Chemical Calculations | General Chemistry 2

The classical chemical calculations are studied in this chapter: the concept of a mole, the atomic, molecular and molar masses, the relation between mass, mole and number of particles, the mass percent of an element, the combustion analysis, the chemical equations (coefficients and stoichiometry), the limiting reagent and the yield calculation.

The Concept of a Mole

To aid in 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 of 12C = 12.0 g
1 atom of 12C = 12.0 amu
1 amu = 1.6605 x 10-24 g

⇒ 1 mole of 12C = 12.0 g x 1 amu1.6605 × 10-24 g x 1 atom of C1212 amu

⇒ 1 mole of 12C = 6.022 x 1023 atoms of 12C


1 mole = 6.022 x 1023 items

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


Isotopic and Atomic Mass

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

Atomic mass (in amu): Average of the masses of the naturally occurring 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 and Molar Masses

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 = sum up the atomic masses of the individual elements that make up each formula unit


Molar mass of CO2:
MCO2 = MC + 2 x MO = 12.0 + 2 x 16.0 = 44.0 g.mol-1

Mass-Mole-Number Relationship

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

n = mM

m = mass of the substance (in g)
M = molar mass of the substance (in g.mol-1)


n = NNA

N = number of particles in the substance
NA = Avogadro’s number (in mol-1)



Number of mole n in 2.0 g of N2:

n = mN2MN2 with MN2 = 2 x MN = 2 x 14.0 = 28.0

⇒ n = 2.028.0 = 7.1 x 10-2 mol

Number of nitrogen atom NN in 2.0 g of N2

n = mN2MN2 = NN2NA ⇒ NN2 = NA x mN2MN2 ⇒ NN2 = 4.3 x 1022 atoms

NN = 2 x NN2 = 8.6 x 1022 atoms


Mass Percent

Mass percent of an element X:

%mass = moles of X in formula x MXMcompound x 100



%mass of aluminum in Al2(SO4)3:

2 atoms of Al in one molecule of Al2(SO4)3 ⇒ moles of Al in formula = 2
MAl = 26.98 g.mol-1
MAl2(SO4)3 = 2 MAl + 3 MS + 12 MO = 342.14 g.mol-1

%mass of aluminum = 2 x 26.98342.14 x 100 = 15.77 %

Combustion Analysis

Combustion analysis: determination of the empirical formulas 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.

Coefficient in Chemical Equations

Coefficients in chemical equations can be interpreted as numbers of molecules or numbers of moles of the entities.


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 about 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

Limiting reagent: reactant that determines how much of the products are made
⇒ limits the amount of formed product(s) based on both molar amounts given and stoichiometric amounts needed

Calculating limiting reagent:

a A + b B → c C

1) Use any one of the starting materials (for example A)
2) Calculate how much of the other starting material (here B) you will need to completely react with it.
Be sure to use stoichiometric ratios from the balanced equation: nAa = nBb
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

Theoretical yield: amount of product predicted from stoichiometry considering limiting reagents
Actual yield: amount of product that is actually formed

% yield = actual yieldtheoretical yield x 100