Atomic weights:
For chemists who need to know the atomic weight of all elements precisely this means the masses of all elements even the mass of every isotope needs to be determined by experiment. This is done in a mass spectrometer which can measure the mass and the abundance of the different isotopes. All masses are compared to the mass of a 12C atom and reported as average relative atomic masses of all naturally occurring isotopes of that element. Magnesium for example has a relative atomic Mass of 24.305 which means it is 24.305/12.00000 = 2.025 times heavier that a 12C atom. To use relative atomic masses in calculations we can do one of two things.We can express the mass of one atom by adding the unit u (atomic mass unit) to the value of the relative atomic mass. E.g. 24.305 u is the mass of one magnesium atom which equals 24.305 x 1.6605E-24 g or we add the unit g/mol to the relative atomic mass and get the mass of a mole of magnesium atoms in g (24.305 g/mol).
The mole:
The mole is an amount of substance, much like a dozen (12) or a gross (144). It is defined as the amount of substance that contains as many entities as there are atoms in exactly 12 g of pure 12C. There are 6.023 × 1023 atoms in 12 g of 12C.By knowing the average atomic mass of each element we can “count” atoms by weighing them. The relative atomic mass of iron is 55.847. This means 1 mole of iron atoms has a mass of 55.847 g. We say the molar mass of iron is 55.847 g/mol. Now we are ready to engage in stoichiometric calculations.
Stoichiometry:
- Stoichiometry refers to the amount of all compounds which react in a balanced chemical equation.
Let’s say we are interested in the chemistry of magnesium with oxygen. We have already seen a balanced reaction equation for this reaction:
2Mg + O2 → 2MgO
How much magnesium oxide will form if 5.73 g magnesium burn in excess oxygen?
3.) Convert moles of MgO to mass MgO: To find the molar mass (or formula weight) of MgO simply add the mass of all atoms in the formula (24.304 g/mol + 16.000 g/mol = 40.304 g/mol):
What if a sample of 1.51 g magnesium is ignited in a sealed 1L flask that contains 1.42 g oxygen gas. How much magnesium oxide can be formed and how much of which reactant is left behind? For this problem you first need to determine which reactant limits the amount of product that can form.
1.) Convert the masses of the reactants to moles of reactant:
2.) Determine which the limiting reactant is:
3.) Calculate the mass of product that can form:
2.50 g MgO can be formed. Some oxygen is left over. Since there are only two reactants which masses we know and one product whose mass we know as well we can simply use the law of conservation of mass to calculate the mass of oxygen left over. The mass of the reactants is:
1.51 g Mg + 1.42 g O2 = 2.93 g reactants.
Since only 2.50 g of product is formed the difference must be O2 left over:
2.93 g – 2.50 g = 0.43 g
0.43 g oxygen are left un-reacted.
Solution Stoichiometry:
In aqueous solution the number of moles of any component dissolved in one litre of solution determines the molar concentration of that component. This molar concentration is referred to as Molarity.
How would you prepare a 1 Molar (M) solution. Well, simply dissolve one mole of the compound in enough water to make 1 L of solution.
How would you prepare a 1 Molar (M) solution. Well, simply dissolve one mole of the compound in enough water to make 1 L of solution.
If 23.4 mL of a 1.200 M solution of NaOH was necessary to neutralize 10.00 mL of a sulphuric acid sample of unknown concentration. What is the molarity of the sulfuric acid.
2NaOH + H2SO4 →Na2SO4 + 2H2O
Since the molar concentration is defined as: 
The number of moles of sodium hydroxide can obtained as follows:
At the equivalence point (neutral) the number of moles of sulfuric acid can be calculated using the coefficients in the balanced chemical equation:
The exact concentration of the sulfuric acid is thus:
Diluting Solutions:
For diluting solutions, you may find the following equation useful:
ci = the initial concentration
Vi = the initial volume
cf = the final concentration
Vf = the final volume
This equation works for all concentrations (not just moles/L), and is based on the assumption that the density of the solution is not changing.
How much of a 6.5 M Ni2+(aq) stock solution is necessary to fill a 220 L nickel-plating bath if the desired concentration of Ni2+(aq) is 0.10 M?
% Yield:
A side problem that may arise is to calculate the % yield of a reaction which is simply the amount of product that you isolated from the reaction mixture divided by the amount of product that in theory could have formed if everything went perfect.
