Chemical Calculations | General Chemistry 2

To facilitate chemical calculations, we use a unit called a mole
mole number n (in mol) is defined as the number of atoms in exactly 12.0 g of ¹²C
 

1 mole = 6.022 x 10²³ items

6.022 x 10²³ is called Avogadro’s number (N_A in items.mol^-1)

Isotopic mass (in amu): mass of an isotope of an element

Atomic mass (in amu): average of the masses of the natural isotopes of an element weighted according to their abundance

Atomic mass of carbon:

Carbon has 2 isotopes: ¹²C (12.000 amu) and ¹³C (13.003 amu)
The abundance of ¹²C is 98.89%
The atomic mass of carbon is: 12.000 x 0.9889 + 13.003 x 0.0111 = 12.011 amu

Molecular mass (in amu): mass of one chemical entity (atom, ion, molecule, formula unit)

Molar mass M (in g.mol^-1): mass of 1 mole of chemical entities (atoms, ions, molecules, formula units) 
M is equal to the sum of the atomic masses of the individual elements that make up each formula unit

Molar mass of CO₂:
M_CO2 = M_C + 2 x M_O = 12.0 + 2 x 16.0 = 44.0 g.mol^-1

 Mole number n (in mol): number of moles in a sample
 

Number of mole n in 2.0 g of N₂:

n = $\frac{{\mathrm{m}}_{\mathrm{N}2}}{{\mathrm{M}}_{\mathrm{N}2}}$ with M_N2 = 2 x M_N = 2 x 14.0 = 28.0

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

N_N = 2 x N_N2 = 8.6 x 10²² atoms

Mass percent %mass of an element X:

%mass of aluminum in Al₂(SO₄)₃:

2 atoms of Al in one molecule of Al₂(SO₄)₃   ⇒   moles of Al in formula = 2
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

%mass of aluminum = 2 x $\frac{26.98}{342.14}$ x 100 = 15.77 %

Combustion analysis: determination of the empirical formula of organic molecules (which contain C, H, and O primarily)

Principle: the sample is combusted in a stream of oxygen gas; all elements present are converted to CO₂ and H₂O. The masses of water and carbon dioxide formed are measured. If oxygen is present in the original sample, it must be determined by mass difference.

Coefficients in chemical equations 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: calculations of the masses, moles, or volumes of reactants and products involved in a chemical reaction

Stoichiometric problems:

- how much of a product can be formed starting from a certain amount of a reactant/reactants
- how much reactant you need to make a desired quantity of product

Stoichiometry problem solving:

1) Balance chemical equation
2) Calculate molar masses of reactants and products of interest
3) Convert any masses given to moles
4) Use balanced equation to determine stoichiometric ratios
5) Calculate moles of desired materials
6) Calculate grams of desired materials

Limiting reagent: reagent which determines the quantity of products formed ⇒ it limits the amount of product(s) formed based on the given molar amounts and the stoichiometric amounts required

Calculating limiting reagent:

a A + b B → c C

1) Use one of the starting materials (for example A)
2) Calculate the mole number of the other reagent (here B) needed for a complete reaction
Be sure to use stoichiometric ratios from the balanced equation: $\frac{{\mathrm{n}}_{\mathrm{A}}}{\mathrm{a}}$ = $\frac{{\mathrm{n}}_{\mathrm{B}}}{\mathrm{b}}$
3) Compare the amount needed for B with the actual amount of B:
- if amount needed > actual amount, B is the limiting reagent
- if actual amount > amount needed, A is the limiting reagent

Percentage yield is defined as: