Molecular formula from
combustion
b ti analysis
l i
In combustion analysis, a sample is burned in a furnace, and the
resulting oxidation products of H2O and CO2 are trapped and
weighed in a device shown on page 96 of your text.
In this section you will learn how to determine the empirical
formula of a compound based on the data obtained from
combustion analysis.
Here is a typical problem. A sample contains only C, H and
N. When .1156 g of this material is burned in combustion
y yyou can isolate .1638 g of CO2 and .1676 g of water.
analysis
What is the empirical formula of this compound?
A sample contains only C, H and N. When .1156 g of this material is
burned in combustion analysis you can isolate .1638 g of CO2 and .1676 g
of water. What is the empirical formula of this compound?
Examine the system for a moment. The amount of CO2 isolated in
the product should give us the amount of C in the original sample.
sample
Similarly the amount of H2O should give us the amount of H in the
material. Notice, however, that the amount of water + the amount
off CO2 is
i actually
t ll greater
t than
th the
th amountt off material
t i l you started
t t d
with! While this may surprise you at first, it makes sense because
both H2O and CO2 contain O that was not present in the original
sample.
A sample contains only C, H and N. When .1156 g of this material is
burned in combustion analysis
y yyou can isolate .1638 g of CO2 and .1676 g
of water. What is the empirical formula of this compound?
Also think about this. How are we going to determine the amount
of N in the material? None of the products contain N! Essentially
we have to determine the amount of C and H first, and then subtract
that from the total amount of sample to find the missing N!
A sample contains only C, H and N. When .1156 g of this material is
burned in combustion analysis you can isolate .1638
1638 g of CO2 and .1676
1676 g
of water. What is the empirical formula of this compound?
Step 1: Determine the amount of products we would have if we
started with a 100g sample.
Just like we did for the empirical formula from % composition, this
is a little mathematical trick that makes later steps easier. We will do
this using simple proportions.
proportions
Original amount of product
Amount of product in 100 g sample
=
;
Original amount of starting material
100 g sample
Original amount of product
100 g ×
= Amount of product in 100 g sample
Original amount of starting material
A sample contains only C, H and N. When .1156 g of this material is
burned in combustion analysis you can isolate .1638
1638 g of CO2 and .1676
1676 g
of water. What is the empirical formula of this compound?
Step 1: Determine the amount of products we would have if we
started with a 100g sample.
(Continued)
CO2
100g x (.1638g
( 1638g CO2/.1156g
/ 1156g product) = 141.7g
141 7g CO2
H2 O
100g x (.1676 g H2O/.1156g product) = 145.0 g H2O
A sample contains only C, H and N. When .1156 g of this material is
burned in combustion analysis you can isolate .1638
1638 g of CO2 and .1676
1676 g
of water. What is the empirical formula of this compound?
Step 2:Determine moles of products and convert to moles of
atoms in sample.
CO2 (Molecular mass = 44.01)
141.7g CO2 x (1 mole/44.01g) = 3.22 mole CO2
Notice that a mole of CO2 contains 1 mole of C.
C You can
use this like another conversion factor: 1 mole CO2= 1 mole C
3.22 moles CO2 x (1 mole C/1 mole CO2) = 3.22 mole C
A sample contains only C, H and N. When .1156 g of this material is
burned in combustion analysis you can isolate .1638
1638 g of CO2 and .1676
1676 g
of water. What is the empirical formula of this compound?
Step 2:Determine moles of products and convert to moles of
atoms in sample.
H2O - A bit trickier (Molar mass = 18.016)
145.0g H2O x (1 mole/18.016g) = 8.05 mole H2O
Notice that a mole of H2O contains 2 moles of H.
H You must
use this additional conversion factor: 1 mole H2O= 2 mole H
8.05 moles H2O x (2 mole H/1 mole H2O) = 16.1 mole H
A sample contains only C, H and N. When .1156 g of this material is
burned in combustion analysis you can isolate .1638
1638 g of CO2 and .1676
1676 g
of water. What is the empirical formula of this compound?
Step 3: Find the missing element.
Right now we have 3.22 mole of C and 16.1 mole of H
3.22 mole C x (12.01g C/1 mole C) = 38.7 g C
16.1 Mole H x (1.008 g/1 mole H) = 16.1 g H
Since this was normalized to a 100 g sample,
sample the missing N is found
by taking 100- g C – g H
= 100-38.7-16.1 = 45.2g N
And converting back moles of N
45.2 g N x ((1 mole/14.01 g) = 3.23 mole N
A sample contains only C, H and N. When .1156 g of this material is
burned in combustion analysis you can isolate .1638
1638 g of CO2 and .1676
1676 g
of water. What is the empirical formula of this compound?
Step 4: Find smallest number of moles and divide all moles by
this number
3.22 mole of C
16.1 mole of H
3.23 mole N
Smallest number is about 3.22 so
C = 3.22/3.22 = 1 C
H = 16.1/3.22 = 5 H
N = 3.23/3.22 = ~ 1 N
So our empirical formula is CH5N
A th Example
Another
E
l
A compound contains only C, O and H. When 1.25 g of this
compound is burned you obtain 2.84 g CO2 and 1.16 g of H2O.
What is the empirical formula of this compound?
Notice this is a little different because it contains O instead of N.
N
It is not that much different however, because our products still only
give us H and C, and we will have to subtract this from the starting
amount to figure out O.
A compound contains only C, O and H. When 1.25 g of this
compound
p
is burned yyou obtain 2.84 g CO2 and 1.16 g of
H2O. What is the empirical formula of this compound?
Step 1: Make it a 100 g sample
100g x (2.84 g CO2/1.25 g sample) = 227.2 g CO2
100g x (1.16 g H2O/1.25 g sample) = 92.8 g H2O
A compound contains only C, O and H. When 1.25 g of this
compound
p
is burned yyou obtain 2.84 g CO2 and 1.16 g of
H2O. What is the empirical formula of this compound?
Step 2:Convert to moles of element
227.2 g CO2 x (1 mole CO2/44.1 g CO2) x (1 mole C/1 mole CO2)
= 5.15 mole C
92.8 g H2O x (1 mole H2O/18.016 g H2O) x (2 mole H/1 mole H2O)
=10 30 mole H
=10.30
A compound contains only C, O and H. When 1.25 g of this
compound
p
is burned yyou obtain 2.84 g CO2 and 1.16 g of
H2O. What is the empirical formula of this compound?
Step 3: Find moles of missing element
5.15 mole C x 12.01 g = 61.85 g C
10.30 mole H x 1.008 g = 10.3 g H
100g total – 61.85 g C – 10.3 g H = 27.85 g missing O
27.85 g x (1 mole/16 g) = 1.74 mole O
A compound contains only C, O and H. When 1.25 g of this
compound
p
is burned yyou obtain 2.84 g CO2 and 1.16 g of
H2O. What is the empirical formula of this compound?
Step 4: Find mole ratios
Smallest number of moles is O at 1.74
5.15 mole C/ 1.74 = 2.96 ~ 3C
10 3 mole H /1.74
10.3
/1 74 = 5.92
5 92 ~ 6H
1.74 mole O /1.74 = 1
Empirical formula C3H6O