In this example, we're gonna look at combustion
analysis where we oxidize a compound containing
carbon hydrogen, and oxygen. We're told
how many grams of water and how many grams of
carbon dioxide form when the reaction is complete
We want to know the empirical formula
So I want to demonstrate how I would solve this problem
Since I have not looked at it before
What I would first want to do is write down
what I know and what I'm trying to determine
And so we have a compound
We don't know, let me call them x, y and z
We don't know the values of x, y and z
We do know we start with 1.00g
and we're gonna react it with oxygen
We don't know the number of moles, but I'm going to
make CO2, and water as my only two products
And we're given that we have 2.818g of CO2
and 1.161g of water
And so what we want to do is convert CO2 grams to moles
and likewise convert the water grams to moles
so we can then determine the moles of carbon
and the moles of hydrogen in our starting compound
and then we're gonna use the total weight to
determine the moles of oxygen beacuse
we can't use the oxygen here to determine that
since we're adding a large amount of oxygen
and then carry out the oxidation
So that means the first thing I want to do is look up
the atomic mass of carbon, hydrogen, and oxygen
Here I have a table. I can then use these to calculate
first the molecular mass of CO2, and that's going to be
one carbon plus two oxygens
and we can calculate that mass.
I'm gonna do that same thing for water
It's going to be two hydrogens plus one oxygen
In both cases it's going to be the number of grams/mole
I'll use my calculator to get the values
And so I can calculate the moles of CO2. I have 2.818g
and 44.008g/mol. Units cancel.
This will be the moles of CO2
And then, likewise I can calcuate the moles of water
1.161g of water, and the molecular mass g/mol
This will be the moles of water
So let me calculate those numbers
And so this tells me I have 0.0640mols of carbon
in my starting compound. And I have, since this is H2O
For every mole of water I have two moles of hydrogen
So two times 0.0644mol of hydrogen
So I need to now go back and calculate
The mass of 0.0640 mol of carbon
and likewise the mass of hydrogen that's in
my original compound. So I have 0.0640mol of carbon
and 12.01g of carbon per mol
So this give me the grams of carbon in my starting compound
And I do the same thing for hydrogen. I have
two times 0.0644mol of hydrogen atoms
Which 1.008 g/mol for hydrogen. And this is gonna be
the grams of hydrogen in my starting compound
So we can calculate these numbers
I start out with 1.00g of my compound
We now know that 0.7690g are carbon so if I
subtract that from the total
And then I substract the mass of hydrogen
This ould be the mass of oxygen in my starting compound
So now have the numbe of grams of oxygen
So I'm gonna have to convert that back to moles
of oxygen to get the empirical formula
15.999g/mol. Moles of oxygen
So i'm almost finished nowI have moles of carbon
So it's here, the moles of carbon
Moles of hydrogen, and moles of oxygen in the compound
So let me re-write those
So what I'm gonna do is divide all the numbers by
.00631. So this becomes 1, and then the other
two numbers are. Then I'm gonna assume within the
accuracy of the measurement, that this is 20, 10 and 1
We have C10H2O1. That's the empirical formula for the starting compound