How do you find the molecular formula from the empirical formula without molar mass?

The empirical formula for a chemical compound is an expression of the relative abundances of the elements that form it. It isn't the same as the molecular formula, which tells you the actual number of atoms of each element present in a molecule of the compound. Different compounds with very different properties may have the same empirical formula. You can derive the molecular formula of a compound from its empirical formula only if you know the molar mass of the compound.

TL;DR (Too Long; Didn't Read)

If you know the empirical formula of a compound, you know the elements present in the compound and their relative proportions. Calculate the molar mass based on the formula and divide this into the mass of the actual compound. The division gives you a whole number. Multiply the subscript of each element in the empirical formula by this number to get the molecular formula for the compound.

How to Find the Empirical Formula

Chemists can determine the elements in a compound and their relative percentages by a chemical reaction with a known compound that produces products that they can collect and weigh. After doing so, they divide the mass of each element by its molar mass to determine the number of moles present in a particular amount – usually 100 grams. The number of moles of each element produces the empirical formula, which is the simplest expression of the elements present in a single molecule of the compound and their relative proportions.

Determining the Molecular Formula

The first step in determining the molecular formula of a compound is to calculate the empirical mass from its empirical formula. To do this, look up the mass of each element present in the compound, and then multiply that number by the subscript that appears after its symbol in the formula. Sum the masses to determine the molar mass represented by the formula.

The next step is to weigh a sample, then divide the empirical mass into the actual mass of the compound. This division produces a whole number. Multiply the subscripts in the empirical formula by this number to determine the molecular formula.

Examples

1. Analysis of a compound reveals it contains 72 g carbon (C), 12 g hydrogen (H) and 96 g oxygen (O). What is its empirical formula?

    Start by dividing the mass of each element present in the compound by the molar mass of that element to find the number of moles. The periodic table tells you the molar mass of carbon is 12 grams (ignoring fractions), that of hydrogen is 1 gram and that of oxygen is 16 grams. The compound therefore contains 72/12 = 6 moles carbon, 12/1 = 12 moles hydrogen and 96/16 = 6 moles oxygen.

    There are 12 moles of hydrogen but only 6 moles of carbon and oxygen, so divide by 6.

    The ratios of carbon to hydrogen to oxygen are 1 : 2 : 1, so the empirical formula is CH2O, which happens to be the chemical formula for formaldehyde.

    2. Calculate the molecular formula for this compound, given that the sample weighs 180g.

    Compare the recorded mass to that of the molar mass expressed by the empirical formula. CH2O has one carbon atom (12g), two hydrogen atoms (2g) and one oxygen atom (16g). Its total mass is thus 30 grams. However, the sample weighs 180 grams, which is 180/30 = 6 times as much. You therefore have to multiply the subscript of each element in the formula by 6 to get C6H12O6, which is the molecular formula for the compound.

    This is the molecular formula for glucose, which has very different properties than formaldehyde, even though they have the same empirical formula. Don't mistake one for the other. Glucose tastes good in your coffee, but putting formaldehyde in your coffee is likely to give you a very unpleasant experience.

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  • Page ID182667
  • Learning Objectives
    • Understand the difference between empirical formulas and molecular formulas.
    • Determine molecular formula from percent composition and molar mass of a compound.

    Below, we see two carbohydrates: glucose and sucrose. Sucrose is almost exactly twice the size of glucose, although their empirical formulas are very similar. Some people can distinguish them on the basis of taste, but it's not a good idea to go around tasting chemicals. The best way to tell glucose and sucrose apart is to determine the molar masses—this approach allows you to easily tell which compound is which.

    How do you find the molecular formula from the empirical formula without molar mass?
    Figure \(\PageIndex{1}\): (A) the molecular structure of glucose and (B) the molecular structure of sucrose.

    Molecular Formulas

    Molecular formulas give the kind and number of atoms of each element present in the molecular compound. In many cases, the molecular formula is the same as the empirical formula. The chemical formula will always be some integer multiple (\(n\)) of the empirical formula (i.e. integer multiples of the subscripts of the empirical formula).

    \[\text{ Molecular Formula} = n (\text{Empirical formula}) \nonumber \]

    therefore

    \[ n = \dfrac{\text{Molecular Formula}}{\text{Empirical Formula}} \nonumber \]

    The integer multiple, n, can also be obtained by dividing the molar mass, \(MM\), of the compound by the empirical formula mass, \(EFM\) (the molar mass represented by the empirical formula).

    \[ n = \dfrac{MM ( molar mass)}{EFM (empirical formula molar mass)} \nonumber \]

    Table \(\PageIndex{1}\) shows the comparison between the empirical and molecular formula of methane, acetic acid, and glucose, and the different values of n. The molecular formula of methane is \(\ce{CH_4}\) and because it contains only one carbon atom, that is also its empirical formula. Sometimes, however, the molecular formula is a simple whole number multiple of the empirical formula. Acetic acid is an organic acid that is the main component of vinegar. Its molecular formula is \(\ce{C_2H_4O_2}\). Glucose is a simple sugar that cells use as a primary source of energy. Its molecular formula is \(\ce{C_6H_{12}O_6}\). The structures of both molecules are shown in Figure \(\PageIndex{2}\). They are very different compounds, yet both have the same empirical formula of \(\ce{CH_2O}\).

    Table \(\PageIndex{1}\): Molecular Formula and Empirical Formula of Various Compounds.
    Name of CompoundMolecular FormulaEmpirical Formulan
    Methane \(\ce{CH_4}\) \(\ce{CH_4}\) 1
    Acetic acid \(\ce{C_2H_4O_2}\) \(\ce{CH_2O}\) 2
    Glucose \(\ce{C_6H_{12}O_6}\) \(\ce{CH_2O}\) 6
    How do you find the molecular formula from the empirical formula without molar mass?
    Figure \(\PageIndex{2}\): Acetic acid (left) has a molecular formula of \(\ce{C_2H_4O_2}\), while glucose (right) has a molecular formula of \(\ce{C_6H_{12}O_6}\). Both have the empirical formula \(\ce{CH_2O}\).

    Empirical formulas can be determined from the percent composition of a compound as discussed in section 6.8. In order to determine its molecular formula, it is necessary to know the molar mass of the compound. Chemists use an instrument called a mass spectrometer to determine the molar mass of compounds. In order to go from the empirical formula to the molecular formula, follow these steps:

    1. Calculate the empirical formula molar mass (EFM).
    2. Divide the molar mass of the compound by the empirical formula molar mass. The result should be a whole number or very close to a whole number.
    3. Multiply all the subscripts in the empirical formula by the whole number found in step 2. The result is the molecular formula.

    Example \(\PageIndex{1}\)

    The empirical formula of a compound of boron and hydrogen is \(\ce{BH_3}\). Its molar mass is \(27.7 \: \text{g/mol}\). Determine the molecular formula of the compound.

    Solution Solutions to Example 6.9.1

    Steps for Problem SolvingDetermine the molecular formula of \(\ce{BH_3}\).
    Identify the "given" information and what the problem is asking you to "find."

    Given:

    Empirical formula \(= \ce{BH_3}\)

    Molar mass \(= 27.7 \: \text{g/mol}\)

    Find: Molecular formula \(= ?\)

    Calculate the empirical formula mass (EFM). \[\text{Empirical formula molar mass (EFM)} = 13.84 \: \text{g/mol} \nonumber \]
    Divide the molar mass of the compound by the empirical formula mass. The result should be a whole number or very close to a whole number.

    \[\dfrac{\text{molar mass}}{\text{EFM}} = \dfrac{27.7 g/mol}{13.84 g/mol} = 2 \nonumber \]

    Multiply all the subscripts in the empirical formula by the whole number found in step 2. The result is the molecular formula. \[\ce{BH_3} \times 2 = \ce{B_2H_6} \nonumber \]
    Write the molecular formula. The molecular formula of the compound is \(\ce{B_2H_6}\).
    Think about your result. The molar mass of the molecular formula matches the molar mass of the compound.

    Exercise \(\PageIndex{1}\)

    Vitamin C (ascorbic acid) contains 40.92 % C, 4.58 % H, and 54.50 % O, by mass. The experimentally determined molecular mass is 176 amu. What are the empirical and chemical formulas for ascorbic acid?

    Answer Empirical FormulaC3H4O3Answer Molecular FormulaC6H8O6

    Summary

    • A procedure is described that allows the calculation of the exact molecular formula for a compound.

    How to find molecular formula when only empirical formula is given?

    STEP 1: Calculate the molar mass of the empirical formula. STEP 2: Divide the given molecular molar mass by the molar mass calculated for the empirical formula. STEP 3: Multiply each subscript by the whole number that resulted from step 2. This is now the molecular formula.

    Is it possible to find the molecular formula from the empirical formula?

    Divide the molar mass of the compound by the empirical formula mass. The result should be a whole number or very close to a whole number. Multiply all the subscripts in the empirical formula by the whole number found in step 2. The result is the molecular formula.

    How do you find the empirical mass if you only have the empirical formula?

    Determine the empirical formula mass by multiplying each element's subscript by its atomic weight on the periodic table and adding them together.