Ratios and Proportions

As chemists study matter, they are often working with ratios. The four statements here all express some sort of ratio.

Density is the ratio of mass to volume: D = m/V.

The units of molar mass are g/mol.

The mole ratio of H2 to O2 is 2:1.

The reaction gave a 78% yield.

Ratios can be written as fractions. For instance, 78% is image . The ratio 2:1 can be written image . Ratios can be reduced like fractions, or they can be added once they have a common denominator. Unit fractions, which equal 1 because the numerator is equivalent to the denominator, can be used to convert between units or to create equivalent fractions.

Whenever you are working with ratios, use the rules for working with fractions.

Example 1

Limiting Reactant

Imagine you start with 10.0 g of Mg(OH)2 and 100.0 mL of 4.0 M HCl. How much MgCl2 will be produced?

Solution

Write the balanced chemical equation.

Mg(OH)2(s) + 2HCl(aq) → MgCl2(aq) + 2H2O(l)

Determine the molar masses of each compound.

Reactants: Mg(OH)2 = 58.3 g/mol HCl = 36.5 g/mol
Products: MgCl2 = 9 5.2 g/mol H2O = 18.0 g/mol

Determine the number of moles of each reactant that you have.

image

Moles of HCl = (0.100 L)(4.0 mol/L) = 0.40 mol

Use the mole ratio to identify the limiting reactant.

A-13

The reactants combine in a 1:2 ratio. So you need 0.17 mol · 2 = 0.34 mol of hydrochloric acid, HCl, to react with the 0.17 mol of magnesium hydroxide, Mg(OH)2.

You have 0.40 mol of HCl, which is plenty, so the limiting reactant is Mg(OH)2. When the reaction is complete, there will be 0.06 mol of HCl left over.

Use the limiting reactant to determine the maximum amount of product.

For every 1 mol of Mg(OH)2, 1 mol of MgCl2 is produced.

So 0.17 mol of MgCl2 is produced. Use the molar mass of MgCl2 to determine the mass of MgCl2.

image

Example 2

Percent Yield

Suppose you ran the reaction from Example 1 again, this time starting with 3 5.0 g Mg(OH)2 and the same amount of HCl as before. In the laboratory, 1 5.4 g of MgCl2 are produced. What is the percent yield of your reaction?

Solution

First, determine the number of moles of each reactant you have. This will make it possible to identify the limiting reactant.

image

y = 0.40 mol HCl

The mole ratio of reactants is 1:2, so you need 1.2 mol of HCl to react with 0.60 mol of Mg(OH)2. You have only 0.40 mol of HCl available, so this time HCl is the limiting reagent.

It takes 2 mol of HCl to produce each mole of MgCl2, so you can expect to produce

0.20 mol of MgCl2: image .

Next, use the molar mass of MgCl2 to calculate the theoretical yield.

image

x = 19 g

Now you can calculate the percent yield for the reaction you ran. Percent yield is yield per 100, or yield/100.

image

Percent yield 581%

Your percent yield was significantly lower than 100%.

Practice Exercises

A-14

  1. What ratios and proportions appear in Examples 1 and 2?

  2. Write each expression as a fraction.

    1. 68%

    2. 3:4

    3. the ratio of volume to temperature

    4. grams per cubic centimetre

  3. Complete these calculations.

    1. image

    2. image

    3. image

    4. image

Show Answers

Answers

1. image

2a. image

2b. image

2c. volume/temperature, or image

2d. image

3a. image

3b. image

3c. image

3d. image