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Nomenclature and Stochiometry. L O S E R S. Be Familiar With…. G A I N E R S. Ionic Compounds- Formed when a ca t ion (metal) and an a n ion (non metal) join together to become neutral (NaCl). Positive ion, have more protons than electrons-loose electrons.
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Nomenclature and Stochiometry
L O S E R S Be Familiar With… G A I N E R S • Ionic Compounds- • Formed when a cation (metal) and an anion (non metal) join together to become neutral (NaCl) Positive ion, have more protons than electrons-loose electrons Negative ion, have less protons than electrons-gain electrons • Molecular Compounds- • 2nonmetal elements held together by covalent bonds. (NO2)
Naming Binary Compounds… • Polyatomic Ions- • Group of atoms that act as a unit and carry a charge 1st cat ion ______ + anion with –ide ending (use roman numerals when necessary) • NaNO3–sodium nitrate • CaSO4–calcium sulfate • CaSO3 –calcium sulfite • NH4NO3- ammonium nitrate • Examples- • Li + F -> LiF -> Lithium Fluoride • Mg + S -> MgS -> Magnesium Sulfide • Na + Br -> NaBr -> Sodium Bromide • Al + F -> AlF3-> Aluminum Fluoride
Using Roman Numerals… • How do you know? • When there is more than one oxidation state • Is not in Groups 1,2, or 13 • Usually in the D block and are transition metals • If transition metals have more than one oxidation state, it will then have one ionic charge to create the ionic charge=oxidation state • *In these cases you must use a roman numeral to state the cations charge • You use roman numerals if the element is a metal and has more than one ionic charge Examples- FeO -> iron (II) oxide Fe2O3 -> iron (III) oxide MnO -> manganese (II) oxide CrCl2-> chromium (II) chloride
Formulas… • Structural Formulas tell us- • Shows the way the atoms are joined together in a molecule. • Chemical Formulas tell us- • Elements that make up the compound • Ratio of the elements in the compound • Subscripts –below the element • Tell you the number of atoms Example- Example- Water-> H2O-> 2 hydrogen atoms and one oxygen atom • Carbon always forms 4 bonds • Molecular Models show orientation in 3 dimensions so are the best molecular models • Molecular Formulas tell us- • Tell you the total number of atoms of each element in 1 molecule of the substance Example- C2H2 ->Acetylene
Formula Mass… • Sum of all the atomic masses in the formula of a substance • Will need to use periodic tables • Atomic masses based on a scale using C-12 • Also known as- • Gram formula mass • Gram molecular weight • Molecular weight
Binary Molecule Compounds… • Naming Rules- • Use prefixes to distinguish how many atoms of each element are present • Ending has an –ide ending • Can omit the prefix of one atom in first element 1=mono 6=hexa 2=di 7=hepta 3=tri 8=octa 4=tetra 9=nona 5=penta 10=deca • Examples- • CO-carbon monoxide • N2O-dinitrogen monoxide • Tetraiodine monoxide-I4O • Carbon tetrafluoride=CF4
Acids… -an acid is a molecular substance that dissolved in water to produce hydrogen ions Ternary Acids Binary Acids • H plus a polyatomic ion • Hydro prefix is DROPPED • If ends with –ate the name of the acid ends with –ic • If ends with –ite the name ends with -ous • Hydrogen and another element • H takes the place of metal therefore the prefix is HYDRO- Hydro+root name of anion+ -ice ending+acid I ateice cream, it was delicious! • Examples- • HCl- hydrochloric acid • HF- hydrofluoric acid • Hydrobromic acid • Hydroiodic acid • Examples- • HNO3- nitric acid • HNO2- nitrous acid
Moles… • Of any element is the number of atom that element which are equal to the atoms in 12.0 grams C-12 6.02 x 10^23 • Number of atoms in 1 mole= 6.02 x 10^23 • Know how to do… • Mass-mole • Particles-mole • Volume-mole • Mixed mole problems • Mass (g) of 1 mol = atomic mass or formula mass
Mass to Mole Particles to Mole • Conversion Factor-Formula Mass • Conversion Factor- 6.02 x 10^23 particles/mol An example of sodium chloride (NaCl) has a mass of 11.2g. How many moles is that? How many molecules are in 2 moles of water? Given: 11.2g NaCl Unknown: # of moles Plan: mass -> mole Formula: g of NaCl/Formula Mass=moles Work: Na 1 x 23 = 23 Cl 1 x 35.5 = 35.5 = 58 g/mol 11.2 g NaCl/ 58.5 g/mol =.191 moles of NaCl Given: 2 moles of water Unknown: molecules Plan: mole -> molecules Formula: 2 moles x 6.02 x 10^23=molecules Work: 2.0 mole water x (6.02 x 10^23) molecules/mole 1.2 x 10^24 molecules water
Volume to Mole Mixed Mole • Conversion Factor-22.4L/mol • Conversion Factor- varies A container with a volume of 893L contains how many moles of air at STP? A sample of NH3 gas occupys 75.0L at STP. How many molecules is this? Given: 893L of air Unknown: # of moles Plan: volume -> mole Formula: 893L/22.4=mole Work: 893L/22.4=39.9 mol of air -or- 893L x 1 mol =39.9 mol of air 1 22.4L Given: 75.0L of NH3 Unknown: molecules Plan: volume -> mole -> molecules Formula: 75.0L/22.4 x 6.02 x 10^23 Work: 75.0L x 1 mol x 6.02 x 10^23 = 22.4L 1 mol = 2.01^24 = 2.02^24 molecules of NH3 • These problems can go any way across the spider diagram This is just one example
Balancing Equations… • Steps • Identify the elements on both sides of the equation • Identify the # of atoms on each side • Pick an element that doesn’t have same number of atoms on both and increase the coefficient on the side where it is smaller than one • Repeat step 3 until all atoms are equal on each side of equation • Check work and be neat • Before moving to Mole-Mole problems use this review for balancing equations • Balanced equations have the same number and kinds of atoms on both sides the equation Example- 2 Ag2O -> 4 Ag+__O2 Ag 2 4 Ag 1 2 3 4 O 1 2 O 2
Mole to Mole Volume to Volume How many moles of water are needed to react completely with 12 moles of sodium according to the balanced equation that follows? 2Na + 2H2O -> 2NaOH + H2 Find volume in liters of hydrogen (H2) that reacts with the 15.5L of N2 Plan: L of N2 ->mol of N2 -> mol of H2 -> L of H2 Work: 15.5L N2 x 1 mol N2 x 3 mol H2 x 22.4L H2 22.4L N2 1 mol N2 1 mol H2 =46.5 L H2 Work: 12 moles of Na x 2 mol H2O 2 mol Na = 12 mol of H2O
Chemical Reactions… Direct Combination or Synthesis -2 or more reactants come together to form a single product A+B -> AB Double Replacement Reaction -2 compounds as reactants -2 compounds as products AB + CD-> AD + CB Decomposition Reaction -breaks down into smaller parts AB -> A + B Single Replacement Reaction -switches partners A + Bx -> Ax + B
Limiting Reactants… • Limiting Reactant- • Reactant that limits the amount of product formed in a chemical reaction • Plan: • g Cu->mole Cu->mol Ag-> g Ag • g AgNO3-> mol ofAgNO3-> mol Ag-> g Ag • Work: • 1. 3.5g Cu x 1 mol Cu x 2 mol Ag x 108g Ag • 63.5g Cu 1 mol Cu 1 mol Ag • = 12 grams of Ag • 2. 6.0g AgNO3 x 1mol AgNO3 x 2mol Ag x 108g Ag • 170g AgNO3 2mol AgNO3 1molAg • = 3.8g Ag • The Limiting Reactant is AgNO3. Ex- Cu + 2AgNO3 -> CO(NO3)2 + 2Ag -Given the balanced equation where copper reactants with silver nitrate to produce copper (II) nitrate plus silver, when 3.5g of Cu is added to a solution containing 6.0g of AgNO3 what is the limiting reactant? -The reactant that produces the least amount is the limiting reactant.
Percent Composition and Formula Mass vs. Density… % = part x 100 whole D=M/V FM=D x 22.4L/mol In water what percent is H and what percent is O2? Density of a gas is 1.96g with volume at STP, what is the formula mass? Formula Mass H 2x1.01=2.02 O 1x16=16 %Composition H: 2.0/18.0 x 100= 11% 0: 16/18 x 100= 89% 18.0grams/mol FM= 1.96g x 22.4L =43.9 g/mol L mol
Empirical Formulas … • Steps for finding the formula… • %-> grams • Grams -> moles • Divide by the smallest until whole • Empirical Formula- • A formula that gives the simplest whole # ratio of the atoms of the compound. • To determine the empirical formula you must know • Mass of each element in sample OR • Percent composition of compound Example- Find empirical formula of a substancethat contains 80% C and 20% H. C 80%->80g x (1 mol/12)=6.7mol C/6.7=1 H 20%->20g x (1 mol/1.0)=20mol H/6.7=3 C1H3 or CH3
Molecular Formulas … Molecular formula = Formula Mass of Molecule Empirical formula’s formula mass • Empirical Formula- • Tells actual number of atoms that each element has • Always whole number multiples of the empirical formula Example-Find the molecular formula for CH2O Experiments show that the formula mass of glucose is 180g/mol (180g/mol)/(30g/mol)=6 Multiply 6 to the empirical formula C6H12O6