If you have cooked food previously, you know even a simple dish can have quite a few ingredients. When following a recipe, you know the exact proportions of each ingredient. For example, in a cake recipe for 250g of flour added, we need 250g of sugar, and this gives us the exact measurements that we need for a perfect cake. But when you’re making your own recipe or trying to recreate something you do not know the proportions, like do you need brown sugar or white sugar or 1 teaspoon of salt or 1 tablespoon. So, how does this relate to chemistry? Well, when we want to react substances together, through different equations we can determine how much of each reactant we need, how much of our desired product is produced and even find out what concentration we require, and that is precisely what we are going to go through in this explanation.
- We will first explore what relative mass is, then explore its relationship with moles.
- After this, we will explore percentage yield, and how to work it out.
- We will then go on to explore atom economy.
- Then we will briefly explore titrations.
- Finally, we will explore some equations related to gases.
Mass and mole calculations in chemical equations
Relative atomic mass, which can also be symbolised by this expression ‘Ar’, is the average mass of a substance compared to 1/12 mass of carbon-12. We measure the atomic mass by comparing it to carbon-12 because, carbon-12 has exactly 12 units, 6 protons and 6 neutrons. In addition, we use the term average because different elements can have different isotopes, so it is the average mass of the isotopes as well.
Isotopes are elements that have the same number of protons and neutrons, but different number of electrons.
From relative atomic mass, we can calculate the relative formula mass, which is symbolised by this expression ‘Mr’. The way we do this is by adding up the relative atomic masses for all the different elements in a compound.
Water is written as H2O, to find the relative formula mass, you first identify the relative atomic mass of each element:
- Hydrogen (H) = 1
- Oxygen (O) = 16
You then identify how much of each element there is:
There are two hydrogens and one oxygen
Then you add everything up, and if there is more than one of each element, you will multiply the relative atomic mass of the elements for how many there are:
Relative formula mass of H2O:
16+1+1 = 18
So our relative formula mass of one molecule of water is 18.
The following section is only required for those studying the higher tier of GCSE Chemistry.
We will also use relative atomic mass and relative formula mass to work out moles. Moles can be described as the amount of substance of a substance. One mole is 6.02·1023 which is absolutely massive and this is known as Avogadro constant.
In order to work out moles, we can use the following equation:
\[moles = mass (g) \div A_r\]
Or if its for a molecule:
\[moles = mass (g) \div M_r\]
Therefore, we can actually do three things with this equation:
- Find out moles.
- Find out the mass of substance in grams.
- Find out the relative atomic mass or relative formula mass.
And this can be done as long as we have two of the three factors.
Chemical calculations percentage yield
In a reaction, reactants react with each other to make a final product. The mass of the product is what we will call the yield. Sometimes the amount of product that is produced, this is known as the actual mass of product that is produced, is different to what we calculated using our chemical equation, which is the theoretical mass produced. When we divide this by each other, then multiply by 100 we find out what our percentage yield is and this can be displayed in the following equation:
\[\text{yield percentage} = \text{ actual yield} \div \text{theoretical yield} \times 100 \]
If we calculate that a reaction will produce 100kg, but after carrying out the reaction we produce 70kg, what is the percentage yield?
First, we identify the actual yield and the theoretical yield:
- Actual yield: 70kg
- Theoretical yield: 100kg
We then plug all these numbers into our equation:
\[\text{yield percentage} = 70 (kg) \div 100(kg) \times 100 \]
Our percentage yield is: 70%
Percentage yield is important, as especially within chemical industries we need to ensure, as chemists we are able to have the highest yield. Sometimes, there are factors that can affect our yield, such as; some product is left in the apparatus and is therefore not measures, reactant may not be pure or we may get additional products.
Atom Economy
Atom economy is another way in which we can analyse our final product. As we learn how to be more sustainable an important way that we need to explore this is within reactions. We need to be able to produce reactions where we do not waste a lot of energy and raw materials and this is where looking at the atom economy can help us. By exploring the atom economy, it allows us to explore how much of the mass of reactants becomes our desired product. It is very similar to how we calculate percentage yield but instead of looking as the mass, we look at relative formula mass and this can be seen in the equation below.
\[\text{percentage atom economy} = \text{ relative formula mass of desired product} \div \text{relative formula mass of reactants} \times 100 \]
Titration chemistry calculations
Titrations are a way for us to monitor when a neutralisation reaction taken place. We use an acid, an alkali and an indicator. The indicator allows us to determine when this reaction has taken place, and we term this as the ‘end point’. If you like at the explanation of Titrations, you will be able to explore how we carry out this reaction, including a step-by-step guide as well as all the apparatus you may need.
Titrations are useful because they allow us to determine the concentration of an unknown substance. We do this by using a balanced equation, having the other reactant with a known concentration and the relative formula mass of the substances.
Chemical calculations gases
There are so many different things that contain gas, we have car wheels, airbags, fizzy drinks and so much more. Each of these have a specific amount, to ensure there is enough for the use and to ensure the object does not explode. To achieve this, scientists needed a way to determine the volume of gas required. They first determined that that different temperature and pressure was a contributing reason, and then they concluded on an equation.
For your GCSE’s you only need to know how to calculate the volume of gas at room temperature which is 20℃and pressure which is 1 atm, whereby the volume of 1 mole of gas is 24cm3
To determine the volume of gas in cm3:
\[\text{moles of gas} = \text{volume of gas (dm^3)} \div 24 dm^3\]
To determine the volume of gas in dm3:
\[\text{moles of gas} = \text{volume of gas (cm^3)} \div 24000 cm^3\]
Let us go through an example to consolidate our knowledge:
The airbag in a car has been inflated by 62g of nitrogen (N2) when it is activated. What is the volume that nitrogen gas occupies at room temperature and pressure, when the Ar of nitrogen is 14.
For this first question, we first need to calculate the moles:
\[moles = mass \div M_r\]
We have the mass which is 62g and as there are two molecules of nitrogen, it means relative formula mass is \(14\times 2 =28)
Now we can put these numbers into our equation:
\[moles = 62 \div 28 = 2.21 mol\]
Now we need to rearrange our gas equation, so we can find the volume:
\[volume = moles \times 24dm^3\]
Finally, we can plug our numbers in to find our answer:
\[volume = 2.21 \times 24dm^3 = 53.14 dm^3\]
Note that as we used the equation with dm3, our final volume is in dm3.
Chemical Calculations - Key takeaways
- Relative atomic mass is the average mass of a substance compared to 1/12 mass of carbon-12.
- Moles can be described as the amount of substance of a substance.
- One mole is 6.02 x 1023
- The relationship between moles, mass and relative atomic mass is seen in this equation: \(moles = mass (g) \div A_r\)
- Percentage yield allows us to determine, the mass of the product actually made, compares to what was calculated.
- Atom economy allows us to explore how much of a reactant was made into product.
- Titrations are a way of exploring neutralisation reactions between an acid and an alkali.
- At room temperature and pressure, we can determine the moles of gas using this equation: \(\text{moles of gas} = \text{volume of gas (dm^3)} \div 24 dm^3\)
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