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Section 7.3 - Percent Composition and Chemical Formulas
Calculating the Percent Composition of a compound
The relative amounts of each element ( Part ) in a compound ( Whole ) are expressed as the Percent Composition or the Percent by Mass of each element in the compound.
%
Part Whole x 100
The percent composition of a compound has as many percent values as there are different elements in the compound.
The percent composition of CaCO 3 is: Ca = 40.1 % C = 12.0 % O = 47.9 %
To calculate the percent composition of a compound, use the chemical formula to determine which elements and how many moles of each are in the compound.
Divide the total mass of one of the elements in the compound by the molar mass of the compound then multiply by 100 to convert to a percent.
Remember that the total mass of the element is the molar mass of the element multiplied by its subscript in the formula.
Example: What is the percent composition of FeSO 4 ?
55.8
151.9
g Fe = g = 36.7 % S = O = 32.1
g 151.9
g x 100 64.0
g 151.9
g x 100 = 21.1 % = 42.2 %
The Percent By Mass in a compound is the number of grams of the element divided by the number of grams of the compound multiplied by 100 .
Sample Problem 7 - 10 p 189 An 8.20 g piece of magnesium combines with 5.40 g of oxygen to form a compound. What is the Percent Composition (Percent By Mass) of this compound?
mass % Mg = mass of 8.20
% Mg = 13.60
g g of Mg compound x 100
x 100 60.3
% % O = mass mass of of O compound x 100 5.40
% O = 13.6
g g x 100
39.7
%
Using Percent Composition as a Conversion Factor
You can use the percent composition to determine the mass of each element in a sample of a compound.
Example: Aluminum makes up 52.9 % of Al 2 O 3 . Calculate the mass of Aluminum in a 50.0 g sample of Al 2 O 3 .
We first assume that we have a 100.0g sample of Al 2 O 3 . 52.9 % of 100.0 g is 52.9 g. We can use 52.9 g divided by 100.0 g sample as a conversion factor.
50.0
g 1 52.9
g x 100.0
g
26.5
g Al
Calculating Empirical Formula
Empirical Formula
EMPIRICAL FORMULA An empirical formula is the smallest number whole ratio of atoms in a formula.
EMPIRICAL FORMULA If the atoms also be in are in a small whole number ratio, then the moles in the substance will a small whole number ratio .
of atoms of each element
Two Sources of Data From Which to Calculate the Empirical Formula Experimental Analysis
-
What is the empirical formula of a compound if a 2.50 gram sample contains 0.900 g of calcium and 1.60 g of chlorine?
Two Sources of Data From Which to Calculate the Empirical Formula Percentage Composition
-
A compound has a percentage composition of 40.0% C, 6.71% H, and 53.3% O. What is the empirical Formula?
The following is an Example of the Empirical Formula Calculation from Experimental Analysis .
What is the empirical formula for a compound if a 2.50 g sample contains 0.900 g of calcium and 1.60 g of chlorine?
Determine the number of moles of each element in the compound.
0.900
1 g Ca x 1 mole 40.1
g Ca Ca
0. 0224 moles Ca 1 .
60 g 1 Cl x 1 mole 35 .
5 g Cl Cl
0.0450 moles Cl
The next step is to determine the smallest whole number ratio.
To obtain the smallest whole number ratio, divide both numbers of moles by the smaller number of moles.
0 .
0 0224 .
moles 0224 moles Ca
1 0 .
0 0450 .
moles 0224 Cl moles
2
The mole ratio is 1 mole of Calcium to 2 moles of Chlorine.
The Empirical formula is therefore CaCl 2 .
Example of Empirical Formula Calculation From Percentage Composition: A compound has a percentage composition of 40.0% C , 6.71% H , and 53.3% O . What is the empirical formula?
Example of Empirical Formula Calculation From Percentage Composition To calculate the ratio of moles of these elements, we assume a 100 g sample . 40.0% of 100 g is 40.0 g.
Example of Empirical Formula Calculation From Percentage Composition
Next we calculate the number of moles of each element present .
Example of Empirical Formula Calculation From Percentage Composition Mol C Mol H Mol O 40.0
g C 1 6.71gH
1 53.3gO
1 x x x 1 mol C 12.0
g C 1molH 1.01gH
1molO 16.0gO
3.33
6.67
3.33
4
Determine the smallest Whole Number Ratio.
The number of moles of each element calculated in step 2 were: 3.33 mol C, 6.64 mol H, and 3.33 mol O. Dividing by the smallest value we obtain the smallest whole number ratio of moles.
For C For H For O 3.33
3.33
mol mol
1 6.64
mol 3.33
mol
2 3.33
mol 3.33
mol
1
With a ratio of 1-C to 2-H to 1-O The empirical formula is therefore -- CH 2 O .
What is the empirical formula for a compound if a 40.0 g sample contains 29.7 g of sodium and 10.3 g of oxygen?
Solution: Step 1 Determine the number of moles of each element.
29.7
1 g Na x 1 mole 23.0
g Na Na
1.29
moles Na 10.3
1 g O x 1 mole 16.0
g O O
.644
moles O
Step 2 Divide each number of moles by the smallest. This will give the smallest whole number ratio.
1.29
0.644
2
0.644
0.644
1
This ratio shows that there are 2 moles of sodium for one mole of oxygen.
The empirical formula is therefore Na 2 O .
Calculating Molecular Formula
The Empirical Formula tells us the ratio of each kind of atom present in a compound.
The Molecular Formula tells us the actual number of each kind of atom in a molecule of a compound.
If the empirical formula and the molar mass are known, the molecular formula can be calculated.
Example: A certain gas, having a molar mass of 78 g, has an empirical formula of CH. What is the molecular formula?
If we multiply the mass of the empirical formula by an integer, the product must equal the molar mass.
(CH) x = 78 g Where x represents an integer .
(13) x = 78 g X = 6
By multiplying each subscript in the empirical formula by the integer (6), we obtain the molecular formula.
(CH) 6 gives C 6 H 6
Problem: The empirical formula of a compound is CHOCl and the molar mass is 129 g. What is the molecular formula?