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Mole / Stoichiometry Calculations

Slide 2 / 157 Table of Contents

· Avagadro's Number · Molar Mass · Emperical Formula · Molar Volume · Percent Composition

Click on the topic to go to that section

Slide 3 / 157

Return to Table of Contents

Avogadro's Number

Slide 4 / 157 Moles

Recall an atom's atomic mass is equal to the number of protons plus the number of neutrons in the atom.

C 6 12.01

Atomic Number

• r number of

protons (Z) Atomic Mass in amu

The unit for atomic mass is amu. Carbon-12 has 6 protons and 6 neutrons. One amu is equal to 1/12 the mass of carbon-12 or approximately the mass of one proton or neutron mass of 1 proton = 1 amu

Slide 5 / 157 Moles

What if you wanted to measure the mass of one atom in the laboratory? Would it be possible? A single atom has a very small mass. One carbon atom has a mass of about 2.0 x 10-23 grams.

Slide 6 / 157

The Mole

It takes a lot of atoms to give us enough material to directly measure in a lab. Hydrogen has a mass of 1 amu. How many atoms of hydrogen would be needed to make a 1 gram sample of hydrogen?

602,200,000,000,000,000,000,000

The amount 6.02x10

23 is called Avogadro's number or a

mole. How big is a mole?

Slide 7 / 157 Calculating Avogadro's Number

The atomic mass of one carbon atom is 12.01 amu or 2x10-23 g. How many carbon atoms would it take to get 12.01 grams of carbon? Givens: mass of 1 carbon atom = 2x10-23 g total mass of carbon atoms = 12.01 g 2x10-23 g x ? of atoms = 12.01 g ? of atoms 12.01 g = 2x10-23 g ? of atoms 6.02 x 1023 atoms =

Slide 8 / 157 Holy Mole-y!

If you were able to count at the rate

• f 1 million numbers a second, it

would take about 20 billion years to count out one mole!

1 mole of pennies could be distributed to all the currently-living people of the world so that they could spend a million dollars per hour every hour (day and night) for the rest of their lives!

One mole of marbles would cover the entire Earth (oceans included) for a depth of two miles!

Slide 9 / 157

The Mole

A mole is just a grouping of numbers...like dozen, ream, etc. A dozen means 12 of something. A mole means 6.02 x 10

23 of something.

Common Grouping Quantities 1 dozen = 12 1 gross =144 1 ream = 500 1 mole = 6.02 x 1023

Slide 10 / 157

1 How many eggs are in two dozen eggs? A 12 B 24 C 0.0833 D 2 E 6.02 x 10

23

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2 How many eggs are in half a dozen eggs? A 12 B 24 C 6 D 0.5 E 6.02 x 1023

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3 How many particles of sand are in 0.5 moles of sand? A 1.5 x 1023 B 3.01 x 1023 C 6.02 x 1023 D 1.2 x 1024 E 6.02 x 1024

1 mole = 6.02 x 1023

Slide 13 / 157

4 How many pieces of gold dust are in 2 moles of gold dust? A 1.5 x 1023 B 3.01 x 1023 C 6.02 x 1023 D 1.2 x 1024 E 6.02 x 1024

1 mole = 6.02 x 1023

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5 How many dozen eggs are in a container of 6 eggs? A 4 B 0.5 C 1 D 2 E 6.02 x 1023

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6 How many dozen eggs are in a container of 18 eggs? A 1.5 B 3 C 0.67 D 2 E 6.02 x 1023

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7 The Milky Way Galaxy may have up to 400 billion (4 x 1014 stars). How many moles of stars does it have? A 1.5 x 10

9

B 6.7 x 10

• 10

C 10 D 0.5 E 6.02 x 1023

1 mole = 6.02 x 1023

Slide 17 / 157

8 Mathematicians estimate Earth's beaches contain nearly 5.6 x 1021 grains of sand. How many moles of sand are on Earth's beaches? A 1.5 B 9.3 x 10

• 3

C 10 D 107.5 E 6.02 x 1023

1 mole = 6.02 x 1023

Slide 18 / 157

Measuring Matter with Moles

The mole is the SI unit for measuring the amount of particles in a chemical substance. 1 mole of Carbon

Slide 19 / 157

One mole (mol) of a substance is 6.02 x 10

23 representative

particles of that substance.

Measuring Matter with Moles

n = NA N Where: n is the number of moles N is the total number of particles NA is Avogadro's Number 6.02 x 10

23

Slide 20 / 157

NA = 6.02 x1023 atoms How many moles of Gold are there in a sample containing 3.01 x 1023 atoms of Gold?

Measuring Matter with Moles Slide 21 / 157

9 How many atoms of titanium are in a sample containing 0.5 mole of titanium? A 1.5 x 1023 B 3.01 x 1023 C 6.02 x 1023 D 1.2 x 1024 n =

NA N NA = 6.02 x 10

23

E 6.02 x 1024

Slide 22 / 157

10 How many atoms of sodium are in a sample containing 2.0 moles of sodium? A 1.5 x 1023 B 3.01 x 1023 C 6.02 x 1023 D 1.2 x 1024 n =

NA N NA = 6.02 x 10

23

E 6.02 x 1022

Slide 23 / 157

11 How many moles of potassium are in a sample

containing 3.01 x 1023 atoms of potassium? A 1.0 B 2.0 C 0.5 D 0.75 n =

NA N NA = 6.02 x 10

23

E 6.02 x 1024

Slide 24 / 157

12 How many moles of potassium are in a sample containing 1.2 x 1024 atoms of potassium? B 0.50 mol C 1.0 mol D 2.0 mol E 3.0 mol A 0.25 mol n =

NA N NA = 6.02 x 10

23

Slide 25 / 157

13 How many moles of tungsten atoms are there in a sample containing 1.8 x 1024 atoms of tungsten? A 0.33 mol B 0.50 mol C 1.0 mol D 1.5 mol E 3.0 mol n =

NA N NA = 6.02 x 10

23

Slide 26 / 157

14 How many moles of silver are there in a pure sample containing 1.5 x 1023 atoms of silver? A 0.10 mol B 0.25 mol C 0.50 mol D 1.0 mol n =

NA N NA = 6.02 x 10

23

E 1.5 mol

Slide 27 / 157

15 How many atoms are there in 5.00 mol of hafnium? A 6.02 x10

23 atoms

B 1.20 x 10

23 atoms

C 1.20 x 10

22 atoms

D 3.43 x 10

23 atoms

E 3.01 x 1024 atoms n =

NA N NA = 6.02 x 10

23

Slide 28 / 157

A mole of ANY substance contains Avogadro’s number of representative particles, or 6.02 x 1023 representative particles. 1 mole of C atoms = 6.02 X 1023 atoms of C 1 mole of pick-up trucks = 6.02 x 1023 pick-up trucks The term representative particle refers to the species or types of particles in the substance For Example: atoms, molecules, formula units, ions

Measuring Matter with Moles

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In 1 mole of water there are 6.02 x 10

23 water molecules.

In 1 mole of NaCl there are 6.02 x 1023 formula units. In 1 mole of carbon there are 6.02 x 10

23 carbon atoms.

molecule of H2O 18.0 amu Avagadro's number of molecules (6.02 x1023) 1 mol H2O (18.0 g) laboratory sample size

H O H

Measuring Matter with Moles

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16 Formula units refer to particles of __________ compounds and molecules refer to particles of __________ compounds. A molecular/covalent, ionic B ionic, molecular/covalent C atoms, molecular D atoms, ionic E ionic, atomic

Slide 31 / 157

17 How many molecules are there in 2.10 mol CO2? A 3.79 x 1024 B 3.49 x 10-24 C 1.05 x 10-23 D 2.53 x 1024 E 1.26 x 1024 n =

NA N NA = 6.02 x 10

23

Slide 32 / 157

18 How many moles of helium atoms are there in a pure sample containing 6.02 x 1024 atoms of helium? A 2.0 mol B 4.0 mol C 6.0 mol D 10.0 mol E 2.4 x 1024 mol n =

NA N NA = 6.02 x 10

23

Slide 33 / 157

19 How many moles of NaCl are there in a pure sample containing 6.02 x 1023 formula units of sodium chloride, NaCl? A 1.0 mol B 2.0 mol C 4.0 mol D 6.0 mol E 6.02 x 1023 mol n =

NA N NA = 6.02 x 10

23

Slide 34 / 157

20 How many formula units of Pb(NO

3)2 are there in

0.5 mole of Pb(NO3)2? A 0.5 formula units B 2.0 formula units C 3.01 x 1023 formula units D 1.2 x 1024 formula units E 6.02 x 1023 formula units n =

NA N NA = 6.02 x 10

23

Slide 35 / 157

Calcium deficiency can cause osteoporosis (weakening of the bones). The minimum amount of calcium in 1 mL of blood should be around 1.3 x 1018 atoms. A patient has her blood tested and the lab finds there are 3 x 10-5 moles of calcium in the blood. Is this patient at risk for

• steoporosis?

(6.02 x 1023 atoms/mole) x (3 x 10-5 moles) = 1.8 x 1023 atoms Ca This exceeds the normal range so they are OK!

Real World Application

slide for answer

Slide 36 / 157

Review: Ionic and Molecular Compounds

Ionic Compounds Molecular Compounds

H2O

2 hydrogen atoms 1 oxygen atom

C6H6

6 carbon atoms 6 hydrogen atoms

NaCl

1 Na+ ion 1 Cl- ion

K2CrO4

2 K+ ions 1 CrO42- ion

Sn(OH)2

1 Sn 2+ ion 2 OH- ions The total number of atoms or ions in a compound depends on electronegativity and bonding.

Slide 37 / 157 Review: Ionic and Molecular Compounds

Chemicals are composed of more than one molecule or formula unit. To indicate more than one molecule or formulat unit, add a coefficient in front of the compound. Example: six molecules of carbon dioxide = 6CO2. 6 atoms of carbon and 12 atoms of oxygen Move to reveal answer How many atoms of carbon and oxygen are in 6CO2?

Slide 38 / 157 Review: Ionic and Molecular Compounds

Ionic Compounds - Fill in Molecular Compounds - Fill in

H2O

__hydrogen atoms __oxygen atom

6 C6H6

__carbon atoms __hydrogen atoms

NaCl

__ Na+ ions __ Cl- ions

3

3 3

K2CrO4

__K+ ions __CrO42- ions

2 Sn(OH)2

__Sn 2+ ions __OH- ions

4 Slide 39 / 157

Converting Moles to Number of Particles

In one mole of water molecules there are:

molecule of H2O 18.0 amu Avagadro's number of molecules (6.02 x10

23 )

1 mol H2O (18.0 g) laboratory sample size

H O H

H2O

2 moles of hydrogen atoms 1 mole of

• xygen atoms

2 x 6.02x1023 atoms

• f hydrogen

1 x 6.02x1023 atoms

• f oxygen

Slide 40 / 157 Converting Moles to Number of Particles

In 3 moles of sodium chloride formula units there are:

3NaCl

3 moles of Na+ ions 3 moles of Cl- ions

Slide 41 / 157

21 How many hydrogen atoms are in six molecules of

ethylene glycol, the major component in antifreeze? The formula for ethylene glycol is: HOCH2CH2OH. A 6 atoms of H B 36 atoms of H C 6 x 6.02 x 10

23 atoms of H

D 36 x 6.02 x 10

23 atoms of H

n = NA N NA = 6.02 x 10

23

E 6.02 x 1023 atoms of H

Slide 42 / 157

22 How many CO32- ions are in one formula unit of

CaCO3? A 1 ion B 3 ions C 6 x 6.02 x 10

23 ions

D 36 x 6.02 x 10

23 ions

n = NA N NA = 6.02 x 10

23

E 6.02 x 1023 ions

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23 How many K+ ions are there in two formula units of potassium hydroxide, 2KOH? A 1 K+ ion B 2 K+ ions C 1 x 6.02 x10

23 ions of K +

D 2 x 6.02 x 10

23 ions of K+

n = NA N NA = 6.02 x 10

23

E 3.12 x 1023 ions of K+

Slide 44 / 157

24 How many sulfide ions (S2- ) are there in 2.0 moles of ammonium sulfide,(NH4)2S? A 2.0 ions B 1.2 x 1024 ions C 2.4 x 1024 ions D 6.02 x 1023 ions

n =

NA N NA = 6.02 x 10

23

E none

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25 How many ammonium ions (NH4 + ) are there in 2.0 moles of ammonium sulfide, (NH4)2S? A 2.0 ions B 1.2 x 1024 ions C 2.4 x 1024 ions D 6.02 x 10 23 ions

n = NA N

NA = 6.02 x 10

23

E 8.0 ions

Slide 46 / 157

Hemoglobin is a protein that carries O2 around your body. The formula for it is approximately C2800H4800N3200O800S8Fe4. If a patient has 2 x 1016 atoms of Fe, how many moles of Hb would be present?

moles of Fe: n = N/Na --> (2 x 1016 atoms)/(6.02 x 1023 atoms/n)= 3.3 x 10-7 moles Fe moles of Hb: There are 1 Hb/4 Fe --> (3.3 x 10-7 moles Fe)/(4 moles of Fe/Hb)

= 8.2 x 10-8 moles Hb

Real World Application

slide for answer

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Return to Table of Contents

Molar Mass

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Mass of Compounds

The total mass of a chemical compound can be calculated by using the masses on the Periodic Table. Example: Calculate the mass of the compound Magnesium Chloride (MgCl2) Mg = 24.305 amu 2Cl = (2)35.453 amu Mass of MgCl2 95.211 amu +

Slide 49 / 157 Masses of Elements/Compounds

Atomic mass Formula mass or Formula weight (FW) Molecular mass or Molecular weight used for elements (like F, V, etc..) only used for ionic compounds (like NaCl, MgO, etc..) only used for molecular compounds (like CO2, H2O, etc..) only amu Units amu amu

Slide 50 / 157

26 What is the formula weight of sodium bromide? A 79.904 amu B 102.894 amu C 205.780 amu D 300.120 amu E 605.102 amu

Slide 51 / 157

27 What is the formula weight of Pb(NO3)2? A 79.90 amu B 102.89 amu C 205.78 amu D 331.34 amu E 605.10 amu

Slide 52 / 157

28 What is the molecular mass of 3H2O2? A 17 amu B 34 amu C 68 amu D 102 amu E 204 amu

Slide 53 / 157 Molar Mass (M)

The mass in grams of one mole of any substance is its molar mass (M). Each of the bars shown below equals

• ne mole of a pure element.

A mole represents the number of atoms it takes to convert from a single atomic mass in amu to the same mass in grams. 1 mole of Aluminum = 26.982 g 1 mole of Copper = 63.546 g

Slide 54 / 157

Molar Mass (M)

One mole of carbon, sulfur and silver are shown.

1 mol of Carbon atoms = 12.0 g 1 mol of sulfur atoms 32.0 g of S 1 mol of silver = 107.9 g of Ag

Slide 55 / 157

Average atomic mass of 1 atom of Kr = 83.8 amu. 1 mole (6.02 x 1023 atoms) of Kr = 83.8 grams. The atomic mass of an element expressed in grams is the mass of one mole or molar mass (M) of the element.

How is the atomic mass of an element related to the molar mass?

1 mole of Kr = 83.8 grams Molar mass of Kr = 83.8 grams or 83.8 g/mol 1 mol

Slide 56 / 157 Molar Mass

Gram atomic mass Gram formula mass or Formula weight (FW) Gram molecular mass or Molecular weight used for elements (like F, V, etc..) only used for ionic compounds (like NaCl, MgO, etc..) only used for molecular compounds (like CO2, H2O, etc..) only grams Units grams grams

The molar mass of an element or compound can be used to convert directly from masses in amu to masses in grams.

Slide 57 / 157

To convert from moles to mass, or vice versa, use the following formula.

Molar Mass

n = M

m

Where: n is the number of moles m is the mass of the sample M is the molar mass of the substance.

Slide 58 / 157

29 How many moles are in a 64-gram sample of

pure sulfur? n = M m

Slide 59 / 157

30 How many moles are in a 72-gram sample of pure

magnesium? n = M m

Slide 60 / 157

31 What is the mass, in grams, of 2 moles of carbon?

n = M m

Slide 61 / 157

32 What is the mass, in grams, of 5 moles of iron?

n = M m

Slide 62 / 157

33 How many grams is 2.0 mol neon atoms?

n = M m

Slide 63 / 157

Diatomic Molecules

(Recall the seven diatomic molecules: HONClBrIF) The molar mass of these molecules will be twice their atomic mass. Examples: M of hydrogen gas: H

2 = (2 x 1) = 2 g/mol

M of bromine liquid: Br

2 = (2 x 79.9) = 159.8 g/mol

M of fluorine molecules: F

2 = (2 x 19) = 38 g/mol

Slide 64 / 157

34 How many grams is 1.0 mole of hydrogen

molecules? n = M m

Slide 65 / 157

35 How many grams is 0.50 mol of oxygen molecules?

n = M m

Slide 66 / 157

The molar mass of a compound is the sum of the molar masses of all the elements in the compound.

Molar Mass of a Compound

To calculate the molar mass of a compound, find the number of grams of each element in one mole of the compound. Then add the masses of the elements in the compound.

32.065 g/mol + 3(16.000) g/mol = 80.650 g/mol

Molar mass of SO3 = 80.650 g/mol

Slide 67 / 157 Molar Mass of a Compound

Note the different molar masses of these two compounds.

18 g of H2O = 1 mole H

2O

180 g of glucose (C

6H12 O6)

= 1 mole glucose

Slide 68 / 157

36 What is the molar mass of K2O, potassium oxide?

[*]

Slide 69 / 157

37 What is the molar mass of CaCO3, calcium carbonate found in eggshells? A 50 amu B 50 grams/mol C 100 amu D 100 grams/mol

Slide 70 / 157

38 What is the molar mass of carbon dioxide? A 28 amu B 28 grams/mol C 44 amu D 44 grams/mol

Slide 71 / 157

39 What is the molar mass of ammonia, NH3? A 17 amu B 17 grams/mol C 45 amu D 45 grams/mol

Slide 72 / 157

Molar mass is the mass of 1 mole of a substance (measured in g/mol). The atomic mass will be the same number as the molar mass (measured in amu). The difference is that the atomic mass refers to only one representative particle and molar mass refers to one mole (6.02 x 1023 ) of representative particles.

Summary of Molar Mass Slide 73 / 157

40 What is the mass of one formula unit of NaCl? A 17 amu B 17 grams/mol C 58.5 amu D 58.5 grams/mol

Slide 74 / 157

41 What is the mass of one molecule of water? A 18 amu B 18 grams/mol C 8 amu D 8 grams/mol

Slide 75 / 157

42 The chemical formula of aspirin is C9H8O4.

What is the mass of 0.200 moles of aspirin? n = M m

Slide 76 / 157

43 How many moles of O are in 2.4 X 1024

molecules of SO3?

Slide 77 / 157

How many grams of iron are in a 68 gram sample of Fe2O3? n = m/M 68 g Fe2O3 = 0.43 mol Fe2O3 x (2 mol of Fe) = 0.86 mol Fe 160 g Fe2O3 then... n = m/M --> m = n*M = (0.86 n Fe)(56 g/n) = 48 grams Fe OR use dimensional analysis M of Fe2O3 = 160 g/mol M of Fe = 56g/mol 69 g Fe2O3 x 1 mol Fe2O3 x 2 mol Fe x 56g Fe = 48 g Fe 160g Fe2O3 1 mol Fe2O3 1 mol Fe

Move to see answer

Slide 78 / 157

44 The molar mass of oxygen (O2) is: A equal to the mass of one mole of oxygen atoms. B 16.0 g/mol C 32.0 g/mol D none of the above n = M m E equal to the mass of one oxygen atom.

Slide 79 / 157

45 There are more moles of CO2 in a 44 g sample of carbon dioxide gas than there are moles of helium atoms in a box containing 2 x 1023 atoms of He.

True False

Slide 80 / 157

Return to Table of Contents

Molar Volume

Slide 81 / 157

The Mole and the Volume of a Gas

The volume of a gas varies with temperature and pressure. But comparisons between gases can be made by designating a standard temperature and pressure (STP).

Slide 82 / 157 STP (Standard Temperature and Pressure)

Standard Temperature (T) is considered 0°C (273 K), the temperature at which water freezes. Standard Pressure (P) is considered 1 atmosphere (101.3 kPa),

• r the pressure of the atmosphere at sea level.

At STP, 1 mole of gas occupies a volume of 22.4 liters (L) Vm = 22.4 L For the time being, we will treat all gases as if they are at STP.

Slide 83 / 157 The Mole and the Volume of a Gas

The volume occupied by one mole of gas is called the molar volume and has the symbol, Vm. It is the same for all gases.

At STP; Vm = 22.4 liters (L).

1 mole of gas = 6.02 x 1023 particles = 22.4 L

Notice that this statement does not depend on the type of gas. It's true of all gases.

Slide 84 / 157

The Mole and the Volume of a Gas At STP; Vm = 22.4 liters (L).

This is true because in a gas the molecules are so far apart that they take up almost no space...the volume of a gas is mostly empty space, regardless of the type of gas. So when it comes to volume (at STP) all gases are created equal. 1 mole of He gas occupies 22.4 L @STP 1 mole of HCl gas occupies 22.4 L @STP

Slide 85 / 157

46 Which of the following must be true about 2 moles of H2 gas vs. 2 moles of CO2 gas at the same temperature and pressure?

A Each sample will have the same density B The same # of atoms will be present in each

sample

C Each sample will have the same mass D Each sample will occupy the same volume E None of these are true

Slide 86 / 157

At STP, 1 mole or, 6.02 x 1023 representative particles, of any gas occupies a volume of 22.4 L. The quantity 22.4 L is called the molar volume of a gas.

The Mole and the Volume of a Gas

n = Vm V

Where: n is the number of moles of gas V is the volume of the gas Vm is 22.4 L, at STP

Slide 87 / 157

47 How many moles are there in 44.8 liters (at STP)

• f fluorine gas?

n = Vm V

Vm = 22.4 L

Slide 88 / 157

48 How many moles of atoms are there in 22.4 liters

(at STP) of Xenon? n = Vm V

Vm = 22.4 L

Slide 89 / 157

49 What is the volume (in liters at STP) of 1.00 mole

• f sulfur dioxide?

n = Vm V

Vm = 22.4 L

Slide 90 / 157

50 What is the volume (in liters at STP) of 2.50 moles

• f carbon monoxide?

n = Vm V

Vm = 22.4 L

Slide 91 / 157

51 What is the volume (in liters at STP) of 4.00

moles of Nitrogen? n = Vm V

Vm = 22.4 L

Slide 92 / 157 Real World Application

An inflated airbag requires 60 L of nitrogen gas (N2) at STP in order to protect the occupant of the vehicle. How many moles of nitrogen gas would need to be created in 40 milliseconds to inflate the airbag?

Slide 93 / 157

Combining the three mole formulas

We now have three formulas for finding the number of moles of a substance. Which one should be used depends on whether you are given (or trying to find): the number of particles the mass of a substance the volume of a gas

n = Vm V

Vm = 22.4 L at STP

n = M m n = NA N NA = 6.02 x 10

23

Slide 94 / 157 Combining the three mole formulas

Sometimes two of these formulas must be used together to solve a problem. Here's an example: How many atoms of carbon are present in a 24 g sample

• f pure carbon?

n = Vm V

Vm = 22.4 L at STP

n = M m n = NA N NA = 6.02 x 10

23

Slide 95 / 157 Combining the three mole formulas

How many atoms of carbon are present in a 24 g sample? Step 1: find the number of moles of C present: Step 2: find the number of atoms in 2.0 moles.

n = m = 24g = M 12g/mol 2.0 mol n = N NA N = n(NA)

rearrange substitute:

2.0mol x 6.02x1023 atoms/mol = 1.2x1024 atoms

Slide 96 / 157

Mole Road Map

n

N

n= N NA

NA=6.03*1023 particles/mole

(Particles: atoms, ions, molecules, etc.)

V

n= V

Vm Vm=22.4 Liters/mole

(Volume)

m

(mass)

n= m

M M: molar mass found using the periodic table

Slide 97 / 157

52 How many molecules are there in 44.8 liters

(at STP) of oxygen gas? n = Vm V

Vm = 22.4 L

Slide 98 / 157

53 How many atoms are there in 11.2 liters (at STP)

• f molecular oxygen ( oxygen gas)?

n = Vm V

Vm = 22.4 L at STP

n = M m n = NA N NA = 6.02 x 10

23

Slide 99 / 157

54 What is the mass of 44.8 liters (at STP) of

molecular oxygen?

n = Vm V

Vm = 22.4 L at STP

n = M m n = NA N NA = 6.02 x 10

23

Slide 100 / 157

55 What is the volume (at STP) of 3.0 x 1023

molecules of fluorine?

n = Vm V

Vm = 22.4 L at STP

n = M m n = NA N NA = 6.02 x 10

23

Slide 101 / 157

56 What is the volume (at STP) of 240 g of

nitrogen gas?

n = Vm V

Vm = 22.4 L at STP

n = M m n = NA N NA = 6.02 x 10

23

Slide 102 / 157

57 How many atoms are present in 30 g of boron ? n = Vm V

Vm = 22.4 L at STP

n = M m n = NA N NA = 6.02 x 10

23

Slide 103 / 157

58 What is the mass of a pure sample of lead

which contains 3.0 x 1024 atoms?

n = Vm V

Vm = 22.4 L at STP

n = M m n = NA N NA = 6.02 x 10

23

Slide 104 / 157

59 How many moles are there in 11.2 liters (at STP)

• f chlorine?

n = Vm V

Vm = 22.4 L

Slide 105 / 157

60 What volume will 6.0x1024 molecules of oxygen gas

• ccupy at STP?

n = Vm V

Vm = 22.4 L at STP

n = M m n = NA N NA = 6.02 x 10

23

Slide 106 / 157

61 How many molecules are in a 32g sample of SO2 ?

n = Vm V

Vm = 22.4 L at STP

n = M m n = NA N NA = 6.02 x 10

23

Slide 107 / 157

62 Determine the volume occupied by 216g of N2O5 at STP?

n = Vm V

Vm = 22.4 L at STP

n = M m n = NA N NA = 6.02 x 10

23

Slide 108 / 157

63 What volume will 5.0x1024 atoms of krypton occupy at

STP?

n = Vm V

Vm = 22.4 L at STP

n = M m n = NA N NA = 6.02 x 10

23

Slide 109 / 157

64 What is the mass of 224 L of hexene (C6H12 ) at STP? n = Vm V

Vm = 22.4 L at STP

n = M m n = NA N NA = 6.02 x 10

23

Slide 110 / 157

65 What is the volume occupied by 6.0x1023 atoms of

Hydrogen at STP?

n = Vm V

Vm = 22.4 L at STP

n = M m n = NA N NA = 6.02 x 10

23

Slide 111 / 157

66 How many atoms of C are in a 30 milligram sample of calcium cyanide. (Hint: you must remember that milli means 1/1000 and you must first write the proper formula for calcium cyanide)

n = Vm V

Vm = 22.4 L at STP

n = M m n = NA N NA = 6.02 x 10

23

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67 What is the mass of 44.8L of Argon gas at STP?

n = Vm V

Vm = 22.4 L at STP

n = M m n = NA N NA = 6.02 x 10

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Percent Composition

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Identify an Unknown Substance

We have been able to calculate the molar mass of a material if we know its formula already. However, what if we encounter an unidentified substance? Using the tools we already have, we are able to determine the composition of a substance if we have certain pieces of information.

Slide 115 / 157 Percent Composition

Potassium chromate, K2 CrO4 Potassium dichromate, K2 Cr2 O7

One such piece of information is the percent composition, or the percentage of a compound's mass, made up by its various elements.

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% element = (number of atoms)(atomic weight) (FW of the compound) x 100

Percent Composition

The percent by mass of an element in a compound is the number of grams of the element divided by the mass in grams of the compound, multiplied by 100%.

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= 80.0% %C = (2)(12.0 amu) 30.0 amu x 100 %H = (6)(1.0 amu) 30.0 amu = 20.0% x 100

The percent composition of carbon and hydrogen in ethane, (C2H6) is…

Percent Composition of Ethane

Since ethane is made of only Carbon and Hydrogen, if the percent composition of Carbon was 80%, the remaining percent must be Hydrogen. A nother method to calculate the %H is: %H = 100% - %C %H = 100% - 80% %H = 20%

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%H = 20% and %C = 80%

The percent composition of ethane, (C2H6) is… Note that even though there are more ATOMS of hydrogen in ethane, there is a much less percentage of mass in the compound than that of carbon. This is because 1 carbon atom (12 amu) is much more massive than 1 hydrogen atom (1 amu). Therefore, carbon accounts for a much greater percentage of the mass of ethane than hydrogen does.

Percent Composition of Ethane Slide 119 / 157

68 In water (H2O), which element do you think accounts for more mass? A Hydrogen B Oxygen C Hydrogen and Oxygen account for the same percent of mass D Carbon

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69 What is the mass percentage of oxygen in water?

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70 What is the percent by mass of carbon in acetone, C3H6O?

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71 What is the percent of Ba in Ba(NO3)2

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72 Iron (II) oxide has a smaller % of iron by mass than iron (III) oxide.

True False

Slide 124 / 157 Real World Application

Aluminum is used in soda cans, aircraft frames, and automobile

• engines. Worldwide aluminum demand

is increasing. Aluminum is isolated from an ore called bauxite which is roughly 12% aluminum by mass. How much bauxite ore (in grams) must be mined to provide enough aluminum to make a car engine requiring 100 kg of aluminum?

bauxite ore Al block engine

100 kg x (1000 g/kg) x (100 g of ore/12 gram of Al) = 830,000 g Al

move for answer

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73 A 2.00 sample of a compound containing only potassium and oxygen is heated. The oxygen gas leaves and the resulting mass of the potassium is 1.66 grams. What is the % by mass of oxygen in the compound?

A 17% B 70% C 30% D 45% E Cannot be determined from the information

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74 What is the % by mass of water in the CaSO4*2H2O crystal?

A 10% B 21% C 19% D 50% E 75%

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Empirical Formula

Slide 128 / 157 Calculating Empirical Formulas

Now that we are able to calculate the percent of a compound by mass of each element, we can begin to identify unknown

• substances. First, we must identify the ratio of the number of

moles of each element in the substance This formula, based on whole-number ratio, is called an empirical formula.

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Calculating Empirical Formulas

Empirical formula: The formula that indicates the molar ratio of elements present in a molecular compound reduced to the least common denominator. For instance, the empirical formula for benzene (C6H6) is CH.

Mass % elements Grams of each element Moles of each element Empirical formula Assume 100g sample use molar mass Calculate mole ratio

Slide 130 / 157 Calculating Empirical Formulas

Empirical formula: There is only one empirical formula for a substance, but two different substances can have the same empirical formula. That is because an empirical formula is always based on the ratio

• f the elements given in their lowest common denominator.

Example: Substance Hydrogen peroxide Hydroxide Molecular Formula Emperical Formula

H2O2 OH OH OH Slide 131 / 157

75 The empirical formula for C6H4(NO3)2 is C3H2N2O3 True False

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76 Which of the following has an empirical formula that is the same as its molecular formula? (1) NH4Cl (2) (NH4)2CO3 (3) CH2Cl2 (4) CHCl2Br A 1 and 3 only B 1, 2 and 4 only C 1 and 4 only D all of them

Slide 133 / 157 Calculating Empirical Formulas

The compound para-aminobenzoic acid (listed as PABA in bottles of sunscreen) is composed of the following elements (by mass): carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA.

Mass % elements

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Step 1: convert the mass percentages to mole amounts in a hypothetical 100 g sample. Assume a 100.00 g amount of PABA to represent the mass of each element: C: 61.31% x 100g = 61.31 g H: 5.14 % x 100g = 5.14 g N: 10.21% x 100g = 10.21 g O: 23.33 % x 100g = 23.33 g

Mass % elements Grams of each element Assume 100g sample

Calculating Empirical Formulas

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Step 2: Now that we know the representative mass, we can use the molar mass of each element to calculate the number

• f moles that would be present.

C: 61.31 g x = 5.105 mol H: 5.14 g x = 5.09 mol N: 10.21 g x = 0.7288 mol O: 23.33 g x = 1.456 mol 1 mol 12.01 g 1 mol 14.01 g 1 mol 1.01 g 1 mol 16.00 g

Calculating Empirical Formulas

Mass % elements Grams of each element Moles of each element Assume 100g sample use molar mass

Slide 136 / 157 Calculating Empirical Formulas

Step 3: Calculate the mole ratio by dividing each mole value by the smallest number of moles. In this case, by that of nitrogen. Divide them all by 0.7288. C: 5.105 mol H: 5.09 mol N: 0.7288 mol O: 1.456 mol 0.7288 mol 0.7288 mol 0.7288 mol 5.105 mol 5.09 mol 0.7288 mol 1.456 mol = 7.005 # 7 = 6.984 # 7 = 1.00 = 1 = 2.001 # 2 0.7288 mol

Mass % elements Grams of each element Moles of each element Empirical formula Assume 100g sample use molar mass Calculate mole ratio

The empirical formula for PABA is C

7H7NO2

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C: 62.1 g x = 5.17 mol = 5.17 mol = 3.01 3 H: 13.8 g x = 13.8 mol = 13.8 mol = 7.96 8 N: 24.1 g x = 1.72 mol = 1.72 mol = 1.00 = 1 1 mol 14.01 g 1 mol 12 .0 g 1 mol 1.0 g 1.72 mol 1.72 mol 1.72 mol # #

The empirical formula is C3H8N

1,6 - diaminohexane is used in making nylon. It is 62.1% C, 13.8% H and 24.1% N. What is the empirical formula?

Calculating Empirical Formulas Slide 138 / 157

77 What is the empirical formula for a compound with

the following percent composition? A C H4 B C2 H5 C C3 H6 D C3 H8

C: 74.9 % H: 25.1 % Slide 139 / 157

78 What is the empirical formula for a compound with

the following percent composition? A C2 H O B C H2 O C C H O2 D C2 H3 O2

C: 40.0 % H: 6.7 % O: 53.0 % Slide 140 / 157

The molar ratio works very well for most compounds. However, in some cases the math leaves us without whole numbers: For Example: iron (?) oxide Fe: 69.92% -> 69.92g = 1.25 mol = 1 O: 30.08% -> 30.08g = 1.88 mol = 1.5

Mass % elements Grams of each element Moles of each element Assume 100g sample use molar mass Calculate mole ratio

55.85 g/mol 16.00g/mol 1.25 mol 1.25 mol

FeO1.5 Special Cases Slide 141 / 157

Special Cases

In these cases, it is necessary to multiply the results by an integer to ensure the ratio is a whole number ratio... The empirical formula is Fe2O3

x 2 = 2 x 2 = 3

The name of the compound is iron (III) oxide

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79 What is the empirical formula for a compound with

the following percent composition? A V O1.67 B V2 O5 C V3 O5 D V3 O8

V: 56.02 % O: 43.98 % Slide 143 / 157

80 A 4.68 gram sample of a sulfur oxide is heated releasing oxygen gas and leaving behind solid sulfur. If the mass of the sulfur left behind was 2.34 grams, what must be the empirical formula of the compound?

A SO B SO2 C S2O D S2O3 E SO3

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Hydrated Crystals

Many ionic compounds have a fixed quantity of water molecules bound within their structure. The mole ratio of water to the dry CuSO4 crystal is 5:1 CuSO4 * 5H2O

Slide 145 / 157 Finding the formula of a hydrate

One can find the formula of a hydrate by heating the crystal to remove the water. Then find the mole ratio between the dry crystal and the water just as you would when finding an empirical formula. g CuSO4

• -> n CuSO4

CuSO4 * ? H2O g H2O --> n H2O

HEAT

Find mole ratio!

Slide 146 / 157 Finding the formula of a hydrate

After heating, a hydrate of MgSO4 was found to be roughly 51.3% water by mass. What is the formula of the hydrate? Step 1: Find the % of the crystal and water and express in grams. 100 - 51.3% water = 48.7 % MgSO4 = 48.7 g MgSO4 & 51.3 g of water Step 2: Convert to moles 48.7 g MgSO4 = 0.405 mol MgSO4 51.3 g H2O = 2.85 mol H2O 120 g/mol 18 g/mol

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Step 3: Find the mole ratio of the water to the dry crystal 0.405 mol MgSO4 = 1 2.85 mol H2O = 7 0.405 mol 0.405 mol .....the formula is MgSO4*7H2O

Finding the formula of a hydrate Slide 148 / 157

81 When a 2.4 g sample of a hydrated crystal of BaCl2 is heated, the dry anhydrous crystal has a mass of 2.08 grams after heating. What is the formula of the hydrate?

A BaCl2*H2O B BaCl2*2H2O C BaCl2*3H2O D BaCl2*4H2O E BaCl2*5H2O

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A molecular formula indicates the number of atoms of each element present in the molecule. The molecular formula of a compound is either the same as its experimentally determined empirical formula, or it is a simple whole-number multiple of its empirical formula

Molecular Formulas

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Molecular Formulas

Formula Name Type of Formula Molar Mass

CH C2H2 Ethyne C6H6 Benzene CH2O Methanal C2H4O2 Ethanoic Acid C6H12O6 Glucose Empirical Molecular Molecular Empirical & Molecular Molecular Molecular 13 (13x2 =) 26 (13x6 =) 78 30 (30x2 =) 60 (30x6 =) 180

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To find the actual molecular formula you need one other piece of information...the molecular weight (mass) of the molecule.

If

you know the ratio of the elements in a molecule

and you know the total mass of the molecule then you can determine the molecular formula. Molecular Formulas Slide 152 / 157

To find the molecular formula: Step 1: Determine the molar mass of the empirical formula. Step 2: Divide the molecular mass by the empirical mass Step 3: Multiply the Empirical Formula by the resulting integer

Molecular Formulas

What is the molecular formula of a compound with an empirical formula of CH

2O that has a mass of 180 g/mol?

Molar mass of CH2O = 30 g/mol Molecular mass = 180 g/mol = 6 Empirical mass 30 g/mol 6 x CH2O = C6H12O6

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82 The molecular mass of Benzene is 78. If the empirical formula of benzene is CH, what is its molecular formula? A C2H2 B CH C C6H6 D C2H4

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Maleic acid is an organic compound composed of 41.39 % carbon, 3.47% hydrogen, and the rest is oxygen. It has a molecular mass of 116 g/mole.

Molecular Formulas

Determine the empirical formula for Maleic acid. Then determine the molecular formula for Maleic acid.

Slide 155 / 157 Real World Application

Clenbuterol is a steroid drug that is illegally used in cattle and in sports like cycling to help the cattle or athlete lose fat and gain lean muscle

• mass. It can be detected by

mass spectroscopy. What is the empirical and molecular formula for clenbuterol if when a 10.0 g sample is combusted in air, it is found it contains 4.60 g of carbon, 0.613 g of H, 0.51 g of oxygen with the rest being chlorine? The molecular weight is 313 g/mol.

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