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Unit 3 (Chp 1,2,3):

Matter, Measurement,

& Stoichiometry

John D. BookstaverSt. Charles Community College

St. Peters, MO 2006, Prentice Hall Inc.

Chemistry, The Central Science, 10th editionTheodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten

Chemistry:The study of matter and the changes it undergoes.

Quantitative

or

Qualitative

(has mass and takes up space)

Ni + HCl

nickel hydrochloric acid

NiCl2nickel(II) chloride

H2 +

hydrogensolid aqueous gas solid crystalsmetal solution

MatterAtom:

Element:

Compound:

H H C C

C C

OO Na

simplest particle retaining properties.

same type of atom (1 or more)

different atoms bonded.

H2O CO2 NaCl

H2 O2

C

molecule

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salt, baking soda,

water, sugaroxygen, iron,

hydrogen, gold

Matter

MixturePure

Substance

ElementsCompoundsHeterogeneous

MixtureHomogeneous

Mixture

separate

physically

cannot

separate

physically

differences or unevenly

mixed

uniform or evenly mixed

separate

chemically

cannot

separate

Chemicalchanges

Physicalchanges

(solutions)

filtering

distillation(boiling)

(suspensions/colloids) NaCl NaHCO3

H2O C12H22O11

O2 Fe

H2 Au

Density:

d =mV

Chemical Property OR Physical Property

Why?

ratio of mass to volume

ormatter to space occupied

Units:

g/mLg/cm3

kg/L

Changes of Matter

• Physical Changes:

� do not change the composition of a substance.

• temperature, changes of state, amount, etc.

• Chemical Changes:

� result in new substances.

• combustion, oxidation, decomposition, etc.

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Chemical Separation:

• Compounds can be decomposed

into elements.

Physical Separation:

Separates heterogeneous

mixtures (solids from liquids).Filtration:

Physical Separation:

Distillation: Separates solution by

boiling point differences.

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Physical Separation:

Chromatography:

Separates homogeneous mixture by

differences in solubility (attractions).

Power of 10 is the number of places

the decimal has been moved.

Examples: 42000 = 4.2 x 104

0.0508 = 5.08 x 10–2

positive power: move decimal right ����

to obtain the original # in standard notation.

negative power: move decimal left to obtain the original # in standard notation.

Scientific Notation

1. Convert the numbers to scientific notation.

(i) 24500

(ii) 0.000985

(iii) 12002

2. Convert to standard notation.

(i) 4.2 x 105

(ii) 2.15 x 10-4

(iii) 3 x 10-3

2.45 x 104

9.85 x 10–4

1.2002 x 104

420,000

0.000215

0.003

Scientific Notation

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Metric Prefixes

BASE UNIT: 1 m 1 L 1 g

0.01 cm

0.001 mL

0.000 001 µg

0.000 000 001 nm

Prefix Symbol Multiplier Examples:

1,000,000,000 GB

1,000,000 MJ

1,000 kg

(atoms)

(light wavelength)

(nuclei)

Uncertainty in MeasurementsMeasuring devices have different uses

and different degrees of precision.

(uncertainty)

% Error = |Accepted – Experimental| x100

Accepted

Uncertainty

5.23 cm

(uncertain)

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Significant Figures• measured digits.

• last digit is estimated, but IS significant.

• do not overstate the precision

5.23 cm

5.230 cm

Significant Zeroes

1. All nonzero digits are significant.

2. Captive Zeroes between two

significant figures are significant.

3. Leading Zeroes at the beginning of a

number are never significant.

4. Trailing Zeroes:

SIG, if at end AND a decimal point.

NOT, if there is no decimal point.

0.0003700400 grams

0’s

round answers to keep the

fewest decimal places

round answers to keep the

fewest significant figures

Significant Figures

+ or –

x or ÷

3.48 + 2.2 =

6.40 x 2.0 =

5.68 5.7

12.8 13

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1. How many sig figs are in each number?

(i) 250.0

(ii) 4.7 x 10–5

(iii) 34000000

(iv) 0.03400

2. Round the answer to the correct sig figs.

(i) 34.5 x 23.46

(ii) 123/3

(iii) 23.888897 + 11.2

(iv) 2.50 x 2.0 – 3

Significant Figures WS 1s

4

2

2

4

809

40

35.1

2

WARM UP (for QUIZ!!!)

• Review WS 1s #1, 3, 10

• Complete WS 1a #1, 2, 8, 9, 10

Law of Definite Proportions

�elemental formulas (composition)

of pure compounds cannot vary.

• 2 H’s & 1 O is ALWAYS water.

• Water is ALWAYS 2 H’s & 1 O.

• 2 H’s & 2 O’s is NOT water.

√ H2O

XXXX H2O2H

O

H

H H

O

O

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Law of Conservation of Mass

The total mass of substances present

at the end of a chemical process is the same as the mass of substances

present before the process took place.

__H2 + __O2 __H2O 22

Balancing Equations!!!

Symbols of Elements

126C

Mass Number

= p’s + n’s

Atomic Number (Z)

= p’s

Element Symbol

All atoms of the same element have the

same number of protons (same Z), but…

can have different mass numbers. HOW?

1

1H2

1H3

1Hprotium deuterium tritium

element:mass:

why?

same or differentsame or differentsame # of protons (& electrons),

but different # of neutrons

Isotopes

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Average Atomic Mass

• average atomic mass: calculated as a

weighted average of isotopes by their relative abundances.

(6.015)(0.0750) + (7.016)(0.925) = 6.94 amu

Avg. Mass = (Mass1)(%) + (Mass2)(%) …

• lithium-6 (6.015 amu), which has a

relative abundance of 7.50%, and

• lithium-7 (7.016 amu), which has a

relative abundance of 92.5%.

WS Atomic

Structure

element sample

atomized,ionized

magnetic field

~75%

~25%

isotopes

separated by

difference

in mass

(35)(~0.75) + (37)(~0.25) = ?

Cl (avg at. Mass) =

Mass Spectrometry

Molecular (Covalent) Compounds

Covalent compounds

contain nonmetals that “share” electrons to

form molecules.

(molecular compounds)

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Diatomic Molecules

These seven elements occur naturally

as molecules containing two atoms.

“H-air-ogens”

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Binary Molecular Compounds

• list less electronegative atom first. (left to right on PT)

• use prefix for the number of atoms of each element.

• change ending to –ide.

CO2: carbon dioxide

CCl4: carbon tetrachloride

N2O5: ________________

CuSO4·5H2Ocopper(II) sulfate pentahydrate

dinitrogen pentoxide

(ionic & covalent)

IonsCations

metals lose e’s

positive (+)

(metal) ion

Anions

nonmetalsgain e’s

negative (–)

(nonmetal)ide

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Ionic Bonds

Attraction between +/– ions formed by

metals & nonmetals transferring e–’s.

Formulas of Ionic Compounds

• Compounds are electrically neutral, so the

formulas can be determined by:

– Crisscross the charges as subscripts (then erase)

– If needed, reduce to lowest whole number ratio.

Pb4+ O2– Pb2O4 PbO2

Naming Ionic Compounds

1) Cation: Write metal name (ammonium NH4+)

For transition metals with multiple charges, write charge as Roman numeral in parentheses.

Iron(II) chloride, FeCl2

Iron(III) chloride, FeCl3

2) Anion: Write nonmetal name with –ide

OR the polyatomic anion name. (–ate, –ite)

Iron(II) sulfide, FeS

Magnesium sulfate, MgSO4

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Common

PolyatomicIons

Name Symbol Charge

*ammonium NH4+ 1+

*acetate

(ethanoate)

C2H3O2–

(CH3COO–)1–

*hydroxide OH– 1–

*perchlorate ClO4– 1–

*chlorate ClO3– 1–

chlorite ClO2– 1–

hypochlorite ClO– 1–

bromate BrO3– 1–

iodate IO3– 1–

*nitrate NO3– 1–

nitrite NO2– 1–

cyanide CN– 1–

*permanganate MnO4– 1–

*bicarbonate

(hydrogen carbonate)HCO3

– 1–

*carbonate CO32– 2–

*sulfate SO42– 2–

sulfite SO32– 2–

*chromate CrO42– 2–

dichromate Cr2O72– 2–

*phosphate PO43– 3–

* these 12 will be on

Quiz 1

- all 20 Polyatomic Ions

will be on Quiz 2

WS 2d

perchlorate ClO4–

chlorate ClO3–

chlorite ClO2–

hypochlorite ClO–

C N O F

Si P S Cl

As Se Br

Te I

nitrate NO3–

nitrite NO2–

In Out Ion Name

–

–

“Oxyanion” Names (elbO’s)

4

3

2

1

4

3

___-ate

___-itesulfate SO4

2–

sulfite SO32–

phosphate PO43–

per-___-ate

hypo-___-ite

nitrate NO3–

nitrite NO2–

sulfate SO42–

sulfite SO32–

perchlorate ClO4–

chlorate ClO3–

chlorite ClO2–

hypochlorite ClO–

Ion Acidadd H+

Name Acids

from these oxyanions:

In Out Ion Name Acid Name

4 – per-___-ate

3 4 ___-ate

2 3 ___-ite

1 – hypo-___-ite

per-___-ic acid

___-ic acid

___-ous acid

hypo-___-ous acid

Naming Acids

WS 2e

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Anatomy of a Chemical Equation

CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)

Anatomy of a Chemical Equation

Reactants appear on the

left side of the equation.

CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)

Anatomy of a Chemical Equation

Products appear on the

right side of the equation.

CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)

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Anatomy of a Chemical Equation

States (s, l, g, aq) written in parentheses next to each compound

CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)

Anatomy of a Chemical Equation

CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)

Subscripts show how many

atoms of each element

Anatomy of a Chemical Equation

Coefficients show the amount of each particle

and are inserted to balance the equation.

CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)

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Reaction

Types

Combination

2 Mg(s) + O2(g) →→→→ 2 MgO(s)

Demo: MgO 2 → 1A + B → AB

2 NaN3(s) ���� 2 Na(s) + 3 N2(g)

Decomposition1 → 2AB → A + B

(50 milliseconds!)

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Replacement Reactions(or “Displacement”)

Single ReplacementAB + C → A + CB

Pb(NO3)2(aq) + KI(aq) → PbI2(s) + KNO3(aq)

Double ReplacementAB + CD → AD + CB

video clip

Demo:

(aq) + (s) → (s) + (aq)

CH4(g) + 2 O2(g) � CO2(g) + 2 H2O(g)

•Often involve hydrocarbons

reacting with oxygen in the air

WS 4a

CxHy + _O2 ���� _CO2 + _H2O

Combustion

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

Formula

Weights

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Formula Weight

(FW)

• Sum of the atomic weightsfor the atoms in a chemical formula

• Formula Weight of calcium chloride, CaCl2, is…

Ca: 1(40.08 amu)

+ Cl: 2(35.45 amu)

110.98 amu

• Sum of the atomic weights for the atoms in a molecule orcompound

• Molecular Weight of ethane, C2H6, is…

Molecular Weight

(MW)

C: 2(12.01 amu)

+ H: 6(1.008 amu)

30.07 amu

Percent Composition

One can find the percent by mass

of a compound of each element in

the compound by using this equation.

% element =(# of atoms)(AW)

(FW)x 100

Percent Composition

So the percentage of carbon in

ethane (C2H6) is…

%C =(2)(12.01)

(30.07)

24.0230.07

= x 100

= 79.88% C

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Moles

Avogadro Constant• One mole of particles

contains the Avogadro

constant of those particles6.022 x 1023

Mole Relationships• One mole of atoms, ions, or molecules contains the

Avogadro constant of those particles 6.022 x 1023

In 1 mol Na2CO3 , how many…

• Na atoms?

• C atoms?

• O atoms?

• How many donuts in 1 mol of donuts?

• How many boogers in 1 mol of boogers?

Which has more atoms, 1 mol CH3 or 1 mol NH3 ?How about CH3CH2OH or H2SO4 ?

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Molar Mass

• the mass of 1 mol of a substance (g/mol)

– molar mass (in g/mol) of an element is the

atomic mass (in amu) on the periodic table

– formula weight (amu) of a compound

same number as the

molar mass (g/mol) of 1 mole of particlesof that compound

Using Moles

Moles are the bridge from

the particle (micro) scale to

the real-world (macro) scale.

Mass(grams)

Particles(atoms)

(molecules)(units)

Moles(groups of 6.022x1023

particles)

molar mass

# g1 mol

1 mol# g

Avogadro constant

6.022x1023

1 mol

1 mol6.022x1023

macro-bridge micro-

Using Moles

1.What is the mass of 1 mole of copper(II)

bromide, CuBr2?

2.How many moles are there in 112 g of

copper(II) bromide, CuBr2?

3.How many particles present in each of the questions #1 & #2 above?

(63.55) + 2(79.90) = 223.35 g

112 g CuBr2 x1 mol CuBr2

223.35 g CuBr2

= 0.501 mol

CuBr2

0.501 mol x6.022 x 1023 particles

1 mol= 3.02 x 1023

particles

= 6.022 x 1023 particles

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•Balanced chemical equations show the amount

of:

Most important are the ratios of reactants and

products in moles, or…

mol-to-mol ratios

Stoichiometry:

calculations of quantities in chemical rxns

–how much reactant is consumed or–how much product is formed

atoms, molecules, moles, and mass

g A

g B mol B

g A1 mol A

g B1 mol B

molar mass A

molar mass B

mol A

Rxn: A(aq) + 2 B(aq) ���� C(aq) + 2 D(aq)

Stoichiometric Calculations

1 mol Ag A

OR mol-to-mol

ratio

2 mol B1 mol A

1 mol A2 mol B

OR

Coefficients of

balanced equation

???

Stoichiometric problems have 1-3 Steps: (usually)

1) Convert grams to moles (if necessary)

using the molar mass (from PT)

2) Convert moles (given) to moles (wanted)

using the mol ratio (from coefficients)

3) Convert moles to grams (if necessary)

using the molar mass (from PT)

grams A x 1 mol A .

grams A=_ mol B

mol Ax x grams B

1 mol B

1) molar mass 2) mole ratio 3) molar mass

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Example : g of A ���� g of B

Solid magnesium is added to an aqueous

solution of hydrochloric acid. What mass of H2

gas will be produced from completely reacting

18.0 g of HCl with magnesium metal?

Mg(s) + 2 HCl(aq) ���� MgCl2(aq) + H2(g)

= ____ g H2

18.0 g HCl xg HCl mol HCl

mol HCl x mol H2

mol H2

g H2

36.46x

2.016

12

1 1

0.498 g H2

molar mass A

molar mass B

g of A

Stoichiometric CalculationsHW p. 114 #58

mole ratio B/A

Finding

Empirical

Formulas

Types of Formulas

• Empirical formulas:

the lowest ratio of atoms of

each element in a compound.

• Molecular formulas:

the total number of atoms of

each element in a compound.

CH3

C2H6

C2H4O

C6H12O3

molecular mass = emp. form.

empirical mass multiple

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Percent to Mass

Mass to Mole

Divide by Small

Times ‘till Whole

Steps (rhyme) ÷ moles by smallest to get mole ratio of atoms

MM from PT

assume 100 g

x (if necessary) to get

whole numbers of atoms

Calculating Empirical Formulas

75 % C 75 g C 6.2 mol C

25 % H 25 g H 24.8 mol H

1 C

4 H

CH4

from Mass % Composition

Butane is 17.34% H and 82.66% C by mass.Determine its empirical formula.

If molecular mass is 58 g·mol–1, what is theMolecular Formula?

82.66 g C

17.34 g H

1) Percent to Mass

2) Mass to Mole

82.66 g C x = 6.883 mol C

17.34 g H x = 17.20 mol H

1 mol C12.01 g C

1 mol H1.008 g H

6.883 mol

6.883 mol

3) Divide by Small

4) Times ’till Whole

= 1 ≈ 1 C

= 2.499 ≈ 2.5 H

x 2= 2 C

C2H5x 2= 5 H

C4H10

58

29.06= 2 2 (C2H5) =

molecular massempirical mass

HW p. 113 #43a, 48

Percent to Mass

Mass to Mole

Divide by Small

Times ‘till Whole

Calculating Empirical Formulas

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• Hydrocarbons with C and H are analyzed

through combustion with O2 in a chamber.

� g C is from the g CO2 produced

� g H is from the g H2O produced

� g X is found by subtracting (g C + g H) from g sample

Combustion Analysis

When 4-ketopentenoic acid is analyzed by

combustion, a 0.3000 g sample produces

0.579 g of CO2 and

0.142 g of H2O.

The acid contains only C, H, and O.

What is the empirical formula of the acid?

Combustion Analysis

Example 1

0.579 g CO2 x1 mol CO2

44.01 g CO2

1 mol H2O

18.02 g H2O

1 mol C

1 mol CO2

x12.01 g C

1 mol Cx

2 mol H

1 mol H2Ox 1.008 g H

1 mol Hx0.142 g H2O x

= 0.158 g C

= 0.0159 g H

0.3000 g sample – (0.158 g C) – (0.0159 g H) =

= 0.126 g O

? g C

? g H

? g O

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0.00788 mol

0.00788 mol

0.00788 mol0.158 g C x

1 mol C

12.01 g C

1 mol H

1.008 g H

0.0132 mol C=

0.0158 mol H=0.0159 g H x

=

=

1.67 C

2 H

0.126 g O x1 mol O

16.00 g O0.00788 mol O= = 1 O

x 3 = 5 C

x 3 = 6 H

x 3 = 3 O

C5H6O3

A sample of a chlorohydrocarbon with a

mass of 4.599 g, containing C, H and Cl,

was combusted in excess oxygen to yield

6.274 g of CO2 and 3.212 g of H2O.

Calculate the empirical formula of the

compound.

If the compound has a MW of 193 g·mol–1,

what is the molecular formula?

Example 2

Combustion Analysis

6.274 g CO2 x1 mol CO2

44.01 g CO2

1 mol H2O

18.02 g H2O

1 mol C

1 mol CO2

x12.01 g C

1 mol Cx

2 mol H

1 mol H2Ox 1.008 g H

1 mol Hx3.212 g H2O x

= 1.712 g C

= 0.3593 g H

4.599 g sample – (1.712 g C) – (0.3593 g H) =

= 2.528 g Cl

? g C

? g H

? g Cl

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0.07131 mol

0.07131 mol

0.07131 mol1.712 g C x

1 mol C

12.01 g C

1 mol H

1.008 g H

0.1425 mol C=

0.3564 mol H=0.3593 g H x

=

=

2 C

5 H

2.528 g Cl x1 mol Cl

35.45 g Cl0.07131 mol Cl= = 1 Cl

C2H5ClIf the compound has a

MW of 193 g·mol–1, what is the molecular formula?

MW

EW193

64.51= 3

C6H15Cl3HW p. 114 #52b

• Which ingredient will run out first?

• If out of sugar, you should stop making cookies.

How Many Cookies Can I Make?

• Sugar is the limiting ingredient, because it will

limit the amount of cookies you can make.

Before After

limiting

2 H2 + O2 2 H2OInitial: ? mol ? mol ? mol

Change:

End:

10 7 0

0 mol 2 mol 10 mol–10 –5 +10

H2

O2

Which is limiting?

excess

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Before After

2 H2 + O2 2 H2OInitial: ? mol ? mol ? mol

Change:

End:

10 7 0

0 mol 2 mol 10 mol–10 –5 +10

H2

O2

O2 is in smallest amount, but…

H2 is in smallest “stoichiometric” amount

Does limiting mean smallest amount of reactant? No!

Solid aluminum metal is reacted with

aqueous copper(II) chloride in solution.

• If 0.030 g Al are reacted with 0.0060 mol

CuCI2, which is the limiting reactant?

• How much product will be produced?

Limiting Reactant

demo

Al(s) + CuCl2(aq) Cu(s) + AlCl3(aq)2 3 3 2

2 Al(s) + 3 CuCl2(aq) 3 Cu(s) + 2 AlCl3(aq)

Limiting Reactant

0.030 g Al

0.0060 mol CuCl2

1 mol Al

26.98 g Al

3 mol Cu

3 mol CuCl2

x

x

x3 mol Cu

2 mol Al= 0.0017 mol Cu

= 0.0060 mol Cu

Al is limiting

HW p. 115 #72

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Theoretical Yield

theoretical yield: the maximum amount of

product that can be formed

– calculated by stoichiometry

(using LR only)

• This is different from the actual yield, the

amount one actually produces and measures (or experimental)

0.030 g Al1 mol Al

26.98 g Alx x

3 mol Cu

2 mol Al= 0.0017 mol Cu

Percent Yield

A comparison of the amount actually obtained

to the amount it was possible to make

%Yield = x 100Actual

Theoretical

(calculate using the LR only)

% Error = |Accepted – Experimental| x100

Accepted

NOT % Error:

Aluminum will react with oxygen gas

according to the equation below

4 Al + 3 O2 2 Al2O3

• In one such reaction, 23.4 g of Al are

allowed to burn in excess oxygen.

39.3 g of aluminum oxide are formed.

What is the percentage yield?

Percent Yield

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101.96 g Al2O3

1 mol Al2O3

4 Al + 3 O2 2 Al2O3

Percent YieldHW p. 116

#79

39.3 g of aluminum oxide are formed.

What is the percentage yield?

23.4 g Al1mol Al

26.98 g Alx

2 mol Al2O3

4 mol Alx

= 44.2 g Al2O3

x

%Yield = x 10039.3 g

44.2 g88.9 %