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CALCULATION of calculating empirical formula from reacting masses FORMULA from reacting masses (NOT using moles)

Advanced UK A/AS level, IB and US grade 11-12 pre-university chemistry

5. How to calculate simple calculating empirical formula from reacting masses formula and formula mass from reacting masses or % composition (easy start, no moles involved !)

Quiz 5 on the basics of calculating empirical formula from reacting masses formula

(easy start, NOT using moles)

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Quiz 5 on the basics of calculating empirical formula from reacting masses formula

(easy start, NOT using moles)

The calculating empirical formula from reacting masses FORMULA of a compound can be worked out by knowing the exact masses of the elements that combine to form a given mass of a compound.

The calculating empirical formula from reacting masses formula of a compound is the simplest whole number ratio of atoms present in a compound. (see section 3. for some simpler examples). Here the word 'calculating empirical formula from reacting masses' means from experimental data.

Do not confuse with molecular formula which depicts the actual total numbers of each atom in a molecule i.e. a summary of the molecular composition of a molecule, element or compound.

The molecular formula and calculating empirical formula from reacting masses formula can be different or the same.

They are the same if the molecular formula cannot be simplified on a whole number basis.

Examples where molecular formula = calculating empirical formula from reacting masses formula

e.g. for sodium sulfate Na₂SO₄  and  propane C₃H₈

You cannot simplify the atomic ratios 2 : 1 : 4 for the salt sodium sulfate

or 3 : 8 for the hydrocarbon propane to smaller whole number (integer) ratios

Similarly for the hydrocarbon pentane , the diagram shows it's molecular formula is C₅H₁₂, which cannot be simplified.

So in these three cases, the molecular formula = calculating empirical formula from reacting masses formula.

Examples of where molecular formula and calculating empirical formula from reacting masses formula are different e.g.

butane molecular formula C₄H₁₀, calculating empirical formula from reacting masses formula C₂H₅

numerically, the calculating empirical formula from reacting masses formula of butane is 'half' of its molecular formula

4 : 10 ==> 2 : 5

glucose molecular formula C₆H₁₂O₆, calculating empirical formula from reacting masses formula CH₂O

numerically, the calculating empirical formula from reacting masses formula of glucose is '¹/₆th' of the full molecular formula

6 : 12 : 6 ==> 1 : 2 : 1

Suppose you start with a molecular structure like octane (left diagram)

If you count the atoms you find the molecular formula is C₈H₁₈.

BUT the simplest ratio formula, that is the calculating empirical formula from reacting masses formula is 'half' of the molecular formula i.e. C₄H₉, but remember, if you have to suggest a structure for a molecule you must know at least its molecular formula.

AND don't forget to be able to think, calculate and deduce the other way round e.g.

(i) Suppose a hydrocarbon molecule has an calculating empirical formula from reacting masses formula of C₂H₅ and a molecular mass of 58 (C = 12, H = 1).

Deduce its molecular formula. The calculating empirical formula from reacting masses formula mass = (2 x 12) + 5 = 29.

Dividing 58 by 29 gives 2. So the molecular formula is 2 x the calculating empirical formula from reacting masses formula = C₄H₁₀

(ii) Suppose a molecule has an calculating empirical formula from reacting masses formula of simply CH, but a molecular mass of 78 (C = 12, H = 1).

The calculating empirical formula from reacting masses formula mass is 12 + 1 = 13. Therefore 78/13 = 6, so the molecular formula is 6 x CH = C₆H₆

Where the calculating empirical formula from reacting masses formula and molecular formula are different, you need extra information to deduce the molecular formula from the calculating empirical formula from reacting masses formula (see link below).

This page is only concerned with calculating empirical formula from reacting masses formula.

For more advanced students see Using moles to calculate calculating empirical formula from reacting masses formula and deduce molecular formula of a compound/molecule (starting with reacting masses or % composition)

The following examples illustrate the ideas using numbers more easily appreciated than in real experiments.

In real laboratory experiments only a fraction of a gram or a few grams of elements would be used, and a more 'tricky' mole calculation method is required than shown here (dealt with later for higher students in section 8).

However the examples below show in principal how formulae are worked out from experiments.

Any calculation method must take into account the different relative atomic masses of the elements in order to get to the actual ratio of the atoms in the formula.

For example, just because 10g of X combines with 20g of Y, it does not mean that the formula of the compound is XY₂ !

If you divide the mass of each element by its atomic mass, you actually get the atomic ratio.

• calculating empirical formula from reacting masses formula calculation Example 5.1 The compound formed between lead and sulfur
  • It is found that 207g of lead combined with 32g of sulphur to form 239g of lead sulphide.
  • From the data work out the formula of lead sulphide. (Relative atomic masses: Pb = 207 and S = 32)
  • In this case it easy to see that by the atomic mass ratio, 239 splits on a 1 to 1 basis of 1 atom of lead to 1 atom of sulphur (1 x 207 to 1 x 32 by mass)
  • so the formula is simply PbS
  • You can set out the calculation in a simple table format, in this case the numbers are very easy to deal with!

• calculating empirical formula from reacting masses formula calculation Example 5.2 The calculating empirical formula from reacting masses formula of a lead oxide
  • It is found that 207g of lead combined with oxygen to form 239g of a lead oxide.
  • From the data work out the formula of the lead oxide. (Relative atomic masses: Pb = 207 and O = 16)
  • In this case, you first have to work out the amount of oxygen combined with the lead.
  • By simple logic from the law of conservation of mass, this is 239 - 207 = 32g
  • In atomic ratio terms, the 207 is equivalent to 1 atom of lead and the 32 is equivalent to 2 atoms of oxygen (1 x 207 to 2 x 16),
  • so the formula is simply PbO₂

  • Note: The mass of oxygen combined with the lead is deduced by subtracting the original mass of lead from final total mass of lead oxide.

• calculating empirical formula from reacting masses formula calculation Example 5.3 The calculating empirical formula from reacting masses formula of aluminium sulfide
  • It is found that 54g of aluminium forms 150g of aluminium sulphide.
  • Work out the formula of aluminium sulphide. (Relative atomic masses: Al = 27 and S = 32).
  • Amount of sulphur combined with the aluminium = 150 - 54 = 96g
  • By atomic ratio, the 54 of aluminium is equivalent to 2 atoms of aluminium and the 96 of sulphur is equivalent to 3 atoms of sulphur.
  • Therefore the atomic ratio is 2 to 3,
  • so the formula of aluminium sulphide is Al₂S₃

• calculating empirical formula from reacting masses formula calculation Example 5.4 From now on, questions just using the table method to work out calculating empirical formula from reacting masses formula from more awkward numbers! In this case a compound formed between copper and chlorine.

  • A compound of copper contained 47.4% copper and 52.6% chlorine.

  • The atomic masses are: Cu = 64 and Cl = 35.5

  • Think of the percentages as masses in grams to solve the calculating empirical formula from reacting masses formula problem.

• calculating empirical formula from reacting masses formula calculation Example 5.5 The calculating empirical formula from reacting masses formula of a compound of carbon and chlorine

  • It was found that 0.39 g of carbon was combined with 4.61g of chlorine.

  • Atomic masses: C = 12 and Cl = 35.5

• calculating empirical formula from reacting masses formula calculation Example 5.6 The formula of a hydrocarbon.

  • It was that 0.75g of carbon was combined with 0.25g of hydrogen.

  • Atomic masses: C = 12 and H = 1

  • Calculate the calculating empirical formula from reacting masses formula of the hydrocarbon

• calculating empirical formula from reacting masses formula calculation Example 5.7 The analysis of sodium sulfate, calculating its calculating empirical formula from reacting masses formula from the % composition by mass.

  • On analysis, the salt sodium sulfate was found to contain 32.4% sodium, 22.5% sulfur and 45.1% oxygen.

  • Atomic masses: Na = 23, S = 32 and O = 16

  • Calculate the calculating empirical formula from reacting masses formula of sodium sulfate

• calculating empirical formula from reacting masses formula calculation Example 5.8 The formula of a hydrocarbon.

  • Analysis of hydrocarbon showed it consisted of 83.3% carbon and 16.7% hydrogen.

  • Atomic masses: C = 12 and H = 1

  • Calculate the calculating empirical formula from reacting masses formula of the hydrocarbon (just think of it as 83.3g C combined with 16.7g H)

See section 8. for more calculating empirical formula from reacting masses/molecular formula calculations involving moles.

Quiz 5 on the basics of calculating empirical formula from reacting masses formula

(easy start, NOT using moles)