Chemical Calculations

In this article, we will be learning how to perform calculations from chemical reactions. These are the learning outcomes:

• Calculate masses of reactants and products by using mole ratio in a chemical equation.
• Calculate volumes of gaseous reactants and products using mole ratio in a chemical equation. 

This article is written keeping in mind the requirements of the students studying the Upper Secondary Chemistry syllabus in Singapore.

What is Stoichiometry?

It is the relationship between the number of moles of reactants and the number of moles of products in a chemical reaction.

Consider the following reaction:

\(\small{\begin{align*} \text{magnesium + oxygen}  &\quad\xrightarrow[\qquad\qquad]{}\quad   \text{magnesium oxide}\\[2ex] \mathrm{2Mg \,(s)+ O_2 \,(g)}   &\quad\xrightarrow[\qquad\qquad]{}\quad   \mathrm{2MgO\,(s)} \end{align*}}\)

In terms of moles, this equation tells us that two moles of magnesium react with one mole of oxygen gas to form two moles of magnesium oxide. 

The coefficients that you see in the balanced chemical equation tells us the ratio of the number of moles of reactants and products that are involved in this chemical reaction. 

Mole ratio of magnesium to oxygen to magnesium oxide is \(\small{2 : 1 : 2}\) 

However, this ratio cannot be applied to mass. 

It is incorrect to say that \(\small{\text{48 g}}\) of magnesium reacts with \(\small{\text{24 g}}\) of oxygen gas to form \(\small{\text{48 g}}\) of magnesium oxide. This does not obey the law of conservation of mass, where it states that matter can neither be created nor destroyed in a chemical reaction. The total mass of the reactants must be equal to the total mass of the products.

So let us see how we can determine the mass of the reactant consumed or the product formed in a reaction using stoichiometry.

Calculating Masses of Reactants and Products

Example 1:

When \(\small{\text{8.4 g}}\) of magnesium carbonate is heated, it decomposes into magnesium oxide and carbon dioxide. Calculate the mass of magnesium oxide formed.

Solution:

Whenever you come across a chemical calculation question, all you need to do is to follow four simple steps.  

• Step 1: Construct the balanced chemical equation (if not provided). 

\(\small{\mathrm{MgCO_3}   \quad\xrightarrow[\qquad\qquad]{}\quad   \mathrm{MgO+ CO_2}}\)
 

• Step 2: Convert the given quantity into moles. 

\(\small{\begin{align} \text{Number of moles} &= \frac {\text{Mass (in g)} }{\text{Molar Mass (g/mol)}}\\[2ex] \text{So, number of moles of }\mathrm{MgCO_3} &= \frac {8.4}{24 + 12 + 3 \times16 }\\[2ex] &= 0.1 \text{ mole} \end{align}}\)
 

• Step 3: Use the mole ratio from the equation to deduce the number of moles of the other substance. 

\(\small{\begin{align} \mathrm{MgCO_3}   \quad\xrightarrow[\qquad\qquad]{}\quad   \mathrm{MgO+ CO_2} \end{align}}\)
 

\(\small{\bbox[5px,border:2px solid #262262] {\begin{align} \mathrm{MgCO_3} &: \mathrm{MgO}\\[2ex] 1 &: 1\\[2ex] 0.1 &: 0.1 \end{align}}}\)
 

• Step 4: Convert mole to the required unit.

\(\small{\begin{align} \text{Mass} &= \text{Number of moles × Molar mass}\\[2ex] \text{Mass of MgO} &= \mathrm{0.1 \times (24 + 16)}\\[2ex] &= \text{4 g} \end{align}}\)
 

Test Your Skill

Question 1:

Magnesium reacts with oxygen as shown below.

\(\small{\mathrm{2Mg+ O_2}   \quad\xrightarrow[\qquad\qquad]{}\quad   \mathrm{2MgO}}\)

\(\small{(A_r \text{ of Mg = 24, O = 16 )}}\)

What is the mass of oxygen required to completely react with \(\small{\text{12 g}}\) of magnesium?

1. \(\small{\text{4 g}}\)
2. \(\small{\text{8 g}}\)
3. \(\small{\text{16 g}}\)
4. \(\small{\text{32 g}}\)

Solution: 

(B) \(\small{\text{8 g}}\)

Explanation:

\(\begin{align} \text{Number of moles of Mg} &= \frac {12}{24}\\[2ex] &= 0.5 \text{ mole} \end{align}\) 

\(\bbox[5px,border:2px solid #262262] {\begin{align} \mathrm{Mg} &:  \mathrm{O_2}\\[2ex] 2 &: 1\\[2ex] 0.5 &: 0.25  \end{align} }\)
 

\(\begin{align} \text{Mass of }\mathrm{O_2} &= 0.25 \times (16 \times 2)\\[2ex] &= \text{8 g} \end{align}\)

Question 2:

When potassium hydrogen carbonate \(\mathrm{(KHCO_3)}\) is strongly heated, the following reaction occurs.

\(\mathrm{2KHCO_3 \,(s)}   \quad\xrightarrow[\qquad\qquad]{}\quad   \mathrm{K_2CO_3 \,(s) + H_2O \,(g) +  CO_2 \,(g)}\)

\((A_r \text{ of K = 39, O = 16, C = 12, H = 1)}\)

What is the mass of potassium carbonate formed when \(\text{40 g}\) of potassium hydrogen carbonate is heated?

1. \(\text{7.2 g}\)
2. \(\text{14.4 g}\)
3. \(\text{27.6 g}\)
4. \(\text{55.2 g}\)

Solution: 

(C) \(\text{27.6 g}\)

Explanation:

Number of moles of KHCO3 = \(\frac {40}{39\; + \;1 + \;12 \; + \;3 \;\times\;16 }\)

                                              = 0.4 mole

KHCO₃ : K₂CO₃

          2 : 1

       0.4 : 0.2

Mass of K₂CO₃  = 0.2 × (2 × 39 + 12 + 3 × 16) 

                           = 27.6g

Calculating volumes of Reactants or Products

Example 2:

Iron(III) oxide reacts with carbon monoxide to give iron and carbon dioxide. Determine the volume of CO₂ released when 5.0 g of Fe₂O₃ reacts completely in the reaction. 

Solution:

• Step 1: Construct the balanced chemical equation.

Fe₂O₃ + 3CO   ———►   2Fe + 3CO₂

• Step 2: Convert the given quantity into moles. 

Number of moles of Fe₂O₃ = \(\frac {5.0}{2\; \times \;56\; + \;3 \;\times\;16 }\) 

                                          = 0.03125 mol

• Step 3: Use the mole ratio from the equation to deduce the number of moles of the other substance.

Fe₂O₃ + 3CO   ———►   2Fe + 3CO₂

    Fe₂O₃ : CO₂

            1 : 3

 0.03125 : 0.09375

• Step 4: Convert mole to the required unit.

Volume of CO₂ = 0.09375 × 24 

                       = 2.25 dm³

Question 3:

Magnesium reacts with hydrochloric acid as shown below. 

                                          Mg (s)+ 2HCl (aq)   ———►   MgCl₂ (aq) + H₂ (g)

(Ar of Mg = 24, Cl = 35.5, H = 1)

What volume of hydrogen gas at r.t.p is produced when an excess of the acid is added to 6 g of magnesium?

1. 1 dm³
2. 6 dm³
3. 12 dm³
4. 24 dm³

Solution:

(B) 6 dm³

Explanation:

Number of moles of Mg = \(\frac {6}{24}\)

                                      = 0.25 mole

  Mg : H₂

     1 : 1

0.25 : 0.25

Volume of H₂ = 0.25 × 24 dm³ 

                      = 6 dm³

Question 4:

The reaction between zinc sulfide and oxygen is as follows:

                                                  2ZnS + 3O₂   ———►   2ZnO + 2SO₂

(A_r of Zn = 65, S = 32, O =16)

What is the volume of sulfur dioxide at r.t.p that will be produced when 9.7 g of zinc sulfide is heated in air?

1. 3.6 dm³
2. 2.4 dm³
3. 1.2 dm³
4. 4.8 dm³

Solution:

(B) 2.4 dm³

Explanation:

Number of moles of ZnS = \(\frac {9.7}{65\;+\;32}\) 

                                        = 0.1 mole

ZnS : SO₂

     2 : 2

  0.1 : 0.1

Volume of SO₂ = 0.1 × 24 

                         = 2.4 dm³

Mole Ratio = Volume Ratio

Recall that one mole of any gas at room temperature and pressure occupies 24 dm³. 

This means that the volume of gas is proportional to the number of moles of gas.

For chemical calculations involving gaseous reactants and products, the mole ratio of the gasses in a balanced chemical equation is the same as the volume ratio of the gasses.

For Gaseous Substances Comparison only

Consider the following reaction involving gasses. 

hydrogen + chlorine   ———►   hydrogen chloride

H₂ (g) + Cl₂ (g)   ———►   2HCl (g)

In terms of mole, 1 mole of hydrogen gas reacts with 1 mole of chlorine gas to form 2 moles of hydrogen chloride gas. 

In terms of volume, 1 volume of hydrogen gas reacts with 1 volume of chlorine gas to form 2 volumes of hydrogen chloride gas. 

Example 3:

Consider the following reaction involving gasses. 

H₂ (g) + Cl₂ (g)   ———►   2HCl (g)

Calculate the volume of hydrogen chloride produced when 50 cm³ hydrogen gas is reacted completely at r.t.p.

Solution:

Find the volume ratio of HCl to H₂ from the equation.

\(\frac {\;\; Volume \; of \;HCl \;\;} {\;\; Volume \;of \;H_2 \;\;}\) = \(\frac { 2 } { 1 }\)

Volume of HCl = 2 × Volume of H₂ 

                        = 2 × 50 cm³ 

                        = 100 cm³

Question 5:

Hydrogen sulfide reacts with oxygen as shown in the following equation.

2H₂S (g) + 3O₂ (g)   ———►   2H₂O (g) + 2SO₂ (g)

(A_r of S = 32, O = 16, H = 1)

If 50 cm³ of hydrogen sulfide is allowed to react completely in an excess of oxygen, what volume of steam would be produced?

1. 20 cm³
2. 30 cm³
3. 40 cm³
4. 50 cm³

Solution: 

(D) 50 cm³

Explanation:

 \(\frac {\;\; Volume \; of \;H_2S \;\;} {\;\; Volume \;of \;H_2O \;\;}\) = \(\frac { 2 } { 2 }\)

    = \(\frac { 1 } { 1 }\)

Volume of H₂O = Volume of H₂S  

                         = 50 cm³

Question 6:

Methane reacts with steam to form hydrogen and carbon monoxide as shown below.

CH₄ (g) + H₂O (g)   ———►   CO (g) + 3H₂ (g)

(A_r of C = 12, O = 16, H=1)

What volume of hydrogen can be obtained from 30 cm³ of methane at r.t.p?

1. 30 cm³
2. 60 cm³
3. 90 cm³
4. 120 cm³

Solution:

(C) 90 cm³

Explanation:

\(\frac {\;\; Volume \; of \;H_2 \;\;} {\;\; Volume \;of \;CH_4 \;\;}\) = \(\frac { 3 } { 1 }\)

   = 3

Volume of H2 = 3 × Volume of CH₄ 

                       = 3 × 30 cm³

                       = 90 cm³

Conclusion

In this article, we have learnt about stoichiometry and how we can use the stoichiometry in a balanced chemical equation to calculate the masses of reactants and products as well as the volume of gaseous reactants and products. 

All these topics have been discussed keeping in mind the Upper Secondary Chemistry syllabus in Singapore.

Keep Learning! Keep Improving!