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Chapter 3: The “Language of Chemistry”

“alphabet”

symbols

(

Know

for the elements, e.g. C, N, F, Mg, Fe, etc.

names/symbols for #1-88 except lanthanides, including spelling!)

“words”

chemical formulas

, e.g. H 2 O, N 2 , Fe 2 (CO 3 ) 3 , etc.

counting atoms in formulas : --1 molecule of H 2 O contains 2 hydrogen atoms and 1 oxygen --the formula Fe 2 (CO 3 ) 3 represents: 2 iron atoms, 3 carbon atoms, and 9 oxygen atoms “sentences” chemical equations ( reactants and products ) Mg(OH) 2(aq) + 2 HCl (aq) MgCl 2(aq) + 2 H 2 O (l) Coefficients Subscripts are used to “balance” the equation indicate states of matter (not always included) Balanced equation : same number of atoms of each element on both sides of arrow

Molecular and Ionic Compounds • Molecular Compounds

– Atoms linked together by “ covalent chemical bonds” in discrete electrically neutral particles called molecules – e.g. H 2 O CO 2 PCl 3

• Ionic Compounds

C 12 H 22 O 11 – Result from transfer of one or more electrons from one atom to another to yield oppositely-charged particles called ions – No discrete molecules; ions held together by electrostatic forces (“ ionic bonds”) in a regular, 3-D pattern called a

crystalline lattice

– e.g. LiF lithium fluoride Li + F Li + + F – MgCl 2 magnesium chloride e Mg e Cl Cl LiF Mg 2+ + 2 Cl MgCl 2

Example Compounds

Molecular CH 4 = = = Ionic

e Na + Cl NaCl Na + Cl -

Formula Unit ratio. (NaCl) : The smallest unit of a compound. Shows the smallest whole-number Ionic compounds must be electrically neutral.

• • •

Types of Chemical Formulas

empirical

formula

present molecular

formula

molecule structural

formula

shows the

simplest ratio

shows the of the elements

actual number

of atoms in one shows how the atoms are connected e.g. for “hydrogen peroxide” the three formulas are: empirical: HO molecular: H 2 O 2 structural:

H O O H

Ionic Compounds

• Usually involve metals and/or polyatomic ions 2- anions 1- anions 1+ cations 2+ cations Other metals may form more than one cation, e.g. Fe 2+ , Fe 3+ , Sn 2+ , Sn 4+

Writing Formulas for Ionic Compounds

• Polyatomic ions--Table 3.5- KNOW formulas and names!!!

– Two or more atoms combined in a single charged unit – e.g. NH 4 + – H 3 O + (ammonium), (hydronium) , – NO 3 – PO 4 3 (nitrate), (phosphate), – HCO 3 (hydrogen carbonate, or bicarbonate) • Look for the simplest combination of cations (+) and anions (-) to yield an electrically neutral formula – e.g. ion combination compound – – – Mg 2+ Na + Fe 3+ and Cl and O 2 and SO 4 2 MgCl 2 Na 2 O Fe 2 (SO 4 ) 3 • Example: What compound should form between sulfur (S) and potassium (K)? • Example: What compound will form between ammonium and phosphate?

Nomenclature for Ionic Compounds

• First, determine if it’s ionic! – metal(s) + nonmetal(s) • Binary ionic compounds (2 different elements) – cation(charge if needed) + anion ide – Know Tables 3.3 and 3.4 (not older names) – e.g. ion combination compound name – – – Mg 2+ Na + Fe 2+ and Cl and O 2 and N 3 MgCl 2 Na 2 O Fe 3 N 2 – Other ionic compounds – With polyatomic ions; cation(charge if needed) + polyatomic ion name – Hydrates; compound name (as above) + prefix hydrate ( Know prefixes, p92) – e.g. ion combination compound name – – – Ca Co Hg 2+ 2+ 2 2+ and Cr and Cl 2 and CN O 7 2 CaCr CoCl 2 2 O 7 •6H 2 O Hg 2 CN 2 •H 2 O

Nomenclature: Molecular Compounds

• First, determine it’s molecular!

– between nonmetals and/or metalloids • Binary molecular compounds (between 2 elements) – prefix element + prefix element ide – First element is most metallic (bottom left of per. table) • Use prefixes to indicate numbers of each atom, e.g.

– PF 3 phosphorus trifluoride – P 2 F 4 diphosphorus tetrafluoride – N 2 O • Exception: hydrogen plus one atom of a nonmetal, see next section!

5 dinitrogen pentoxide

Nomenclature; Binary Acids

• First, determine it’s a binary acid!

– hydrogen + nonmetal – Hydro element ic + acid – e.g. compound name – HCl – HBr

Nomenclature: Oxoacids and Their Salts

• oxoacid H x EO y • Removal of H + (E = nonmetal) yields polyatomic anions oxoacid H 2 SO 4 sulfur ic acid polyatomic ions SO 4 2– sulf ate salt example Na 2 SO 4 sodium sulf ate polyprotic acids H 2 SO 3 Sulfur ous acid HSO 4 – Hydrogen sulf ate SO 3 2– sulf ite NaHSO Sodium hydrogen sulf ate 4 CaSO 3 Calcium sulf ite HSO 3 – Hydrogen sulf ite Ca(HSO Calcium hydrogen sulf ite 3 ) 2 Series of chlorine oxoacids and their salts: HClO x (x = 1,2,3,4) acid salts

Stoichiometric Equivalence

A chemical formula shows the ratio by atoms and by moles the elements in the formula.

of e.g. in the compound N 2 O 5 : ratio by atoms: 2 atoms N : 5 atoms O ratio by moles: 2 moles N : 5 moles O

in N

2

O

5

,

2 moles N 5 moles O (a chemical equivalence) Problem: How many moles of N atoms are combined with 15 moles of O in N 2 O 5 ?

Use the mole ratio as a conversion factor!

(15 moles O) x (2 moles N/5 moles O) = 6.0 moles N

• •

Formula Mass and Molecular Mass

formula mass formula = sum of all atomic masses of elements in a (remember that atomic mass = the mass of a single atom) molecular mass = formula mass of a

molecular

substance {“formula weight” and “molecular weight” are often used instead} Problem What is the molecular mass of N 2 O 5 ? (add the atomic masses!) N 2 O 5 = 2 N + 5 O = 2(14.0) + 5(16.0) = 108.0

What are the units? For 1 molecule: amu For 1 mole: grams 1 mole of a substance = its formula mass in grams e.g. 1 mole of N 2 O 5 = 108.0 g N 2 O 5 (just another conversion factor!)

Example Problems

• What is the mass of 0.65 moles of N 2 O 5 ?

(0.65 moles N 2 O 5 ) x (108 g N 2 O 5 /1 mol N 2 O 5 ) = 70 g N 2 O 5 • What mass of iron combines with 5.00 g of oxygen to make Fe 2 O 3 ?

Method: grams A --> moles A --> moles B --> grams B (5.00 g O) x(1 mol O)/16.0 g O) x (2 mole Fe/3 mol O) x (55.85 g Fe/mole Fe) = 11.6 g Fe

Sample Problems

• The label on my water bottle says there are 5.0 mg of sodium in it; if that were pure sodium, how many atoms of sodium would that be?

• The same water bottle contains 16 oz of water. If I drink it all, how many moles of water did I drink?

•

Percentage Composition

percentage composition -- mass % of elements in a compound Theoretical % composition -- from given formula Example Problem What is the percentage composition of H 2 CO 3 ?

mole ratio = 2 mol H : 1 mole C : 3 mol O molecular mass = 2(1.01) + 1(12.01) + 3(15.99) = 62.00 g/mol % composition: % H = [mass H / mass H 2 CO 3 ] x 100% = [2(1.01)/62.00] x 100% = 3.36% % C = (12.01/62.00) x 100% = 19.36% % O = [3 (16.00)/62.00] x 100% = 77.38% Total 100.00%

Empirical Formula Determination

Example Problem A certain fluorocarbon is found to be 36.52% C, 6.08% H, and 57.38% F. What is the empirical formula for this compound?

We’re looking for the mole ratio of the elements.

In 100 g of the compound, there are: (36.52 g C) x (1 mol C/12.01 g C) = 3.041 mol C (6.08 g H) x (1 mol H/1.008 g H) = 6.02 mol H (57.38 g F) x (1 mol F/19.00 g F) = 3.020 mol F So, the mole ratio is: C 3.041

H 6.02

F 3.020

Now reduce to the simplest ratio (divide by the smallest number): C 3.041/3.020

H 6.02/3.020

F 3.020/3.020

= C 1.007

H 1.99

F = CH 2 F (the empirical formula )

Molecular Formula

Empirical formula formula combined with molecular mass = molecular Problem The above fluorocarbon is found to have a molecular mass of 66.08 g/mole. What is the molecular formula?

n x (mass of empirical formula) = molecular mass (n = ?) Empirical formula = CH 2 F Formula mass = 1 C + 2 H + F = 33.03 g/mole n x (33.03 g/mole) = 66.08 g/mol so, n = 2 molecular formula is C 2 H 4 F 2

Sample Problem

Carboranes are an interesting class of compounds that contain carbon, hydrogen, and boron. One such carborane is found to have the following percentage composition: 28.18% C, 63.45% B, and 8.26% H. Determine the empirical formula of this carborane.

• •

Balancing Chemical Equations I

Adjust coefficients to get equal numbers of each kind of element of both sides of arrow.

Use smallest, whole number coefficients.

e.g. start with unbalanced equation (for the combustion of butane): C 4 H 10 + O 2 CO 2 + H 2 O

Hint

-- first look for an element that appears only once on each side; e.g. C C 4 H 10 + 13/2 O 2 4 CO 2 + 5 H 2 O Multiply through by 2 to remove fractional coefficient: 2 C 4 H 10 + 13 O 2 8 CO 2 + 10 H 2 O

Nomenclature: Organic Compounds • Compounds of carbon--organic chemistry

– If they have only C and H, hydrocarbons – e.g. alkanes: methane CH 4 , ethane C 2 H 6 , propane C 3 H 8 • general formula: C n H 2n+2 – Know Table 3.7

• Functional Groups

– R = hydrocarbon group – e.g. alcohols: methanol CH 3 OH, ethanol C 2 H 5 OH – e.g. amines: propyl amine, butylamine – Recognize functional groups in Table 3.8

Methanol (wood alcohol), CH 3 OH, is related to methane, CH 4 , by replacing one H with OH.