S1.4 Flashcards
the mole
1 mole = ?
- = 6.02⋅10^-23 elementary entities (atoms, ions, …)
- equal to the number of atoms in a 12g of 12C ⇒ M(C) = 12 g/mol
Avogadro constant
NA = 6.02 ⋅ 10^-23 mol^-1
= number of constituent particles in one mole of a substance
amount of substance
fixed number of particles ⇒ n = [mol]
relative atomic mass
- weighted average of the atomic masses and their relative abundances of an isotope
- ratio of the mass of a certain atom to 1/12 of the mass of a 12C atom
- Ar = [/]
relative molecular mass
- combination of Ar values for individual atoms or ions
- the ratio of the mass of a molecule or other multiatomic species to 1/12 of the mass of a 12C atom
- Mr = [/]
why was the 12C isotope used to find Ar?
because it is highly stable and C does not appear in many isotopes
what is 1 amu?
= atomic mass unit
= m(C)/12
molar mass
mass of one mole of a substance
M = [g/mol]
calculating amount of substance
formulas (mass, number of particles)
n = m/M = N/NA
empirical vs molecular formula
- empirical formula ⇒ the simplest ratio of atoms of the different elements that are present in the substance
- molecular formula ⇒ simple multiple of the empirical formula, specifies the actual number of atoms of each element in one molecule or formula unit of the substance
- molecular formula = n(empirical formula)
hydrate vs anhydrate
Hydrate = an ionic compound that contains water molecules in its structure (X ⋅ yH2O)
Anhydrate = the substance that remains after the water is removed from a hydrate (X)
solution
def
= homogeneous mixtures of two or more components, consisting of a solvent (dissolver) and one or more solutes (dissolves)
concentrated vs diluted solution
concentrated solution ⇒ ratio of solute to solvent is high
dilute solution ⇒ ratio of solute to solvent is low
molar concentration
[solute] = n(solute) : V(solution) = [mol/dm3 = M]
mass concentration
γ(solute) = c(solute) = m(solute) : V(solution) = [g/dm3]
