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CHEM- Empirical and Molecular Formulae Calculations

Chemistry, Empirical and Molecular Formula Calculations- IBDP | DSE | GCE | IAL | AP Chemistry

· chemistry,question analysis,Empirical,Molecular Formula,Calculations

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 ❗️ ❕

#Surprised

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

  •  .25 × 4
  • .5 × 2
  • .33 × 3

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 Mg3N2

Example 2👇​​

The empirical formula and the molecular mass of a colorless liquid are CH2 O and 60. Find its molecular formula. (Relative atomic masses: H=1, C=12, O=16)

📝​​Hint: Let (CH2O)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 (CH2O)2 = C2H4O2

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 CH2

Molecular Formula

📝​Let ( CH2)n be the molecular formula

So the relative molecular mass = n (12+2) = 14n

Answer: 14n = 56 *n=4

Thus, the molecular formula is ( CH2)4 = C4H8

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