C4: Stoichiometry

Compounds are written in formulae which tell you its structure of atoms

The letters show the symbols of the elements in the compound (Note that capital letters make a difference, eg. Na and NA are different things)
The subscript (small number at the bottom right of the element symbol) shows the amount of that type of atom present in the molecule

Determining Formulae of Ionic Compounds

Through knowing the charges of the ions present, we are able to know how many of each ion needs to be present
This is because ionic compounds always have a neutral charge, or a charge of 0. Therefore, the positive and negative charges from the ions have to be equal
For example, if we needed to find the equation of magnesium chloride, we would work it out by the following

A magnesium (Mg) ion has a charge of 2+
A chloride (Cl) has a charge of 1-
To balance out the charges, we need to make the negative charge to -2 to match the negative charge of +2
From this, we know that we need two chloride ions and one magnesium ion for a neutral charge
Therefore, the equation of sodium carbonate is MgCl2

For example, if we needed to find the equation of sodium carbonate, we would work it out by the following

A sodium (Na) ion has a charge of 1+
A carbonate (CO3) ion has a charge of 2-
To balance out the charges, we need to make the positive charge to +2 to match the negative charge of -2
From this, we know that we need two sodium ions and one carbonate ion for a neutral charge
Therefore, the equation of sodium carbonate is Na2CO3

Determining equations from diagrams

Equations can be determined after identifying the number and separate components in the compound

Let's try this with the example of ethanol, which has a structure like the image above
Ethanol has 5 hydrogen (H) atoms, 2 carbon (C) atoms, and 1 hydroxide (OH) molecule
In organic compounds, Carbon and Hydrogen are always written first. Let's write our equation in that order
With this order and the numbers for each atom, we can write our equation
Therefore, ethanol is C2H5OH

Writing word equations (with state symbols)

First of all, we need to know what state symbols are. Well, they are simply symbols which tell you what physical state a substance is

(s) means that the substance is solid
(l) means that the substance is liquid
(g) means that the substance is gas
(aq) means that the substance is aqueous (dissolved in water)

Ionic equations are equations that show only the ions that react in the equation. Let's use an example to see the different steps needed to find the ionic equation

Lets try this out with the example: BaCl₂(aq) + Na₂SO₄(aq) —> 2NaCl (aq) + BaSO₄(s). These are the steps...

Write the separate ions in the aqueous compounds (but not the solid, liquids, and gases)

Ba²⁺(aq) + 2Cl^-(aq) + 2Na⁺(aq) + SO₄^2-(aq) —> 2Na⁺(aq) + 2Cl^-(aq) + BaSO₄(s)
When you see the same ion on both sides, cross them out. Whatever remaining is your ionic equation

Ba²⁺(aq) + SO4^2-(aq) —> BaSO₄(s)

Balancing equations

When given word equations, sometimes we need to balance the equations so that they make sense and both sides are equal

Let's try this with the example Methane(g)+ Oxygen(g) —> Carbon dioxide(g) + Water (l). The unbalanced equation of this reaction is CH₄ + O₂ —> CO₂ + H₂O

First we have to count the amount of atoms on both side to see which atoms are unbalanced

On the left side, there are 1 C atoms, 4 H atoms, and 2 O atoms
On the right side, there is 1 C atom, 2 H atoms, 3 O atoms

We can see that the number of carbon is already balanced, and that we need to balance the hydrogen and oxygen atoms. We cannot change the reactants or products of the reaction, but can only add a number in front to increase the number of a specific component.

As we have 4 H atoms on the left and 2 H on the right, we need to put a 2 in front of the H2O on the right side. This leaves us with CH₄ + O₂ —> CO₂ + 2H₂O which has

1 C atom, 4 H atoms, and 2 O atoms on the left side
1 C atom, 4 H atoms, 4 O atoms on the right side

All that is left is the Oxygen (O). To make this balanced, we simply needed to add a 2 in front of the O2, to balance out both sides. That leaves us with our final equation: All that is left is the Oxygen (O). To make this balanced, we simply needed to add a 2 in front of the O₂, to balance out both sides. That leaves us with our final equation CH₄ + 2O₂ —> CO₂ + 2H₂O which has

1 C atom, 4 H atoms, and 4 O atoms on the left side
1 C atom, 4 H atoms, 4 O atoms on the right side

Therefore, we can deduce the balanced equation through balancing the amount of atoms on each side

Relative atomic and molecular mass

The relative atomic mass is the average mass of natural atoms of an element. This is equivalent to adding the amount of neutrons and protons of an atom (eg. The relative atomic mass of carbon is 12 as it has 6 protons and neutrons each)
The relative molecular mass is the addition of the atomic masses of the atoms in the molecule. For example, the molecular mass of H2O is 18 (1+1+16)

The Mole

A mole is defined as 6.02 x 10^23 particles of a substance
It is a unit to measure the amount of particles present
A mole of an element is equivalent to the relative atomic mass of the element in grams (eg. a mole of oxygen would have a mass of 16g)
To work out the molecular mass, we can add the atomic masses (for example, a mole of carbon dioxide would have a mass of 12g+16g+16g=44g)

Molar Volume

The molar volume is the volume that one mole of any gas will have
The molar volume is 24dm³ or 24,000 cm³ (at room temperature and pressure)

Calculating reacting masses

With certain information, we can calculate the masses of products and reactants in a chemical reaction
Let's see how we do this with the example of this question: Calculate the mass of produced magnesium oxide from 12g of magnesium
Balanced equation: 2Mg + O2 —> 2MgO

Work out the amount of moles of magnesium present. The molar mass of Mg is 24g. As 24g is the the mass of one mole of Mg, 12g is therefore the mass of the half a mole of Mg
As 2Mg of reactants will produce 2MgO, the ratio of Mg to MgO is one. Therefore, when half a mole of Mg reacts, half a mole of MgO will be formed.
Now we know that the answer is equivalent to the mass of half a mole of MgO
As the mass for one mole of MgO is 40g, the mass for half a mole will be 20g
Therefore, when 12g of magnesium reacts, 20g of magnesium oxide is produced

Calculating volumes of gases

The molar volume of one mole of any gas is equivalent to 24dm³ or 24,000 cm³
The volume of a gas can be worked out by multiplying the molar volume by the amount of gas in moles
For example, 5mol of a gas would have the mass of 120dm³ or 120,000 cm³
Equation: Volume (cm3)= Amount of gas (mol) x Molar volume (24dm3 or 24,000 cm3)

Calculating concentrations of solutions

The concentration of a solution shows how common a substance is in a solution
To show the different steps let's use the following example
Equation: Concentration (mol / dm³) = Amount of substance (mol) ÷ Volume of solution (dm³)
If 0.25mol of solute dissolved into a solution of 500cm3, what is the concentration of the solution in mol/dm3?

First, we have to convert cm3 to dm3 to make sure we are working in the correct units. To do this we need to divide by 1000.

500cm³ ÷ 1000 = 0.5dm³
Now, we can use the equation below to work out the concentration with the information we have

Concentration (mol / dm³) = 0.25mol ÷ 0.5dm³
Therefore, the concentration of the solution is 0.5mol / dm³

Limiting reactants

Limiting reactants are the reactants that run out before the others, and therefore cause the reaction to stop
The reactant that runs out first is the limiting reactant
To show the different steps let's use the following example
3g of magnesium reacts with 7.3g of hydrochloric acid. Which is the limiting reactant? (Equation of reaction: Mg + 2HCl —> MgCl2 + H2)

First, we need to find the amount of moles each reactant has. To work this out, we need to divide the amount of grams with the molar mass (mass of the reactant if it had one mole). The molar mass of Mg is 24g and of HCl is 36.5g

Mg: 3g/24g=0.125g

HCl: 7.3g/36.5g=0.2g
Now we need to consider the ratio of the reactants. One Mg reacts with 2HCl. Therefore, 0.125g needs to react with 0.25g of HCl. However, there are only 0.2g of HCl in this scenario, which is not enough.

Therefore, the limiting reactant of this equation is HCl, as it has less moles and will run out before Mg does

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