# Chemical Calculations | General Chemistry 2

## Mass-Mole-Number Relationship

**Mole number n (in mol):**

The number of moles in a sample

$\mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}}$

m = mass of the substance (in g)

M = molar mass of the substance (in g.mol^{-1})

$\mathrm{n}=\frac{\mathrm{N}}{{\mathrm{N}}_{\mathrm{A}}}$

N = number of particles in the substance

N_{A} = Avogadro’s number = 6.022 x 10^{23} mol^{-1}

Number of mole n in 2.0 g of N_{2}: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 N_{2}: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 10^{22}atomsN

_{N}= 2 x N_{N2}= 8.6 x 10^{22}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

M_{x} = atomic mass of X (in amu or g.mol^{-1})

M_{compound} = molecular mass of the compound (in amu or g.mol^{-1})

Determine %Al in Al_{2}(SO_{4})_{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 Al

_{2}(SO_{4})_{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 %

## 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 H

_{2}+ N_{2}→ 2 NH_{3}Molecular interpretation: 3 molecules of H

_{2}react with 1 molecule of N_{2}to form 2 molecules of NH_{3}

Molar interpretation: 3 moles of H_{2}react with 1 mole of N_{2}to form 2 moles of NH_{3}

**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:**

- Balancing the chemical equation
- Calculate the molar masses of reactants and products of interest
- Convert all given masses to moles
- Use the balanced equation to determine the stoichiometric ratios
- Calculate the number of moles of desired materials
- 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

- Consider one of the starting reagents as the limiting reactant (for example A)
- 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: $\frac{{\mathrm{n}}_{\mathrm{A}}}{\mathrm{a}}$ = $\frac{{\mathrm{n}}_{\mathrm{B}}}{\mathrm{b}}$ - 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 = $\frac{\mathrm{actual}\mathrm{yield}}{\mathrm{theoretical}\mathrm{yield}}$ x 100