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Chemistry calculations - EdexcelAn empirical formula experiment

An empirical formula of a substance is found using the masses and relative atomic masses of the elements it contains. The law of conservation of mass applies to closed and non-enclosed systems.

Part of Chemistry (Single Science)Key concepts in chemistry

An empirical formula experiment

The of a simple can be found using experiments. This page outlines one common experiment. Eye protection should be worn when carrying out experiments like this.

Aim

To determine the empirical formula of magnesium oxide.

Method

  1. Measure and record the of an empty with its lid.
  2. Put a length of magnesium ribbon into the crucible.
  3. Measure and record the total mass of the crucible, its lid and contents.
  4. Place the crucible on a tripod with a pipe clay triangle. Strongly heat the crucible for several minutes using a Bunsen burner.
  5. When the magnesium has stopped glowing, turn off the Bunsen burner and allow the crucible to cool down.
  6. Repeat step 3.

Results

The table shows some example results.

Mass (g)
Mass of crucible and lid (step 1)50.00
Mass of crucible, lid and magnesium (step 3)50.24
Mass of crucible, lid and magnesium oxide (step 7)50.40
Mass of crucible and lid (step 1)
Mass (g)50.00
Mass of crucible, lid and magnesium (step 3)
Mass (g)50.24
Mass of crucible, lid and magnesium oxide (step 7)
Mass (g)50.40

Analysis

  1. calculate the mass of magnesium used.
    • mass = 50.24 g - 50.00 = 0.24 g
  2. calculate the mass of oxygen gained during heating
    • mass = 50.40 g - 50.24 = 0.16 g
  3. calculate the empirical formula of magnesium oxide (Ar of Mg = 24 and Ar of O = 16)
StepActionResultResult
1Write the element symbolsMgO
2Write the masses0.24 g0.16 g
3Write the Ar values2416
4Divide masses by Ar0.24 梅 24 = 0.010.16 梅 16 = 0.01
5Divide by the smallest number0.01 梅 0.01 = 10.01 梅 0.01 = 1
6Write the formulaMgOMgO
Step1
ActionWrite the element symbols
ResultMg
ResultO
Step2
ActionWrite the masses
Result0.24 g
Result0.16 g
Step3
ActionWrite the Ar values
Result24
Result16
Step4
ActionDivide masses by Ar
Result0.24 梅 24 = 0.01
Result0.16 梅 16 = 0.01
Step5
ActionDivide by the smallest number
Result0.01 梅 0.01 = 1
Result0.01 梅 0.01 = 1
Step6
ActionWrite the formula
ResultMgO
ResultMgO

Empirical formula is MgO.

Question

In an experiment, 1.27 g of hot copper reacts with iodine vapour to form 3.81 g of copper iodide. Calculate the empirical formula of copper iodide. (Ar of Cu = 63.5, Ar of I = 127)