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Chemical Formulas

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    Learning Objectives
    • Explain the meaning of various chemical formulas (empirical, molecular, and structural formulas).
    • Calculate weight and mole percentages from a formula.
    • Derive chemical formulas from a given set of information.
    Key Concepts
    • Chemical formulas: empirical formula, molecular formula, and structural formula
    • Formula weight, molecular weight and molar mass
    • Weight percentage and mole percentage
    • Elemental analysis

    Chemical formulas such as \(\ce{HClO4}\) can be divided into empirical formula, molecular formula, and structural formula. Chemical symbols of elements in the chemical formula represent the elements present, and subscript numbers represent mole proportions of the proceeding elements. Note that no subscript number means a subscript of 1.

    From a chemical point of view, an element contained in the substance is a fundamental question, and we represent the elemental composition by a chemical formula, such as \(\ce{H2O}\) for water. This formula implies that the water molecules consist of 2 hydrogen and 1 oxygen atoms. The formula \(\ce{H2O}\) is also the molecular formula of water. For non-molecular substances such as table salt, we represent the composition with an empirical formula. Sodium chloride is represented by \(\ce{NaCl}\), meaning that sodium and chlorine ratio in sodium chloride is 1 to 1. Again, the subscript 1 is omitted. Since table salt is an ionic compound, the formula implies that numbers of \(\ce{Na+}\) ions and \(\ce{Cl-}\) ions are the same in the solid. The subscript numbers in an empirical formula should have no common divisor.

      H H
      | |
      | |
      H H
    Structural of
    A 3-Dimensional
    structure of \(\ce{C6H12}\)

    A structural formula reflects the bonding of atoms in a molecule or ion. For example, ethanol can be represented by \(\ce{CH3CH2OH}\). This is a simple way of representing a more elaborated structure shown on your left. Molecular structures are often beautiful, but the representation is an artwork. For example, a 3-dimensional structure of cyclohexane is shown on the right. This is a chair form, and another structure has a boat form. You will learn more about it in organic chemistry. The molecular formula of benzene is \(\ce{C6H6}\), and its empirical formula is \(\ce{CH}\).

    You may refer to a substance by its name, and recognize it by its properties. Properties are related to the structure and the composition of the molecules. Knowing the chemical formula is a giant step towards understanding a substance.

    Formula Weights, Molecular Weights and Molar Masses

    The formula weight is the sum of all the atomic weights in a formula. The evaluation of formula weight is illustrated in this example.

    Example 1

    What is the formula weight of sufuric acid \(\ce{H2SO4}\)?


    The formula also indicates a mass as the sum of masses calculated this way

    \(\mathrm{2\times1.008 + 32.0 + 4\times16.0 = 98.0}\)

    where 1.008, 32.0 and 16.0 are the atomic weights of \(\ce{H}\), \(\ce{S}\), and \(\ce{O}\) respectively.


    If the formula is a molecular formula, the mass associated with it is called molecular mass or molecular weight. As an exercise, work out the following problem.

    What is the molecular weight of caffeine, \(\ce{C8H10N4O2}\)?

    The diagram shown here is a model of the caffeine molecule.

    With the aid of a table of atomic weights, a formula indirectly represents the formula weight. If the formula is a molecular formula, it indirectly represents the molecular weight. For simplicity, we may call these weights molar masses, which can be formula weights or molecular weights.

    A chemical formula not only represents what a substance is made of, it provides a great deal of information about the substance. Do you know that chemical formulas are used all over the world, regardless of the language? Chinese, Russian, Japanese, African, and South Americans use the same notations we do. Thus, \(\ce{H2S}\) is recognized as a smelly gas all over the world. Chemical formula is an international or universal language.

    Weight Percentage and Mole Percentage

    A chemical formula not only gives the formula weight, it accurately represents the percentages of elements in a compound. On the other hand, if you know the percentage of a compound, you may figure out its formula. Percentage based on weights is called weight percentage, and percentage based on the numbers of atoms or moles is called mole percentage.

    Example 2

    What are the weight and mole percentages of \(\ce{S}\) in sufuric acid?


    From example 1, we know that there are 32.0 g of \(\ce{S}\) in 98.0 g of sulfuric acid. Thus the weight percentage is

    \(\mathrm{Weight\: percentage = \dfrac{32}{98} = 32.7\%}\)

    From the formula, there is one \(\ce{S}\) atom among 7 atoms in \(\ce{H2SO4}\)

    \(\mathrm{Mole\: percentage = \dfrac{1}{7} = 14.3\%}\)


    You have learned what weight and mole percentages are and how to evaluate them in this example. As an exercise, work out the following problem:

    What are the weight and mole percentages of \(\ce{C}\), \(\ce{H}\), \(\ce{N}\), and \(\ce{O}\) for caffeine, \(\ce{C8H10N4O2}\)?

    Determination of Chemical Formulas

    How would you find the chemical formula of a substance? If you know the substance, its formula and other information is usually listed in a handbook. Handbooks such as the CRC Handbook of Chemistry and Physics contain information on millions of substances.

    If you are a researcher and you made a new compound that no one has ever made before, then you need to determine its empirical or molecular formula. For an organic compound, you burn it completely to convert all carbon (\(\ce{C}\)) to \(\ce{CO2}\), and all hydrogen (\(\ce{H}\)) to \(\ce{H2O}\).

    \(\mathrm{C_xH_{2y} \xrightarrow{(burned\: in\: O_2)} x\, CO_2 + y\, H_2O}\)

    Thus, from the weight of \(\ce{CO2}\) and \(\ce{H2O}\) produced by burning a definite amount of the substance, you can figure out the percent of \(\ce{C}\) and \(\ce{H}\) in the compound.

    Nitrogen is determined by converting it to \(\ce{NH3}\). The amount of \(\ce{NH3}\) can be determined by titration, and the percentage can also be determined.

    Percentage of \(\ce{O}\) is usually obtained by subtracting all percentages of \(\ce{C}\), \(\ce{H}\), and \(\ce{N}\), if the compound does not contain any other element.

    Example 3

    A compound contains 92.3 weight percent of carbon and 7.7 weight percent of \(\ce{H}\). What is the empirical formula?


    Assume that you have 100 g of the compound, then you have 92.3 g of carbon and 7.7 g of hydrogen. Thus the mole ratio of \(\ce{C}\) to \(\ce{H}\) should be

    \(\mathrm{\dfrac{92.3}{12}:\dfrac{7.7}{1.008} = 7.7 : 7.7 = 1 : 1}\)

    Thus, the empirical formula is \(\ce{CH}\).

    You have learned how to determine a chemical formula if the percentages of various elements present in the compound are known in this example. To test your skill, you may be asked to work out the empirical formula of any compound. Try this problem:

    Exercise \(\PageIndex{1}\)

    Aspartic acid contains 36.09% \(\ce{C}\), 5.30% \(\ce{H}\), 10.52% \(\ce{N}\), and 48.08 \(\ce{O}\) by weight. What is the empirical formula for aspartic acid?
    Aspartic acid is one of the non-essential amino acids, usually present in young plants. It is obtained by hydrolysis of asparagine, which is abundant in asparagus

    Example 4

    A compound with an empirical formula of \(\ce{CH}\) has a molecular weight of 78 g/mol. What is the molecular formula?


    The formula weight of \(\ce{CH}\) is 13.0.

    Since 78/13 = 6, the molecular formula is \(\ce{C6H6}\), the formula for benzene.


    This example illustrates the difference between empirical and molecular formula, for which the molecular weight must be known.

    Example 5

    When 1.00 g of benzene is burned, how much \(\ce{CO2}\) and \(\ce{H2O}\) should be produced?


    \(\mathrm{1\:g\:CH \times \dfrac{1\:mol\:C}{13\:g\:CH}\times\dfrac{1\:mol\:CO_2}{1\:mol\:C}\times\dfrac{44.0\:g\:CO_2}{1\:mol\:CO_2} = 3.38\:g\:CO_2}\)

    Use the same method to calculate the amount of \(\ce{H2O}\) produced (Ans. 0.692 g).

    Example 6

    When 1.00 g of a compound containing only carbon and hydrogen is burned completely, 3.14 g of \(\ce{CO2}\) and 1.29 g of \(\ce{H2O}\) is produced. What is the empirical formula?


    Amounts of carbon and \(\ce{H}\) in 3.14 g of carbon dioxide,

    \(\mathrm{3.14\: g\: CO_2\times \dfrac{12\: g\: C}{44\: g\: CO_2}\times\dfrac{1\: mol\: C}{12\: g\: C}= 0.0714\: mol\: C}\)

    \(\mathrm{1.29\: g\: H_2O\times \dfrac{2\: g\: H}{18\: g\: H_2O}\times\dfrac{1\: mol\: H}{1\: g\: H}= 0.143\: mol\: H}\)

    Thus, mole ratio of \(\ce{C}\) : \(\ce{H}\) is 0.0714 : 0.143 = 1 : 2. Therefore, the empirical formula is \(\ce{CH2}\)


    The molecular formula for ethylene is \(\ce{C2H4}\) and cyclohexane is \(\ce{C6H12}\). What are their molecular weights?

    Example 7

    Chloroform is a common solvent used in chemical labs. It has a molecular formula of \(\ce{CHCl3}\). What is the weight percentage of chlorine (\(\ce{Cl}\))? (Atomic weight, \(\ce{Cl}\), 35.453; \(\ce{H}\), 1.00794; \(\ce{C}\), 12.0110)

    You should understand the reason for using this formula to calculate it:

    \(\mathrm{\dfrac{3\times 35.453}{3\times 35.453 + 12.011 + 1.00794}= 89.094\%\: (weight\: percentage)}\)


    What is the weight percentage of \(\ce{C}\) in \(\ce{CCl4}\)?

    These examples illustrate some fundamental types of problems related to the understanding of the chemical formula. Their solutions are related to the skills outlined at the start of this page.

    Skill Developing Problems

    1. What is the molecular weight of sugar \(\ce{C12H22O11}\)?


    Skill - Know where to find atomic weights and evaluate molecular weights for:
    caffeine, \(\ce{C8H10N4O2}\) (194.2);
    calcium aspirin, \(\ce{Ca(OOCC6H4OCOCH3)2}\) (398.4);
    ascorbic acid (vitamin C), \(\ce{C6H8O6}\);
    quartz (silicon), \(\ce{SiO2}\);
    ruby, nearly all \(\ce{Al2O3}\).

    1. What are the weight and mole percentages of \(\ce{C}\) in sugar \(\ce{C12H22O11}\)?

    12*12/(12*12+22+11*16) and (12/45)

    Skill - Evaluate weight and mole percentages of any element in any substance.

    1. The oil of cinnamon contains almost 90% of cinnamaldehyde. By weight, this compound contains 81.79% \(\ce{C}\), 6.10% \(\ce{H}\) and 12.11% \(\ce{O}\). What is the empirical formula for cinnamaldehyde? The empirical formula is also the molecular formula in this case.


    Skill - Find a chemical formula when weight percentages are known.

    1. If 1.80 kg glucose \(\ce{C6H12O6}\) is burned completely, how many kg of \(\ce{CO2}\)is produced?

    2.64 kg

    Skill - Calculate the amount of water produced in this case.

    1. Calculate the amount (g) of \(\ce{H}\) present in 1.0 L of water. Assume density of water to be 1 g/mL. (\(\ce{H}\), 1; \(\ce{O}\), 16).

    111 g

    Skill - Calculate the amount (kg) of \(\ce{C}\) in a block of dry ice weighing 88 kg.

    1. A sample of a compound containing only \(\ce{C}\) and \(\ce{H}\) gave 0.090 g of \(\ce{H2O}\) and 0.440 g of \(\ce{CO2}\). (\(\ce{H}\), 1; \(\ce{C}\), 12; \(\ce{O}\), 16). Find the empirical formula.


    Skill - Describe elemental analysis for the determination of chemical formula.

    1. How many moles is 0.130 g of \(\ce{C6H6}\)?

    0.0017 mol

    Skill - Determination of chemical formula from elemental analysis.

    1. How many hydrogen atoms are present in a water molecule?

    Two \(\ce{H}\) atoms in an \(\ce{H2O}\) molecule.

    Skill: Explain the meaning of chemical formula. How many hydrogen atoms are there in an ethanol molecule, \(\ce{C2H5OH}\)?

    1. In an experiment, 12.0 g of carbon gives 44.0 g of carbon dioxide. What is the weight percentage of carbon in \(\ce{CO2}\)?


    Skill: The weight percentage and mole percentage can be calculated if you know the chemical formula.
    If you reduce 44.0 grams of \(\ce{CO2}\) to carbon, how many grams of \(\ce{C}\) will you get?

    1. The combustion of 1.00 g of carbon gives 2.33 g of carbon oxide. The atomic weight of \(\ce{C}\) is 12.011 g/mole and that of \(\ce{O}\) is 15.9994 g/mole. What is the mole percent of \(\ce{C}\) in the carbon oxide?

    50% by mole

    Skill: Differentiate mole percentage and weight percentage.

    1. The empirical formula for benzene is \(\ce{CH}\). Its molecular weight was determined to be 78. What is the molecular formula?


      Skill: Differentiate empirical and molecular formulas.

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