Always assume that significant numbers are applied to all problems even if it isn’t stated.
Non-zero digits are always significant numbers
Ex. 1288 is 4 significant numbers
Zero Significant rules
If zero is sandwiched between two non-zero digits, then it is always significant
Ex. 10002 has 5 significant numbers
Leading zeroes (from the left) are never significant
Ex. 0.004 has only 1 significant number
Trailing zeroes (from the right) after a nonzero digit is significant only if there is a
decimal place.
Ex. 60000. is 5 significant numbers
● Mass relationships between substances in a chemical relationship
● Based on the mole ratio
○ Indicated by coefficients in a balanced equation
■ 2Mg + 1O2 → 2MgO
● Calculation steps
○ Write a balanced equation
○ Identify the known and unknown
■ What you already know / given information?
■ What you want to figure out / what’s not given?
○ Line up Conversion Factors
■ Mole Ratio - moles to moles
■ Molar Mass - moles to grams
■ Molarity - moles to liters solution
■ Molar Volume - moles to liters gas
● *Molar volume at STP: 1 mol of a gas is 22.4 L
■ *this is a core step in all stoichiometry problems
○ Example: How many moles of KClO3 must decompose in order to produce 9 moles of oxygen gas?
■ 2KClO3 → 2 KCL + 3O2
■ 9 moles of O2 * 2 moles of KClO3 / 3 moles of O2 = ?
■ Answer = 6 moles of KClO3
● Percent Yield
○ Chemical reactions are never perfect
○ You won’t get the amount of product you calculate
○ Actual Yield: Amount of product obtained experimentally
○ Theoretical Yield: Expected amount of product (using stoichiometry)
○ Formula: Actual yield/ Theoretical Yield * 100
○ Example: What’s the percent yield if 13.1 grams of CaO is actually produced with 24.8 grams of CaCO3
■ 24.8 g CaCO3 * 1 mol CaCO3/100.1 g CaCO3 * 1 mol CaO / 1 mol CaCO3 * 56.08g CaO/ 1 mol CaO = 13.9 g CaO
■ 13.1/ 13.9 *100 = 94.2%
● Limiting Reagent
○ The reactant that runs out first
○ We care about it bc it determines how much product will be made
● Excess Reactant
○ Added to ensure that the other reactant is completely used up
○ Cheaper and easier to recycle
● How to find the Limiting Reagent?
○ Assume all of the first reactant gets used up
○ Determine how much of one of the products will be made
○ Do the same for the second reactant
○ The reactant that makes less product is the limiting reagent
○ Example: If you have 18 mol SO2 and 7.0 mol O2, what is the limiting reagent?
■ 2SO2 + O2 → 2SO3
■ 18 mol O2 * 2 mol SO3/ 2 mol SO2 = 18 mol SO3
■ 7.0 mol O2 * 2 mol O2/ 1 mol O2 = 14 mol SO3
■ Limiting Reagent = 14 mol SO3
● Convert amounts of elements into moles
● Find the mole ratio: Divide all numbers by the lowest number of moles
● If necessary, use a multiple of your ratio to get whole numbers
○ Ex. Find the empirical formula of a compound containing 10.4 grams P and 35.7 grams Cl.
■ Phosphorus: 10.4 g P * 1 mol P/ 31.0 g P = 0.335 mol P
■ Chlorine: 35.7 g Cl * 1 mol Cl/ 35.5 g Cl = 1.01 mol Cl
● Phosphorus has the lowest amount of moles, so Cl will also be divided by that
■ 0.335 mol/ 0.335 mol = 1
■ 1.01/ 0.335 mol = 3
■ Answer = P to Cl ratio is 1:3, formula is PCl3
● Physical changes are when objects and substances undergo a change that doesn’t change their chemical composition
○ Ex. Boiling Point, Melting Point, Freezing Point, dissolving, bending or cutting
● Chemical changes are when the compositions of a substance changes or one or more substances combine or break up to form new substances
○ Ex. Color change, bubbles, odorous, gas bubbles, production of heat
● When Chemical reactions occur, a change in energy is noticed as there is a change in temperature of the reaction
○ Exothermic reactions release energy → Increase in temperature
○ Endothermic reactions absorb energy → Decrease in temperature
● Oxidation: Loss of electrons but increase in oxidation numbers
● Reduction: Gain of electrons but decrease in oxidation numbers
○ Assigning Oxidation Numbers
■ Elements: ox. number = zero
■ Monatomic Ions: ox. number = charge of ion
■ F in a compound is always -1
■ O in a compound is usually -2 (exception: O in peroxide)
■ H is +1 in covalent bonds
■ Determine other elements from there
● Compounds add up to zero
● Polyatomic Ions add up to their charge
● Redox reactions
○ Identify them when any element’s ox. number changes
○ The element that’s oxidized is the reducing agent and provides the electrons
○ The element reduced is the oxidizing agent
● Balancing Redox Equations
○ “Half Reaction method,” use when equation is too complicated to balance conventionally
○ If reaction occurs in acid:
■ Write separate half-reactions for oxidation and reduction
■ Balance the elements (except H and O) in each half
■ Balance O using H2O
■ Balance H using H+ ions (available from the acid solution)
○ Balance electrons (e-) in each half (e- are products in oxidation and reactants in reduction)
○ Use multiples of half-reactions (if necessary) to get the numbers of electrons equal
○ Add half-reactions together
○ Add spectator ions back in
● If reaction occurs in base:
○ Write half-reactions and balance H and O as if its an acid
○ To both sides of the balanced equation, add OH- (equal to the number of H+ ions)
○ Form water on the sides that have H+ and OH-. Reduce H2O from both sides if possible
○ Balance the rest as in acid solution