# 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}