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Lesson 3.6: Quantitative Information from Balanced Equations

← Lesson 3.5: Lesson 3.7: →

When reactants turn into products we often want to know how much product can be made from a given amount of reactant. In chemistry, problems of this sort are organized around the concept of a balanced chemical equating. We can make up a somewhat fanciful problem to illustrate the point.

Imagine that you wanted to start a fertilizer business because you have found an inexpensive way to manufacture ammonia from nitrogen and hydrogen. Every work station can process 1.200 kg of hydrogen per hour. The cost of the hydrogen is $5/kg, the cost of nitrogen $4/kg, and you can sell the ammonia for $9/kg. The processing costs are $10 for every 1.0 kg of hydrogen used. Is it profitable for you to manufacture ammonia under these conditions? If so, how much profit can you make per hour with a single work station?

How can we make sense of this kind of problem? At first sight it may appear that no profit can be made. If we imagine that the ingredients, hydrogen and nitrogen are combined to produce ammonia without any change in mass, we may think that 1 kg of hydrogen ($5) will combine with 1kg of nitrogen ($4) to produce 2kg of ammonia worth 2kg x $9/kg = $18. Since it costs $10 to process the 1 kg of hydrogen, the total production costs appear to be $19 for product that sells for $18. It does not appear profitable.

But is our reasoning sound? If not, where did we go wrong? We can start to answer this by writing a balanced chemical equation for the reaction between hydrogen and nitrogen.

N2 + 3H2 → 2NH3

Does the balanced equation show us directly how many grams of ammonia can be made from a given mass of hydrogen? Not directly. Balanced equations give number not mass ratios. In this case we can see that 2 molecules of ammonia can be made for every 3 molecules of hydrogen (assuming that we have enough nitrogen) but the mass ratio involved is less obvious. How can we work out the mass ratio of interest?

Start by organizing the information in a particular way. To avoid confusing mass and number ratios, let's agree to write the masses involved above the substance in the chemical equation and the numbers below. We can start with the mass to per hour for our (imaginary made up) process – 1.2 kg of hydrogen.

mass (g) 1200
N2 + 3H2 2NH3
number (mol)

To use the fact that 2 ammonia molecules can be made from every 3 hydrogen molecules, we first ask how many molecules are in 1200 g of hydrogen. It is convenient to find the number of molecules in the unit moles.

$$ 1200 \, g \times \frac{1 \, mol}{ 2.0 \, g} = 600 \, mol $$

We place this number of hydrogen molecules below the hydrogen symbol to remind us hat it is a number not a mass.

mass (g) 1200
N2 + 3H2 2NH3
number (mol) 600

We can now use the 3:2 ratio to find the number of ammonia molecules that can be produced from the 600 moles of hydrogen needed (per hour) and place that value in the chart.

$$ 600 \, mol \, H_2 \times \frac{2 \, mol \, NH_3}{ 3 \, mol \, H_2 } = 400 \, mol \, NH_3 $$

mass (g) 1200
N2 + 3H2 2NH3
number (mol) 600 400

What is the mass of this many ammonia molecules?

$$ 400 \, mol \times \frac{17 \, g}{ 1 \, mol} = 6800 \, g $$

Placing the mass in the table gives:

mass (g) 1200 6800
N2 + 3H2 2NH3
number (mol) 600 400

Our profit margin is looking more promising. Before considering the cost of the nitrogen we see that 6800 g of ammonia selling for 6.8 kg x $9/kg = $61.2 , requires only 1.2 kg x $5/kg = $6 for the hydrogen and 1.2 kg x $10/kg = $12 for processing. That means $18 of costs per hour for $61.2 of product.

To get a complete a complete picture we need to find the cost of nitrogen involved. From the balanced equation we see that 1 nitrogen molecule is needed for every 3 hydrogen molecules. To process 600 moles of hydrogen we need

$$ 600 \, mol \, H_2 \times \frac{1 \, mol \, N_2}{ 3 \, mol \, H_2 } = 200 \, mol \, N_2 $$

mass (g) 1200 6800
N2 + 3H2 2NH3
number (mol) 200 600 400

What is the mass of this many nitrogen molecules?

$$ 200 \, mol \times \frac{28 \, g}{ 1 \, mol} = 5600 \, g $$

Placing the mass in the table gives:

mass (g) 5600 1200 6800
N2 + 3H2 2NH3
number (mol) 200 600 400

What is the cost of this much nitrogen? 5.6 kg x $4/kg = $22.4

That means that our total cost to process 1200 of hydrogen is $18 + $22.4 = $40.4 to produce $61.2 of product a significant profit per hour! More than $20 per hour for a single work station!

So what was wrong with our original logic? We did not take into account the ratio of reactants in the reaction. This may not seem obvious. The problem is with the ratio of the reactants to each other, not the ratio of the reactants to products. In both cases we found the same mass of reactants and products. In the final analysis 6.8 kg of reactants turned into 6.8 kg of product, and in our original (but faulty for other reasons) logic 2kg of reactants turned into 2kg of product – the mass of the products is equal to the mass of the reactants. So far so good. The mistake comes in when we further add that 2 kg of reactants implies 1 kg of hydrogen and 1 kg of nitrogen. The mass ratio of reactants is not 1:1 but 14:3!

Those working in the discipline have found that the best way to keep track of these ratios (and thus avoid logical mistakes) is to base our reasoning on the balanced chemical equation involved in the reaction. Here is a procedural approach to this type of problem.

Example 1: How much ammonia can be produced from 112 g of nitrogen?

Step 1: Write the balanced chemical equation, indicating the mass of reactants and products above each formula, and the number of particles in a given mass below.

mass (g) 112 ?
N2 + 3H2 2NH3
number (mol)

Step 2: Divide the mass by the molar mass to find out how many particles are in the nitrogen. Since we are finding how many nitrogen molecules are present we use the molar mass of the molecule, 28.0 g, not the molar mass of the atom, 14.0 g.

$$ 112 \, g \times \frac{1 \, mol}{ 28.0 \, g} = 4.00 \, mol $$

mass (g) 112 ?
N2 + 3H2 2NH3
number (mol) 4.00

Step 3: Multiply the number of molecules of nitrogen by the ammonia : nitrogen ratio to find out how many ammonia molecules will form.

$$ 4.00 \, mol \, N_2 \times \frac{2 \, mol \, NH_3}{ 1 \, mol \, N_2 } = 8.00 \, mol \, N_2 $$

mass (g) 112 ?
N2 + 3H2 2NH3
number (mol) 4.0 8.0

Step 4: Multiply the number of molecules of ammonia by the molar mass of ammonia.

$$ 8.00 \, mol \times \frac{17.0 \, g}{ 1 \, mol} = 136 \, g $$

mass (g) 112 136
N2 + 3H2 2NH3
number (mol) 4.0 8.0

Summary: When we are given the mass of one substance involved in chemical reaction and want to know about the mass of another,

  1. Divide the mass by the molar mass.
  2. Multiply by the number ratio of the substances involved – the ratio of their coefficients in the balanced equation
  3. Multiply the number found in step 2 by the molar mass.

Example 2: Use the procedure to find the mass of hydrogen required in the above example.

Since mass is conserved in chemical reactions we can tell by inspection that 24.0 g of hydrogen are involved.

Step 2: We already know the number of nitrogen molecules involved, so we start with step 2. The hydrogen : nitrogen ratio is 3:1.

$$ 4.00 \, mol \, N_2 \times \frac{3 \, mol \, H_2}{ 1 \, mol \, N_2 } = 12.0 \, mol \, H_2 $$

mass (g) 112 136
N2 + 3H2 2NH3
number (mol) 4.0 12.0 8.0

Step 3: Now we multply by the molar mass of hydrogen.

$$ 12.0 \, mol \times \frac{2.00 \, g}{ 1 \, mol} = 24.0 \, g $$

mass (g) 112 24 136
N2 + 3H2 2NH3
number (mol) 4.0 12.0 8.0