ENG 023 Chap 1

Chapter 1 Chapter 1 Matter Matter

CHEMISTRYCHEMISTRY The Central Science The Central Science

10th Edition10th Edition

The Molecular Perspective of Chemistry• Matter is the physical material of the universe.• Matter is made up of relatively few elements.• On the microscopic level, matter consists of atoms and

molecules.• Atoms combine to form molecules.• Molecules may consist of the same type of atoms or

different types of atoms.

The Study of ChemistryThe Study of Chemistry

The Molecular Perspective of Chemistry

The Study of ChemistryThe Study of Chemistry

States of Matter• Matter can be a gas, a liquid or a solid.• These are the three states of matter.• Gases have no fixed shape or volume.• Gases can be compressed to form liquids.• Liquids have no shape, but they do have a volume.• Solids are rigid and have a definite shape and volume.

Classification of MatterClassification of Matter

Pure Substances and Mixtures• Atoms consist only of one type of element.• Molecules can consist of more than one type of element.

– Molecules can have only one type of atom (an element).– Molecules can have more than one type of atom (a

compound).• If more than one atom, element, or compound are found

together, then the substance is a mixture.

Classification of MatterClassification of Matter

Pure Substances

and Mixtures

Pure Substances and Mixtures• If matter is not uniform throughout, then it is a

heterogeneous mixture.• If matter is uniform throughout, it is homogeneous.• If homogeneous matter can be separated by physical

means, then the matter is a mixture.• If homogeneous matter cannot be separated by physical

means, then the matter is a pure substance.• If a pure substance can be decomposed into something

else, then the substance is a compound.

Classification of MatterClassification of Matter

Elements• If a pure substance cannot be decomposed into

something else, then the substance is an element.• There are 114 elements known.• Each element is given a unique chemical symbol (one or

two letters).• Elements are building blocks of matter.• The earth’s crust consists of 5 main elements.• The human body consists mostly of 3 main elements.

Classification of MatterClassification of Matter

Elements

Classification of MatterClassification of Matter

Elements

• Chemical symbols with one letter have that letter capitalized (e.g., H, B, C, N, etc.)

• Chemical symbols with two letters have only the first letter capitalized (e.g., He, Be).

Classification of MatterClassification of Matter

Compounds• Most elements interact to form compounds.• The proportions of elements in compounds are the same

irrespective of how the compound was formed.• Law of Constant Composition

(or Law of Definite Proportions):– The composition of a pure compound is always the

same.

Classification of MatterClassification of Matter

Compounds• If water is decomposed, then there will always be twice

as much hydrogen gas formed as oxygen gas.• Pure substances that cannot be decomposed are elements.

Classification of MatterClassification of Matter

Mixtures• Heterogeneous mixtures are not uniform throughout.• Homogeneous mixtures are uniform throughout.• Homogeneous mixtures are called solutions.

Classification of MatterClassification of Matter

If matter is uniform throughout and cannot be separated into other substances by physical processes, but can be decomposed into other substances by chemical processes, it is called a (an) __________. A) heterogeneous mixture B) element C) homogeneous mixture D) compound E) mixture of elements

Quiz

If matter is uniform throughout and cannot be separated into other substances by physical processes, but can be decomposed into other substances by chemical processes, it is called a (an) __________. A) heterogeneous mixture B) element C) homogeneous mixture D) compound E) mixture of elements

Quiz

Answer: D

The law of constant composition applies to __________.

A) solutions B) heterogeneous mixtures C) compounds D) homogeneous mixtures E) solids

Quiz

The law of constant composition applies to __________.

A) solutions B) heterogeneous mixtures C) compounds D) homogeneous mixtures E) solids

Quiz

Answer: C

Dalton’s Atomic TheoryDalton’s Atomic Theory

• Each element is composed of extremely small particles called atoms

• All atoms of a given element are identical; the atoms of different elements are different and have different properties

• Atoms of an element do not changed into different types of atoms by chemical reactions; atoms are neither created nor destroyed in chemical reactions

• Compounds are formed when atoms of more that one element combine; a given compound always has the same relative number and kind of atoms

Fundamental Particles Of An Fundamental Particles Of An Atom – Atom –

Proton, Neutron And ElectronProton, Neutron And Electron• 3 subatomics particles – proton, neutron and electron• Charge of an electron is -1.602 x 10-19 C• Charge of a proton is +1.602 x 10-19 C• Neutrons are uncharged• Atoms have an equal number of electrons and protons• Protons and neutrons reside together in the nucleus of

the atom• Atoms has extremely small masses• Atoms are extremely small (1 – 5 Ǻ)• One angstrom (Ǻ) = 10-10 m

IsotopesIsotopes

• All atoms of an element have the same number of protons in the nucleus

• The number of protons is different for different elements

• Atoms of a given element that differ in the number of neutrons and mass are called isotopes

IsotopesIsotopes

• Atomic number (Z) = no. of proton• Mass number (A) = total no. protons + neutrons• By convention, for element X, we write X.• Isotopes have the same Z but different A.• Eg. C, C. • Naturally occurring 98.93% 12C, 1.07% 13C• We can find Z from the periodic table.

Atomic Mass ScaleAtomic Mass Scale

• Hydrogen, has a mass of 1.6735 x 10-24 g• 1 amu = 1.66054 x 10-24 g

and 1 g = 6.02214 x 1023 amu• 12 amu = an atom of the 12C isotope of carbon• Average atomic masses = atomic weight• Eg Atomic weight of carbon =

(0.9893)12 amu + (0.0107)13.00335 amu = 12.01 amu

Chemical Substances – Chemical Substances – Formulas And NamesFormulas And Names

• Periodic table• Molecular formulas – chemical formula indicate the

actual no. and types of atoms in a molecule• Empirical formulas – only relative number of atoms

of each type in a molecule• Example – H2O2, HO and C2H4, CH2

• Structural formulas – show which atoms are attached to which within the molecule– Perspective drawing– Ball and stick model– Space – filling model

• Structural formulas

– show which atoms are attached to which within the molecule

– Perspective drawing– Ball and stick model– Space – filling model

Chemical Substances – Chemical Substances – Formulas And NamesFormulas And Names

Physical and Chemical Changes • When a substance undergoes a physical change, its

physical appearance changes. – Ice melts: a solid is converted into a liquid.

• Physical changes do not result in a change of composition.• When a substance changes its composition, it undergoes a

chemical change:– When pure hydrogen and pure oxygen react completely,

they form pure water. In the flask containing water, there is no oxygen or hydrogen left over.

Properties of MatterProperties of Matter

Physical and Chemical Changes

Properties of MatterProperties of Matter

Physical and Chemical Changes

• Intensive physical properties do not depend on how much of the substance is present.– Examples: density, temperature, and melting point.

• Extensive physical properties depend on the amount of substance present.– Examples: mass, volume, pressure.

Properties of MatterProperties of Matter

Separation of Mixtures• Mixtures can be separated if their physical properties are

different.• Solids can be separated from liquids by means of

filtration.• The solid is collected in filter paper, and the solution,

called the filtrate, passes through the filter paper and is collected in a flask.

Properties of MatterProperties of Matter

Separation of Mixtures• Homogeneous liquid mixtures can be separated by

distillation.• Distillation requires the different liquids to have different

boiling points.• In essence, each component of the mixture is boiled and

collected.• The lowest boiling fraction is collected first.

Properties of MatterProperties of Matter

Separation of Mixtures

Separation of Mixtures• Chromatography can be used to separate mixtures that

have different abilities to adhere to solid surfaces.• The greater the affinity the component has for the surface

(paper) the slower it moves.• The greater affinity the component has for the liquid, the

faster it moves.• Chromatography can be used to separate the different

colors of inks in a pen.

Properties of MatterProperties of Matter

Which one of the following is an intensive property?

A) mass B) temperature C) heat content D) volume E) amount

Quiz

Which one of the following is an intensive property?

A) mass B) temperature C) heat content D) volume E) amount

Answer: B

Quiz

Chemical equations: descriptions of chemical reactions.• Two parts to an equation: reactants and products:

2H2 + O2 2H2O, CH4 + O2 CO2 + H2O

Chemical EquationsChemical Equations

• Combination reactions, A + B C

– 2Mg(s) + O2(g) 2MgO(g)

• Decomposition reactions, C A + B

– CaCO3(s) CaO(s) + CO2(g)

• Combustion

– C3H8(g) + 5O2(g) 3CO2 + 4H2O

• Metathesis

– AgNO3(aq) + KCl(aq) AgCl(s) + KNO3 (aq)

• The chemical equation for the formation of water can be visualized as two hydrogen molecules reacting with one oxygen molecule to form two water molecules:

2H2 + O2 2H2O

Chemical EquationsChemical Equations

• Stoichiometric coefficients: numbers in front of the chemical formulas; give ratio of reactants and products.

Chemical EquationsChemical Equations

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

Chemical EquationsChemical Equations

• Law of conservation of mass: matter cannot be lost in any chemical reactions.

Chemical EquationsChemical Equations

Combination and Decomposition Reactions

• Combination reactions have fewer products than reactants:

2Mg(s) + O2(g) 2MgO(s)• The Mg has combined with O2 to form MgO.• Decomposition reactions have fewer reactants than

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

(the reaction that occurs in an air bag)• The NaN3 has decomposed into Na and N2 gas.

Chemical EquationsChemical Equations

Combination and Decomposition Reactions

Chemical EquationsChemical Equations

Combustion in Air

Chemical EquationsChemical Equations

Combustion is the burning of a substance in oxygen from air:

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

Which of the following are chemical processes? 1. rusting of a nail 2. freezing of water 3. decomposition of water into hydrogen and oxygen gases 4. compression of oxygen gas

A) 2, 3, 4 B) 1, 3, 4 C) 1, 3 D) 1, 2 E) 1, 4

Quiz

Which of the following are chemical processes? 1. rusting of a nail 2. freezing of water 3. decomposition of water into hydrogen and oxygen gases 4. compression of oxygen gas

A) 2, 3, 4 B) 1, 3, 4 C) 1, 3 D) 1, 2 E) 1, 4

Answer: C

Quiz

Formula and Molecular Weights

• Formula weights (FW): sum of atomic weights (AW) for atoms in formula.

FW (H2SO4) = 2AW(H) + AW(S) + 4AW(O) = 2(1.0 amu) + (32.0 amu) + 4(16.0 amu)

= 98.0 amu

• Molecular weight (MW) is the weight of the molecular formula.

MW(C6H12O6) = 6(12.0 amu) + 12(1.0 amu) + 6(16.0 amu) = 180.0 amu

Formula WeightsFormula Weights

Percentage Composition from Formulas

• Percent composition is the atomic weight for each element divided by the formula weight of the compound multiplied by 100:

Formula WeightsFormula Weights

100

Compound ofFW

AWElement of atoms of NoElement %

Mole: convenient measure of chemical quantities.• 1 mole of something = 6.0221367 1023 of that thing.• Experimentally, 1 mole of 12C has a mass of 12 g.

Molar Mass• Molar mass: mass in grams of 1 mole of substance (units

g/mol, g·mol-1).• Mass of 1 mole of 12C = 12 g.

The MoleThe Mole

The MoleThe Mole

The MoleThe Mole

The MoleThe Mole

This photograph shows one mole of solid (NaCl), liquid (H2O), and gas (N2).

Interconverting Masses, Moles, and Number of Particles

• Molar mass: sum of the molar masses of the atoms:molar mass of N2 = 2 (molar mass of N).

• Molar masses for elements are found in the periodic table.

• Formula weights are numerically equal to the molar mass.

The MoleThe Mole

• Start with mass % of elements (i.e. empirical data) and calculate a formula, or

• Start with the formula and calculate the mass % elements.

Empirical Formulas from Empirical Formulas from AnalysesAnalyses

Molecular Formula from Empirical Formula

• Once we know the empirical formula, we need the MW to find the molecular formula.

• Subscripts in the molecular formula are always whole-number multiples of subscripts in the empirical formula

Eg. Ascorbic acid contains 40.92% C, 4.58% H and 54.50% O by mass. What is its empirical formula?

What is the molecular formula if the molar mass is 176 amu?

Empirical Formulas from Empirical Formulas from AnalysesAnalyses

Molecular Formula from Empirical Formula

• Once we know the empirical formula, we need the MW to find the molecular formula.

• Subscripts in the molecular formula are always whole-number multiples of subscripts in the empirical formula

Eg. Ascorbic acid contains 40.92% C, 4.58% H and 54.50% O by mass. What is its empirical formula?

What is the molecular formula if the molar mass is 176 amu?

Ans: C3H4O3, (88amu) Molecular formula: C6H8O6 (176 amu)

Empirical Formulas from Empirical Formulas from AnalysesAnalyses

Empirical Formulas from Empirical Formulas from AnalysesAnalyses

C:H:O = 1:1.33:1 = 3:4:3 i.e. C3H4O3 = 88 amu Molecular formula is C6H8O6 = 176 amu

An empirical formula always indicates __________.

A) which atoms are attached to which in a molecule

B) how many of each atom are in a molecule

C) the simplest whole-number ratio of different atoms in a compound

D) the isotope of each element in a compound

E) the geometry of a molecule

Quiz

An empirical formula always indicates __________.

A) which atoms are attached to which in a molecule

B) how many of each atom are in a molecule

C) the simplest whole-number ratio of different atoms in a compound

D) the isotope of each element in a compound

E) the geometry of a molecule

Answer: C

Quiz

• Balanced chemical equation gives number of molecules that react to form products.

• Interpretation: ratio of the number of moles of reactant required to give the ratio of number of moles of product.

• These ratios are called stoichiometric ratios.

NB: Stoichiometric ratios are ideal proportions• Real ratios of reactants and products in the laboratory

need to be measured (in grams and converted to moles).

Quantitative Information Quantitative Information from Balanced Equationsfrom Balanced Equations

• If the reactants are not present in stoichiometric amounts, at end of the reaction some reactants will still be present (in excess).

• Limiting Reactant: one reactant that is consumed

Limiting ReactantsLimiting Reactants

Limiting ReactantsLimiting Reactants

Theoretical Yields• The amount of product predicted from stoichiometry

taking into account limiting reagents is called the theoretical yield.

• The percent yield relates the actual yield (amount of material recovered in the laboratory) to the theoretical yield:

Limiting ReactantsLimiting Reactants

100yield lTheoretica

yield ActualYield %

A 3.82g sample of magnesium nitride (Mg3N2) is reacted with 7.73g of water. Mg3N2 + 3H2O 2NH3 + 3MgO

The yield of MgO is 3.60g. What is the percent yield in the reaction?

A) 94.5 B) 78.7 C) 46.6 D) 49.4 E) 99.9

Quiz

A 3.82g sample of magnesium nitride (Mg3N2) is reacted with 7.73g of water. Mg3N2 + 3H2O 2NH3 + 3MgO

The yield of MgO is 3.60g. What is the percent yield in the reaction?

A) 94.5 B) 78.7 C) 46.6 D) 49.4 E) 99.9

Quiz

Answer: B

Solution:

1 mole of Mg3N2 produces 3 moles of MgO

1mole of Mg3N2 = (3x24.3+2x14)g1mole of MgO = (24.3+16)g

Therefore 100.9g of Mg3N2 produces 120.9g of MgOand theoretical yield of,

3.82g of Mg3N2 produces 4.577g of MgO

But actual yield = 3.60g of MgO

Therefore %yield = (3.60/4.577)x100 = 78.65%

End of Chapter 1 End of Chapter 1 Matter Matter