# CSEC Chemistry: Calculations Involving Equations and Moles

Updated: Apr 25, 2020

In a balanced and complete equation, you are given a lot of information. Apart from what the products are and what the reactants are, you get the** quantities** of each involved. This gives you the relationship between all of the substances in the reaction. With the quantities, you can convert moles, volumes, masses or number of molecules of one reactant into those of the products.

Take, for example, the following equation:

**N2(g) + 3H2(g) --> 2NH3(g)**

This gives us the following information about the **quantitative relationships **in the equation:

1 mol N2(g) + 3 mol H2(g) --> 2 mol NH3(g)

6.0 x 10^23 molecules N2(g) + 3 x 6.0 x 10^23 molecules H2(g) --> 2 x 6.0 x 10^23 molecules NH3(g)

28 g N2(g) + 6 g H2(g) --> 34 g NH3(g)

22.4 dm^3 N2(g) + 67.2 dm^3 H2(g) --> 44.8dm^3 NH3(g) (at s.t.p)

Therefore, if you are given a chemical equation and the **number of moles**, **volumes**, **masses** or **number of molecules** of one reactant or product, you can find the corresponding information for the other substances in the reaction.

__Example:__ Calcium carbonate decomposes on heating according to the equation: **CaCO3(s) ---> CaO(s) + CO2(g)**

a) What is the mass of Calcium Oxide formed when 0.025 mol of Calcium Carbonate is heated this way?

The equation shows the reactant and products in a 1:1:1 ratio. So, 0.025 mol of CaCO3 gives 0.025 mol of CaO and 0.025 mol of CO2.

Therefore, we must find the mass of 0.025 mol of CaO:

**mass of CaO formed = number of mol x molar mass**

** = 0.025 mol x (16 g/mol + 40 g/mol)**

** = 0.025 mol x 56 g/mol**

** = 1.4 g**

b) What is the mass and volume (at s.t.p.) of the CO2 formed?

To find the mass, we take the same procedure:

**mass of CO2 formed = number of mol x molar mass**

** = 0.025 mol x (12 g/mol + [2 x 16 g/mol])**

** = 0.025 mol x 44 g/mol**

** = 1.1 g**

Since we already know that 1 mole of any gas at s.t.p. is 22.4 dm^3, we simply multiply the number of moles by this known volume.

**volume of CO2 (at s.t.p) formed = number of moles x 22.4 dm^3/mol**

** = 0.025 mol x 22.4 dm^3/mol**

** = 0.56 dm^3**

__Example:__ On heating, KClO3(s) decomposes as follows:

**2KClO3(s) --> 2KCl(s) +3O2(g)**

If 9.8 g of KClO(s) is heated:

a) What is the mass of KCl(s) formed?

The equation gives the reactant and products in the ratio of 2:2:3. This equates to 2 mol KClO3(s) gives 2 mol KCl(s) and 3 mol O2(g).

So, we convert the mass given to moles:

**number of moles of KClO3(s) = mass/molar mass**

** = 9.8 g/(39+35.5+48)**

** = 9.8 g/122.5 gmol^-1**

** = 0.08 mol**

The equation rewritten for this, then, is:

**0.08 mol KClO3(s) --> 0.08 mol KCl(s) + 0.12 mol O2(g)**

Now, we find the mass of 0.08 mol of KCl(s):

**mass = number of moles x molar mass**

** = 0.08 mol x (39 + 35.5)g/mol**

** = 0.08 mol x 74.5 g/mol**

** = 5.96 g**

b) What is the mass and volume of O2(g) formed at s.t.p.?

**mass of O2 = number of moles x molar mass**

** = 0.12 mol x 32 g/mol**

** = 3.84 g**

**volume of O2 = number of moles x 22.4 dm^3**

** = 0.12 mol x 22.4 dm^3**

** = 2.688 dm^3**

In some cases, you will need to calculate the** volumes **of gases consumed or produced in reactions. If the **mass** or **volume at s.t.p.** is given, first **convert** these quantities to **moles**, then **use the coefficients** in the balanced equation to find the number of moles of the gas consumed or produced, and hence the **volume at s.t.p.**

__Example:__ **2NO(g) +O2(g) --> 2NO2(g)**

If 1.5 g of NO(g) is used:

a) find the volume of oxygen required at s.t.p.

**number of moles of NO(g) = mass/molar mas**

** = 1.5 g/ (14+16) gmol^-1**

** = 1.5 g/30 gmol^-1**

** = 0.05 mol**

From the equation, we know that 2 mol NO(g) + 1 mol O2(g) gives 2 mol NO2(g) (2:1:2 ratio).

Applying our known number of moles of NO(g) to this equation, we get:

**0.05 mol NO(g) + 0.025 mol O2(g) --> 0.05 mol NO2(g)**

So, the **volume of oxygen required (at s.t.p)= 0.025 mol x 22.4 dm^3/mol**

** = 0.56 dm^3**

b) What is the volume of NO2(g) produced?

Since we already know the number of moles of NO2(g) to be 0.05 mol:

So, the **volume of NO2(g) produced (at s.t.p)= 0.05 mol x 22.4 dm^3/mol**

** = 1.12 dm^3**

What is you are asked to find the mass of a product given the concentration of the reactants?

You can split up the calculation into two steps:

Find the number of moles of solute with the volume of solution of known concentration:

**number of moles of solute= volume in dm^3 x molar concentration***(Please note that you can convert cm^3 to dm^3 by dividing by 1000: cm^3/1000=dm^3)*Use the coefficients of the balanced equation to work out the number of moles of product formed and hence, the mass of the product.

__Example:__ 50 cm^3 of 0.2 mol/dm^3 BaCl2(aq) reacts according to the following equation:

**BaCl2(aq) + MgSO4(aq) --> BaSO4(s) + MgCl2(aq)**

Find:

a) the volume, in cm^3, of 0.4 moldm^-3 MgSO4(aq) needed.

We first find the number of moles of BaCl2(aq) used:

**number of moles of BaCl2(aq) = (50 cm^3/1000) x 0.2 moldm^-3**

** = 0.01 mol BaCl2**

From the equation, we know that **1 mol BaCl2(aq) + 1 mol MgSO4(aq) --> 1 mol BaSO4(s) + 1 mol MgCl2(aq) **(1:1:1:1 ratio)

Hence, **0.01 mol BaCl2(aq) + 0.01 mol MgSO4(aq) --> 0.01 mol BaSO4(s) + 0.01 mol MgCl2(aq)**

Now, to find the volume of 0.4 moldm^-3 MgSO4(aq) needed, we just need to calculate the volume of 0.4 moldm^-3 MgSO4(aq) which contains 0.01 mol:

**Volume of 0.4 moldm^-3 MgSO4(aq) needed = 0.01 mol/0.4 moldm^-3**

** = 0.025 dm^3**

** = 25 cm^3**

b) the mass of BaSO4(s) formed

We already know that 0.01 mol BaSO4(s) is produced in this reaction. So, we just find the mass:

mass of BaSO4(s) produced = 0.01 mol x (137 + 32 + [4 x 16])

= 0.01 mol x 233 gmol^-1

= 2.33 g