# Chemical Calculations | General Chemistry 2

## 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 ^{12}C

1 mole of ^{12}C = 12.0 g

1 atom of ^{12}C = 12.0 amu

1 amu = 1.6605 x 10^{-24} g

⇒ 1 mole of ^{12}C = 12.0 g x $\frac{1\mathrm{amu}}{1.6605\times {10}^{-24}\mathrm{g}}$ x $\frac{1\mathrm{atom}\mathrm{of}{}^{12}\mathrm{C}}{12\mathrm{amu}}$

⇒ 1 mole of ^{12}C = 6.022 x 10^{23} atoms of ^{12}C

**1 mole = 6.022 x 10 ^{23} items**

6.022 x 10^{23} is called Avogadro’s number (N_{A} 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 CO

_{2}:

M_{CO2}= M_{C}+ 2 x M_{O}= 12.0 + 2 x 16.0 = 44.0 g.mol^{-1}

## Mass-Mole-Number Relationship

** **(in mol): 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 (in mol^{-1})

Number of mole n in 2.0 g of N

_{2}:n = $\frac{{\mathrm{m}}_{\mathrm{N}2}}{{\mathrm{M}}_{\mathrm{N}2}}$ 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 N_{2}: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 10^{22}atomsN

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

_{2}(SO_{4})_{3}:2 atoms of Al in one molecule of Al

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

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 CO_{2} and H_{2}O. 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 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

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 = $\frac{\mathrm{actual}\mathrm{yield}}{\mathrm{theoretical}\mathrm{yield}}$ x 100

actual yield = amount of product that is actually formed

theoretical yield = amount of product predicted from stoichiometry