Moles. One mole of substance contains 6.02 x 10^23 (L) of particles.
1. Write the balanced Equation, 2. List the
measurements that is given and a symbol for the
one to be calculated. 3. Determine the conversion
factors (conversion to moles) and list them under
the correct formula. 4. Calculate n. 5. Multiply n by
the mole ratio (unknown coefficient over known
coefficient). 6. Calculate what you want.
Limiting Reactant: Limits the amount of product that can be formed. The
reaction will stop when all of the lim reactant is consumed.
Excess Reactant: Remains when a reaction stops when the lim reactant is
1. Calculate n for both reactants. 2.
Divide n by stoichiometry coefficients
and identify the smaller number. 3.
Perform all calculations with initial n
(not the n divided by the coefficients) of
the limiting reactant.
Gas Stoichiometry: When the gas is at STP, 1 mol = 22.7 dm^3
When the gas is not at STP, use PV = nRT
N (number of particles)
n = N / L
N = n x L
m (Molar Mass)
m = n x M
n = m / M
Solid: Particles tighlty
packed, strong inter-particle
forces, particles vibrate in
position, fixed shape and volume.
Liquid: Particles are more
spaced, inter-molecular forces are
weaker, particles can slide over
each other, no fixed shape, fixed
Gas: particles spread out,
inter-particle forces are
negligible, particles move more
freely, no fixed shape or volume.
Temperatura is a measure of the average kinetic
energy of the particles of a substance. States of matter
at a given T and P are determined by the strength of
the forces that may exist between particles. Higher
Temperature (T) means higher Kinetic Energy (KE)
Empirical Formula: the simplest whole number ratio of the elements in a compounds.
Finding the empirical formula: 1. Start
with the number of grams of each element
(if percentages are given, assume total
mass is 100). 2. Convert the mass of each
element to moles using its molar mass. 3.
Divide each mole value by the smallest
number of moles calculated. 4. Round to
the nearest whole number.
Molecular Formula: shows all the atoms in a molecule, it is a multiple of the empirical formula.
Calculating molecular formula: the empirical formula and
the molar mass of the compound are needed. Find the
molar mass of the molecule with the empirical formula.
Divide the larger molar mass (the one from the molecular
formula) by the smaller one (the one from the empirical
formula) and you will find a whole number. Multiply all
the subscripts of the empirical formula by that number.
Relative Atomic Mass (Ar): (weighted average of one atom of the element) / (1/12 mass of one atom of Carbon - 12). Ratio, therefore unit less.
Relative Formula Mass (Mr): (sum of the weighted average of the masses of the atoms in a formula unit) / (1/12 mass of one atom of Carbon - 12)
Percent Composition: (mass of element) / (total mass of compound) x 100%
Balancing: Determine the correct
formulas for all the reactants. 2. Being
balandinc, if possible wtith the simplest
metal, or the most complex species.
Leave elements like H2 or O2 last. Check
when two or more simple
compounds combine to
form a complicated one.
Mg + O --> MgO
Decomposition Reaction: opposite of
a synthesis reaction, complex
molecules breaks down to makes
simples ones. They require heat or
energy to decompose.
2NaCL --> 2NA + Cl2
ReactionL one element
trades place with another
element in a compound.
Metal and H:
Mg + 2HCl --> MgCl2 + H2
F2 + KCl --> FCl + K
Activity Series of Metals: top ones are more reactive so they replace the ones below it.
Reaction: when the
anions and cations of two
switch places, forming
two entirely different
Pb(NO3)2 + K2CrO4 --> PbCrO4 + 2KNO3
Combustion Reaction: generally with
oxygen (oxidation), products are CO2
(complete combustion) and H2O, when
there is not enough O2, an incomplete
combustion happesn, products are CO,
C, and H20.
2CH4 + 3O2 --> 2CO2 + 4H2O
Percentage Uncertainty = absolute uncertainty / measured value x 100%
Percentage Error: (Accepted Value - Experimental Value) / Accepted Value x 100%
%U > %E --> Random Error
%U < %E --> Systematic Error
Solute: less abundant component. Solvent: more
abundant component in which the solute is
Concentration is in mol dm^-3 and it is C = n/V. n is in
moles and V is in dm^3
Parts per million = mass of solute /
total mass of solution
C1V1 = C2V2
n1 + n2 = n3 --> C1V1 + C2V2 = C(V1+V2)
Concentrated solutions have
relatively large quantities of
dissolved solute. Dilute solution has
small quantity os dissolved solute.
Solubility of solids increases with
temperature and solubility of gases
decreases with increasing