Lesson 3: Conservation of Mass

In any chemical change, mass is conserved. This simply means that the total mass of the substances you begin with must equal the total mass of the substances you end up with.

In other words, in ordinary chemical reactions, mass is neither created nor destroyed. This is not true in nuclear reactions, where mass can be converted into energy and energy can be converted into mass. But in ordinary experience, mass is conserved. Early measurements confirmed that where mass appeared to have been lost, in fact the missing mass was in the form of a gas, and so escaped from the reaction vessel. When such reactions were carried out in closed vessels, no mass was lost. This came to be known as the law of conservation of mass. Dalton's atomic theory addressed this when he said that chemical reactions involve the rearrangement of atoms and atoms are neither created nor destroyed in chemical reactions.

In this section we'll practice solving some problems which apply the law of conservation of mass.

Example 1 | Example 2 | Example 3 | Example 4

Problem-solving with Conservation of Mass

Example 1

(Ex. 7 in your workbook)

When 10 grams of calcium are burned in oxygen, 14 grams of calcium oxide are formed. Conservation of mass says that the sum of the masses of the reactants (the substances we start with) must be the same as the sum of the masses of the products (the substances we produce):

10 g + x = 14 g

This means that 4 grams of oxygen must have combined with the calcium. A chemist might indicate this schematically as you see on the screen.

We also have to remember to put the units (g is for grams) by each number; this helps us keep track of what's going on and also expresses to the reader that we are dealing with mass and not some other property.

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Example 2

(Ex. 8 in your workbook)

That was an example of a simple stoichiometry problem. Stoichimetry problems ar those that involve weight relationships in chemical reactions. Here is a similar problem that’s a little more complicated.

When 33 g of wood was burned in air, it combined with 32 g of oxygen to form 3 g of ash and a gas mixture of carbon dioxide and water. When the gas mixture was cooled, 18 g of water condensed. How much carbon dioxide was formed?

As in the previous problem, it's helpful to set up an equation to focus on the information we know and how it relates to our problem.

wood	+	oxygen		ash	+	water	+	carbon dioxide
33 g	+	32 g		3 g	+	18 g	+	? g

 

If we add up the masses of the reactants (the wood and the oxygen), we get 65 g. That means that the total mass of the products (the ash, water, and carbon dioxide) must also be equal to 65 g.

We can set the problem up algebraically as:	3 g + 18 g + x = 65 g
	21 g + x = 65 g
And solve it by subtracting 21 g from both sides of the equation:	X = 65 g – 21 g
	X = 44 g

The mass of the carbon dioxide produced in this reaction was 44 g.

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Example 3

When 16.0 g of methane react with chlorine, 40.5 g of methyl chloride and 36.5 g of hydrogen chloride form. How much chlorine was used?

methane	+	chlorine		methyl chloride	+	hydrogen chloride
16.0g	+	? g	=	40.5 g	+	36.5 g

 

Remember that the law of conservation of mass says that the sum of the mass of the reactants (the methane and the chlorine in this problem) must be the same as the sum of the masses of the products (the methyl chloride and the hydrogen chloride). The sum of the masses of the products is 40. 5 g + 36.5 g = 77.0 g

Since the reactants must also weigh a total of 77.0 g, and since one of them, the methane, weights 16.0 g, what must the other, the chlorine, weigh?

We can set the problem up algebraically as:	16.0 g + x = 77.0 g
And solve it by subtracting 16.0 g from both sides of the equation:	X = 77.0 g – 16.0 g
	X = 61.0 g

Out of the 77.0 g of methane and chlorine, 16.0 g are the methane. The rest, 61.0 g, must therefore be the mass of the chlorine.

Notice that we are consistent in the precision with which we express the masses – using the rules for significant digits in calculations that you learned in lesson 1 – and that we always include the units with any number that we write down. It can be tedious, sometimes, to always write down the units, but as the problems get more complex, the habit will prevent you from making many mistakes. You must always have both the correct significant digits (the precision) and the units in your final answers for problems that you solve in chemistry.

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Example 4

Here is a similar problem to try on your own.

When 32.0 g of methane react with chlorine, 81.0 g of methyl chloride and 73.0 g of hydrogen chloride form. How much chlorine was used?

Answer: You should have found that 122.0 g of chlorine was used. If you had trouble coming up with that answer, review the previous examples or get some help from the instructor.

Try Ex. 9 now, before moving on; the answers are posted below.

Answers to Ex. 9:

a. 8 g oxygen
b. 2.75 g carbon dioxide
c. 1.64 g water

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