Chemical Calculations | General Chemistry 2

 Mole number n (in mol): 

The number of moles in a sample
 

Number of mole n in 2.0 g of N₂:

n = $\frac{{\mathrm{m}}_{{\mathrm{N}}_2}}{{\mathrm{M}}_{{\mathrm{N}}_2}}$

m_N2 = 2.0 g
M_N2 = 2 x M_N = 2 x 14.0 = 28.0 g.mol^-1 

⇒ n = $\frac{2.0}{28.0}$ = 7.1 x 10^-2 mol

Number of nitrogen atom N_N in 2.0 g of N₂:

n = $\frac{{\mathrm{m}}_{{\mathrm{N}}_2}}{{\mathrm{M}}_{{\mathrm{N}}_2}}$ = $\frac{{\mathrm{N}}_{{\mathrm{N}}_2}}{{\mathrm{N}}_{\mathrm{A}}}$

m_N2 = 2.0 g
M_N2 = 28.0 g.mol^-1
N_N2 = 2 x N_N
N_A = 6.022 x 10^-23 mol^-1

⇒ N_N2 = N_A x $\frac{{\mathrm{m}}_{\mathrm{N}2}}{{\mathrm{M}}_{\mathrm{N}2}}$ = 4.3 x 10²² atoms

N_N = 2 x N_N2 = 8.6 x 10²² atoms

The percent composition by mass is the percent of the total mass contributed by each element of a compound. It is calculated as follows:

Determine %Al in Al₂(SO₄)₃:

%Al = n x $\frac{{\mathrm{M}}_{\mathrm{Al}}}{{\mathrm{M}}_{{\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3}}$ x 100%

n = 2 [2 atoms of Al in one molecule of Al₂(SO₄)₃]
M_Al = 26.98 g.mol^-1
M_Al2(SO4)3 = 2 M_Al + 3 M_S + 12 M_O = 342.14 g.mol^-1

%Al = 2 x $\frac{26.98}{342.14}$ x 100 = 15.77 %

Stoichiometric coefficients:

The numeric values written to the left of each species in a chemical equation to balance the equation. The stoichiometric coefficients can be interpreted as the number of molecules or the number of moles of a substance produced or consumed during the reaction
 

3 H₂ + N₂ → 2 NH₃

Molecular interpretation: 3 molecules of H₂ react with 1 molecule of N₂ to form 2 molecules of NH₃
Molar interpretation: 3 moles of H₂ react with 1 mole of N₂ to form 2 moles of NH₃

Stoichiometry:

The calculations of the masses, moles, or volumes of reactants and products involved in a chemical reaction. Reactants are said to be in stoichiometric amounts when their are combined in the same relative amounts as those represented in the balanced chemical equation

Stoichiometry problems:

• How much of a product can be formed starting from a certain amount of reactants
• How much of one reactant is necessary to react with a given amount of another
• How much reactant is required to produce a desired amount of product

How to solve stoichiometry problems:

1.  Balancing the chemical equation
2. Calculate the molar masses of reactants and products of interest
3. Convert all given masses to moles
4. Use the balanced equation to determine the stoichiometric ratios
5. Calculate the number of moles of desired materials
6. Calculate the masses of desired materials

Limiting reactant:

The reactant that is consumed completely in a chemical reaction. The limiting reactant determines the maximum amount of product that can be formed during a reaction
 

How to determine the limiting reactant:

a A + b B → c C

1. Consider one of the starting reagents as the limiting reactant (for example A)
2. Calculate the mole number of the other reagent, B, required for a complete reaction of A
   Be sure to use the stoichiometric ratios of the balanced equation: $\frac{{\mathrm{n}}_{\mathrm{A}}}{\mathrm{a}}$ = $\frac{{\mathrm{n}}_{\mathrm{B}}}{\mathrm{b}}$
3. Compare the amount of B needed for a complete reaction with the actual amount of B:
   If amount of B needed > actual amount of B, B is the limiting reactant
   If actual amount of B > amount of B needed, A is the limiting reactant

Theoritical vs. actual yield:

Theoritical yield: the amount of product that will form if all the limiting reactant is consumed
Actual yield: the amount of product actually recovered​​​​​
 

Percent yield:

A measure of the efficiency of a chemical reaction. The percent yield is calculated as follows: