CHEMISTRY 2202 UNIT 1 STOICHIOMETRY · STOICHIOMETRY: - the calculation of quantities USED IN or PRODUCED by a chemical reaction. 3 Sections: Part 1 - Mole calculations (ch. 2 & 3

CHEMISTRY 2202UNIT 1

STOICHIOMETRY

Introduction

Chemistry is both a qualitative (describing with words) and a quantitative (describing with numbers) science.

Chemical equations are ‘recipes’ that show chemists what we need and what will be produced in a reaction.

eg. H2SO4(aq) + 2 NaOH(aq) Na2SO4(aq) + 2 H2O(l)

Introduction

This balanced equation helps us answer questions like:● How much sulfuric acid is needed to neutralize

30 mL of 5.00 M sodium hydroxide? ● What volume of water will be produced when

30 mL of sodium hydroxide reacts with 10 mL of

sulfuric acid?

Chemical equations help us “count” molecules or QUANTIFY chemical reactions.

Unit Outline

STOICHIOMETRY:

- the calculation of quantities USED IN or PRODUCED by a chemical reaction.

3 Sections:

Part 1 - Mole calculations

(ch. 2 & 3 in text)

Part 2 - Using mole ratios in chemical equations (ch. 4)

Part 3 - Solution stoichiometry

(ch. 6)

PART 1 – MOLE CALCULATIONS ● Isotopes and Atomic Mass● Average Atomic Mass ● Molar Mass ● Mole Calculations ● Molar Volume ● Percent Composition ● Empirical Formulas ● Molecular Formulas ● Formula of a Hydrate

ISOTOPES and ATOMIC MASS Atomic number:

▫The # of protons in an atom or ion.

▫In a neutral atom: #p+ = #e-

eg.

● Atomic number of C is _____

ISOTOPES and ATOMIC MASS •Mass number:

▫The # located in the bottom of each box on the periodic table (has decimal places). ▫Mass Number = #p+ + #no

ie. mass # - #p+ = #no

eg. ● mass number of C = _____ ● # of no in C = _____

Your Turn!!!

Atom Atomic # Atomic Mass # #p

+#e

-#n

o

Carbon (C) 6 12

Carbon (C) 6 13

Carbon (C) 6 14

ISOTOPES AND ATOMIC MASS . . .

•ISOTOPE

▫Atoms that have the same number of protons (ie. same element) but a different NUMBERS OF NEUTRONS are called isotopes

ISOTOPES AND ATOMIC MASS . . . ▫Isotopes have different mass numbers.

eg. Carbon has 3 naturally occurring isotopes:

● carbon-12 98.9 % of all carbon in the world

● carbon-13 1.1 % of all carbon in the world

● carbon-14 0.0000000001 % of all carbon

▫When these isotope values are averaged, with their percent abundance in nature, we get the number that appears on the periodic table (12.01 g/mol).


ISOTOPE NOTATION:

ISOTOPES and ATOMIC MASS Example

Carbon-12 Carbon-13 Carbon-14

How many neutrons are in each isotope?


Name each isotope of magnesium.

How many neutrons are in each isotope?

YOUR TURN!!!

A. hydrogen-2

B.

YOUR TURN!!!

Write the isotope notation for an isotope of:

A. Copper that has 29 p+ and 35 no

B. Vanadium that has 23 p+

and 28 no.

AVERAGE ATOMIC MASS (AAM)

ANALOGY: ● The average mark you receive in this course is a

weighted average of all your scores. ● Your final grade depends on how well you do in

different pieces of work and their % of the total grading scheme.

● Your final mark is NOT a straight average. ● On the periodic table, the atomic mass numbers

are also an average, based on different types of isotopes and their relative percent abundance.

AVERAGE ATOMIC MASS . . .

● This average is based on the concept of the ATOMIC MASS UNIT (AMU)

● The symbol for the AMU is μ. ● 1 μ = 1/12 the mass of a carbon-12 atom

(from IUPAC conference, 1961) ie. 1 carbon-12 atom has a mass of 12 μ. ● On the periodic table, carbon’s atomic mass is

an average, 12.01, expressed in g/mol.

AVERAGE ATOMIC MASS . . . To calculate average atomic mass (AAM):

1.Change percent abundance value to decimal value (divide by 100).

2. Multiply the atomic mass by its % abundance for each isotope. 3. Add these numbers together. 4. Use significant digits.

In other words: AAM = (exact atomic mass x % abundance as a decimal) +

(exact atomic mass x % abundance as a decimal) +

(exact atomic mass x % abundance as a decimal) + .…


ISOTOPE ATOMIC MASS % ABUNDANCE

Lithium-6 6.01513 u 7.420

Lithium-7 7.01601 u 92.58

Calculate the average atomic mass of lithium from the given data:


Calculate the average atomic mass of Mg given that it has three isotopes: ● magnesium-24 → 23.98 μ; 78.60% abundance ● magnesium-25 → 24.99 μ; 10.11 % abundance ● magnesium-26 → 25.98 μ; 11.29 % abundance

Naturally occurring magnesium exists as a mixture of three isotopes. Mg-24 has an atomic mass of 23.985 amu and a relative abundance of 78.70 %. Mg-25 has an atomic mass of 24.985 amu and a relative abundance of 10.13%. The average atomic mass of magnesium is 24.31 amu.

Calculate the atomic mass of the remaining isotope.

For you to do!

Page 45-#2

Page 46-#2

The Mole

•INTRODUCTION:

▫Chemists face the

difficult task of trying to

measure extremely small

particles, such as: ● FORMULA UNITS and IONS for ionic compounds ● MOLECULES for molecular compounds ● ATOMS for elements

The Mole

eg.

1 teaspoon of water → 1.67 x 1023

molecules

(5.0 mL)

•A device that could count molecules at a rate of 1 million per second would take over 5 billion years to count the molecules in a teaspoon of water!!!

The mole

Item Quantity Amount

Gloves Pair 2

Soft-drinks six-pack 6

eggs dozen 12

pens Gross (12 dozen) 144

paper ream 500

Names to represent some common quantities:

The Mole

IUPAC to the RESCUE!!

•Chemists use the concept of the MOLE as a convenient way to deal with counting huge numbers of particles.

One mole is the number of atoms contained in exactly 12 g of carbon-12.

The Mole

1 mol = 6.02 x 1023 particles

(atoms, ions, molecules, formula units)

•This number is referred to as Avogadro’s Number (in honour of scientist Amadeo Avogadro who discovered it), and is written:

NA = 6.02 x 1023 particles/mol

The letter n is used to represent

# of moles

The Mole

•Just how big is this number?

•Put it this way . . .

▫If we had 6.02 x 1023

marbles ( 1 mol of marbles), it would be enough to cover all of Earth to a depth of 80 km!

▫If you won 6.02 x 1023

dollars, it would be enough to spend a billion dollars a day for over a trillion years before you had to worry about running out!

▫(In book, how big? p. 49)

The Mole•One mole always contains 6.02 x 10

23 particles.

ie.

● 1 mol C = 6.02 x 1023 atoms

● 1 mol O2 = 6.02 x 1023 molecules of O2

● 1 mol CH4 = 6.02 x 1023 molecules of CH4

= 2.41 x 1024 atoms of H

● 1 mol NaCl = 6.02 x 1023 formula units

= 6.02 x 1023 Na+

ions

= 6.02 x 1023 Cl- ions

Mole Calculations – Mole to Particle

● Where:

n = number of moles (mol)

N = number of particles

(formula units, ions, molecules, or atoms)

NA = Avogadro’s number

(6.02 x 1023 particles/mol)


eg #1:

How many moles in 1.21 x 1035

molecules of CO2?


eg #2: How many atoms in 0.35 mol of Fe?


eg #3:

How many formula units in 5.69 mol of Na2SO4?


eg #4:

How many Na+

ions in 5.69 mol of Na2SO4?

HOMEWORK!!!

Pg. 52 #'s 7,8,9,10

MOLAR MASS (M)

INTRODUCTION:

•Moles cannot be measured directly.

•Chemists measure chemical amounts using

mass for solids and volume for gases & liquids.

•To convert between moles and mass we need

the mass of one mole or the MOLAR MASS.

Molar Mass (M) of Elements •The mass of 1 mol of a substance is referred to as its molar mass (M) and is measured in g/mol.

eg.

1 mol C → 12.01 g/mol

1 mol H → 1.01 g/mol

1 mol Pb → 207.19 g/mol

1 mol Na → 22.99 g/mol

Molar Mass (M) of Compounds

•The molar mass of a compound is the sum of the molar masses of each element in the compound.

eg.

Calculate the molar mass of: ● H2O ● Na2SO4

● Ba(MnO4)2 .8 H2O

For you to do!

● Pg. 57 # 17 &19 Answers on page 77

Mole Calculations – Mole to Mass

Rearrange:

Where: n = number of moles (mol) m = mass (g) M = molar mass (g/mol)


eg #1:

How many moles in 4.67 g of copper ?


eg # 2:

What is the mass of 0.487 mol of H2O ?


eg # 3:

Calculate the mass of 2.00 mol of H2(g).


eg # 4:

How many moles are in 80.0 g of NH4Cl ?

For you to do!!

Pg. 59 #20 & Pg. 60 #25

Answers on page 77

Note● Molar amount = number of moles of whatever

● 1 kg = 1000 g ● 1 g = 1000 mg

Mole Calculation – Mass to Particles

• We do not have a direct quantity that relates mass (m) and number of particles (N), but they are related through number of moles (n).


eg.

How many molecules are in 26.9 g of H2O(l)?

m = 26.9 g

Mwater= 18.02 g/mol

NA = 6.02 x 1023 molecules/mol

Find N


eg.

A sample of Sn contains 4.69 x 1028 atoms. Calculate

its mass.

N = 4.69 x 1028 atoms


MSn = 118.69 g/mol

Find m


eg.

How many molecules are in 4.78 g of glucose?

m = 4.78 g

Mglucose= 180.18 g/mol


Find N


Particles(N)

Moles(n)

Mass(m)

X NA

÷ NA

x M

÷ M


Practice: p. 63 #’s 28 & 33

p. 64 #’s 34 & 37

MOLAR VOLUME (gases)

Pressure (P)

•The force exerted per unit of surface area.

•SI unit is the Pascal (Pa).

(Used to be the atmosphere - atm)

Temperature (T)

•measures the kinetic energy of the particles.

•SI unit is Celsius degrees (oC).

•Other unit is Kelvin (K).


•The volume of a gas is dependent on P and T.

A ‘COOL’ balloon

Pressure & Volume

https://www.youtube.com/watch?v=5doUIdm0jTU

https://www.youtube.com/watch?v=DcnuQoEy6wA


Standard Conditions

(needed to measure gas volumes)

STP – Standard Temperature and Pressure

T = 0oC (273 K)

P = 101.3 kPa (1 atm)

SATP – Standard Ambient Temperature and Pressure

T = 25oC

P = 100 kPa


•Experimental evidence shows that the volume of one mole of ANY GAS at STP is 22.4 L/mol.

ie. MV = 22.4 L/mol OR VSTP = 22.4 L/mol

Mole Calculations – Molar Volume

Where:

n = number of moles, in mol

v = volume, in L

VSTP

= molar volume of a gas at STP

22.4 L/mol


eg 1:

What is the volume, in L, of 0.600 mol of SO2 gas at STP?


eg 2:

What is the number of moles in 33.6 L of He gas at STP?


eg 3:

What is the mass of 44.8 L of methane at STP?


eg 4:

A chemist isolates 2.99 g of a gas. The sample occupies 800.0 mL at STP. Is the gas most likely to be O2(g), Kr(g), Ne(g), or F2(g)?

(Hint: Determine the molar mass of the gas.)

HOMEWORK!!!

p. 73 #’s 38 - 43

SUMMARY

PERCENT COMPOSITION

The composition of many substances is expressed in terms

of percentages.

The London 2012 Olympic medals weigh between 375 - 400g.

Gold medal - 92.5% silver, 1.34% gold, with the remainder copper.

Silver medal - 92.5% silver, with the remainder copper.

Bronze medal - 97.0% copper, 2.5% zinc and 0.5% tin.

3.5 % of sea water is dissolved salt

EXCEPT

the Dead Sea which is 33% salt.

PERCENT COMPOSITION

Two types of calculations:

1.Percent Composition given mass

% element = mass of element x 100

mass of compound

Percent Composition given mass (g)

eg. 1

8.20 g of Mg combines with 5.40 g of O to form MgO. Determine the percent composition of this compound?

Percent Composition given mass (g)

eg. 2

29.0 g of Ag combines with 4.30 of S to

form Ag2S. Calculate the mass percent for S

in this compound?

Your Turn!

p. 82 #’s 2 & 4

PERCENT COMPOSITION •Percent composition is the percent by mass of each element in a compound

•Should add up to 100%.

•AKA: Mass Percent

eg.

C8H

8O

3 – the vanillin molecule

C-->63.1 %

H-->5.3 %

O-->31.6 %

TOTAL: 100 %

PERCENT COMPOSITION

2.Percent Composition given the formula

% element = M of element x 100 M = molar mass

M of compound

Percent Composition given the Formula

eg. 3

Calculate the percent composition of C2H6.

79.85 %

20.2 %

Based on your answer, calculate the mass

of C in an 82 g sample of C2H6.

Percent Composition given the Formula

eg. 4

Calculate the mass percent of NaHSO4.

M = 120.07 g/mol

19.15 % 0.841 % 26.71 % 53.30 %

Based on your answer, calculate the mass of H in a 20.2 g sample of NaHSO4.

For you to do!

Pg. 85 #'s 5 & 8

EMPIRICAL and MOLECULAR FORMULAS

● An empirical formula gives the simplest ratio of elements in a compound.

● A molecular formula shows the actual number of atoms in a molecule of a compound.

Ex: glucose: molecular formula-->

empirical formula-->● Ionic compounds are always wrote as empirical

formulas – simplest ratio● A molecular formula is only used for molecular

compounds

EMPIRICAL and MOLECULAR FORMULAS Compound Molecular Formula Empirical Formula

Butane

C4H

10 C

2H

5

Glucose

C6H

12O

6 C

6H

12O

6

Water

H2O H

2O

Benzene

C6H

6

CH

EMPIRICAL FORMULAS

The empirical formula of a compound may be determined from the % composition of a compound.

Method:

1.Change % to mass out of 100 g.

2.Convert mass to moles.

3.Divide by the smallest number of moles to get the ratio for the empirical formula.

4.Multiply to get a whole number ratio.

(if needed)

EMPIRICAL FORMULAS

Page 90 in book. Helpful for EF calculations.

A FUN WAY TO REMEMBER EF calculations . . .

Percent to mass

Mass to Mole

Divide by small

Multiply till whole

By Joel Thompson

EMPIRICAL FORMULAS

eg. 1

A compound was analyzed and found to contain 87.4% N and 12.6 % H by mass. Determine the empirical formula of the compound.

EMPIRICAL FORMULAS

eg. 2

What is the empirical formula of a compound that is 25.9 % N and 74.1 % O ?

EMPIRICAL FORMULAS

eg. 3

A compound contains 89.91% C and 10.08 % H by mass. Determine the empirical formula of the compound.

MOLECULAR FORMULA CALCULATIONS

•We can determine the molecular formula of a compound using the molar mass and the empirical formula of a compound.

METHOD:

1.Determine empirical formula (may be given).

2.Divide the molar mass by the empirical formula mass.

3.Round off the result to the nearest whole number.

4.Multiply empirical formula by this whole number to get the molecular formula.


eg. 1

Determine the molecular formula of a compound with a molar mass of 60.00 g/mol and an empirical formula of CH4N.


eg. 2

The empirical formula of hydrazine is NH2. The molar mass of hydrazine is 32.06 g/mol. What is the molecular formula for hydrazine?

For You to Try

p. 97 #’s 17 – 20

Unit 1 Part 2

CALCULATIONS AND CHEMICAL EQUATIONS

I know the answer!

TEE HEE!

UNIT 1 – PART 2 (CH 4)

Topics: ● Stoichiometry

● definition – what it is ● types of stoichiometry ● mole ratios ● calculations using n & m

● other calculations using v & VSTP

● Theoretical yield, actual yield and percent yield ● Limiting and Excess Reagent ● Core Lab 2 – % Yield in a Chemical Reaction

STOICHIOMETRY

● Stoichiometry comes from the Greek words stoikheion, meaning “element” and metron, meaning “to measure”.

● Stoichiometry is the determination of quantities needed for, or quantities produced by, chemical reactions.

STOICHIOMETRY

3 types of stoichiometry

1. Gravimetric stoichiometry ● stoichiometry using mass (g) of solids(s).

2. Solution stoichiometry ● stoichiometry using concentration & volume of

solutions.

3. Gas stoichiometry ● stoichiometry using volume (L) of gases (g) at

STP.

STOICHIOMETRY

Q: How do we calculate quantities used

or quantities produced in chemical reactions?

A: Mole Ratios

• The coefficients from a balanced chemical equation provide the mole ratios of reactants

and products in a chemical reaction.

• Coefficients from a balanced chemical equation, work like a sandwich recipe . . .

MOLE RATIOS

Clubhouse sandwich recipe

MOLE RATIOS

Slices of Toast

Slices of Turkey

Strips of Bacon

# of Sandwiches

12

27

66

100

Fill in the missing quantities

MOLE RATIOS

● Mole ratios from balanced chemical equations work in the same way

eg: equation for an air bag exploding

sodium azide sodium + nitrogen gas 2 NaN3(s) 2 Na(s) + 3 N2(g)

2 UNITS or

2 MOLESNOT 2 g

MOLE RATIOS

2 NaN3(s) 2 Na(s) + 3 N2(g)

6

10

150

10

MOLE RATIOS

● A mole ratio is a mathematical expression that shows the relative amounts of two species involved in a chemical change.

● A mole ratio ● comes from a balanced chemical equation ● show the relative amounts of reactants &

products in moles

looks like ---> coefficient required

coefficient given

MOLE RATIOS

● Mole ratios from balanced chemical equations represent relative number of particles (atoms, molecules, ions, formula units or moles).

C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g)

15 mol

18 molecules

35 molecules

7.38 mol

MOLE RATIOS

NOTE:

Mole ratios DO NOT REPRESENT

MASS

P4 + 5 O2 → 2 P2O5

1 g of phosphorus reacts with 5 g of oxygen

MOLE TO MOLE STOICHIOMETRY

STEPS:

1.Write a balanced chemical equation with subscripts

(Identify given and required values).

2.Use MOLE RATIOS from balanced chemical equation .

nrequired = ngiven x coefficient required

coefficient given


eg.1

How many moles of ammonia gas are produced when 0.60 mol of nitrogen gas reacts with hydrogen gas?

(1.2mol)


eg.2

How many moles of CO2 are produced when 31.5 mol of C3H8 is burned? (94.5 mol)


eg.3

Calculate the number of moles of magnesium needed to produce 0.586 mol of Mg3N2. (1.76 mol)


eg.4

How many moles of sulfur are produced when 4.25 mol of SO3 decomposes? (0.531 mol)

FOR YOU TO TRY

● Pg. 114-117 #'s 2,5,9

MORE STOICHIOMETRY CALULATIONS

● Other types of Stoichiometry: ● Mole to Mass (or Mass to Mole) ● Mass to Mass ● Volume to Volume (for gases at STP) ● SolutionStoichiometry (Part 3 of Unit 1...)

MOLE TO MASS STOICHIOMETRY

● Mole to Mass Calculations are the same as Mole to Mole but also require conversion between mass and moles:

● Use n=m or m= n x M

M


STEPS:

1.Balanced chemical - identify given and required

2.Convert mass to moles

3.Mole Ratio


coefficient given


eg. 1:

Calculate the number of moles of oxygen that will react with 6.49 g of aluminum. (0.180 mol)


eg. 2:

How many moles of water are produced when 20.6 g of CH4 burns? (2.57 mol)


eg. 3: Calculate the mass of aluminum oxide that can be produced from the reaction between 0.1804 mol of oxygen gas and excess aluminum. (18.39g)


eg. 4: What mass of Ca3P2 is produced when

4.38 mol of Ca reacts with phosphorus? (266g)

For you to Try

Pg. 149 #'s 6 & 7

Mass to Mass Stoichiometry

STEPS:

1.Balanced chemical (given!! required!! subscripts!!)

2.Convert mass to moles.

3.Mole Ratio


coefficient given

4.Convert moles to mass


NOTE: The coefficients from balanced equations

represent MOLE RATIOS.

MASSES are NOT to be used

in the MOLE RATIO step.


eg1:

How many grams of NH3 gas are produced when 5.40 g of hydrogen gas reacts with nitrogen gas? (30.4g)


eg 2:

What mass of nitrogen gas is needed to react completely with hydrogen gas and produce 30.6 g of ammonia gas?

(24.72g)


eg 3:

What mass of CaCl2(aq) is produced when Ca(NO3)2(aq)

reacts with 4.39 g of NaCl(s)? (4.17g)


eg 4:

Given that 45 g of sulfur (S8) gas reacts with aluminum solid, find the mass of product. (7.0x101g)

For you to try

Page 121 #'s 11-14

LAW OF CONSERVATION OF MASS

● Mass is conserved in chemical reaction.

OR

total mass of = total mass of

the reactants the products. ● We can verify this law by using the mole ratios

and molar masses of reactants and products from balanced chemical equations.

Law of Conversation of Mass

Quantity 2 Mg(s) + O2(g) 2 MgO(s)

n (mol) 2 1 2

m (g) 2(24.31) 32.00 2(40.31)

The total mass of the reactants is 2(24.31) + 32.00 = 80.62gThe total mass of the products is 2(24.31+16.00) = 2(40.31) = 80.62g

The mass of the products and reactants are the same. So mass is conserved!!

Law of Conservation of Mass

Ex:

The decomposition of 500.00 g of Na3N produces

323.20 g of N2. How much Na is produced in this

decomposition?

Mole to Volume Stoichiometry

If gas volume at STP is GIVEN or REQUIRED you must use VSTP in your calculations.

eg 1:

What volume of oxygen at STP is needed to react with 2.34 g of NO2 ?

2 N2O5(s) 4 NO2(g) + O2(g)

Mole to Volume Stoichiometry

eg 2:

Calculate the mass of N2H4(g) needed to produce 157 L of nitrogen gas at STP in the reaction below. 2 N2H4(l) + N2O4(l) 3 N2(g) + 4 H2O(l)

For you to try

p. 123 # 16

LIMITING AND EXCESS REAGENTS

Sandwich Time Again!


Here’s what we find in our kitchen:

◦How many sandwiches can we make??

6 Slices of Toast

12 Pieces of Turkey

20 slices of bacon

What limits the # of sandwiches??


Chemical reactions with 2 or more reactants, stop when 1 reactant is completely used up.

Limiting reagent ● AKA limiting reactant ● The reactant that is completely used in a

chemical reaction, thereby limiting the amount of product that can form.


● Excess reagent ● AKA excess reactant ● the reactant that is left over after the reaction

is complete. ● there is more than is needed for complete

reaction. ● We can predict which reactant is the LR by

using mole ratios from balanced chemical equations.

DETERMING THE LIMITING REAGENT

How to Solve:

1.Write balanced chemical equation.

2.Calculate n(mols) for EACH reactant.

3.Use the mole ratio for each reactant. ● The LR is the reactant that produces less moles

of product

4.Calculate the amount of product using the LR.

DETERMINE THE LIMITING REAGENT

eg 1:

Determine the Limiting Reagent and Excess Reagent if 14.8 g of C3H8(g) reacts with 2.14 g of O2(g).

NOTE:

When you are given the quantity (mass, volume or moles) of 2 or more reactants in a stoichiometry question, YOU MUST ALWAYS determine the LR and use it to determine the amount of product.

Limiting Reagent

eg 2.

What mass of NaCl is produced when 6.70 mol of Na reacts with 3.20 mol of Cl2?

Limiting Reagent

eg 3.

What mass of CaCO3 will be produced when 20.0 g of Ca3(PO4)2 reacts with 15.0 g of Na2CO3 ?

Limiting Reagent

eg 4.

What volume of H2 at STP will be produced when 10.0 g of Zn reacts with 20.0 g of HCl(aq)?

For you to try

p. 134

#’s 27ab & 30a

Percent Yield Calculations

● The term YIELD refers to the mass of product formed in a chemical reaction.

eg. Burning 99.5 g of propane in sufficient oxygen produces 298 g of carbon dioxide gas.

THEORETICAL YIELD ● what you are supposed to get ● The amount of product predicted or calculated.


● A question that asks you to find the theoretical yield is asking you to calculate the amount expected.

Actual yield ● what you actually get in an experiment ● The amount actually produced in a chemical

reaction in the LAB


For most reactions, your actual yield is less than your theoretical yield, for many reasons, including:

Reaction is slow and was incomplete. Impurities are present. Reaction goes to equilibrium (never really

finishes). Sample is wet when weighed. Experimental error.



eg 1.

Calculate the theoretical yield of copper if 1.87 g of Al reacts with excess aqueous copper (II) sulfate?

2 Al + 3 CuSO4 → Al2(SO4)3 + 3 Cu


eg 1.(cont’d)

If the reaction produces 3.74 g of copper, what is the percent yield of copper?


eg 2.

20.0 g of bromic acid, HBrO3 (aq) , is reacted with excess HBr.

HBrO3(aq) + 5 HBr(aq) 3 H2O(l) + 3 Br2(aq)

(a) What is the theoretical yield of Br2 for this reaction? (74.4 g)

(b) If 47.3 g of Br2 are produced, what is the percentage yield of Br2? (63.6%)


eg 3.

Barium sulfate forms a precipitate in this reaction:

Ba(NO3)2 (aq) + Na2SO4 (aq) BaSO4(s) + 2 NaNO3(aq)

When 35.0 g of Ba(NO3)2 reacts with excess Na2SO4,

29.8 g of BaSO4 are recovered by the chemist.

(a) Calculate the theoretical yield of BaSO4. (31.3 g)

(b) Calculate the percentage yield of BaSO4. (95.2 %)


eg 4.

Yeasts can act on a sugar, such as glucose, C6H12O6, to produce ethyl alcohol, C2H5OH, and carbon dioxide.

C6H12O6 (s) 2 C2H5OH(l) + 2CO2(g)

If 223 g of ethyl alcohol are recovered after 1.63 kg of glucose react, what is the percentage yield of the reaction? (834 g, 26.7 %)

Solutions Chemistry

Chapter 7

Solutions

● Terms ● Molar Concentration (mol/L)● Dilutions● % Concentration (pp. 255 – 263)● Solution Process● Solution Preparation● Solution Stoichiometry● Dissociation

Terms

solutionsolventsoluteconcentrateddiluteaqueousmiscibleimmisciblealloysolubility

molar solubilitysaturatedunsaturatedsupersaturateddissociationelectrolytenon-electrolytesolubleinsoluble

limiting reagentexcess reagentactual yieldtheoretical yielddecantingpipettingfiltrateprecipitatedynamic equilibrium

Terms

solution - is a homogeneous mixture solute - the substance that dissolves OR the

substance in lesser quantitysolvent - the substance which dissolves the

solute OR the substance in greater quantityconcentrated - a large amount of solute relative

to the amount of solvent dilute - a small amount of solute relative to the

amount of solvent

Terms

saturated – contains the maximum amount of dissolved solute at a given temperature and pressure

unsaturated – contains less than the maximum amount of dissolved solute at a given temperature and pressure

supersaturated – contains more than the maximum amount of dissolved solute for a given temperature and pressure

Terms

miscible – liquids that dissolve in each other immiscible – liquids that do not dissolve in each

otheraqueous - the solvent is water alloy - a solid solution of two or more metals

Terms

Solubility - the maximum amount of solute that can be dissolved under specific temperature and pressure conditions

eg. the solubility of HCl at 25 °C is 12.4 mol/L

eg. 100.0 mL of water at 25°C dissolves 36.2 g of sodium chloride

Terms

soluble – solubility is greater than 1 g per 100 mL of solvent.

insoluble - solubility is less than 0.1 g per 100 mL of solvent.

9 types of solution

Rate of dissolving

● for most solids, the rate of dissolving is greater at higher temperatures

● stirring a mixture or by shaking the container increases the rate of dissolving.

● decreasing the size of the particles increases the rate of dissolving.

Solubility

● small molecules are often more soluble than larger molecules.

● solubility of solids increases with temperature.● the solubility of most liquids is not affected by

temperature.● the solubility of gases decreases as

temperature increases● an increase in pressure increases the solubility

of a gas in a liquid.

“Like dissolves like”

● ionic solutes and polar covalent solutes both dissolve in polar solvents

● non-polar solutes dissolve in non-polar solvents.

● compounds with very strong ionic bonds, such as AgCl, tend to be less soluble in water than compounds with weak ionic bonds, such as NaCl.

Surprise Review

● Find the molar mass of Ca(OH)2

● How many moles in 45.67 g of Ca(OH)2?

● Find the mass of 0.987 mol of Ca(OH)2.

Molar Concentration

The terms concentrated and dilute are qualitative descriptions of solubility.

A quantitative measure of solubility uses numbers to describe how much solute is dissolved or the concentration of a solution.

Molar Concentration

The MOLAR CONCENTRATION of a solution is the number of moles of solute (n) per litre of solution (v).

Molar Concentration

FORMULA:Molar Concentration = number of moles volume in litres

C = n V

mol/L OR M (molar)

C

n V orVxCn Rearranged Formulae

Molar Concentration

Calculate the molar concentration of:

1). 4.65 mol of NaOH is dissolved to prepare 2.83 L of solution. (1.64 mol/L)

2). 15.50 g of NaOH is dissolved to prepare 475 mL of solution.

( 0.3875 mol → 0.816 mol/L)

Molar Concentration

3). the volume of 6.00 mol/L HCl(aq) that can be made using 0.500 mol of HCl. (0.083L)

4). the volume of 1.60 mol/L HCl(aq) that can be made using 20.0 g of HCl. (0.343L)

Dilution (p. 272)When a solution is diluted:- The concentration decreases- The volume increases- The number of moles remains the

same

ni = nf Number of moles after dilution

Number of moles before dilution

Dilution (p. 272) ni = nf

Ci Vi = Cf Vf

eg. Calculate the molar concentration of a vinegar solution prepared by diluting 10.0 mL of a 17.4 mol/L solution to a final volume of 3.50 L. (0.0497 mol/L)

p. 273 #’s 25 – 27

p. 276 #’s 1, 2, 4, & 5

I will actually be checking these questions. So please do them!!

June 2010 # 12.

A lab technician prepares a dilute solution of hydrochloric acid. If 50.0 mL of 2.50 mol/L hydrochloric acid is added to 450.0 mL of water, what is the new concentration?

(A) 0.250 mol/L (C) 3.60 mol/L

(B) 0.278 mol/L (D) 4.00 mol/L

Answer: (B)

Percent Concentration● Concentration may also be given as a %.● The amount of solute is a percentage of the

total volume or mass of solution. liquids in liquids - % v/v solids in liquids - % m/v solids in solids - % m/m

Percent Concentration

100x(mL) solutionofvolume

(g) soluteofmass(m/v)Percent

100x(g) solutionofmass

(g) soluteofmass(m/m)Percent

100x(mL) solutionofvolume

(mL) soluteofvolume(v/v)Percent

Concentration in ppm and ppb

Parts per million (ppm) and parts per billion (ppb) are used for extremely small concentrations

610solution

solute

mxppmm

910solution

solute

mxppbm

eg. 5.00 mg of NaF is dissolved in 100.0 kg of solution. Calculate the concentration in:

a) ppm

b) ppb

ppm = 0.005 g x 106

100,000 g

= 0.05 ppm

ppb = 0.005 g x 109

100,000 g

= 50.0 ppb

For you to do as homework!!

● p. 265 #’s 15 – 17

Solution Preparation & Dilution● standard solution – a solution of known

concentration● volumetric flask – a flat-bottomed glass vessel that

is used to prepare a standard solution● delivery pipet – a pipet that accurately measures

one volume● graduated pipet – pipets that have a series of lines

that can be used to measure many different volumes

To prepare a standard solution:

1. calculate the mass of solute needed

2. weigh out the desired mass

3. dissolve the solute in a beaker using less than the final volume

4. transfer the solution to a volumetric flask (rinse the beaker into the flask)

5. add water until the bottom of the meniscus is at the etched line

To dilute a standard solution:● 1. Rinse the pipet several times with deionized

water.● 2. Rinse the pipet twice with the standard

solution.● 3. Use the pipet to transfer the required

volume.● 4. Add enough water to bring the solution to

its final volume.

Solution Stoichiometry

1. Write a balanced equation

2. Calculate moles given

n=m/M

3. Mole ratios

4. Calculate required quantity

m = n x M

V

nCOR

C

nVOR

OR n = CV

Solution Stoichiometry

eg. 45.0 mL of a HCl(aq) solution is used to neutralize 30.0 mL of a 2.48 mol/L NaOH solution.

Calculate the molar concentration of the HCl(aq) solution.

Sample Problems1. What mass of copper metal is needed to

react with 250.0 mL of 0.100 mol/L silver nitrate solution? (0.794g)

2. Calculate the volume of 2.00 M HCl(aq) needed to neutralize 1.20 g of dissolved NaOH. (0.0150L)

3. What volume of 3.00 mol/L HNO3(aq) is needed to neutralize 450.0 mL of 0.100 mol/L Sr(OH)2(aq)? (30.0mL)

Sample Problem Solutions

Cu(s) + 2 AgNO3(aq) → 2 Ag(s) + Cu(NO3)2(aq)

Step 2 n = 0.02500 mol AgNO3

Step 3 n = 0.01250 mol Cu

Step 4 m = 0.794 g Cu


HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

Step 2 n = 0.0300 mol NaOH

Step 3 n = 0.0300 mol HCl

Step 4 V = 0.0150 L HCl


2 HNO3(aq) + Sr(OH)2(aq) →

2 H2O(l) + Sr(NO3)2(aq)

Step 2 n = 0.04500 mol Sr(OH)2

Step 3 n = 0.0900 mol HNO3

Step 4 V = 0.0300 mol/L HNO3

The Solution Process (p. 299)

● Dissociation occurs when an ionic compound breaks into ions as it dissolves in water.

● A dissociation equation shows what happens to an ionic compound in water.

eg. NaCl(s) → Na+(aq) + Cl-(aq)

K2SO4(s) → 2 K+(aq) + SO4

2-(aq)

Ca(NO3)2(s) → Ca2+(aq) + 2 NO3

-(aq)


● Solutions of ionic compounds conduct electric current.

● A solute that conducts an electric current in an aqueous solution is called an electrolyte.


● Acids are also electrolytes.● Acids form ions when dissolved in water.

eg. H2SO4(aq) → 2 H+(aq) + SO4

2-(aq)

HCl(s) → H+(aq) + Cl-(aq)

● Molecular Compounds DO NOT dissociate in water.

eg. C12H22O11(s) → C12H22O11(aq)

● Because they DO NOT conduct electric current in solution, molecular compounds are non-electrolytes.


The molar concentration of any dissolved ion is calculated using the ratio from the dissociation equation.

eq. What is the molar concentration of each ion in a 5.00 mol/L MgCl2(aq) solution:

5.00 mol/L 5.00 mol/L 10.00 mol/L


What mass of calcium chloride is required to What mass of calcium chloride is required to prepare 2.00 L of a solution in which the Clprepare 2.00 L of a solution in which the Cl--

(aq)(aq) concentration is 0.120 mol/L ? concentration is 0.120 mol/L ?

Documents

CHEMISTRY 2202 UNIT 1 STOICHIOMETRY · STOICHIOMETRY: - the calculation of quantities USED IN or PRODUCED by a chemical reaction. 3 Sections: Part 1 - Mole calculations (ch. 2 & 3