# Chemical Calculations | General Chemistry 2

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

## The Concept of a Mole

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 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  x

⇒ 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)

## 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 is equal to the sum of 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

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 (in mol-1)

Number of mole n in 2.0 g of N2:

n = $\frac{{\mathrm{m}}_{\mathrm{N}2}}{{\mathrm{M}}_{\mathrm{N}2}}$ with MN2 = 2 x MN = 2 x 14.0 = 28.0

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

Number of nitrogen atom NN in 2.0 g of N2:

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

NN = 2 x NN2 = 8.6 x 1022 atoms

## Mass Percent

Mass percent %mass of an element X:

%mass = moles of X in formula x $\frac{{\mathrm{M}}_{\mathrm{X}}}{{\mathrm{M}}_{\mathrm{compound}}}$ 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 $\frac{26.98}{342.14}$ x 100 = 15.77 %

## Combustion Analysis

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

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

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

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

Percentage yield is defined as:

% yield =  x 100

actual yield = amount of product that is actually formed​​​​​
​​​​theoretical yield = amount of product predicted from stoichiometry