1 mole of magnesium chloride is decomposed to form 1 mole of magnesium and 1 mole of Chlorine.
Chemical reactions involve moles of substance. Mass (grams) are not used because the relationships are too complex. If the above reaction was discussed using grams, it would have to say:
95 grams of magnesium chloride are decomposed to form 24 g of Mg and 71 grams of chlorine.
As you can see the ratios would be 95:24:71 and these values bear no resemblance to the coefficients of the chemical equation. There is no way to simplify these numbers. By dealing with moles the ratios are 1:1:1 and the calculations are basic.
When the products of a reaction like this are measured, it is easiest to measure the weight (mass) of the products.
For example, a chemist decomposes some MgCl2. After the reaction is over, the chemist weighs the Mg and finds that 24 g were produced. The question is how much MgCl2 was needed to produce the 24 g of Mg.
Since chemical reactions are about moles, we first need to convert the 24 g of Mg into an equivalent number of moles of Mg. The following calculation does that:
1 mole Mg 24 g Mg x ----------- 24 g Mg
Looking at the reaction, it can be seen that for 1 mole of Mg to be produced, it takes 1 mole of MgCl2 to be decomposed. A close look at the reaction above shows the ratios of MgCl2:Mg:Cl2 to be 1:1:1. This means that 1 mole MgCl2 makes 1 mole Mg and 1 mole Cl2.
Once we know how many moles of MgCl2 were need we can then convert the number of moles into an equivalent mass through the atomic mass of MgCl2 using the following calculation:
First the atomic mass of MgCl2 is 24 g + 35.5 g + 35.5 g = 95 g.
95 g MgCl2 1 mole MgCl2 x -------------- 1 mole MgCl2
Thus it requires 95 g of MgCl2 to produce 24 g of Mg.
Some other reactions:
A chemist decomposes sucrose according to the following reaction:
Suppose that the chemist measured the mass of C produced and found the value to be 144 g. How much sucrose was needed to produce 144 g of C?
First, determine how many moles of C is equivalent to 144 g like so:
1 mole C 144 g C x ---------- 12 g C
The ratio of sucrose to C is 1:12. Since we have 12 moles of C produced, the reaction and the ratios show that we must have had 1 mole of sucrose since we need a ratio of 1:12.
Last we calculate the mass of 1 mole of sucrose.
342 g C12H22O11
1 mole C12H22O11 x ------------------
1 mole C12H22O11
Another example is the decompostion of water.
A chemist measures 64 g of O2 being produced. How much water was decomposed to produce the 64 g of O2?
First calculate how many moles of O2 are equivalent to 64 g of O2.
1 mole O2
64 g O2 x -----------
32 g O2
The ratio is 2:1 for H2O:O2. That is 2 moles of H2O are needed to produce 1 mole of O2. We calculated above that 2 moles of O2 were produced. So if 2 moles H2O produce 1 mole O2 then it must take 4 moles H2O to produce 2 moles O2.
Last convert the moles of H2O to grams of H2O using the atomic weight.
The atomic weight or atomic mass of H2O is 1g + 1g + 16g = 18 g.
18 g H2O
4 mole H2O x ------------
1 mole H2O