Unit 3- Equations, The Mole, and Chem Formulas Flashcards
chem equations
a shorthand notation to describe a chem reaction
reactants
substances consumed
products
substances formed
coefficient
s before the formula of a substance in equation
Balancing equations
- start with most complex molecule and balance on either side. Move to next element. The last element will look at the product to balance the reactants
- if fraction, multiply WHOLE equation by the denom
neutralization
the reaction of an acid with a base to form a SALT and WATER
(acid + base —> salt + water)
acid
compound that dissolves water forming hydrogen ions
base
dissolves in water forming hydroxide ions.
- usually soluble metal hydroxide
salt
ionic compound consisting of the cation of a base and the anion of an acids
balancing neutralization reactions
the # of hydrogen ions provided by he acid must equal # of hydroxide ions provided by the base (= to # of water molecules formed)
combustion reaction
process of burning
- when hydrocarbon burns in oxygen, carbon forms CO2 and hydrogen forms H2O
oxidation-reduction reactions
reaction in which electrons are transferred from 1 species to another
assigning oxidation #s
- an atom in its elemental state has an oxidation # of zero
- The oxidation # of monoatomic ions is the charge of the ion
- In combination with other elements, the oxidation # of F is -1, O generally -2, H generally +1 with nonmetal or -1 with metal. Other halogens are generally -1.
- Sum of oxidation #s in any species must = charge of species
oxidation
loss of electrons in reaction (ox # inc)
reduction
gain of electrons in reaction (ox # dec)
mole
amount of substance that contains as many entities as the # of atoms in exactly 12g of the 12C isotope of Carbon
Avogadro’s #
the experimentally determined # of 12C atoms in 12g (= 6.022 x 10^23)
converting moles to entities
and vice versa
moles x avogadro’s # = # atoms/molec
atoms/ avogardo’s # = moles
molar mass (M)
the mass of one molec of that substance (g/mol) (= atomic mass)
- used to convert grams to moles and vice versa
mass % from formula
atomic mass / mass of compound x 100%
combustion
mass CO2(MM CO2) = molesCO2(CO2: C) = moles of C (MM C) = mass C
*same with H2O to H and O
Calc Empirical Formula
mass or % mass(MM) = moles/ smallest mol = # become subscripts
*mass% –> assume 100g sample. % = g
molec formula
whole # multiple of empirical formula
(empirical)n
n= MM of cpd/ MM of empirical formula
stiochiometry
study of the quantitative relationships involving the substances in chem reactions
formula for reaction stoichiometry
mass A(MM A) = moles A(coefficients A:B) = moles B(MM B) = mass B
theoretical yield
max quantity of product that can be obtained from a chem reaction based on amounts of starting materials
(stoichiometry)
limiting reactants
the reactant that is completely consumed when a chem reaction occurs
( stoichiometry but choose smaller mole to get mass of product)
actual yield/ % yield
mass of the product in a reaction
% yield= actual yeild/ theoretical yield x 100%