Mass Relationships in Chemical Reactions

Chapter 3HW 39-53 odd, 59, 66, 69,75, 78, 83, 86, 93-94

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 is the mass of an atom in atomic mass units (amu)

Micro-Worldatoms & molecules

Macro-Worldgrams

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

12121212121212121213

1212121212121212121212121212121212121212

12121212121212121212121212121212121212121212121212121212121212121212121212121212121212121212121212121212121212121212121212121212121212121212

Average Atomic Mass CalculationA random sample of Magnesium is composed of

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

If 100 atoms of Mg were randomly selected, 79 atoms would have 12 neutrons (N0), 10 atoms would have 13 N0, and 11 atoms would have 14 N0

• Find the weighted average for the mass of Magnesium by writing percentage as a decimal (e.g. 75% = 0.75)

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

Average Atomic Mass Calculation

A random sample of chlorine is

• 75.8% Cl-35 (34.97 amu) and

• 24.2% Cl-37 (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 3.1Copper, a metal known since ancient times, is used in electrical cables and pennies, among other things.

The atomic masses of its two stable isotopes: (62.93 amu) 69.09%

(64.9278 amu) 30.9%

Calculate the average atomic mass of copper. The relative abundances are given in parentheses.

Example 3.1 Solution

First the percents are converted to fractions:

69.09 percent to 69.09/100 or 0.6909

30.91 percent to 30.91/100 or 0.3091.

We find the contribution to the average atomic mass for each isotope, then add the contributions together to obtain the average atomic mass.

(0.6909) (62.93 amu) + (0.3091) (64.9278 amu) = 63.55 amu

*Always average atomic mass shown of 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

~0.6 Septillion6.02 x 1023

1 mole = NA = 6.02 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 miles2

6.02*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 Moon

7.3 × 1022 kg

Molar mass is the mass of 1 mole of in grams

eggsshoes

marblesatoms

1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g

1 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 molecules

58.443 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

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

1 12C atom

12.00 amux

12.00 g

6.022 x 1023 12C atoms=

1.66 x 10-24 g

1 amu

M = molar mass in g/mol

NA = Avogadro’s number

One Mole of:

C S

Copper Iron

Mercury

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

Particles AtomsMoleculesFormula 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 Factors

MM = molar mass in g/molConverts between mass (g) & moles

NA = Avogadro’s number = 6.02 x 1023

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

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

Example

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

ZincThe conversion factor is the molar mass of Zn.

From the periodic table we find that 1 mol Zn = 65.39 g Zn

Example

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

More Practice ConversionsConvert 58 grams Calcium to moles

58 grams Ca= 1.45 moles Ca1 mol Ca

40.1 grams Cax

Convert 0.2 moles Nickel to grams

Convert 54 grams C to moles atoms

0.2 moles Ni= 11.7 grams Ni58.7 grams Ni

1 moles Nix

54 g C = 2.7 x 1024 C atoms

1 mol C

12 grams Cx

6.02 x 1023 C atoms

1 mol Cx

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

1S 32.07 amu

2O + 2 x 16.00 amu SO2 64.07 amu

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

1 molecule SO2 = 64.07 amu

1 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 amu

1Cl + 35.45 amuNaCl 58.44 amu

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

1 formula unit NaCl = 58.44 amu

1 mole NaCl = 58.44 g NaCl

NaCl

Example 3.5 Determine Molar Mass

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 3.5 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”

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*(1.008 g) = 16.04 g

Example 3.7

How 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 hydrogen atoms in each Urea molecule

urea

= 1.03 × 1024 H atoms

grams of urea → moles of urea → molecules of urea → atoms of H

6.02 x 1023 molecules1 mol of urea

4 Hydrogen atoms1 molecule of urea

1 mol of urea60.06 g urea

25.6 g of urea x xx

Practice Conversions

How many atoms of P are found in 6.30 mg of Mg3(PO4)2?

What is the mass of 3.4 x 1025 atoms of Boron?

What is the % composition of Aluminum in Galaxite (MnAl2O4)? How many grams in 35.6 g galaxite?

Hea

vy

Hea

vyL

ight

Lig

ht

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 abundancies. 1922 Nobel Prize in Chemistry

Sample gets ionized

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 3.8 Percent Composition

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

Calculate the percentcomposition by mass of H, P, and O in this compound.

Example 3.8 Solution

Solution The molar mass of H3PO4 is 97.99 g. The percent by mass of each of the elements in H3PO4 is calculated as follows:

*Always make sure your numbers add back up to 100%

Example 3.10 Compound Mass Percentages

Chalcopyrite (CuFeS2) is a principal mineral of copper.

Calculate the number of kilograms of Cu in 3.71×103 kg of chalcopyrite using percent compositions.

Chalcopyrite.

How do we calculate mass percent of an element?

mass % = (Element Mass)*(Quantity)

Compound Massx 100%

Example 3.10 Solution

The 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 × 103 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 × 103 kg)

= 1.28 × 103 kg

Example Compound Mass PercentagesTalc 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

Percent Composition and Empirical Formulas

Early analysis of many molecules originally only

yielded % composition

Elemental analysis of compounds reveals the mass

% of each element

This allows us to derive the empirical formula

(atom to atom ratio)

Step 1&2

Step 3

Step 4

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 based on experimental

error, Round only if within one tenth.

When the element % composition is known:

Empirical Formula determination

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 ~ 2 O

2.29/2.29 = 1 C

CO2

72.55 g O

27.45 g C

Example 3.9

Determine 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.

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

Example 3.9

Solution If we have 100 g of ascorbic acid, then each percentage can be converted directly to grams and from grams to moles of each element.

Example 3.9

However, chemical formulas are written with whole numbers. Try to convert to whole numbers by dividing all the subscripts by the smallest subscript (3.406):

where the sign means “approximately equal to.”

This gives CH1.33O as the formula for ascorbic acid. Next, we need to convert 1.33, the subscript for H, into an integer.

Multiply by the smallest whole number to obtain all integers.

Example 3.9

This can be done by a trial-and-error procedure:1.33 × 1 = 1.331.33 × 2 = 2.661.33 × 3 = 3.99 ~ 4

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 C0.5H1.5O0.25

Divide by smallest subscript (0.25)

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

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

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

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.

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 Symbols

Symbol 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 Equations

2 Mg + O2 2 MgO

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

Or 2 moles Mg + 1 mole O2 makes 2 moles MgO

Or 48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgONOT

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

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 Equations

2. 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 Glucose

__C6H12O6 + __O2 → __CO2 + __H2O

Reactants6 Carbon12 Hydrogen8 Oxygen

Products1 Carbon2 Hydrogen3 Oxygen

==

6

13

6

6

18

6

18

The balanced equation shows it take 6 molecules of O2 to react with 1 molecule glucose to produce 6 molecules of CO2 and H2O each.

6 mole of O2 reacts with 1 mole glucose to produce 6 moles of CO2 and H2O each.

OR

12

Balance these Chemical equations:

• H3PO4 + NaOH → Na3PO4 + H2O

• Mg + O2 → MgO

• NH4NO2 → N2 + H2O

• P2O5 + H2O → H3PO4

• 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 3.12

When 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.

Stoichiometry Stoichiometry is the quantitative relationship

found between reactants and 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 Stoichiometry is the "recipe of a reaction".

1x2x 2x

1x

From the equation we see that 1 mole CH4 2 mole O2 = “Stoichiometric equivalents”

Mole Ratio1C2H6O + 3O2 → 2CO2 + 3H2O

• 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 + 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 water

x

1. Write balanced chemical equation

2. Convert quantities of known substances into moles

3. Use coefficients in balanced equation to calculate the number of moles of the sought quantity

4. Convert moles of sought quantity into desired units

Amounts of Reactants and Products

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?

Example 3.14

Lithium reacting with water to produce hydrogen gas.

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

9.89 g of H2? g of Li

Example 3.14 Solution

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

35 g C3H81 mol C3H8

44 g C3H8

x x 4 mol H2O1 mol C3H8

x 18 g H2O1 mol H2O

= 57.3 g H2O

35 g C3H8 1 mol C3H8

44 g C3H8

3 mol CO2

1 mol C3H8

44 g CO2

1 mol CO2

= 105 g CO2

x x x

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

*Note: Different amounts of each product are made by the same amount of reactant

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 → 2KOH

SiCl4 + 4H2O → H4SiO4 + 4HCl

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

? g of Al 100 g of Fe

Example 3.13

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).

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

(1 mol C6H12O6 6 mol CO2)

86 g Glucose 1 mol Glucose180 g Glucose

x x 6 mol CO2

1 mol Glucosex

44 g CO2

1 mol CO2

= 126 g CO2

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

Stoichiometry: Chemistry for Massive Creatures - Crash Course Chemistry #6www.youtube.com/watch?v=UL1jmJaUkaQ

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.

% 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 only ended up with 20.3 grams MgO, what is our percent yield?

10 g O21 mol O2

32 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 MgO

25.2 MgOx 100% = 80.6%

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

Limiting ReagentsThe reaction can only proceed as far as the limiting reagent 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 Reagents: 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)

1P2O5 + 3H2O → 2H3PO4

If we react 44 grams of P2O5 with 25 grams of water, which is the limiting reagent 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 P2O51 mol P2O5

141 g P2O5

x x 2 mol H3PO4

1 mol P2O5

x 98 g H3PO4

1 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 reagent and there is excess water.

Example: Limiting Reagent and Excess

Urea [(NH2)2CO] is prepared by reacting ammonia with CO2:

637.2 g of NH3 are reacted with 1142 g of CO2.

(a) Which of the two reactants is the limiting reagent?

(b) Calculate the mass of product (NH2)2CO formed.

(c) How much excess reagent (in grams) is left at the end of the reaction?

Example

Solution We carry out two separate calculations. First, starting with 637.2 g of NH3, we calculate the number of moles of (NH2)2CO that could be produced if all the NH3 reacted according to the following conversions:

Combining these conversions in one step, we write

3.15 Solutions

Example

Second, for 1142 g of CO2, the conversions are

The number of moles of (NH2)2CO that could be produced if all the CO2 reacted is

It follows, therefore, that NH3 must be the limiting reagent because it produces a smaller amount of (NH2)2CO.

3.15 Solutions

Example

Solution The molar mass of (NH2)2CO is 60.06 g. We use this as a conversion factor to convert from moles of (NH2)2CO to grams of (NH2)2CO:

3.15 Solutions

Example 3.15

(c) Strategy Working backward, we can determine the amount of CO2 that reacted to produce 18.71 moles of (NH2)2CO.

The amount of CO2 left over is the difference between the initial amount and the amount reacted.

Solution Starting with 18.71 moles of (NH2)2CO, we can determine the mass of CO2 that reacted using the mole ratio from the balanced equation and the molar mass of CO2. The conversion steps are

3.15 Solutions

Example 3.15 Solutions

Combining these conversions in one step, we write

The amount of CO2 remaining (in excess) is the difference between the initial amount (1142 g) and the amount reacted (823.4 g):

mass of CO2 remaining = 1142 g − 823.4 g = 319 g

40 g of NO are mixed with 8 g O2 to form NO2:2NO + O2 → 2NO2

Which reactant is limiting the reaction?

How much NO2 is produced?

How much of the other reactant is left?

If only 18 g is recovered from the reaction, what’s the percent yield?

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

Example 3.17

Titanium is a strong, lightweight, corrosion-resistant metal that is used in rockets, aircraft, jet engines, and bicycle frames. It is prepared by the reaction of titanium (IV) chloride with molten magnesium between 950°C and 1150°C:

In a certain industrial operation 3.54 × 107 g of TiCl4 are reacted with 1.13 × 107 g of Mg.

• Calculate the theoretical yield of Ti in grams.

• Calculate the percent yield if 7.91 × 106 g of Ti are actually obtained.

Example 3.17

Solution Carry out two separate calculations to see which of the two reactants is the limiting reagent. First, starting with 3.54 × 107 g of TiCl4, calculate the number of moles of Ti that could be produced if all the TiCl4 reacted. The conversions are

so that

Example 3.17

Next, we calculate the number of moles of Ti formed from1.13 × 107 g of Mg. The conversion steps are

And we write

Therefore, TiCl4 is the limiting reagent because it produces a smaller amount of Ti.

Example 3.17

The mass of Ti formed is

(b) Strategy The mass of Ti determined in part (a) is the theoretical yield. The amount given in part (b) is the actual yield of the reaction.

Example 3.17

Solution The percent yield is given by

Check Should the percent yield be less than 100 percent?

Fun Practice 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 ~120?

Fun? Practice 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

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

2) Find the Molecular formula PxOy : 43.7% P & 56.3% O

with a molar mass between 280 – 290 g/mol.

3) 5.8 g of PxOy react with 15 g water to form H3PO4.

Balance Rxn and find the theoretical yield.

Mass Relationships Exam Review