Download Percent Composition

Percent Composition - Formulas
This program demonstrates how to find percentage composition as well as empirical and
molecular formulas from that data. Please read each section carefully. You might wish to
take notes. Remember that all atomic masses are rounded to 3 significant figures.
Percentage Composition Calculations
To calculate the correct percentage composition of each element in some compound you
must correctly follow several steps. It is not very difficult, once you get the hang of it.
Please take notes on the steps to follow.
Several problems follow this one for you to work.
Lets look at a sample problem: magnesium chloride --> MgCl2
step 1: calculate formula mass
Mg: 1 x 24.3 = 24.3
Cl: 2 x 35.4 = 70.8
---formula mass = 95.1
step 2: divide each component mass by the formula mass and multiply by 100
Mg: 24.3/95.1 x 100 = 25.6%
Cl: 70.8/95.1 x 100 = 74.4%
step 3: make certain the percentages add up to 100 (+/- 0.1)
25.6 + 74.4 = 100
This problem shows all the steps to follow. Be careful to use three significant figures at
all times. Remember to first find the correct formula, then find formula mass, then
calculate the percentage composition.
The next two problems allow you some practice. Work these out on paper and then
check your answer.
Practice Problem #1
Find the percentage composition for each element in: iron (III) silicate
iron (III) silicate
Fe: 32.9%
--> Fe2(SiO3)3
Si: 24.8%
O: 42.4%
Remember that with significant figures your answer can be one above or below the final
digit. Example: iron could have been reported as 32.8%, 32.9%, or even 33.0%.
Work: Fe: 2 x 55.8 = 112
Si: 3 x 28.1 = 84.3
O: 9 x 16.0 = 144
-----340.
112/340. x 100 = 32.9%
84.3/340. x 100 = 24.8%
144/340. x 100 = 42.4%
----100.1 (ok)
Practice Problem #2
Find the percentage composition for each element in: tin (IV) arsenate . To see the correct
answers and the method to solve the problem, please continue.
tin (IV) arsenate
Sn: 39.1%
--> Sn3(AsO4)4
As: 32.8%
O: 28.0%
Work: Sn: 3 x 119 = 357
As: 4 x 74.9 = 300.
O: 16 x 16.0 = 256
----913
357/913 x 100 = 39.1%
300./913 x 100 = 32.8%
256/913 x 100 = 28.0%
----99.9 (ok)
Empirical Formula Calculations
Finding the empirical formula is somewhat the reverse of finding percentage
composition. First you will be given the percentages of each element in a particular
compound.
Assuming that you will always be given a 100 gram sample we will convert these
percentages to grams. The next step is to find the ratio of moles of each element and then
calculate the simplest ratio of subscripts that maintains that ratio.
It is much easier done than said! Take notes as we work through this sample problem.
Lets start with this data: 36.5% Na, 25.4% S, and 38.1% O
First, convert each percentage to grams: 36.5 g Na, 25.4 g S, 38.1 g O.
Next, divide each by the grams/mole of that element:
Na:
36.5 g
----= 1.58 mol Na
23.0 g/mol
S:
25.4 g
-----= 0.791 mol S
32.1 g/mol
0:
38.1 g
= 2.38 mol O
-----16.0 g/mol
Now we can set up the ratio of moles of each element:
Na
S
O
1.58 0.791 2.38
To convert these decimal numbers into whole numbers and maintain the same ratio
between them, just divide each by the smallest of the subscripts.
Na
S
O
1.58 0.791 2.38
----------0.791 0.791 0.791
Our final formula would look like: Na2SO3 --> sodium sulfite
Following will be two problems for you to work.
Practice Problem #1
Given the following data, find the correct empirical formula:
49.0% C, 2.70% H, 48.2% Cl
The correct formula would be: CH3Cl2
To solve:
C: 49.0g
----- = 4.08 mol
12.0 g/mol
H: 2.70g
----- = 2.67 mol
1.01 g/mol
Cl: 48.2g
----- =
1.36 mol
35.5 g/mol
C
H
Cl
4.08
2.57
1.36
---------1.36
1.36
1.36
gives the final formula: CH3Cl2
Practice Problem #2
Given the following data find the empirical formula:
N = 26.2%,
H = 7.50%,
Cl = 66.4%
The correct formula would be: NH4Cl
(ammonium chloride)
The work:
N: 26.2g
----- = 1.87 mol
14.0 g/mol
N
1.87
---1.87
H
7.50
---1.87
Cl
1.87
--1.87
H: 7.50g
----- = 7.50 mol
1.01 g/mol
Cl: 66.4g
----- =
35.5 g/mol
1.87 mol
--> NH4Cl
Molecular Formulas
To find the molecular formula, first find the empirical formula for the data you have
been given. Then compare the formula mass for your empirical formula with the formula
mass for the molecular formula. You may have to multiply the subscripts in your empirical
formula by some factor. Please take notes as we work through a sample problem.
Sample Problem: Molecular Formula Calculation
Given: 38.7% C, 9.70% H, 51.6% O and a molecular formula mass of 62.0 find the true
molecular formula:
First, find the empirical formula: --> CH3O
Next find the formula mass: C
1 x 12.0 = 12.0
H 3 x 1.01 = 3.03
O 1 x 16.0 = 16.0
-------formula mass =
31.0
Now divide the molecular mass by this formula mass:
62.0/31.0 = 2
Multiply each subscript in your formula by this factor (2 here):
Your final molecular formula: C2H6O2
To check your answer find the formula mass of your final formula. It should be the same
as the molecular formula mass.
Practice Problem #1
Given the following data, find the correct molecular formula:
24.3% C, 4.1% H, 71.6% Cl molecular formula mass = 99.0
To see the correct answer and the method to solve the problem, please continue.
The correct formula would be: C2H4Cl2
The empirical formula would be CH2Cl with a formula mass of 49.5
Dividing 99.0 by 49.5 gives a factor of 2. Each subscript is doubled.
Again, check the formula mass of your final answer to confirm your answer.
Practice Problem #3
Given the following data, find the correct molecular formula:
54.6% C, 9.00% H, 36.4% O and a molecular mass of 176
The correct answer would be: C8H16O4
The empirical formula would be: C2H4O
with a formula mass of 44
The multiplication factor would be 4 (176/44.0)