Download Pre-AP Chemistry Chemical Quantities Review Sheet

Pre-AP Chemistry
Chemical Quantities Review Sheet
 PART 1: THE MOLE AND CONVERSION
- A conversion factor allows for the conversion between units.
- Representative particle – species present in a substance
o Include atoms, molecules, ion, formula units
- Conversion Factor 1: 1 mole = 6.02 x 1023 items
o Avogadro’s number = 6.02 x 1023 items
o This conversion factor is used to go from moles to items (or vice versa) ONLY!
o You must have units and chemical species throughout your calculations.
o Remember that 1 mole of anything is equivalent to an Avogadro’s number.
o 1 mole of any element has the same number of particles.
- Conversion Factor 2: 1 mole = (Molar Mass) grams
o For an element, the Molar Mass (MM) comes directly from the periodic table.
o For a compound or molecule, the Molar Mass (MM) is the sum of the number of
particles that make up the compound/molecule.
 MM of CaCl2 is found by adding the mass of Ca plus 2 times the mass of Cl
o The units of Molar Mass are grams per mole (g/mol)
o While 1 mole of any element has the same number of particles, 1 mole of each element
has a different mass as dictated by the periodic table.
- Conversion Factor 3: mole X = mole Y
o The chemical formula serves as a conversion factor.
o This helps to convert between one chemical species to another chemical species.
o This is a mole conversion factor NOT a gram conversion factor
o For the compound K2C2O4
 Correct: 1 mole K2C2O4 : 2 mole K : 2 mol C : 4 mole O
 WRONG: 1 gram K2C2O4 : 2 gram K : 2 grams C : 4 grams O
- You must be able to perform all the conversion factors listed above.
- All work must have correct units and chemical species listed.
- Be sure that your answers are in the correct
number of significant digits.
- You must be able to combine conversion factors
correctly to come to the correct answer
o Always make sure your units cancel.
- Use the flow chart provided to help determine the
number of steps required and the position of units
and numbers.
- Practice all the different types of conversion over
and over. Build speed and accuracy.
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 PART 2: PERCENT COMPOSITION
- The percent composition of a molecule describes the relative amount of each element.
- This is like any percentage: a part out of the whole thing.
o
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Percent by Mass X =
All of the percentages must total to about 100%.
The percent composition can be found from a chemical formula only.
o The molar mass of the formula is the total mass.
o The mass of each element is found by: Subscript x Molar Mass.
 PART 2: EMPIRICAL AND MOLECULAR FORMULA
- Each chemical formula has a basic ratio of elements.
- Empirical Formula – formula with lowest whole number ratio of elements.
- Molecular Formula – formula that shows the actual number of elements.
o This can be the same as the empirical formula.
o This can be a whole number multiple of the empirical formula.
- Steps to determining the empirical and molecular formula from percent composition:
1. Convert the percents to grams.
2. Convert the grams to moles by dividing by the Molar Mass.
3. Divide all the moles by the smallest number of moles.
4. Use the lowest, whole-number of moles (you may need to multiply by two if you do not
have whole numbers) to determine the EMPIRICAL FORMULA.
5. Find the Molar Mass of the empirical formula.
6. Calculate the ratio: Ratio =
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(the molecular molar mass is given).
7. Multiply the Ratio by all the subscripts of the empirical formula to determine the
MOLECULAR FORMULA.
You must maintain units throughout all of these steps and the chemical species.
 PART 3: MOLARITY AND DILUTIONS
- Concentration is the measure of the amount of solute in a given volume of solvent.
o Solute = particle of lesser quantity ; Solvent = particle of greater quantity.
o Dilute = small amount of solute ; Concentrated = large amount of solute.
- Molarity is a quantitative unit of measurement for concentration of solutions
o
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or
or
o The equation requires LITERS of solution.
o This is specific for the volume of a solution NOT solvent in the denominator.
Dilutions are used to prepare dilute solutions form highly concentrated (stock) solutions.
o The number of moles of solute per unit volume decrease, but the amount of moles does
not change.
o Dilution Equation:
or
o M = Molarity, V = Volume, 1 and C = concentrated solution, 2 and D = dilute solution.
Accurate/precise volume measure equipment: buret, volumetric flask (beaker is not accurate.)
The volume of concentrated solution must be added to enough water to reach desired volume.
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 PART 4: NOMENCLATURE
Type I: Ionic Bonding with FIXED oxidation states (charge)
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How to identify: Metal is in the main groups (Group 1, 2, 3) and Ag+, Zn2+, Cd2+
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SYMBOL  NAME
1) Write full name of the metal/cation.
2) Write name of the nonmental/anion with –ide ending.
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NAME  SYMBOL
1) Write symbol for the metal/cation.
2) Write symbol for the nonmetal/anion.
3) Use the charge from the periodic table (based on element’s group #) to determine the
subscripts of the metal and nonmetal to give an overall charge of zero (pos = neg).
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Examples:
1) Na2S
2) Calcium fluoride
sodium sulfide
CaF2
Type II: Ionic Bonding with VARIABLE oxidation states (charge)
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How to identify: Metal is a transition metal except Ag, Zn, Cd
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SYMBOL  NAME
1) Write full name of the metal/cation.
2) Write name of the nonmental/anion with –ide ending.
3) Use the fixed charge of the anion (from the periodic table) to determine the total
positive charge to give an overall compound charge of zero.
4) Divide the overall positive charge by the number of metal atoms to determine the
charge of EACH metal atom. Write this number as a Roman Numeral between the metal
and nonmetal in the chemical name.
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NAME  SYMBOL
1) Write symbol for the metal/cation.
2) Write symbol for the nonmetal/anion.
3) The charge of the EACH metal cation comes from the Roman Numeral in the given
name.
4) The charge of the anion comes from the periodic table.
5) Determine the subscripts of the metal and nonmetal to give an overall charge of zero
(pos = neg).
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Examples:
1) Sn3P4
2) Cobalt(II) nitride
tin(IV) phosphide
Co3N2
Type III: Ionic Bonding with Polyatomic Ions
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How to identify: A polyatomic ion will be present in the name or formula.
The list of polyatomic ions on Page 257 (textbook) MUST be memorized (name, formula, charge)
Consider the polyatomic ion as one large unit that does not change
NEVER touch or change the polyatomic ion --- use parentheses to denote multiple quantities
The charge of a polyatomic ion is the charge for the WHOLE unit/group of atoms. The sulfate ion
(SO42-) has a charge of 2- on the entire sulfate molecule…one SO4 has a charge of 2-.
Do NOT change the ending of the polyatomic anion to an –ide ending. Write the full name of the
ion.
1) If the metal/cation has a fixed charge, follow rules for Type I (except do not use –ide
ending).
2) If metal/cation has a variable charge, follow rules for Type II (except do not use –ide
ending.)
Examples:
1) Mn3(PO4)7
2) Barium acetate
permanganate(VII) phosphate
Ba(C2H3O2)2
Type IV: Covalent Bonding
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How to identify: 1st element is a nonmetal.
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SYMBOL  NAME
1) Write full name of the 1st nonmetal.
2) Write name of the 2nd nonmetal with –ide ending.
3) Put Greek prefix before both nonmetal based on the number of each nonmetal atom.
4) Remove “mono-” if placed in front of the first nonmetal. NEVER put “mono-” at
beginning of the chemical name.
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NAME  SYMBOL
1) Write symbol for the 1st nonmetal.
2) Write symbol for the 2nd nonmetal.
3) Use prefix before each metal to give the subscript of the corresponding nonmetal.
Greek Prefixes
1 = mono- 6 = hexaExamples:
2 = di7 = hepta1) BrO3
bromine trioxide
3 = tri8 = octa2) Diphosphorus pentoxide
P2O5
4 = tetra9 = nona5 = penta- 10 = decaPage 4 of 4
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