CHEM- Empirical and Molecular Formulae Calculations
Chemistry, Empirical and Molecular Formula Calculations- IBDP | DSE | GCE | IAL | AP Chemistry
Jumping straight into the examples
Example 1 👇
In a compound formed by magnesium and nitrogen, it is found that 2.34 g of magnesium combine with 0.91 g of nitrogen. Find the empirical formula of the compound.
(Relative atomic masses: N= 14, Mg= 24)
There are just three steps to solve this question ❗️ ❕
Step 1- Find the number of moles of atoms of Magnesium and Nitrogen
Moles of Mg = 2.34 / 24 = 0.0975 mol
Moles of N = 0.91 / 14 = 0.065 mol
Moles = Mass ÷ Molar Mass
Step 2-Calculate the relative numbers of moles of atoms
Divide each mole value by the smallest number of moles calculated❗️
For the example, 0.065 is the smallest number of moles calculated
For Mg : 0.0975 / 0.065 = 1.5
For N : O.065 / 0.065 = 1
Step 3- Make the calculated numbers into whole numbers
📝Tips: When the numbers end with
For the above example, it's .5 so we need to multiply ALL the calculates values by 2 to get a whole number
For Mg : 1.5 × 2 = 3
For N : 1 × 2 = 2
Therefore, the empirical formula is Mg_{3}N_{2}
Example 2👇
The empirical formula and the molecular mass of a colorless liquid are CH_{2} O and 60. Find its molecular formula. (Relative atomic masses: H=1, C=12, O=16)
📝Hint: Let (CH_{2}O)_{n} be the molecular formula
So the relative molecular mass = n (12+2+16) = 30n
Answer: 30n = 60 *n =2
Thus, the molecular formula is (CH_{2}O)_{2} = C_{2}H_{4}O_{2}
Example 3👇
A compound X contains 85.7% carbon and 14.3% hydrogen. Its relative molecular mass is 56. Find its empirical and molecular formula.
(Relative atomic masses: H=1, C=12)
Empirical formula
Step 1- Find the number of moles of atoms of Carbon and Hydrogen
❗️ ❕For some questions, if percentages or fractions by mass of elements are provided, assume these values to mass in grams
Moles of C: 85.7 / 12 = 7.14 mol
Moles of H: 14.3 / 1 = 14.3 mol
Step 2-Calculate the relative numbers of moles of atoms
For C: 7.14 / 7.14 = 1
For H: 14.3 / 7.14 = 2.00
*The numbers are already given in whole numbers so STEP 3 is not required for this example
Hence, the empirical formula of the compound is CH_{2}
Molecular Formula
📝Let ( CH_{2})_{n} be the molecular formula
So the relative molecular mass = n (12+2) = 14n
Answer: 14n = 56 *n=4
Thus, the molecular formula is ( CH_{2})_{4} = C_{4}H_{8}
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