3.3.1 Chemical Formulae
Chemical Formulae
- Each element is represented by its own chemical symbol.
- In an element, the chemical formula represents the atoms in the molecule.
- The chemical formula tells us:
- the types of atoms or ions in the compound,
- the number of atoms or ions in the compound,
- For example, the chemical formula for ethene is C2H4. This shows that ethene is the result of the combination of the elements carbon and hydrogen, and there are 2 carbon atoms and 4 hydrogen atoms in each molecule of ethene.
- Generally, chemical formula can be divided into
- Empirical formula
- Molecular formula
- The empirical formula of a substance is the chemical formula that gives the simplest whole-number ratio of atoms of each element in the substance.
- The molecular formula of a substance is the chemical formula that gives the actual number of atoms of each element in the substance.
Empirical Formula
- The empirical formula of a substance is the chemical formula that gives the simplest whole-number ratio of atoms of each element in the substance.
- Empirical = information gained by means of observation, experience, or experiment.
Example:
| Chemical Substances | Molecular Formula | Empirical Formula |
Glucose | C6H12O6 | CH2O |
Water | H2O | H2O |
Carbon Dioxide | CO2 | CO2 |
Benzene | C6H6 | CH |
Butane | C4H8 | CH2 |
Finding Empirical Formula
Steps to determine the empirical formula of a compound
STEP 1: Find the mass
STEP 2: Find the mole
STEP 3: Find the simplest ratio
Example:
In a chemical reaction, 4.23g of iron reacts completely with 1.80g of oxygen gas, producing iron oxide. Calculate the empirical formula of iron oxide. [Relative atomic mass: Iron = 56; Oxygen = 16]
Answer:
| Element | Fe | O |
| Mass | 4.23g | 1.80g |
| Number of mole | 4.23/56 =0.0755 | 1.80/16 =0.1125 |
| Simple ratio | 0.0755/0.0755 =1 | 0.1125/0.0755 =1.5 |
| Ratio in round number | 2 | 3 |
The empirical formula of iron oxide = Fe2O3
Example:
Determine the empirical formula of a compound which has a percentage of composition Mg: 20.2%, S: 26.6%, O: 53.2%. [Relative atomic mass: Mg = 24; S = 32; O = 16]
Answer
| Element | Mg | S | O |
| Percentage | 20.2% | 26.6% | 53.2% |
| Mass in 100g | 20.2g | 26.6g | 53.2g |
| Number of mole | 20.2/24 =0.8417mol | 26.6/32 =0.8313mol | 53.2/16 =3.325mol |
| Simple ratio | 0.8417/0.8313 =1 | 0.8313/0.8313 =1 | 3.325/0.8313 =4 |
The empirical formula of the compound is MgSO4
Example:
From an experiment, a scientist found that a hydrocarbon contains 85.7% of carbon according to its mass. Find the empirical formula of the hydrocarbon. [Relative atomic mass: Carbon = 12; Hydrogen = 1]
Answer:
| Element | C | H |
| Percentage | 85.7% | 14.3% |
| Mass in 100g | 85.7g | 14.3g |
| Number of mole | 85.7/12 =7.142mol | 14.3/1 =14.3mol |
| Simple ratio | 7.142/7.142 =1 | 14.3/7.142 =2 |
The empirical formula of the hydrocarbon = CH2
Molecular Formula
- The molecular formula of a substance is the chemical formula that gives the actual number of atoms of each element in the substance.
- A molecular formula is the same as or a multiple of the empirical formula.
- For example, the empirical of carbon dioxide is CO2 and the molecular formula is also CO2.
- Whereas, the empirical formula of ethane is CH3 while the molecular formula of ethane is C2H6.
Finding Molecular Formula
Given that the empirical formula of benzene is CH and its relative molecular mass is 78. Find the molecular formula of benzene. [Relative Atomic Mass: Carbon: 12; Hydrogen: 1]
Answer:
Let’s say the molecular formula of benzene is CnHn.
The relative molecular mass of CnHn
= n(12) + n(1)
= 13n
13n = 78
n = 78/13 = 6
Therefore, the molecular formula of benzene
= C6H6
Example:
What is the mass of metal X that can combine with 14.4g of oxygen to form X oxide with molecular formula X2O3. (RAM: O = 16; X = 56 )
Answer:
Number of mole of oxygen = 14.4/16 =0.9 mol
From the molecular formula, we learn that the ratio of element X to oxygen X:O = 2:3
Therefore, the number of mole of X = 0.9 × 2 3 =0.6 mol
Number of mole,
n = mass/Molar mass = 0.6
mass
= 0.6 x Molar mass*
= 0.6 x 56
= 33.6g
The mass of element X = 33.6g
*Molar mass of a substance = Relative atomic mass of the substance
Percentage of Composition of a Compound
- To find the percentage of composition of a substance means to find the percentage of mass of each element in the molecule of the substance to the mass of the molecule.
- The percentage of mass of an element can be determined by the following equation:
RAM = Relative atomic mass
RMM = Relative molecular mass
Example
Calculate the percentage of composition of DDT (C14H9Cl5). [Relative atomic mass: Carbon = 14; Hydrogen = 1; Chlorine = 35.5]
Answer:
Relative molecular mass of DDT =14(12)+9(1)+5(35.5) =354.5
