the elemental makeup

of a compound.

• The percent composition for a compound is the percent by mass

of each element in the compound.

• What is the formula mass for ammonia (NH3)?

•

•

•

One nitrogen atom: 14.01 amu

Three hydrogen atoms: 3 x 1.008 amu = 3.024 amu

The formula mass for NH3 is therefore 17.03 amu

What percentage of an ammonia

sample is nitrogen?

%𝑁 =

14.01 𝑎𝑚𝑢

× 100 = 82.27%

17.03 𝑎𝑚𝑢

What percentage of an ammonia

sample is hydrogen?

%𝐻 =

(3 × 1.008) 𝑎𝑚𝑢

× 100 = 17.76%

17.03 𝑎𝑚𝑢

18

3.2 EMPIRICAL AND MOLECULAR FORMULAS

Percent Composition

• In the previous section, the molecular mass of aspirin was

calculated. What is the percent composition of aspirin?

Carbon?

Hydrogen?

Oxygen?

19

Example – Composition of Amethyst

Amethyst is a purple form of quartz, SiO2. Amethyst gets its

purple color from tiny amounts of iron and manganese in the

quartz crystals.

a. What is the percent (by mass) of silicon in amethyst?

b. What is the percent (by mass) of oxygen in amethyst?

c. If you wanted to obtain 250.0 grams of silicon, how

much amethyst would you need?

20

3.2 EMPIRICAL AND MOLECULAR FORMULAS

Empirical Formulas

• The empirical formula is the chemical formula with the smallest

whole number ratio of atoms.

• The molecular formula is the formula that represents the entire

• The empirical formula is not necessarily the molecular formula!

• Percent composition can be used to determine the empirical

formula of a substance.

Ethane has the structural formula:

The molecular formula is C2H6.

The simplified formula is the

empirical formula, CH3.

21

3.2 EMPIRICAL AND MOLECULAR FORMULAS

Empirical Formulas

Benzene is made up of six carbon atoms and six hydrogen atoms.

a. What is the molecular formula for benzene?

b. What is the empirical formula for benzene?

c. Determine the percent composition of benzene from its

molecular formula.

d. Determine the percent composition of benzene from its

empirical formula.

22

3.2 EMPIRICAL AND MOLECULAR FORMULAS

Example 3.11 – Grams to Empirical Formula

A sample of the black mineral hematite, an oxide of iron found in

many iron ores, contains 34.97 g of iron and 15.03 g of oxygen.

What is the empirical formula for hematite?

Solution on the next slide.23

3.2 EMPIRICAL AND MOLECULAR FORMULAS

Example 3.11 – Empirical Formula

A sample of the black mineral hematite, an oxide of iron found in

many iron ores, contains 34.97 g of iron and 15.03 g of oxygen.

What is the empirical formula for hematite?

Know:

34.97 g Fe

15.03 g O

g

Fe: 55.85 mol

g

O: 15.9994 mol

𝑚𝑜𝑙

34.97 𝑔 𝐹𝑒 ×

= 0.6261 𝑚𝑜𝑙 𝐹𝑒

55.85 𝑔

15.03 𝑔 𝑂 ×

𝑚𝑜𝑙

= 0.9395 𝑚𝑜𝑙 𝑂

15.9994 𝑔

Want:

empirical formula

Convert grams à moles.

Divide the smallest # of moles.

0.6261

= 1.000 𝑚𝑜𝑙 × 2 = 𝟐

0.6261

Fe2O3

0.9395

= 1.501 𝑚𝑜𝑙 𝑂 × 2 = 𝟑

0.6261

24

3.2 EMPIRICAL AND MOLECULAR FORMULAS

Empirical Formula Summary

1. Determine the number of moles of each element.

2. Divide the number of moles by the smallest number.

3. If necessary, multiply by an integer value so that all the values

end up as whole numbers.

25

3.2 EMPIRICAL AND MOLECULAR FORMULAS

Empirical Formula: Process

1. Determine the number of moles of each element.

2. For each element, divide the number of moles by the smallest

number of moles.

3. If all the resultant values are whole numbers, those are the

subscripts in the empirical formula.

•

•

If necessary, multiply by an integer value so that all the values

end up as whole numbers.

In this course, subscripts in a chemical formula must be

whole numbers.

Round numbers that are very

close to whole numbers.

2.04 à 2

3.98 à 4

2.95 à 3

Other numbers must be multiplied by integer values

to get a whole number.

2.50 x ____?

3.25 x ____?

3.80 x ____?

1.33 x ____?

1.20 x ____?

2.75 x ____?

26

3.2 EMPIRICAL AND MOLECULAR FORMULAS

Example 3.12 – Percent Composition to Empirical Formula

The bacterial fermentation of grain to produce ethanol forms a

gas with a percent composition of 27.29% C and 72.71% oxygen.

What is the empirical formula for this gas?

1. Assume 100 g of substance.

2. Convert grams to moles.

3. For each element, divide the number of moles by the smallest

number of moles.

4. Complete as in previous example.

Solution on the next slide.27

3.2 EMPIRICAL AND MOLECULAR FORMULAS

Example 3.12 – Percent Composition to Empirical Formula

The bacterial fermentation of grain to produce ethanol forms a

gas with a percent composition of 27.29% C and 72.71% oxygen.

What is the empirical formula for this gas?

Know:

Assume 100 grams.

C: 27.29% à 27.29 g

O: 72.71% à 72.71 g

g

C: 12.011 mol

g

O: 15.9994 mol

𝑚𝑜𝑙

27.29 𝑔 𝐶 ×

= _______ 𝑚𝑜𝑙 𝐶

12.011 𝑔

Want:

empirical formula

Convert grams à moles.

Divide the smallest # of moles.

à?

𝑚𝑜𝑙

72.71 𝑔 𝑂 ×

= _______ 𝑚𝑜𝑙 𝑂 à ?

15.9994 𝑔

28

3.2 EMPIRICAL AND MOLECULAR FORMULAS

Empirical Formula à Molecular Formula

Benzene has a molecular mass of 78.11 g/mol. The empirical

formula for benzene is CH. What is the molecular formula?

Does the molecular formula make sense given the molecular

structure of benzene?

29

3.2 EMPIRICAL AND MOLECULAR FORMULAS

Molecular Formula from Percent Composition

Example 3.13

% Composition à Molecular Formula

Nicotine is a poisonous compound found in tobacco leaves. It is

74.02% carbon, 8.710% hydrogen, and 17.27% nitrogen. Its

molar mass is 162.3 g/mol. Determine both the empirical formula

and the molecular formula for nicotine.

30

Example – Eugenol

Eugenol is the major component in clove oil, but it can also be

extracted from nutmeg, cinnamon, and basil. It has a molar mass

of 164.2 g/mol and is 73.14% carbon and 7.37% hydrogen. The

remainder is oxygen.

a. What is the molecular formula?

b. What is the empirical formula?

31

Example – Composition of Striker Flints

The strikers used to light Bunsen burners are made from a mix

of iron, magnesium, and several rare earth metals. The

substance in strikers is sometimes referred to as ferrocium or

fire-steel. If ferrocium is 42.55% cerium, 23.35% lanthanum,

21.35% iron, 4.25% neodymium, 4.25% praseodymium, and

4.25% magnesium, what is the empirical formula?

32

Example – Chemistry and Art

In art of the Ancient Near East (Assyrian Art), sculptures often had

carved eyes that were accented with shells set in bitumen, a black

tar-like substance. Chemical analysis of bitumen by using a

combustion reaction yields 6.575 g CO2 and 1.346 g H2O.

a. What is the empirical formula for bitumen (CxHy)?

b. The molecular weight of CxHy is 78.1134. What is the

molecular formula?

33

3.2 EMPIRICAL AND MOLECULAR FORMULAS

Water of Hydration

Hydrates

• Solid chemical compounds exposed to the atmosphere often

have water adsorbed to the surface of crystals.

• The number of water molecules attached to a salt is written

after a dot at the end of the formula.

• Chemists can drive off the water by gentle heating.

• In a method similar to that used to determine empirical

formulas, moles of water lost can be used to determine

hydrate formulas.

• Example hydrates:

CoCl2 . 6H2O

SnCl2 . 2H2O

KAl(SO4)2 . 12H2O

34

3.2 EMPIRICAL AND MOLECULAR FORMULAS

Water of Hydration

Example

A 2.5640 g sample of a magnesium sulfate hydrate was found to

contain 1.3125 g of water. Determine the complete formula for the

hydrate.

MgSO4 . x H2O

35

Example – Water of Hydration

A student was given a sample of calcium chloride hydrate.

Determine the number of water molecules in the hydrate given

the following data:

Mass Crucible:

13.1381 g

Mass Crucible + Sample

before heating 14.8205 g

Mass Crucible + Sample

after heating

13.9897 g

36

3.3 MOLARITY

Solutions

• A solution is a homogeneous mixture of a solute dissolved in a

solvent.

• The solute is typically the substance present in a relatively small

amount.

• The solvent is the substance that is used to dissolve the solute.

solute

solvent

37

3.3 MOLARITY

Solutions

• The concentration of a solution can be expressed as a molarity.

• Molarity (M) is the amount of solute (mol) dissolved in some

volume (L) of solution.

• Molarity can be used as a conversion factor in calculations.

𝑚𝑜𝑙 𝑠𝑜𝑙𝑢𝑡𝑒

𝑀=

𝑙𝑖𝑡𝑒𝑟𝑠 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

What is the molarity of the solution?

solvent

solute

38

3.3 MOLARITY

Example 3.17 – Molarity to Mass

How many grams of sodium chloride are contained in 0.250 L of a

5.30 M aqueous solution?

Square brackets can be used to denote concentration. [NaCl] = 5.30 M.

Know:

NaCl: 58.44

g

mol

0.250 L of a 5.30

volume

[NaCl] = 5.30 M

mol

L

molarity

Want:

grams of NaCl

solution

à

# moles

5.30 𝑚𝑜𝑙

0.250 𝐿 ×

= 1.325 𝑚𝑜𝑙 𝑁𝑎𝐶𝑙

1𝐿

58.44 𝑔

1.325 𝑚𝑜𝑙 𝑁𝑎𝐶𝑙 ×

= 77.4 𝑔 𝑁𝑎𝐶𝑙

𝑚𝑜𝑙

5.30 mol

L

250 mL

39

3.3 MOLARITY

Example 3.18 – Molarity to Volume

The concentration of a vinegar sample is 0.839 M. What volume

of vinegar contains 75.6 g of acetic acid (CH3COOH)?

Know:

75.6 g CH3COOH

Want:

liters of vinegar

g

CH3COOH: 60.05 mol

[CH3COOH] = 0.839 M = 0.839 mol

L

75. 6 𝑔 𝐶𝐻! 𝐶𝑂𝑂𝐻 ×

𝑚𝑜𝑙

= 1.259 𝑚𝑜𝑙 𝐶𝐻! 𝐶𝑂𝑂𝐻

60.05 𝑔

1.259 𝑚𝑜𝑙 𝐶𝐻! 𝐶𝑂𝑂𝐻 ×

𝐿

= 1.50 𝐿 𝑣𝑖𝑛𝑒𝑔𝑎𝑟

0.839 𝑚𝑜𝑙

40

Example – Molarity of a Solution

A solution was made by dissolving 10.00 g of potassium sulfate

in 250.0 mL of water.

a. What is the concentration (molarity) of the solution?

b. What is the concentration of potassium ions?

c. What is the concentration of sulfate ions?

41

3.3 MOLARITY

Dilution

• A concentrated stock solution of known concentration can be

used to make a dilute solution of a desired concentration.

42

3.3 MOLARITY

Dilutions

• In this course, use dimensional analysis and the number of

moles to perform dilution calculations.

M1V1 = M2V2

Note: The formula M1V1 = M2V2 works for dilution but not titration. To avoid

confusion, please use dimensional analysis for all calculations in this course.

43

3.3 MOLARITY

Example: Dilutions

A stock solution of 2.00 M NaOH is available. An experiment requires

250.0 mL of a 1.50 M solution. Describe how to prepare such a solution.

2.00 M NaOH

sodium

hydroxide

250 mL

44

3.3 MOLARITY

Example: Dilutions

If 0.850 L of a 5.00 M copper(II) nitrate solution is diluted to a volume of

1.80 L by the addition of water, what is the molarity of the diluted solution?

5.00 M

Cu(NO3)2

1.80 L

copper(II)

nitrate

45

3.3 MOLARITY

Example: Serial Dilutions

A sample of blue dye with a concentration of 0.36 M is diluted twice. In the

first step, 10.0 mL of the 0.36 M solution is diluted to 100.00 mL to make

solution #2. In the next step, 5.00 mL of solution #2 is diluted to 100.00 mL.

What is the concentration of dye in the resulting solution?

46

3.4 CONCENTRATION

Other Concentration Units

Mass Percentage

• is the ratio of a component’s mass to the solution’s mass

expressed as a percentage.

• is also known as %mass, %weight, and (w/w)%.

• is used for many consumer products.

• The active component of bleach is sodium hypochlorite,

NaOCl.

• Bleach has an NaOCl concentration of 7.4 (w/w)%.

• 100 g of bleach therefore contains 7.4 g of NaOCl.

47

3.4 CONCENTRATION

Other Concentration Units

Example 3.22 – Spinal Fluid

A 5.0 g sample of spinal fluid contains 3.75 mg of glucose. What is

the percent by mass of glucose in spinal fluid?

scripps.org

48

3.4 CONCENTRATION

Other Concentration Units

Volume Percentage

• is used to represent the concentration of a solution formed by

dissolving a liquid solute in a solvent.

• %vol or (v/v)%

• is also used in many consumer products.

Example 3.24 – Volume/Volume Percentage

Rubbing alcohol, also called isopropyl alcohol, is

usually sold as a 70 (v/v)% solution. If this density of

an isopropyl alcohol solution is 0.785 g/mL, how

many grams of the alcohol are present in a 355 mL

bottle?

49

3.4 CONCENTRATION

Other Concentration Units

Mass-Volume Percentage

• is a ratio of solute mass to solution volume expressed as a

percentage: (m/v)%

• is commonly used in medical settings.

Example – Mass/Volume Percentage

Physiological saline solution used for

intravenous fluids is typically 0.9 (m/v)%. If

an IV bag contains 100 mL solution, what

mass of solute (NaCl) does it contain?

50

3.4 CONCENTRATION

Other Concentration Units

Parts per Million and Parts per Billion

• are units used to express concentrations of extremely dilute

solutions.

• are often used to refer to trace contaminant levels in water, air,

or other solvents.

𝑚𝑎𝑠𝑠 𝑠𝑜𝑙𝑢𝑡𝑒

𝒑𝒑𝒎 =

× 10%

𝑚𝑎𝑠𝑠 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

𝑚𝑎𝑠𝑠 𝑠𝑜𝑙𝑢𝑡𝑒

𝒑𝒑𝒃 =

× 10&

𝑚𝑎𝑠𝑠 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

51

3.4 CONCENTRATION

Flint Water Crisis 2017

Example – Lead Levels in Water

The graphic on the next slide shows some

data from a study conducted by Virginia

Tech scientists.

a. In one sample, the lead concentration

was 158 ppb. How many micrograms of

lead would be present in a glass of water

that contains 355 mL?

b. In another sample, the lead concentration

was 13,000 ppb. How many micrograms

of lead would be present in a glass of

water that contains 355 mL?

cbsnews.com

52

michiganlivemedia.org

53

Challenge Problem

Two unknown elements A and Z combine to produce two

different compounds: A2Z3 and AZ2. If 0.15 mol A2Z3 has a

mass of 15.90 g and 0.15 mol AZ2 has a mass of 9.30 g, what

are the atomic masses of A and Z?

54

END OF CHAPTER PROBLEMS – CHAPTER 3

Atomic Mass and Molar Mass: #1, 5, 11, 15, 17, 19, 23, 29, 31

Chemical Formulas and

The Mole Concept: #25, 33, 35, 37, 39, 41, 43

Molarity: #47, 49, 51, 53, 59, 61

For detailed solutions to these problems, go to the OpenStax

Chemistry website and download the Student Solution Guide.

VIDEOS – CHAPTER 3

Gram to Mole to Atoms Conversions

http://screencast.com/t/ofLlIvVxC

Chemical Formula and Mole Conversion

http://screencast.com/t/V5nTAsvDJd

Calculating Empirical and Molecular Formulas

(not an MC video – posted on YouTube by Step-by-Step Science)

Calculating Molarity

http://screencast.com/t/w4R7NcDjfa4

Dilution Calculations

http://screencast.com/t/6MphEryfyQlG

*All videos were created by MC Chemistry faculty unless otherwise indicated.

SIMULATIONS – CHAPTER 3

Sugar and Salt Solutions (Solutions, Ionic, Covalent)

https://phet.colorado.edu/en/simulation/legacy/sugar-and-salt-solutions

Molarity

https://phet.colorado.edu/en/simulation/molarity

Dr. Praveen Vadapally, Barton Community College – Adapted from Patricia Takahara, Montgomery College – “OpenStax Chemistry

Enhanced Lecture Slides” – CC BY 4.0 BY 4.0. Takahara, Patricia. “OpenStax Chemistry Enhanced Lecture Slides (Chapters 1 to 11)”.

OER Commons. Institute for the Study of Knowledge Management in Education, 06 Apr. 2018. Web. 07 Jan. 2020. Slides by Montgomery

College licensed under CC BY 4.0.

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