What is a chemical formula?
A compound's chemical formula gives the mole ratio of the elements in that compound.
Water has the formula H2O
The formula tells us that there are two moles of hydrogen atoms for every mole of oxygen atoms in a mole of water molecules (18g/mol)
Or take sodium chloride formula NaCl
The formula tells us that for every mole of sodium ions Na+ there is also one mole of chloride ions Cl- in a mole of sodium chloride (58.5g/mol)
These are the simplest mole ratios of elements in the compound.
The simplest mole ratio of elements in a compound is called an empirical formula.
But how do we calculate these and other formulae like them?
School experiments to determine simple formulae tends to focus on binary or two element compounds.
The methods are usually to either take away an element from the compound and find the masses of each element in the sample by difference.
Or the other method is to add an element to another and again by weighing determine the masses of the two elements in a given mass of the compound and hence from that data find the formula.
In the first method a common approach is to remove oxygen from copper oxide using natural gas.
In the second it is to burn magnesium in air and add oxygen to the metal to create magnesium oxide.
If you have done any kind of useful practical chemistry you will have carried out at least one of these experiments in a laboratory yourself.
In the UK they can be found in most GCSE and A level courses.
Here is a link to heating magnesium in air if you have never done this practically.
Here is link to the experiment to remove copper from copper oxide to find the formula of copper oxide
Let's have a look at some results from a typical experiment and see how to use the mole to arrive at the empirical formula for the compound.
Let's begin with finding the formula of magnesium oxide.
Here's the data ( it's not "real" data but imagined to serve the purpose of the calculation!! if ever you see results like these in a real experiment you ought to be suitably suspicious!!):
Mass of magnesium(g) Mass of magnesium oxide(g) Mass of oxygen(g)
0.0 0.0 0.0
0.6 1.0 0.4
1.2 2.0 0.8
1.8 3.0 1.2
2.4 4.0 1.6
There are several things we can see from these results.
First if we were to plot a graph of mass of oxygen against mass of magnesium we'd have straight line.
The mass of oxygen added is directly proportional to the mass of magnesium burned.
This illustrates the law of constant composition for a compound i.e. the mole radio of the elements in the compound never changes no matter how many moles of the compound you look at and wherever it is found in the Universe (though having never been to Mars or Jupiter i'm taking the last bit of that definition on trust).
We can calculate the amount of each substance used.
We use the equation from Mole(2)
So if n=m/M then amt magnesium = mass of magnesium(g)/molar mass of magnesium (g/mole)
Therefore Amt magnesium = 0.6g /24 g/mol = 0.025 mol
The answers are below but you ought to just try a few calculations from memory to check you can actually come up with the same results as here.
Amt magnesium(mol) Amt magnesium oxide(mol) Amt oxygen(mol)
0.0 0.0 0.0
0.025 0.025 0.025
0.050 0.050 0.050
0.075 0.075 0.075
0.10 0.10 0.10
What do you notice about the numbers of moles?
In every example they are the same !!
Why is that?
Let's look at the mole ratio of elements in magnesium oxide:
magnesium oxygen
mass (g) 1.2 0.8
molar mass (g/mol) 24 16
amount (mol) 0.05 0.05
simplest
mole ratio 1.0 : 1.0
empirical
formula Mg O
You'll see here the table method I have used to set out the calculation
Try some examples for yourself and see how easy this stuff is.
Then try some of these examples below.
If you want the answers you'll have to send in an email comment for them
Happy calculating
Example empirical formula calculations:
A compound of carbon, hydrogen and oxygen contains 40.0% carbon, 6.6% hydrogen and 53.4% oxygen. Calculate its empirical formula.
Determine the formula of a mineral with the following mass composition: Na = 12.1%, Al = 14.2%, Si = 22.1%, O = 42.1% and H2O = 9.48%.
A 10.00g sample of a compound contains 3.91g pf carbon , 0.87g of hydrogen and the rest is oxygen. Calculate the empirical formula of this compound.
The next examples calculate the formula of hydrated salts i.e x in this type of formula MO.x H2O
A sample of a hydrated compound was analysed and found to contain 2.10g of cobalt, 1.14g of sulphur, 2.28g of oxygen and 4.5g of water. Calculate its empirical formula.
Mole (1)
Mole (2)
The Mole (4) Using the mole to determine equation stoichiometry
Mole (5)
A compound's chemical formula gives the mole ratio of the elements in that compound.
Water has the formula H2O
The formula tells us that there are two moles of hydrogen atoms for every mole of oxygen atoms in a mole of water molecules (18g/mol)
Or take sodium chloride formula NaCl
The formula tells us that for every mole of sodium ions Na+ there is also one mole of chloride ions Cl- in a mole of sodium chloride (58.5g/mol)
These are the simplest mole ratios of elements in the compound.
The simplest mole ratio of elements in a compound is called an empirical formula.
But how do we calculate these and other formulae like them?
School experiments to determine simple formulae tends to focus on binary or two element compounds.
The methods are usually to either take away an element from the compound and find the masses of each element in the sample by difference.
Or the other method is to add an element to another and again by weighing determine the masses of the two elements in a given mass of the compound and hence from that data find the formula.
In the first method a common approach is to remove oxygen from copper oxide using natural gas.
In the second it is to burn magnesium in air and add oxygen to the metal to create magnesium oxide.
If you have done any kind of useful practical chemistry you will have carried out at least one of these experiments in a laboratory yourself.
In the UK they can be found in most GCSE and A level courses.
Here is a link to heating magnesium in air if you have never done this practically.
Here is link to the experiment to remove copper from copper oxide to find the formula of copper oxide
Let's have a look at some results from a typical experiment and see how to use the mole to arrive at the empirical formula for the compound.
Let's begin with finding the formula of magnesium oxide.
Here's the data ( it's not "real" data but imagined to serve the purpose of the calculation!! if ever you see results like these in a real experiment you ought to be suitably suspicious!!):
Mass of magnesium(g) Mass of magnesium oxide(g) Mass of oxygen(g)
0.0 0.0 0.0
0.6 1.0 0.4
1.2 2.0 0.8
1.8 3.0 1.2
2.4 4.0 1.6
There are several things we can see from these results.
First if we were to plot a graph of mass of oxygen against mass of magnesium we'd have straight line.
The mass of oxygen added is directly proportional to the mass of magnesium burned.
This illustrates the law of constant composition for a compound i.e. the mole radio of the elements in the compound never changes no matter how many moles of the compound you look at and wherever it is found in the Universe (though having never been to Mars or Jupiter i'm taking the last bit of that definition on trust).
We can calculate the amount of each substance used.
We use the equation from Mole(2)
So if n=m/M then amt magnesium = mass of magnesium(g)/molar mass of magnesium (g/mole)
Therefore Amt magnesium = 0.6g /24 g/mol = 0.025 mol
The answers are below but you ought to just try a few calculations from memory to check you can actually come up with the same results as here.
Amt magnesium(mol) Amt magnesium oxide(mol) Amt oxygen(mol)
0.0 0.0 0.0
0.025 0.025 0.025
0.050 0.050 0.050
0.075 0.075 0.075
0.10 0.10 0.10
What do you notice about the numbers of moles?
In every example they are the same !!
Why is that?
Let's look at the mole ratio of elements in magnesium oxide:
magnesium oxygen
mass (g) 1.2 0.8
molar mass (g/mol) 24 16
amount (mol) 0.05 0.05
simplest
mole ratio 1.0 : 1.0
empirical
formula Mg O
You'll see here the table method I have used to set out the calculation
Try some examples for yourself and see how easy this stuff is.
Then try some of these examples below.
If you want the answers you'll have to send in an email comment for them
Happy calculating
Example empirical formula calculations:
A compound of carbon, hydrogen and oxygen contains 40.0% carbon, 6.6% hydrogen and 53.4% oxygen. Calculate its empirical formula.
Determine the formula of a mineral with the following mass composition: Na = 12.1%, Al = 14.2%, Si = 22.1%, O = 42.1% and H2O = 9.48%.
A 10.00g sample of a compound contains 3.91g pf carbon , 0.87g of hydrogen and the rest is oxygen. Calculate the empirical formula of this compound.
The next examples calculate the formula of hydrated salts i.e x in this type of formula MO.x H2O
A sample of a hydrated compound was analysed and found to contain 2.10g of cobalt, 1.14g of sulphur, 2.28g of oxygen and 4.5g of water. Calculate its empirical formula.
Mole (1)
Mole (2)
The Mole (4) Using the mole to determine equation stoichiometry
Mole (5)
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