Lesson 10: Formulas & Composition
From the name of a compound you can determine its formula and from that you can determine the formula weight of the compound (the molecular weight if it is a molecular material) and its composition in weight percent. The pages listed here give you information and examples of how to do this. Determining the formula of a compound from its name is crucial to each of the tasks remaining in this lesson.
Objective 8. From the name of a compound, determine the formula and formula weight.
To determine the formula of a compound from its name, first consider what type of name it is. (More information on this is available near the beginning of this lesson in "Types of Names" section.)
If it is a prefix name, use the prefixes to determine how many atoms of each element are contained in the formula.
If it is a Stock name, several steps are needed. Use the Roman numeral to determine the charge (or oxidation state) of the first element. From the second part of the name, determine whether it is a simple anion or a polyatomic ion. If it is a simple anion, determine its charge (or oxidation state) from its position on the periodic table. If it is a polyatomic ion, determine its charge from memory. Use the charges on the ions to determine the formula of the compound.
If it is a Latin name, several steps are needed. Use the -ous or -ic suffix to determine whether the charge (or oxidation state) of the first element is the lower or higher of its common charges. You might remember the common charges of the element, or you may have to look them up. From the second part of the name, determine whether it is a simple anion or a polyatomic ion. If it is a simple anion, determine its charge (or oxidation state) from its position on the periodic table. If it is a polyatomic ion, determine its charge from memory. Use the charges on the ions to determine the formula of the compound.
If it is a simple name, first determine whether the compound is ionic or covalent. If it is ionic, determine the charges on the ions from their postitions on the periodic table. Then use the charges on the ions to determine the formula of the compound. If the compound is covalent, use the periodic table to determine how many electrons each element needs to gain. Reverse (and simplify) the "need ratio" to get the "atom ratio" and the formula of the compound.
If it is a common name you will need to determine the formula from memory (or by looking up the compound).
Once you have the formula, the formula weight is determined by adding up the atomic weights of each element, using each one as many times as is indicated by the subscripts in the formula.
Exercises
Determine the formula and formula weight for each of the following compounds:
a. ammonia
b. magnesium chloride
c. chlorine(V) oxide
d. cuprous oxide
e. carbon tetrabromide
f. sodium sulfate
h. barium nitride
Worked-Out Examples (a,b)
(a) Ammonia, you should have learned, is NH3. Even if all you remember is that ammonia contains nitrogen and hydrogen, you should be able to figure out the formula. Because they are both nonmetals you know that each wants to gain electrons. Nitrogen wants to gain three electrons; hydrogen wants to gain one electron; therefore, it will take three hydrogens to satisfy the needs of the one nitrogen atom. So the formula will be NH3.
Knowing that the formula is NH3, you can come up with the formula weight by adding up the atomic weights. There are different degrees of precision to which you can do this. For simplicity's sake in these examples, we're using one decimal place in the atomic weights. Nitrogen has an atomic weight of 14.0; hydrogen has an atomic weight of 1.0; so three of them would be 3.0. That gives the sum of 17.0. So the formula weight of ammonia is 17.0, in either grams per mole or amu per molecule.
(b) Magnesium chloride is an ionic compound because it has a metal, magnesium, and a nonmetal, chlorine. Magnesium will lose two electrons and form a +2 charge. Chlorine will gain one electron to form a chloride ion with a -1 charge. The formula for the compound is MgCl2.
To get the formula weight, find the atomic weights and add them together taking the subscripts into account. Magnesium is 24.3; chlorine is 35.5; so two would be 71.0. The total gives 95.3 as the formula weight.
Answers to Exercises
Determine the formula and formula weight for each of the following compounds: (in either grams per mole or amu per molecule)
a. ammonia is NH3, and its formula weight is 17.0
b. magnesium chloride is MgCl2 and its formula weight is 95.3
c. chlorine(V) oxide is Cl2O5 and its formula weight is 151.0
d. cuprous oxide is Cu2O, and its formula weight is 143.0
e. carbon tetrabromide is CBr4, and its formula weight is 331.6
f. sodium sulfate is Na2SO4, and its formula weight is 142.1
h. barium nitride is Ba3N2, and its formula weight is 439.9
Objective 9. From the name of a compound, determine the composition (weight percent).
To determine the composition of a compound you need to expand on what you just did in the previous exercise. That involves figuring out the formula, then the formula weight and then the percentage of each one.
Exercises
What is the composition (in weight percent of each element) of the following compounds?
a. calcium nitrate
b. sodium sulfide
c. carbon dioxide
d. iron(III) chloride
e. iron(III) sulfide
f. aluminum oxide
g. ammonium carbonate
Worked-Out Example (a)
Calcium nitrate consists of calcium ion (+2) and nitrate ion, which has a -1 charge. So the formula for calcium nitrate, is Ca(NO3)2. That formula contains one calcium, two nitrogens, and six oxygens. The atomic weight for calcium is 40.1; the atomic weight for nitrogen is 14.0, so times two gives 28.0; and the atomic weight for oxygen is 16.0 times six of them gives 96.0. The sum of all of those atomic weights is 164.1. So, one mole of calcium nitrate weighs 164.1 grams. Of that, 40.1 grams is calcium; 28.0 grams is nitrogen; and 96.0 grams is oxygen. |
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To calculate the percentages, take each of those parts and divide by the total. So the percentage for calcium is 40.1 grams of calcium over 164.1 grams total and that comes out to be 24.4%. For nitrogen, it's 28.0 grams of nitrogen divided by 164.1 grams total and that comes out to be 17.1%. For oxygen, we have 96.0 grams of oxygen divided by 164.1 grams total and that comes out to be 58.5%. As a double check on the mathematics, add up the percentages to make sure they come up to 100%. |
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Answers to Exercises
What is the composition (in weight percent of each element) of the following compounds?
a. calcium nitrate - Ca(NO3)2
24.4% Ca
17.1% N
58.5% Ob. sodium sulfide - Na2S
58.9% Na
41.1% Sc. carbon dioxide - CO2
27.3% C
72.7% Od. iron(III) chloride - FeCl3
34.4% Fe
65.6% Cle. iron(III) sulfide - Fe2S3
53.7% Fe
46.3% Sf. aluminum oxide - Al2O3
52.9% Al
47.1% Og. ammonium carbonate - (NH4)2CO3
29.1% N
8.4% H
12.5% C
49.9% O