Unit 3 (Chp 1,2,3): Matter, Measurement, & Stoichiometry John D. Bookstaver St. Charles Community College St. Peters, MO 2006, Prentice Hall Inc. Chemistry,

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.

Quantitativeor

Qualitative

(has mass and takes up space)

Ni + HClnickel hydrochloric acid

NiCl2nickel(II) chloride

H2 +

hydrogen solid 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

salt, baking soda, water, sugar

oxygen, iron, hydrogen, gold

Matter

MixturePure

Substance

ElementsCompoundsHeterogeneous

MixtureHomogeneou

s 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 FeH2 Au

Density:

d =mV

Chemical Property OR Physical Property

Why?

ratio of mass to volumeor

matter 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.

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.

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

420,0000.000215

0.003

Scientific Notation

Metric Prefixes

BASE UNIT: 1 m 1 L 1 g

0.01 cm

0.001 mL

0.000 001 µg0.000 000 001 nm

Prefix Symbol Multiplier Examples:

1,000,000,000 GB

1,000,000 MJ1,000 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)

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

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

224

80940

35.12

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

X H2O2

HO

H

H H

O

O

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?

11H

21H

31H

protium deuterium tritium

element:mass:

why?

same or differentsame or differentsame # of protons (& electrons), but different # of neutrons

Isotopes

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)

Diatomic Molecules

These seven elements occur naturally as molecules containing two atoms.

“H-air-ogens”

7

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)

IonsCationsmetals lose e’spositive (+)(metal) ion

Anionsnonmetals

gain e’snegative (–)

(nonmetal)ide

Ionic BondsAttraction 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, FeCl2Iron(III) chloride, FeCl3

2) Anion: Write nonmetal name with –ide

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

Iron(II) sulfide, FeS

Magnesium sulfate, MgSO4

Common Polyatomic

Ions

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 FSi P S Cl

As Se BrTe I

nitrate NO3–

nitrite NO2–

In Out Ion Name

–

–

“Oxyanion” Names (elbO’s)

4321

43

___-ate___-ite

sulfate SO42–

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-___-ate3 4 ___-ate2 3 ___-ite1 – hypo-___-ite

per-___-ic acid ___-ic acid ___-ous acid hypo-___-ous acid

Naming Acids

WS 2e

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)

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)

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!)

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: PbI2

AgNO3(aq) + Cu(s) Ag(s) + CuNO3(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

Formula Weight (FW)

• Sum of the atomic weights for 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 or compound

• 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

Moles

Avogadro Constant• One mole of particles

contains the Avogadro constant of those particles 6.022 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 ?

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 particles of that compound

Using MolesMoles 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 Moles1.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

•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

Example : g of A g of BSolid magnesium is added to an aqueoussolution 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

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 H 1.008 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

C2H5 x 2= 5 H

C4H10

5829.06

= 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

• 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 H2O18.02 g H2O

1 mol C1 mol CO2

x12.01 g C1 mol C

x

2 mol H1 mol H2O

x 1.008 g H1 mol H

x0.142 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

0.00788 mol

0.00788 mol

0.00788 mol0.158 g C x

1 mol C12.01 g C

1 mol H1.008 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 OC5H6O3

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 2Combustion Analysis

6.274 g CO2 x1 mol CO2

44.01 g CO2

1 mol H2O18.02 g H2O

1 mol C1 mol CO2

x12.01 g C1 mol C

x

2 mol H1 mol H2O

x 1.008 g H1 mol H

x3.212 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

0.07131 mol

0.07131 mol

0.07131 mol1.712 g C x

1 mol C12.01 g C

1 mol H1.008 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?

MWEW

19364.51

= 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 H2O

Initial: ? mol ? mol ? molChange: End:

10 7 0

0 mol 2 mol 10 mol–10 –5 +10

H2

O2

Which is limiting?

excess

Before After

2 H2 + O2 2 H2O

Initial: ? mol ? mol ? molChange: 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

Theoretical Yieldtheoretical 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

101.96 g Al2O3

1 mol Al2O3

4 Al + 3 O2 2 Al2O3

Percent Yield HW 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 g44.2 g

88.9 %