Moles in Equations

Empirical Formula

The empirical formula is the simplest whole number ratio of atoms of each element present in a compound.

Worked Example

A compound is found to contain $0.6075g$ of magnesium, $Mg$ and $3.995g$ of bromine, $Br$. What is the empirical formula?

  1. First, find the molar ratio of atoms:
    $$ Mg: \frac {0.6075}{24.3} = 0.025
    \hspace7ex
    Br: \frac {3.995}{79.9} = 0.050 $$
  2. Divide by the smallest number:
    $$ Mg: \frac {0.025}{0.025} = 1
    \hspace7ex
    Br: \frac {0.050}{0.025} = 2 $$
  3. Therefore, the empirical formula is:
    $$ MgBr_{2} $$

Molecular Formula

Molecular formulae are used for compounds that exist as simple molecules. A molecular formula tells you the number of each atom that make up a molecule.

Moles in Gases

Avogadro's Hypothesis states that equal volume of gases at the same temperature and pressure contain equal number of molecules. One mole of gas occupies $24.0\,dm^{3}$ or $(24,000\,cm^{2})$, therefore the volume per mole of a gas is $24\,dm^3mol^{-1}$. The number of moles of a gas can be calculated with:
$$ n = \frac {v}{24} $$

Moles in Solutions

In a solution, a solute is dissolved into a solvent. The concentration of a solution is the amount of solute dissolved per unit volume. Concentration is measured in $mol\,dm^{-3}$. For example, a solution with a concentration of $4\,mol\,dm^{-3}$ has 4 moles of solute in every $dm^{3}$ of solution.
$$ n = c \times v $$

Standard Solutions

Standard solutions have known concentrations. They are often used in titrations in order to determine unknown information about another substance.

Reacting Quantities and Stoichiometry

In chemistry, stoichiometry studies the amount of substances involved in a chemical reaction.

Reacting Masses

Potassium reacts with chlorine to form 3.925g of potassium chloride:
$$ 2K_{(s)} + Cl_{2(g)} \rightarrow 2KCl_{(s)} $$
1. First, calculate the amount in mol of $KCl$:
$$ mol = \frac {mass}{M_{r}}
\hspace3ex
\frac {3.925}{39.1+35.5} = 0.05\,mol $$
2. Calculate the moles of K and Cl that would form 3.925g of $KCl$.
$$ \eqalign{
2K_{(s)} &+\,Cl_{2(g)} \rightarrow 2KCl_{(s)} \\
2\,mol &+\,1\,mol \rightarrow 2\,mol \\
0.05 &+\,0.025 \rightarrow 0.05} $$
3. Finally calculate the masses of $K$ and $Cl_{2}$.
$$ K: 0.05 \times 39.1 = 1.95g \\
Cl: 0.025 \times 71 = 1.775g $$

Reacting Masses and Gas Volumes

A student heated $2.55g$ of potassium nitrate, $KNO_{3}$, which decomposes in the equation:
$$ 2KNO_{3(s)} \rightarrow 2KNO_{2(s)} + O_{2(g)} $$

What volume (in $cm^{3}$) of oxygen is formed?

  1. First calculate the amount, in mol of $KNO_{3}$ that decomposed.
    $$ mol = \frac {mass}{M_{r}}
    \hspace3ex
    \frac {2.55}{101.1} = 0.025\,mol $$
  2. Calculate the amount of $O_{2}$ that formed.
    $$ \eqalign{
    2KNO_{3(s)} &\rightarrow 2KNO_{2(s)} + O_{2(g)} \\ 0.025 &\rightarrow 0.025 + 0.0125} $$
  3. Calculate the volume of $O_{2}$ that formed.
    $$ \eqalign{ volume &= mol \times concentration \\
    volume &= 0.0125 \times 24000 \\
    &= 300\, cm^{3}} $$

Reacting Masses, Gas and Solutions

A chemist reacted $0.23g$ of potassium with water to form $250cm^{3}$ of aqueous potassium hydroxide. Hydrogen gas was also produced. The equation is shown below:
$$ 2K_{(s)} + 2H_{2}O_{(l)} \rightarrow 2KOH_{(aq)} + H_{2(g)}$$
1. Calculate the amount, in mol of $K$ that reacted.
$$ mol = \frac {mass}{M_{r}}
\hspace3ex
\frac {0.23}{39.1} = 0.0058\,mol $$
2. Calculate the moles of $H_{2}$ formed.
$$ \eqalign{ 2K_{(s)} + 2H_{2}O_{(l)} &\rightarrow 2KOH_{(aq)} + H_{2(g)} \\
0.0058 + 0.0058 &\rightarrow 0.0058 + 0.0029
}$$
3. Calculate the volume of $H_{2}$ formed at RTP.
$$ \eqalign{
volume &= mol \times 24 \\
volume &= 0.0029 \times 24 \\
&= 0.096\,dm^{3}} $$
4. Calculate the concentration, in $(moldm^{-3})$ of $KOH_{aq}$.
$$ volume = 0.25dm^{3} \\
\eqalign{
concentration &= \frac {mol}{volume} \\
&= \frac {0.0058}{0.25} \\
&= 0.235 mol} $$

© Andrew Deniszczyc, 2018