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

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

- First, find the molar ratio of atoms:

$$ Mg: \frac {0.6075}{24.3} = 0.025

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Br: \frac {3.995}{79.9} = 0.050 $$ - Divide by the smallest number:

$$ Mg: \frac {0.025}{0.025} = 1

\hspace7ex

Br: \frac {0.050}{0.025} = 2 $$ - Therefore, the empirical formula is:

$$ MgBr_{2} $$

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.

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:

$$ n = \frac {v}{24} $$

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

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

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}}

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\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 $$

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?

- 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 $$ - 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} $$ - Calculate the volume of $O_{2}$ that formed.

$$ \eqalign{ volume &= mol \times concentration \\

volume &= 0.0125 \times 24000 \\

&= 300\, cm^{3}} $$

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, 2021