chemistry t hird e dition gilbert | kirss | foster | davies © 2012 by w. w. norton & company...

CHEMISTRYTHIRD EDITIONGilbert | Kirss | Foster | Davies

© 2012 by W. W. Norton & Company

CHAPTER 3-C

Percent Composition,Molecular Formulae,Combustion Analysis

2

• Percent Composition: Identifies the elements

present in a compound as a mass percent of the

total compound mass.

• The mass percent is obtained by dividing the mass

of each element by the total mass of a compound

and converting to percentage.

Percentage CompositionPercentage Composition

3

Empirical FormulaEmpirical Formula

• The empirical formula gives the ratio of the number

of atoms of each element in a compound.

Compound Formula Empirical FormulaHydrogen peroxide H2O2 OH

Benzene C6H6 CH

Ethylene C2H4 CH2

Propane C3H8 C3H8

4


• Glucose has the molecular formula C6H12O6. What

is its empirical formula, and what is the percentage

composition of glucose?

Empirical Formula = smallest whole number ratio

CH2O

5


CH2O

Total mass = 12.01 + 2.02 + 16.00 = 30.03

%C = 12.01/30.03 x 100% = 39.99%%H = 2.02/30.03 x 100% = 6.73%

%O = 16.00/30.03 x 100% = 53.28%

6


Saccharin has the molecular formula C7H5NO3S. What is its empirical formula, and what is the percentage composition of saccharin?

Empirical Formula is same as molecular formula

MW = 183.19 g/mole

%C = (7 x 12.011)/183.19 x 100% = 45.89% etc.

7


• A compound’s empirical

formula can be determined

from its percent composition.

• A compound’s molecular

formula is determined from

the molar mass and empirical

formula.

8


• A compound was analyzed to be 82.67% carbon and 17.33%

hydrogen by mass. What is the empirical formula for the

compound?

• Assume 100 g of sample, then 82.67 g are C and 17.33 g are

H

9


• Convert masses to moles:

82.67 g C x mole/12.011 g = 6.88 moles C

17.33 g H x mole/1.008 g = 17.19 mole H

Find relative # of moles (divide by smallest number)

10


Convert moles to ratios:

6.88/6.88 = 1 C

17.19/6.88 = 2.50 H

Or 2 carbons for every 5 hydrogens

C2H5

11


Empirical Formula is: C2H5

Formula weight is: 29.06 g/mole

If the molecular weight is known to be 58.12 g/mole

Then the molecular formula must be:

C4H10

12


• Combustion analysis is one of the most common methods for determining empirical formulas.

• A weighed compound is burned in oxygen and its products analyzed by a gas chromatogram.

• It is particularly useful for analysis of hydrocarbons.

13

Combustion AnalysisCombustion Analysis

Combustion Analysis: the technique of finding the mass composition of an unknown sample (X) by examining the products of its combustion.

X + O2 → CO2 + H2O

0.250 g of compound X produces:

0.686 g CO2 and 0.562 g H2O

14


X + O2 → CO2 + H2O

1. Find the mass of C & H that must have been present in X

(multiply masses of products by percent composition of the

products).

C: 0.686 g x 12.01 g/44.01 g = 0.187 g C

H: 0.562 g x (2 x 1.008 g)/18.02 g = 0.063 g H

15


X + O2 → CO2 + H2O

0.187 g C + 0.063 g H = 0.250 g total

So compound X must contain only C and H!!

2. Find the number of moles of C and H

C: 0.187 g x mole/12.01 g = 0.0156 moles C

H: 0.063 g x mole/1.008 g = 0.063 moles H

16


X + O2 → CO2 + H2O

3. Find the RELATIVE number of moles of C and H in whole

number units (divide by smallest number of moles)

C: 0.0156/0.0156 = 1

H: 0.063/0.0156 = 4

If these numbers are fractions, multiply each by the same

whole number.

17


X + O2 → CO2 + H2O

3. Write the Empirical Formula (use the relative numbers as

subscripts)

CH4

18


Summary

1. Find the mass of C and H in the sample.

2. Find the actual number of moles of C and H in the

sample.

3. Find the relative number of moles of C and H in whole

numbers.

4. Write the empirical formula for the unknown compound.

19


NOTE!

In step # 1 always check to see if the total mass of C and H adds

up to the total mass of X combusted. If the combined mass of

C and H is less than the mass of X, then the remainder is an

unknown element (unless instructed otherwise).

If a third element is known, calculate the mass of that element by

subtraction (at the end of step 1), and include the element in

the remaining steps.

20


Combustion Analysis provides the Empirical Formula. If a

second technique provides the molecular weight, then the

molecular formula may be deduced.

1. Calculate the empirical formula weight.

2. Find the number of “formula units” by dividing the known

molecular weight by the formula weight.

3. Multiply the number of atoms in the empirical formula by

the number of formula units.

21


The molecular weight of glucose is 180 g/mole and its

empirical formula is CH2O. Deduce the molecular

formula.

1. Formula weight for CH2O is 30.03 g/mole

2. # of formula units = 180/30.03 = 6

3. Molecular formula = C6H12O6

chemistry t hird e dition gilbert | kirss | foster | davies © 2012 by w. w. norton & company...

Documents