Ch. 3: Mass Relationships in Chemical Reactions

2

By definition and International agreement: 1 atom 12C “weighs” 12 amu

On this scale1H = 1.008 amu - which is 8.400% as massive

16O = 16.00 amu - which is 133.33% as massive

Atomic mass: mass of an atom in atomic mass units (amu)

Micro-World(atoms & molecules)

Macro-World(grams)

where 1 amu = 1.66 x 10-24 g

The average atomic mass is the weighted average of all of the naturally occurring isotopes of the

element.

Carbon average atomic mass = 0.9890 * 12 amu + 0.0110 * 13.00335 amu =

12.01 amu*98.90% = 0.9890

121212121212121212 13

1212121212121212121212121212121212121212

12121212121212121212121212121212121212121212121212121212121212121212121212121212121212121212121212121212121212121212121212121212121212121212

A random sample of Magnesium is composed of

• 79.0% Mg-24 , 10.0% Mg-25 , and 11.0% Mg-26

Find the weighted average for the mass of Magnesium

(0.790 x 24) + (0.100 x 25) + (0.11 x 26) = 24.3 amu

79.0% of 24 10.0% of 25 11.0% of 26+ + = Average atomic mass

*Do Not Divide by the total

Use the percentage in the decimal form (e.g. 75% = 0.75)

Example: Average Atomic Mass Calculation #1

A random sample of chlorine is

75.8% 35Cl (34.97 amu) & 24.2% 37Cl (36.97 amu)

• Find the weighted average for the mass of chlorine by writing percentage as a decimal (e.g. 50% = 0.50)

(0.758 x 34.97) + (0.242 x 36.97) = 35.45 amu

Example: Average Atomic Mass Calculation #2

Average atomic mass for relative abundant isotopes

The mole (mol) is the amount of a substance that contains as many elementary units as there are atoms in exactly 12.00 grams of 12C

Dozen = 12

Pair = 2

The Mole (mol): A grouping unit to count numbers of particles

The mass of a single carbon atom in grams:

0.0000000000000000000000199

Avogadro’s Constant (number)602,214,150,000,000,000,000,000

602 Septillion6.022 x 1023

1 mole = NA = 6.022 x 1023

Some BIG Numbers…World population:

Cells in our body:

Stars in our galaxy:

~7,000,000,000 ~300,000,000,000~50,000,000,000,000

Jean Baptist Perrin

268,820 miles26.02*1023 grains of sand could cover Texas in almost 3 feet of sand

6.02*1023 Starbursts© would cover the

surface of the entire US stacked and be

24.2 miles tall

6.02*1023

moles (mammal)

The Moon7.3 × 1022 kg

1 amu = 1.66 x 10-24 g or 1 g = 6.022 x 1023 amu

1 12C atom12.00 amu

x12.00 g

6.022 x 1023 12C atoms=

1.66 x 10-24 g1 amu

M = molar mass in g/mol

NA = Avogadro’s number

Molar mass is the mass of 1 mole of in gramseggsshoesatoms

1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g1 12C atom = 12.00 amu

12.00 g 12C = 1 mole 12C atoms 65.39 g of Zn = 1 mole Zinc atoms

18.015 g of water = 1 mole water molecules58.443 g of NaCl = 1 mole NaCl formula units

1 mol = NA = 6.02 x 1023

*Read the masses on the periodic table as grams/mole

How to “Read” Chemical Equations

2 Mg + O2 2 MgO

2 atoms Mg + 1 molecule O2 makes 2 units of MgO

or... (Scale everything up by NA)

2 moles Mg + 1 mole O2 makes 2 moles MgO so (using the Periodic Table we have)

48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO

One Mole of:

C S

Copper Iron

Mercury

Ted-Ed: How big is a mole? www.youtube.com/watch?v=TEl4jeETVmg

Particles Atoms

MoleculesFormula units

MM: Molar mass (grams/mol)

Avogadro’s constant NA = 6.02 x 1023

NA

NA

Remember Train Track Unit Conversions

How many inches are in 3 miles?

3 miles 1

5280 feet 1 mile

Multiply all numbers on the topDivide all numbers on the bottom

3 x 5280 x 12 1 x 1x 1

12 inches 1 foot

Need conversion factors: 1 mile = 5,280 ft; 1 foot = 12 in

= 190,080 inches

Chemistry Conversion FactorsMM = molar mass in grams/mole

Converts between mass (g) & moles

NA = Avogadro’s number = 6.02 x 1023

Converts between moles & particles(atoms / molecules / formula units)

Grams ↔ Moles ↔ Atomsx NA ÷ Molar Mass

x Molar Mass ÷ NA

Grams to Moles example

How many moles of Helium atoms are in 6.46 g of He?

Molar mass is the conversion factor and is found on the periodic table to be 4.003 g/mol.

This can be written as 1 mol He = 4.003 g He

6.46 g He = 1.61 moles He1 mol He4.00 g He

x

More Practice: 45.5 grams of Gold (Au) to moles

How many grams of zinc are in 0.356 mole of Zn?

Zinc

The conversion factor is the molar mass of Zn. From the periodic table we find that

1 mol Zn = 65.4 g Zn

Moles to Grams example

More Practice: 0.56 mol Potassium to grams

0.356 mol Zn= 23.3 gram Zn

65.4 gram Zn1 mol Zn

x

How many atoms of Carbon are in 2.54 moles of C?

Carbon

The conversion factor is Avogadro’s Number1 mole Carbon = 6.02 x 1023 atoms of Carbon

Moles to Atoms example

More Practice: 0.074 mol U to atoms of U

= 1.53 x 1024 atoms of C

2.54 mol C 6.02 x 1023 atoms of C1 mol C

x

How many atoms are in 16.3 g of Sulfur?

Grams of S → moles of S → # atoms S

MM: 1 mol S = 32.07 g 1 mol S = 6.02 x 1023 S atoms

Grams to Atoms Example

16.3 g S

= 3.06 x 1023 atoms of S

1 mol S32.1 g S

x 6.02 x 1023 atoms of S1 mol S

x

More Practice: 0.01 g of Carbon to atoms

What would be the mass in grams of 2.4 x 1025 atoms of Copper?

# atoms of Cu → moles of Cu → grams CuMM: 1 mol Cu = 63.55 g Cu6.02 x 1023 Cu atoms = 1 mol Cu

Atoms to Grams Example

2.4 x 1025 atoms Cu

= 2.5 x 103 g = 2.5 kg Cu

1 mol Cu 63.55 g Cux

6.02 x 1023 atoms 1 mol Cux

More Practice: 9.8 x 1021 atoms of Uranium to grams

xxx

More Practice ConversionsConvert 24 grams Potassium to moles

24 grams K = 0.61 moles K1 mol K39.1 grams K

x

Convert 1.25 moles Gold to grams

Convert 16 grams Argon to atoms

1.25 mol Au= 246 grams Au197.0 g Au

1 mol Aux

16 g Ar = 2.4 x 1023 Ar atoms

1 mol Ar39.9 g Ar

x 6.02 x 1023 Ar atoms1 mol Ar

x

Bozeman Science Mole conversion tutorial: www.youtube.com/watch?v=xPdqEX_WMjo

Convert 3.4 x 1025 atoms of Boron to grams of Boron

Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a molecule.

1S 32.07 amu2O + 2 x 16.00 amu SO2 64.07 amu

For any molecule molecular mass (amu) = molar mass (grams)

1 molecule SO2 = 64.07 amu1 mole SO2 = 64.07 g SO2

SO2

Formula mass is the sum of the atomic masses (in amu) in a formula unit of an ionic compound.

1Na 22.99 amu1Cl + 35.45 amuNaCl 58.44 amu

For any ionic compound formula mass (amu) = molar mass (grams)

1 formula unit NaCl = 58.44 amu1 mole NaCl = 58.44 g NaCl

NaClx

Determining Molar Mass Examples

Calculate the molecular or formula masses (in g/mol) of the following compounds:

(a) Water (H2O)

(b) Isopropanol (C3H8O1) (rubbing alcohol)

(c) (NH4)2S

(d) Mg(NO3)2

(e) caffeine (C8H10N4O2)

Example Molar Mass Solutions

a) H2O: 2*(1.01 g) + 1*(16.00 g) = 18.02 grams

b) C3H8O1: 3*(12.01 g) + 8*(1.01 g) + 1*(16.00 g) = 60.05 grams

c) (NH4)2S : 2*(14.01 g) + 8*(1.01 g) + 1*(32.07 g) = 68.17 grams

d) Mg(NO3)2: 1*(24.31 g) + 2*(14.01 g) + 6*(16.00 g) = 148.33 grams

e) C8H10N4O2: 8*(12.01 g) + 10*(1.01 g) + 4*(14.01 g) + 2*(16.00 g) = 194.20 grams

Each Molar mass can be read with units “grams/mole”

Compound Mass to Mole Example

Methane (CH4) is the principal component of natural gas.

How many moles of CH4 are present in 6.07 g of CH4?

Molar mass of CH4 = 12.01 g + 4(1.0 g) = 16.04 g

More Practice: 32 grams of Isopropanol (C3H8O) to moles

6.07 g CH4 = 0.38 moles CH41 mol CH4

16.0 g CH4

x

Example Atoms in a CompoundHow many hydrogen atoms are present in 25.6 g of urea [(NH2)2CO]

The molar mass of urea is 60.06 g.

*Note there are 4 H atoms per 1 Urea molecule So there are 4 moles of H per 1 mole urea

urea

= 1.03 × 1024 H atoms

grams of urea → moles of urea → moles of H → atoms of H4 mol of H atoms

1 mol of urea6.02 x 1023 H atoms

1 mole of H1 mol of urea60.06 g urea

25.6 g of urea x xx

More Practice: How many Fluorine atoms in 4.5 grams of AlF3?

Example Compound Conversions

Alanine(C3H7NO2)

= 5.1 × 1022 C atoms

3 mol of C . 1 mol C3H7NO2

6.02 x 1023 C atoms1 mol of C

1 mol of C3H7NO2

89.0 g C3H7NO2

c) 2.5 g of C3H7NO2

Alanine is one of the 20 amino acids that form proteins.

In 2.5 grams of Alanine, determine the number of: a) moles of Alanine b) molecules of Alanine

c) atoms of Carbon

= 5.1 × 1022 C atoms

3 atoms of C . 1 molecule C3H7NO2

6.02 x 1023 molecules1 mol C3H7NO2

1 mol of C3H7NO2

89.0 g C3H7NO2

*c) 2.5 g of C3H7NO2

= 1.7 × 1022 molecules

6.02 x 1023 molecules1 mole of C3H7NO2

1 mol of C3H7NO2

89.0 g C3H7NO2

b) 2.5 g of C3H7NO2

1 mol of C3H7NO2

89.0 g C3H7NO2

a) 2.5 g of C3H7NO2 = 0.028 moles C3H7NO2

*Last 2 conversions can be used in either order

Or

Bell Ringer Practice Conversions

How many atoms of Nitrogen (N) are found in 6.30 g of Ca(NO3)2?

Heavy

Heavy

Light

Light

Mass Spectrometer

Mass Spectrum of Ne• Invented by Francis Aston in the 1920’s• Separates particles by e/m (charge/mass)• First demonstration of isotopes (Ne 20 & 22)• Led to precise determination of atomic

masses and relative isotope abundances. 1922 Nobel Prize in Chemistry

Sample gets ionized

Fuse School: Mass Spectrometry; https://www.youtube.com/watch?v=sTi--ixdAME

Percent composition of an element in a compound =

n x molar mass of elementmolar mass of compound

x 100%

n is the number of moles of the element in 1 mole of the compound

Ethanol: C2H5OH

%C =2 x (12.01 g)

46.07 gx 100% = 52.14%

%H =6 x (1.008 g)

46.07 gx 100% = 13.13%

%O =1 x (16.00 g)

46.07 gx 100% = 34.73%

52.14% + 13.13% + 34.73% = 100.0%Molecular weight: 46.07 g

Example Percent Composition #2Phosphoric acid (H3PO4) is a colorless, syrupy liquid used in detergents, fertilizers, toothpastes, and in carbonated beverages for a “tangy” flavor.

More Practice: What is the % composition of Chromium in Chromite (FeCr2O4)?

Calculate the percent composition by mass of H, P, and O

Molar Mass of H3PO4

H: 3 x 1.008P : 1 x 30.97O: 4 x 16.00

= 97.99 g

Using Percent Compositions Example #1Talc is a mineral composed of hydrated magnesium silicate with the chemical formula H2Mg3(SiO3)4

Determine the number of grams of pure Magnesium we could extract from 45.0 grams of the talc mineral using percent composition.

% Mg = (Mg Mass)*(3) Total Mineral Mass x 100%

% Mg = 72.93 g Mg 379.29 g Mineral x 100% = 19.23% Mg

g Mg = % Mg 45.0 gram compound

= 0.1923 x 45.0 g = 8.65 g Mg

More Practice: What is the % composition of Aluminum in Galaxite (MnAl2O4)? How many grams Al in 35.6 g of Galaxite?

Example Using Percent Compositions #2 Chalcopyrite (CuFeS2) is a principal mineral of copper.

Calculate the number of kilograms of Cu in 3.71 kg of chalcopyrite using percent compositions. Chalcopyrite.

How do we calculate mass percent of an element?

mass % = (Element Mass)*(Quantity)

Compound Massx 100%

Example #2 SolutionThe molar masses of Cu and CuFeS2 are 63.55 g and 183.5 g, respectively. The percent composition of Cu is therefore:

To calculate the mass of Cu in a 3.71 kg sample of CuFeS2, we need to convert the percentage to a fraction (that is, convert 34.63 percent to 34.63/100, or 0.3463) and write:

mass of Cu in CuFeS2 = 0.3463 × (3.71 kg) = 1.28 kg Copper

Percent Composition → Empirical Formulas

Early analysis of many compounds only yielded % mass composition

of each element

This allows us to derive the empirical formula (atom to atom ratio)

Step 1 & 2

Step 3

Step 4

Molecular Formula: C6H12O6

Empirical Formula:C1H2O1

Determining Empirical Formulas

1. Assume 100g sample and change element percents to grams

2. Convert each to moles by dividing by molar mass of each atom

3. Divide each number by the smallest number

If all Numbers are integers (or close) you are done; Put in formula

4. If not, multiply all numbers by the SAME smallest whole number so that all numbers are whole.

We don’t expect numbers to be perfect integers due to some experimental error, Generally round only if within one tenth.

When the element % composition is known:

An analysis of an unknown gas has determined that it contains 72.55% Oxygen and 27.45% Carbon by mass. What is the empirical formula of the compound? C#O#

1 mol C12 g C

1 mol O16 g O

= 4.53 mol O

= 2.29 mol C

4.53/2.29 = 1.98 mol

2.29/2.29 = 1 mol C

CO2

72.55 g O

27.45 g C

Example: Empirical Formula Determination #1

C(1 mol)O(2 mol) C(1 atom)O(2 atom)

~ 2 mol O

Example: Empirical Formula #2Determine the empirical formula of Ascorbic acid (vitamin C: cures/prevents scurvy).

It is composed of 40.92% C, 4.58% H, and 54.50% O by mass.

40.92 g C 4.58 g H 54.50 g O

To determine the empirical formula, we will first assume a 100 g sample of Ascorbic Acid

2nd: convert each mass to mole

Example: Empirical Formula #2 Solution3rd:We then take the smallest value of moles and divide each number by that same smallest value

• This gives CH1.33O as the formula for ascorbic acid.• If any # is more than 0.1 away from whole, it is too far to round.• Multiply all values by integers until all values are whole numbers

1.33 × 1 = 1.331.33 × 2 = 2.661.33 × 3 = 3.99 ~ 4

More Practice: CxHyNz : 42.1% C, 8.8% H & 49.1% N

Because 1.33 × 3 gives us an integer (4), we multiply all the subscripts by 3 and obtain

C3H4O3 as the empirical formula for ascorbic acid.

g O = (11.5 g sample) – (6 g C + 1.5 g H) =

Combust 11.5 g ethanol

Collect 22.0 g CO2 and 13.5 g H2O

6.0 g C 0.5 mol C

1.5 g H 1.5 mol H

4.0 g O 0.25 mol O

Empirical formula C2H6O

g C = % C x 22.0 g CO2

g H = % H x 13.5 g H2O

27.3% x 22.0 g = 6.0 g C

11.1% x 13.5 g = 1.5 g H

“desiccant” SiO2 “scrubber” LiOH

4.0 g O

SciShow: What's in Those Packets That Say 'Do Not Eat'?https://www.youtube.com/watch?v=3NYHycpgk6E

A sample of a compound contains 30.46% Nitrogen and 69.54% Oxygen by mass

In a separate experiment, the molar mass of the compound is estimated to be between 90 g and 95 g.

Determine the molecular formula and the accurate molar mass of the compound.

The molecular formula can be determined with both % Composition and Molar Mass

Example: Molecular Formula

We first assume 100 g and convert % to grams. We then convert each element from grams to moles.

To convert to whole numbers we divide each mole by the smaller subscript (2.174). After rounding off, we obtain NO2 as the empirical formula.

The Molecular formula must have a Molar Mass that is a multiple of the Empirical formula’s Molar Mass

(NO2, N2O4, N3O6, N4O8…)

Example: Molecular Formula Solution

Empirical molar mass = 14.01 g + 2(16.00 g) = 46.01 g

Next, we determine the ratio between the molar mass and the empirical molar mass

The molar mass is twice the empirical molar mass. This means that there are two NO2 units in each molecule of the compound, and the molecular formula is N2O4.

More practice: Find the Molecular formula PxOy : 43.7% P & 56.3% Owith a molar mass between 280 – 290 g/mol.

Example: Molecular Formula Solution

Aconitic acid has a mass composition of: 41.4% Carbon

3.4% Hydrogen 55.2% Oxygen

It also has a molar mass of 174 grams/mole

What is the molecular formula (CxHyOz)of Aconitic acid?

Bell Ringer: Molecular Formula

A chloro-hydrocarbon a mass composition of: 38.4% Carbon

4.8% Hydrogen 56.7% Chlorine

It also has a molar mass of 187.5 grams/mole

What is the molecular formula (CxHyClz)?

Bell Ringer: Molecular Formula

3 ways of representing the reaction of H2 with O2 to form H2O

Chemical Reaction: A process in which one or more substances is changed into one or more new substances.

A chemical equation uses chemical symbols to show what happens during a chemical reaction:

reactants products

Equations SymbolsSymbol Meaning

+ “and”

→ “yields” or “produces”

↔ Reversible

(g) Gas

(l) Liquid

(s) Solid

(aq) Aqueous (dissolved in H2O)

or

Phase State :represent whatstate of matter asubstance is in.

Water, H2O(l)

Ice, H2O(s)

Vapor, H2O(g)

NaCl(s) + H2O(l) Na+(aq) + Cl-

(aq) + H2O(l)

How to “Read” Chemical Equations2 Mg + O2 2 MgO

2 atoms Mg + 1 molecule O2 makes 2 formula units MgOOr 2 moles Mg + 1 mole O2 makes 2 moles MgO Or 48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO

NOT

2 grams Mg + 1 gram O2 makes 2 g MgO

We can not compare mass to mass, it must be in particles or Moles

In 1 mole of HBr there are 80 grams of Br to 1 gram H, Yet still 1 atom Br to 1 atom H

Law of Conservation of Matter/Mass• Chemical reactions only rearrange bonded atoms

with no atoms created or destroyed• Every atom from the reactants must be equal in

the products

2C2H6 + 7O2 → 4CO2 + 6H2O

Reactants4 Carbons12 Hydrogens14 Oxygens

Products4 Carbons12 Hydrogens14 Oxygens

=

Fire converts/Not destroys

SciShow: When You Burn Fat, Where Does it Go? www.youtube.com/watch?v=C8ialLlcdcw

Balancing Chemical Equations

1. Write the correct formula(s) for the reactants on the left side and the product(s) on the right side of the equation.

Ethane reacts with oxygen to form carbon dioxide and water

C2H6 + O2 CO2 + H2O

We only change the coefficients (number in front) to make the number of atoms of each element the same on both sides.

Never change the subscript (that would change the chemicals)

2C2H6 NOT C4H12

Balancing Chemical Equations2. Count the number of atoms on each side.

Check to see if it’s already balanced.LiBr + KNO3 → LiNO3 + KBr

• On the reactant side, there is 1 Li, 1 Br, 1 K,1 N, and 3 O atoms.

• On the product side, there is 1 Li, 1 Br, 1 K,1 N, and 3 O atoms.

• This equation is balanced.

Balancing Chemical Equations3. Leaving Hydrogen and Oxygen till last,

place a coefficient to balance one element. Continue balancing the others elements with

coefficients__H2SO4 + __NaOH → __Na2SO4 + __H2O

Reactants1 Sulfur1 Sodium3 Hydrogen5 Oxygen

=Products1 Sulfur2 Sodium2 Hydrogen5 Oxygen

2 2

=Reactants1 Sulfur2 Sodium4 Hydrogen6 Oxygen

Products1 Sulfur2 Sodium4 Hydrogen6 Oxygen

Combustion of propane__C3H8 + __O2 → __CO2 + __H2O

Reactants3 Carbon8 Hydrogen2 Oxygen

Products1 Carbon2 Hydrogen3 Oxygen

==

3

7

3

4

10

5

10

The balanced equation shows it take 5 molecules of O2 to react with 1 molecule propane to produce 3 molecules of CO2 and 4 of H2O.

5 mole of O2 reacts with 1 mole propane to produce 3 moles of CO2 and 4 of H2O.

OR

8

__P2O5 + ___H2O → ___H3PO4More Practice:

Balance these Chemical equations:

• H3PO4 + NaOH → Na3PO4 + H2O

• Mg + O2 → MgO

• NH4NO2 → N2 + H2O

• NaHCO3 → Na2CO3 + CO2 + H2O

• Li + H2O → LiOH + H2

Balancing Difficult Chemical EquationsEquations can be difficult if the reactants don’t react in a simple

1:2, 1:3, 1:4 etc. ratio.

C2H6 + O2 CO2 + H2O *multiply CO2 by 2

C2H6 + O2 2CO2 + H2O *multiply H2O by 3

C2H6 + O2 2CO2 + 3H2O

7 oxygen atoms2 oxygen atoms

Because it’s O2, no whole number coefficient can give us 7 atoms on the reactants (only multiples of 2)

Balancing Chemical Equations

C2H6 + _O2 2CO2 + 3H2O

*multiply O2 by to balance 72

Because each particle must be an integer, remove the fraction by multiplying the whole reaction by the denominator

Balance using the smallest fraction possible that would give the desired quantity

C2H6 + O2 2CO2 + 3H2O72

7 oxygen atoms

2C2H6 + 7O2 4CO2 + 6H2Ox2

Example 3.12When aluminum metal is exposed to air, a protective layer of aluminum oxide (Al2O3) forms on its surface. This layer prevents further reaction between Al and O2, and it is why Al does not corrode.

An atomic scale image of aluminum oxide.

Write a balanced equation for the formation of Al2O3.

Use the simplest fraction to balance Oxygen

Example 3.12Multiplying both sides of the equation by 2 gives whole-numbercoefficients.

Check

The equation is balanced. Also, the coefficients are reduced to the simplest set of whole numbers.

More Practice: _C3H6O2 + _O2 → _CO2 + _H2O

Stoichiometry is the quantitative relationship found between reactants

and/or products in a reaction.

• 2 hydrogen molecules and 1 oxygen molecules produce 2 water molecules

• 2 moles hydrogen and 1 mole oxygen produce 2 moles water

Multiply all by Avogadro's #

Stoichiometry is the "recipe of a reaction".

1x 2x 2x1x

From the equation we see that 1 mole CH4 2 mole O2

= “Stoichiometric equivalents”

TedEd: The law of conservation of mass https://www.youtube.com/watch?v=2S6e11NBwiw

Mole Ratio

• The coefficients in a balanced reaction tell us the mole ratio.

• It takes 1 mole of ethanol to react with 3 moles of oxygen. This produces 2 moles of carbon dioxide and 3 moles of water.

• The mole ratio will act as our conversion factor for calculations.

1C2H6O + 3O2 → 2CO2 + 3H2O1C2H6O + 3O2 → 2CO2 + 3H2O

Mole ratio conversion factor

Mole ratio conversion factor

Combustion of Ethanol (C2H6O)1C2H6O + 3O2 → 2CO2 + 3H2O

• If we have 4 moles of ethanol, how many moles oxygen are needed for the reaction?

• How many moles CO2 are made from reacting 0.6 moles ethanol?

4 mol ethanol = 12 mol oxygen3 mol oxygen1 mol ethanol

x

0.6 mol ethanol = 1.2 mol CO22 mol CO2

1 mol ethanolx

2C2H6 + 7O2 → 4CO2 + 6H2OHow many moles of oxygen does it take to produce 3.5 moles of water (with excess ethane)?3.5 mol water = 4.1 mol oxygen7 mol oxygen

6 mol waterx

Calculations can go forward or backward, always looking at the mole to mole ratio

How many moles of ethane (C2H6) does it take to produce 3.5 moles of water (with excess oxygen)?

3.5 mol water = 1.2 mol ethane2 mol ethane6 mol waterx

More Practice: 1) moles of CO2 from 1.5 moles C2H6

2) moles of H2O from 0.34 moles of O2

1. Write balanced chemical equation2. Convert quantities of known substances into moles3. Use coefficients in balanced equation to calculate the

number of moles of the sought quantity4. Convert moles of sought quantity into desired units

Amounts of Reactants and Products

Stoichiometry Example #1 The food we eat is broken down, in our bodies to provide energy. The overall equation for this complex process represents the degradation of glucose (C6H12O6).

(1 mol C6H12O6 6 mol CO2)

86 g Glucose 1 mol Glucose180 g Glucose

x x 6 mol CO21 mol Glucose

x44 g CO21 mol CO2

= 126 g CO2

(180 g/mol) (44 g/mol)

More Practice: Grams of O2 needed for 86 grams C6H12O6

If 86 g of C6H12O6 is consumed, what is the mass of CO2 produced?

Conversion Factors:

C3H8 + 5O2 → 3CO2 + 4H2O How many grams of water are produced from the combustion of 35.0 g of propane (C3H8)?

35.0 g C3H8 1 mol C3H844 g C3H8

x x 4 mol H2O1 mol C3H8

x 18 g H2O1 mol H2O

= 57.3 g H2O

35.0 g C3H8 1 mol C3H844 g C3H8

3 mol CO21 mol C3H8

44 g CO21 mol CO2

= 105 g CO2

x x x

How many grams of CO2 are produced from the combustion of 35 g of propane (C3H8)?

Stoichiometry Example #2

35.0 g

C3H8 + 5O2 → 3CO2 + 4H2O 57.3 g105 g? g

Mass is conserved from reactants to products

35.0 + g O2 = 105 g + 57.3 gMass O2 = 127.3 g

All alkali metals react in water to produce hydrogen gas and the corresponding alkali metal hydroxide.

How many grams of Li are needed to produce 9.89 g of H2?

Lithium reacting with water to produce hydrogen gas.

Mass to Mole to Mole to Mass (Products) to (Reactants)

Stoichiometry Example #3

2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g)

9.89 g of H2? g of Li

More Practice: Grams of LiOH produced from 2.9 grams Li

Stoichiometry Example #3 Solution

Crash Course: Stoichiometry: www.youtube.com/watch?v=UL1jmJaUkaQ

2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g)

How many grams of Barium phosphate are produced when adding 1.3 grams of Barium hydroxide to a

solution of phosphoric acid?

__Ba(OH)2 + __ H3PO4 → __ Ba3(PO4)2 + __H2O

1.3 grams Excess ? grams

More Practice

Second problem: How many grams of water from 2.5 grams of H3PO4?

Practice problems• How many grams are produced of each product in the

following decomposition reactions?

• How many grams of H2O are required for complete reaction with 37 g of the other reactant?

• How many grams of Aluminum are required to obtain 100 g Iron? 2Al + 3FeO → Al2O3 + 3Fe

K2O + H2O → 2KOHSiCl4 + 4H2O → H4SiO4 + 4HCl

CaCO3 → CaO + CO212.0 g2KClO3 → 2KCl + 3O22.8 g

? g of Al 100 g of Fe

Theoretical Yield is the amount of product that wouldresult if all the limiting reagents reacted. This is

what you have already been calculating.

Actual Yield is the amount of product actually obtainedfrom a reaction. This is an observed value and you

must be told this amount.

Percent (%) Yield = Actual Yield

Theoretical Yieldx 100%

Reaction Percent YieldNot every reaction goes to 100% completion every time

Or we can’t always collect (purify) all of the product.

2Mg + O2 → 2MgOIf we react 10 grams of O2 with excess Magnesium, how much MgO would be theoretically expect to obtain?

If we actually ended up with 20.3 grams MgO, what is our % yield?

10 g O2 1 mol O232 g O2

x x 2 mol MgO1 mol O2

x 40.3 g MgO1 mol MgO

= 25.2 g MgO

Theoretical yield = 25.2 g MgO

% Yield = Actual Yield

Theoretical Yieldx100% =

20.3 g MgO25.2 MgO

x 100% = 80.6%

We obtained a reaction % yield of 80.6% MgO from this reaction.

Example: Percent Yield

More Practice: React 25 g Mg with excess O2; recovered 32 g MgO.What is the percent yield?

Limiting ReactantThe reaction can only proceed as far as the limiting reactant will allow it.

Once it is gone, the other reactant has nothing to react with so the reaction stops.

To determine, we must calculate the product yield from both reactants and choose the smaller amount.

Limited by the number of car bodies so production stops

Limiting Reactant: Reactant used up first in the reaction to result in its completion.

N2 + 3H2 → 2NH3

H2 is limiting

N2 Excess

Unless all reactants are in exact stoichiometric equivalents, one will limit the reaction (i.e. 2 moles of N2 and 3 moles of H2)

Bozeman Science Stoichiometry Tutorial:www.youtube.com/watch?v=LQq203gyftA

1P2O5 + 3H2O → 2H3PO4

If we react 44 grams of P2O5 with 25 grams of water, which is the limiting reactant and how many grams of H3PO4 will be produced?

We first find out how many grams of product could be made in each case?

44 g P2O5 1 mol P2O5141 g P2O5

x x 2 mol H3PO4

1 mol P2O5

x 98 g H3PO41 mol H3PO4

= 61.2 g

25 g H2O 1 mol H2O18 g H2O

x x 2 mol H3PO4

3 mol H2Ox 98 g H3PO4

1 mol H3PO4

= 90.7 g

We then choose the smaller amount of product: 61.2 g of H3PO4

P2O5 is the limiting reactant and there is excess water.

Example #1: Limiting Reactant and Excess

Example #2: Limiting Reactant and Excess

Urea [CH4N2O2] is prepared by reacting ammonia with CO2:

24 g of NH3 are reacted with 65 g of CO2.

(a) Determine the limiting reactant and calculate the mass of product urea formed.

(b) How much excess reactant (in grams) is left at the end of the reaction?

2NH3(g) + CO2(g) → CH4N2O2(s) + H2O(l)

Example #2: Limiting Reagent and Excess24 g 65 g ? grams Urea

2NH3 + CO2 → CH4N2O2 + H2O

24 g NH3 1 mol NH317 g NH3

x x 1 mol Urea2 mol NH3

x 76 g Urea1 mol Urea

= 54 g Urea

65 g CO2 1 mol CO244 g CO2

x x 1 mol Urea 1 mol CO2

x 76 g Urea1 mol Urea

= 112 g Urea

a) NH3 is the limiting reactant and 54 g of urea is produced.

24 g NH3 1 mol NH317 g NH3

x x 1 mol CO2

2 mol NH3

x 44 g CO21 mol CO2

= 31 g CO2 used

b) Started with 65 g CO2 and used 31 g. 65 – 31 = 34 g CO2 left

35 g of NO are mixed with 18 g O2 to form NO2:

2NO + O2 → 2NO2

• Which reactant is limiting the reaction and how much NO2 is produced? • If only 16 g is recovered from the reaction, what’s the percent yield? • How much of the other reactant is left?

Practice: Limiting Reactant and Excess

How much Iron was added to Copper (II) Sulfate to actually obtain 32.1 g Copper (a 73.1% reaction yield)?

Fe + CuSO4 → FeSO4 + Cu

% Yield = Actual Yield

Theoretical Yieldx 100% =

32.1 g Cu

? g Cux 100% = 73.1%

The theoretical yield must have been 43.9 g Cu

43.9 g Cu 1 mol Cu63.5 g Cu

x x 1 mol Fe1 mol Cu

x 55.8 g Fe1 mol Fe

= 38.6 g Fe

38.6 grams of Iron was added to CuSO4

Fun Practice Exam Problems

How many oxygen atoms are found in 20 g of CaCO3?

A compound analysis yields the following data:44.4% C, 6.21% H,

39.5% Sulfur, 9.86% Oxygen

What is the molecular formula if the molar mass ~162?

Fun? Practice Exam ProblemsCO + O2 → CO2

Balance the above reaction and then calculate the mass of oxygen needed to produce 27 g CO2 with excess CO

If 10 g methane CH4 is combusted (uses O2) to form CO2 and H2O. How many grams of each product are formed by the reaction and how many grams of oxygen were used?

40 g of NO are mixed with 8 g O2 to form NO2, which reactant is limiting the reaction, how much NO2 is produced, and how much of the other reactant is left? (Write equation first) If only18 g is recovered from the reaction, what’s the % yield?

__H3PO4 + __NaOH → __Na3PO4 + __H2O

What is the theoretical yield of water if 12.5 grams of H3PO4 reacts with 18.4 grams of NaOH?

The actual yield was 5.2 grams water, what is the percent yield?

1) Average Atomic Mass: 80.1% 11B and 19.9% 10B

2a) How many Phosphorous atoms are in 2.4 g Mg3(PO4)2

2b) What is the % Composition of Mg in Mg3(PO4)2?

3a) Find the Molecular formula PxOy : 43.7% P & 56.3% Owith a molar mass between 280 – 290 g/mol.

3b) 5.8 g of PxOy react with 15 g water to form H3PO4.Balance Rxn and find the theoretical yield.

3c) What is the % yield if only 5.4 grams of product is recovered?

Mass Relationships Exam Review