# CHEM 1806 Barton Community College Atomic Mass and Molar Mass Chemistry Questions

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)
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)
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
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)
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
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)
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)
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
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?
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.
• 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
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
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
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
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)
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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