# Chemical Calculations | General Chemistry 2

Typical chemical calculations are studied in this chapter: relationship between mass, mole and number of particles, percent composition by mass, stoichiometry, limiting reactant and yield calculation

## Mass-Mole-Number Relationship

Mole number n (in mol):

The number of moles in a sample

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

N = number of particles in the substance
NA = Avogadro’s number = 6.022 x 1023 mol-1

Number of mole n in 2.0 g of N2:

n = $\frac{{\mathrm{m}}_{{\mathrm{N}}_{2}}}{{\mathrm{M}}_{{\mathrm{N}}_{2}}}$

mN2 = 2.0 g
MN2 = 2 x MN = 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 NN in 2.0 g of N2:

n = $\frac{{\mathrm{m}}_{{\mathrm{N}}_{2}}}{{\mathrm{M}}_{{\mathrm{N}}_{2}}}$ = $\frac{{\mathrm{N}}_{{\mathrm{N}}_{2}}}{{\mathrm{N}}_{\mathrm{A}}}$

mN2 = 2.0 g
MN2 = 28.0 g.mol-1
NN2 = 2 x NN
NA = 6.022 x 10-23 mol-1

⇒ NN2 = NA x $\frac{{\mathrm{m}}_{\mathrm{N}2}}{{\mathrm{M}}_{\mathrm{N}2}}$ = 4.3 x 1022 atoms

NN = 2 x NN2 = 8.6 x 1022 atoms

## Percent Composition by Mass

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

%X = n x $\frac{{\mathrm{M}}_{\mathrm{X}}}{{\mathrm{M}}_{\mathrm{compound}}}$ x 100%

%X = percent composition of X
n = number of atoms X in a molecule of the compound
Mx = atomic mass of X (in amu or g.mol-1)
Mcompound = molecular mass of the compound (in amu or g.mol-1)

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]
MAl = 26.98 g.mol-1
MAl2(SO4)3 = 2 MAl + 3 MS + 12 MO = 342.14 g.mol-1

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

## Stoichiometry

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

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

## Percent Yield

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:

% yield =  x 100