1.2 Formulas
1.2.1
Relative Atomic Mass ( Ar ): The weighted mean of all the naturally occuring isotopes of the element relative to Carbon-12.Because of this the Ar is not a whole number. Ar has no units
Relative Molecular Mass (Mr) : The wieghted mean of all atomic masses of the atoms in one molecule relative to Carbon-12. Because of this Mr in not a whole nu,ber. Mr has no units.
1.2.2: Calculate the mass of one mole of a species from its formula
Number of moles =mass/molecular mass
Molarity = number of moles/ volume
1.2.3: Solve problems involving the relationship between the amount of substance in moles, mass and molas mass.
1.2.4
Empiricqal Formula: The simples way to write the formula of a compound
- Can either be obtained by knowing the mass of teh percent composition of the elements
Molecular Formula: The actual number of atoms of each element in a molecule of a substance
1.2.5: Determine the empirical formula from the percentage composition or from other experimental data.
Ex: 344g of a compound contains 0.365g of Na, 0.221g of N, and 0.758g of O. Find the empirical formula:
Na 0.365/23 = 0.0159/0.0158 =1
N 0.221/14 = 0.0158/0.0158 = 1 NaNO3
O 0.0758/16 = 0.474/0.0158 = 3
1.2.6: Determine the molecular formula when guven both the empirical formula and experiemntal data.
Relative Atomic Mass ( Ar ): The weighted mean of all the naturally occuring isotopes of the element relative to Carbon-12.Because of this the Ar is not a whole number. Ar has no units
Relative Molecular Mass (Mr) : The wieghted mean of all atomic masses of the atoms in one molecule relative to Carbon-12. Because of this Mr in not a whole nu,ber. Mr has no units.
1.2.2: Calculate the mass of one mole of a species from its formula
Number of moles =mass/molecular mass
Molarity = number of moles/ volume
1.2.3: Solve problems involving the relationship between the amount of substance in moles, mass and molas mass.
1.2.4
Empiricqal Formula: The simples way to write the formula of a compound
- Can either be obtained by knowing the mass of teh percent composition of the elements
Molecular Formula: The actual number of atoms of each element in a molecule of a substance
1.2.5: Determine the empirical formula from the percentage composition or from other experimental data.
Ex: 344g of a compound contains 0.365g of Na, 0.221g of N, and 0.758g of O. Find the empirical formula:
Na 0.365/23 = 0.0159/0.0158 =1
N 0.221/14 = 0.0158/0.0158 = 1 NaNO3
O 0.0758/16 = 0.474/0.0158 = 3
1.2.6: Determine the molecular formula when guven both the empirical formula and experiemntal data.
