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```NS F1 (pg ! 1 of 1! )
Mass % Composition for Compounds
(aka Elemental Analysis)
Name____________________
Elemental Analysis
The chemical formula of a compound contains essential information about its composition. To determine a chemical formula, a
chemist would like to count the atoms of each element in one molecule. But of course, atoms are too small to count, so we have to
count the atoms by the number of mole of each element present in a mole of the compound. Unfortunately there is no direct
experimental method for measuring moles. Instead, laboratory experiments can give the masses of the various elements contained
in the total mass of the compound. This common practice is called elemental analysis or mass percent composition or more
simply percent composition.
Theoretical % Composition Using Molar Masses
Using water as an example, H2O the mass percentages of this compound from the known formula is illustrated below.
H
2 ( 1.0 g/mole) + 16.0 g/mole = 18 g/mole total
H !
2.0
× 100 = 11%
18
!
(these numbers are the molar masses from the periodic
chart.)
16
× 100 = 89%
18
Of course it is more likely that a chemist does not know the chemical formula, but would like to determine it. A compound can be
analyzed for the exact % of each element present from which empirical formulas can be determined.
Experimental % Composition Using Lab Data
You may recall Lab C6− Law of Constant Composition in Unit C, in which we burned magnesium in the crucible to produce
Mg
magnesium oxide. In that Lab we used the results to calculate a mass ratio of !
. Here we will use the same data to calculate the
O
mass percent composition of each element.
The following data represents typical results from this Lab.
mass of empty crucible (g)
12.47
mass of magnesium (g)
0.34
mass of crucible and magnesium oxide (g)
13.03
•
Subtraction will allow calculation of magnesium oxide:
•
A second subtraction will allow calculation of the oxygen in the product:
•
This will allow % calculations of both the magnesium and the oxygen:
Mg
!
0.34g
× 100 = 60.7%
0.56g
O
!
0.22g
× 100 = 39.3%
0.56g
13.03 – 12.47 g = 0.56 g of magnesium oxide
0.56 g – 0.34 g = 0.22 g of oxygen
alternatively, you could calculate this by subtraction.
Another example with a twist
If you were given 2.85 g of magnesium oxide, what mass of magnesium would be in this sample?
From the problem above, we know that MgO is 60.7 % Mg
Convert the 60.7% to a decimal and then multiply by the mass of the sample: ! 0.607 × 2.85 = 1.73g
```