http://www.ck12.org Chapter 10. The Mole
% Zn=16 .07 g Zn
20 .00 g
×100%= 80 .35% Zn% O=
3 .93 g O
20 .00 g×100%= 19 .65% O
The calculations make sense because the sum of the two percentages adds up to 100%. By mass, the compound is
mostly zinc.
Determining Masses from Percent Composition Data
We can also perform the reverse calculation, determining the mass of an element in a given sample, if we know the
total mass of the sample and its percent composition.
Example 10.14
You have a 10.0 g sample of a metal alloy that contains only aluminum and zinc. If the sample is 36% aluminum by
mass, what masses of Al and Zn are present?
Answer:
We are told that the sample is 36% aluminum by mass. Because the only other component is zinc, it must make up
the remaining 64% of the mass. We can multiply each of these percentages by 10.0 grams to find the masses of each
element.
10.0 g sample×0.36 = 3.6 g Al
10.0 g sample×0.64 = 6.4 g Zn
Empirical Formulas
Recall that anempirical formulais one that shows the lowest whole-number ratio of the elements in a compound.
Because the structure of ionic compounds is an extended three-dimensional network of positive and negative ions,
only empirical formulas are used to describe ionic compounds. However, we can also consider the empirical formula
of a molecular compound. Ethene is a small hydrocarbon compound with the formula C 2 H 4 (Figure10.5). While
C 2 H 4 is its molecular formula and represents its true molecular structure, it has an empirical formula of CH 2. The
simplest ratio of carbon to hydrogen in ethene is 1:2. In each molecule of ethene, there is 1 carbon atom for every
2 atoms of hydrogen. Similarly, we can also say that in one mole of ethene, there is 1 mole of carbon for every 2
moles of hydrogen. The subscripts in a formula represent the molar ratio of the elements in that compound.
FIGURE 10.5
Ball-and-stick model of ethene, C 2 H 4.