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?
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 $KCl$:
$$ mol = \frac {mass}{M_{r}}
\hspace3ex
\frac {3.925}{39.1+35.5} = 0.05\,mol $$
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 $Cl_{2}$.
$$ K: 0.05 \times 39.1 = 1.95g \\
Cl: 0.025 \times 71 = 1.775g $$
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?
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 $KOH_{aq}$.
$$ volume = 0.25dm^{3} \\
\eqalign{
concentration &= \frac {mol}{volume} \\
&= \frac {0.0058}{0.25} \\
&= 0.235 mol} $$