Hi everyone, welcome to IGCSE Study Buddy where you can revise chemistry topics from the Cambridge IGCSE syllabus. If you are enjoying our videos so far, please don't forget to hit the like button and subscribe to our channel. In this video, you are going to learn part 1 of topic 3, Stoichiometry.
A formula is a way to represent the elements and their proportions in a compound. For example, the formula for table salt is NaCl or sodium chloride, indicating one sodium atom and one chlorine atom in the compound. The symbols of elements are shown in the periodic table. Each element is represented by a unique chemical symbol such as H for hydrogen, O for oxygen and Na for sodium. Where an element symbol has two letters, the first letter is always capitalized while the second is simple.
For example, magnesium is represented as capital M, simple G. not capital M capital G. We use numbers to indicate the number of atoms of the element. For example, O2 represents two oxygen atoms and N2 represents two nitrogen atoms.
When there is only one atom of an element, you typically do not write one. For example, H is used instead of H1. The molecular formula of a compound is the number and type of different atoms in one molecule. Let's look at the structure of methane.
Methane is a molecule consisting of one carbon atom bonded to four hydrogen atoms. So, its structural formula is CH4. This picture shows how atoms are arranged in the molecule and it's also a way of displaying the structural formula.
Let's look at another example. So the chemical formula for a compound is determined by looking at how many of each type of atom it contains. This molecule has two carbon atoms, six hydrogen atoms, and one oxygen atom, so its formula would be C2H5OH, which is ethanol. The empirical formula of a compound is the simplest whole number ratio of the different atoms or ions in a compound.
So, molecular formulas show the actual number of atoms of each element in a compound, while empirical formulas represent the simplest ratio of elements in that compound. The molecular formula is often a multiple of the empirical formula. Here are some examples of molecular formulas and their corresponding empirical formulas.
As you can observe, In the cases of carbon dioxide and water, their molecular formula is already presented in its simplest ratio form, which happens to be the same as their empirical formula. To find the empirical formula from the molecular formula, just divide all the subscripts by their lowest common denominator. It's easy to find the formulae of either molecular compounds or ionic compounds if we understand the valency of atoms.
What is valency? Valency is how much an atom can make bonds with other atoms. It tells us how many electrons an atom needs to gain, lose or share to achieve. a stable electron configuration.
For example, if an atom's valency is 1, it means it can lose or share one electron with another atom to achieve a stable electron configuration. Here is a table displaying the valency of elements within each group. The group of an element on the periodic table shows how many outer shell electrons it has. This helps us know if it needs to gain or lose electrons to have a full outer shell. Groups 1 to 3, metals, lose electrons.
Groups 5 to 7, non-metals, gain electrons. And group 4 can do either. Let's work out the formula for aluminium oxide. Aluminium or Al has a valency of 3 and oxygen or O has a valency of 2. Crisscross each atom's valency with the valency of the combining atom to determine the formula of a compound. So in this case, the formula will be Al2O3.
If valencies are equal, you must cancel them out. Ionic Ionic compound formulas are always empirical. When atoms form ions through ionic bonding, their valency also tells you the charge of the ion. For example, sodium ion has a plus one charge and chloride ion has a minus one charge and so on. Here's a table that displays the charges of ions determined by their respective groups on the periodic table.
Here are some common ions you will come across and their associated charges. Example, the ion of copper 2 is Cu2+. The roman numeral 2 in the bracket stands for the charge.
Let's now deduce the formula of an ionic compound. In an ionic compound, the positive and negative charges of the ions cancel each other out, resulting in a total charge of zero for the compound. This balance of charges is essential for ionic compounds.
Let's go through some examples to help you understand how to do this. First example, potassium chloride. So if we look at the electron configurations of potassium and chlorine, we know that potassium must lose one electron to gain a stable electron configuration and chlorine must gain one electron. So potassium ion will be K+. We don't write the 1 on charges, only the sign.
And chloride ion will be K+. Cl-. We always write the positive ion first. Since both have the same number of charges, their ratio is simply 1 is to 1. So in potassium chloride or KCl, for every potassium ion, you need one chloride ion, Cl-, to balance the charges and create a neutral ionic compound. Let's consider boron oxide.
In boron oxide, boron B typically forms a plus 3 cation, so it has a charge of plus 3. Oxygen O typically forms a minus 2 anion, so it has a charge of minus 2. To create a neutral ionic compound, we need to ensure that the total positive charge from the boron ions balance the total negative charge from the oxygen ions. The ratio required for this is 2 boron ions to 3 oxygen ions, resulting in a 2 is to 3 ratio. So for boron oxide B2O3, the ratio of ions is 2 boron ions to 3 oxygen ions to balance the charges and create a neutral compound.
So for a detailed understanding, consider this diagram. Boron donates two of its electrons to oxygen, resulting in a stable electron configuration for oxygen. However, boron still possesses an extra electron to give away, leading to the involvement of another oxygen atom. Now, both boron and the first oxygen have stable electron configurations. But the second oxygen still needs one more electron to complete its shell.
Another boron ion steps in to contribute one electron, leaving boron with two more to give away. Finally, a third oxygen atom accepts those two electrons. The result is a neutral ionic compound with no net charge. In essence, To create a neutral compound, we need an equal number of positive and negative charges. In this example, when we multiply boron's charge by 2 and the oxide's charge by 3, both ions will have equal charges of plus 6 and minus 6, effectively cancelling each other out.
This balancing of charges is how we arrive at the simplest ratio of 2 to 3. Now for exam purposes, here's a simple trick to work out the formula of an ionic compound. We simply use the criss-cross method where we criss-cross the charges of the ions to determine the simplest ratio needed to form a neutral ionic compound. If the charges on the ions are the same, they combine in a 1 is to 1 ratio to create a neutral compound. In this case, you do not write the number 1 in the chemical formula because it's understood.
If the charge is plus or minus 1, we do not write the number 1. We only write if it's a positive or a negative charge. Let's try the criss-cross method with the example calcium nitrate. Calcium has a charge of plus 2 and nitrate has a charge of minus 1. When creating the formula of a compound ion, it's important to enclose the compound ion in brackets when more than one of that ion is needed in the formula. So in this example, the compound ion is nitrate.
So, calcium nitrate is represented as CaNO3 twice, where the brackets emphasize the presence of multiple nitrate ions. One more example, ammonium carbonate. Once again, once you cross-multiply the charges, we enclose the compound ion that is needed more than once in the brackets.
In this case, carbonate is not enclosed in a bracket because only one of it is needed. That concludes Part 1 of Topic 3, Stoichiometry. Are you enjoying our videos?
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