AP chemistry
Unit 1 reference sheet
Formula conventions
2+
Superscripts:
used to show the charges on ions Mg the 2 means a 2+ charge (lost 2 electrons)
Subscripts:
used to show numbers of atoms in a formula unit
H2SO4 two H’s, one S, and 4 O’s
–
Coefficients
used to show the number of formula units 2Br the 2 means two individual bromide ions
Hydrates
CuSO4 • 5 H2O some compounds have water molecules included
Empirical formula: formula reduced to lowest terms
Molecular formula: the true number of atoms of each element in the formula of a compound
A mole (abbreviated mol) is a certain number of things. It is sometimes called the chemist’s dozen. A dozen is 12 things, a mole is
23
23
Avogadro’s Number: 1 mole of any substance contains 6.02 x 10 molecules
6.02 x 10 things.
Percentage Composition quantifies what portion (by mass) of a substance is made up of each element.
Set up a fraction: mass of element
round to the tenth’s place. The percentage compositions of
mass of molecule
each element should add up to 100% or very close
Change to percentage by multiplying by 100
Given the Percentage Composition of a formula, you can calculate the empirical formula of the substance.
Step 1 assume you have 100 g of substance so the percentages become grams
Step 2 change grams of each element to moles of atoms of that element
Step 3 simplify the formula by dividing moles by the smallest value
Step 5 If ratio becomes… 1:1.5 multiply by 2
1:1.33 or 1:1.66 multiply by 3
Combustion analysis: turn grams of CO2 to moles of carbon and grams of carbon (using stoichiometry)
Turn grams of water to moles of hydrogen and grams of hydrogen (using stoichiometry)
Add the grams of carbon and grams of hydrogen and subtract them from the total grams = grams of O
Turn grams of O to moles of O and divide all of the moles by the smallest value of moles
Mass spectrometry: This gives proof that there are isotopes. The height of the peak tells the abundance of the isotope and the x
axis shows the m/z ratio (assume the charge is +1 so it is numerically equal to the mass of the isotope).
To find the average mass of the isotope: mass of isotope A x abundance + mass of isotope B x abundance, etc.
Balancing chemical reactions
Write the formula equation of...sodium metal + water → sodium hydroxide + hydrogen gas
Na + H2O → NaOH + H2
• some compounds have common names that you should just know... water, H2O; ammonia, NH3; methane, CH4
• remember the seven diatomic elements so they can be written as diatomic molecules when they appear in their
elemental form. Other elemental substances are written as single atoms
The balanced equation represents what actually occurs during a chemical reaction. Since atoms are not created
or destroyed during a normal chemical reaction, the number and kinds of atoms must agree on the left and
right sides of the arrows.
_2_Na2CO3 + _4_HCl → __4 NaCl + _2_H2O + _2_CO2
To balance the equation, you are only allowed to change the coefficients in front of the substances... not change the
formulas of the substances themselves.
Reduce the coefficients to the lowest terms.
Fractions may be used in front of diatomic elements.
Predicting products
Synthesis: A+ X AX
Decomposition: AX A + X
Single replacement: A + BY  B + AY
Double replacement: AX + BY  AY + BX (these happen if a precipitate, water, or a gas is formed)
Combustion: burns in air to react with oxygen
Memorize this type of combustion: Hydrocarbon + Oxygen  CO2 + H2O
If you see the following substances formed during Double replacement, realize that they will breakup into gases and
NH4OH  NH3(g) + H2O
leave the system (preventing re-formation of the reactants). H2CO3  CO2(g) + H2O and
Learn: a metal oxide + water  base (metal hydroxide)
Make sure products are written with the correct subscripts (charges must balance out for ionic). They don’t always have the same
subscript that they had on the reactant side
Stoichiometry study of the quantitative relationships in chemical formulas and equations.
Conversion factors in stoichiometry relates the Given and the Desired compounds using the coefficients from the balanced equation.
Example: N2 + 3 H2 → 2 NH3
could be used
3 moles H2
2 moles NH3
…which means that every time 2 moles of NH3 is formed, 3 moles of H2 must react.
The units of the given and desired will guide you as to which conversion factor to use:
Mass: grams or kg (use molar mass to convert to moles)
Particles: molecules or atoms (use Avogadro’s number to convert to moles)
The answer you calculate from a stoichiometry problem can be called the Theoretical Yield. Theoretically , you should get this
amount of product. In reality , you often get less than the theoretical amount due to products turning back to reactants or side
reactions. The amount you actually get is called the Actual Yield.
Percentage Yield = Actual Yield
Theoretical Yield x 100
Limiting reactant
In a problem with two Given values, one of the Given’s will limit how much product you can make. This is called the
limiting reactant. The other reactant is said to be in excess. Solve the problem twice using each Given… the reactant
that results in the smaller amount of product is the limiting reactant and the smaller answer is the true answer.
Example: N2 + 3 H2 → 2 NH3
When 28.0 grams of N2 reacts with 8.00 grams of H2, what mass of NH3 is produced?
(in this case, the N2 is the limiting reactant)
To find out how much H2 is left over, do another line equation
Given: 28.0 g N2 (make sure you always start with the limiting reactant once you know what it is)
Desired: ? g H2
Subtract the answer of this problem from 8.00 g H2
It is difficult to simply guess which reactant is the limiting reactant because it depends on two things:
(1) the molar mass of the reactant and
(2) the coefficients in the balanced equation
The smaller mass is not always the limiting reactant.
Example: N2 + 3 H2 → 2 NH3
1 mole (28 g N2) will just react with 3 moles (6.06 g H2)
so, if we react 28.0 g N2 with 8.0 g H2, only 6.06 g H2 will be used up and 1.94 g of H2 will be left over.
In this case, N2 is the L.R. and H2 is in X.S.
Solutions:
Solutions of acids, bases, and salts contain mobile ions and conduct electricity. These solutions are called electrolytes. Salts
are ionic compounds that dissociate in water. Acids are actually molecular compounds (covalently bonded) the become
ions when dissolved in water. Only 7 acids are strong electrolytes (H2SO4, HClO4, HClO3, HNO3, HI, HCl, HBr) and completely
dissociate when dissolved. All others dissolve completely, but only partially dissociate into ions.
Only 8 bases are strong electrolytes (alkali metals with OH and Ca, Sr, Ba with OH) because they dissolve completely. All
others have low solubility and remain solids rather than dissolve. One common exception is the weak base NH4OH . It
dissolves, but partially dissociates
strong electrolytes: completely ionize or dissociate in water (soluble ionic, strong acids)
weak electrolytes: partially ionize or dissociate in water (insoluble ionic, weak acids)
nonelectrolytes: have no ions in solution (covalent compounds)
+
+
+
+
+
Always Soluble compounds with alkali metal ions(Li , Na , K , Cs , Rb )
Usually Soluble
Usually NOT Soluble
Never Soluble
+
–
–
–
–
NH4 , NO3 , C2H3O2 , ClO3 & ClO4
–
– –
+
2+,
2+
Cl , Br , I [except ... Ag , Pb Hg2 ]
2–
2+
2+
2+
2+
SO4 [except “: Ca , Ba , Sr & Pb ]
2–
–
2+
2+
2+
O , OH [except alkali and Ca , Ba , Sr ]
2–
2– 2–
3–
+
CO3 , SO3 , S , PO4 [except NH4 & alkali metals]
Net ionic reactions
Shows what is chemically happening in a reaction. Write all strong electrolytes as ions (because in they completely turn into
ions in solution) and keep weak and nonelectrolytes as compounds. Take out any spectator ions-don’t chemically change so
they look exactly the same on both sides of the reaction.
Molecular equation: NaCl + AgNO3  NaNO3 + AgCl
+
+
+
Ionic equation: Na + Cl + Ag + NO3  Na + NO3 + AgCl
+
Net ionic equation: Ag + Cl  AgCl