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Mole unit - Example calculation

Mole unit - Example calculation: Encyclopedia II - Mole unit - Example calculation

In this example, moles are used to calculate the mass of CO2 given off when 1 g of ethane is burnt. The equation for this chemical reaction is: 7 O2 + 2 C2H6 → 4 CO2 + 6 H2O that is, 7 molecules of oxygen react with 2 molecules of ethane to give 4 molecules of carbon dioxide and 6 molecules of water. The first thing is to figure out how many molecules of ethane were burnt. We know that it was just enough to make 1 g, ...

Mole unit: Encyclopedia II - Mole unit - Example calculation

Mole unit - Example calculation

In this example, moles are used to calculate the mass of CO₂ given off when 1 g of ethane is burnt. The equation for this chemical reaction is:

7 O₂ + 2 C₂H₆ → 4 CO₂ + 6 H₂O

that is,

7 molecules of oxygen react with 2 molecules of ethane to give 4 molecules of carbon dioxide and 6 molecules of water.

The first thing is to figure out how many molecules of ethane were burnt. We know that it was just enough to make 1 g, so we now need the molecular mass of ethane. This can be calculated : the mass in grams of one mole of a substance is by definition its atomic or molecular mass; The atomic mass of hydrogen is 1, and the atomic mass of carbon is 12, so the molecular mass of C₂H₆ is (2 × 12) + (6 × 1) = 30. One mole of ethane is 30 g. So 1 g of ethane is 1/30th of a mole : the amount burnt was 1/30th of a mole (remember that it is a number, quite like "half a dozen").

Now we can calculate the number of molecules of CO₂ given off. Since for 2 molecules of ethane we obtain 4 molecules of CO₂, we have 2 molecules of CO₂ for each molecule of ethane. So, for 1/30th of a mole of ethane, 2 × 1/30th = 1/15th of a mole of CO₂ were produced.

Next, we need the molecular mass of CO₂. The atomic mass of carbon is 12 and that of oxygen is 16, so one mole of carbon dioxide is ) is 12 + 2 × 16 g = 44 g.

Finaly, the mass of CO₂ is 1/15 mole × 44 g per mole = 2.93 g of carbon dioxide.

Notice that the number of moles does not need to balance on either side of the equation (remember also the simple equation for water: a mole of oxygen and two moles of hydrogen give a two moles of water). This is because a mole does not count mass or the number of atoms involved, but the number of particules involved (each of them composed of a variable number of atoms). However, we could likewise calculate the mass of oxygen consumed, and the mass of water produced, and observe that the mass of products (carbon dioxide and water) is equal to the mass of dioxygen plus ethane :

7/2 x 1/30th mole of dioxygen x (2 x 16) g/mole = 7 * 16 / 30 g = 3,73 g
6/2 x 1/30th mole of water x ((2 x 1) + 16) g/mole = 1,8 g
3,73 g + 1 g = 2,93 + 1,8 g

(nota : actually, according to the mass-energy relationship, there is a very slim difference between the mass of carbon, hydrogen and oxygen separated, on one side, and on the other side the mass of the molecules made of them ; this has not to be cared of here)