(C) - Copyright, 1996 F.W. Boyle, Jr.

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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 MgCl_{2}. After the reaction
is over, the chemist weighs the Mg and finds that 24 g were
produced. The question is how much MgCl_{2} 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:

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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 MgCl_{2} to be
decomposed. A close look at the reaction above shows the ratios
of MgCl_{2}:Mg:Cl_{2} to be 1:1:1. This means
that 1 mole MgCl_{2} makes 1 mole Mg and 1 mole
Cl_{2}.

Once we know how many moles of MgCl_{2} were need we can
then convert the number of moles into an equivalent mass through
the atomic mass of MgCl_{2} using the following
calculation:

First the atomic mass of MgCl_{2} is 24 g + 35.5 g + 35.5
g = 95 g.

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95 g MgCl_{2}1 mole MgCl_{2}x -------------- 1 mole MgCl_{2}

Thus it requires 95 g of MgCl_{2} to produce 24 g of Mg.

Some other reactions:

A chemist decomposes sucrose according to the following reaction:

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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:

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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.

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342 g C_{12}H_{22}O_{11}

1 mole C_{12}H_{22}O_{11}x ------------------

1 mole C_{12}H_{22}O_{11}

Another example is the decompostion of water.

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A chemist measures 64 g of O_{2} being produced. How much water was
decomposed to produce the 64 g of O_{2}?

First calculate how many moles of O_{2} are equivalent to 64 g of
O_{2}.

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1 mole O_{2}

64 g O_{2}x -----------

32 g O_{2}

The ratio is 2:1 for H_{2}O:O_{2}. That is 2
moles of H_{2}O are needed to produce 1 mole of
O_{2}. We calculated above that 2 moles of O_{2}
were produced. So if 2 moles H_{2}O produce 1 mole
O_{2} then it must take 4 moles H_{2}O to produce
2 moles O_{2}.

Last convert the moles of H_{2}O to grams of
H_{2}O using the atomic weight.

The atomic weight or atomic mass of H_{2}O is 1g + 1g +
16g = 18 g.

So:

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18 g H_{2}O

4 mole H_{2}O x ------------

1 mole H_{2}O