How to Find the Molecular Formula (with Pictures)

2024 Author: Samantha Chapman | [email protected]. Last modified: 2024-01-15 13:47

If you need to find the molecular formula of a mysterious compound within an experiment, you can do the calculations based on the data you get from that experiment and some key information available. Read on to learn how to proceed.

Steps

Part 1 of 3: Finding the Empirical Formula from Experimental Data

Step 1. Review the data

Looking at the data from the experiment, look for the percentages of mass, pressure, volume, and temperature.

Example: A compound contains 75.46% carbon, 8.43% oxygen and 16.11% hydrogen by mass. At 45.0 ° C (318.15 K) and at 0.984 atm of pressure, 14.42 g of this compound has a volume of 1 L. What is the molecular compound of this formula?

Step 2. Change the percentage masses to masses

Look at the mass percentage as mass of each element in a 100g sample of the compound. Instead of writing the values as percentages, write them as masses in grams.

Example: 75, 46 g of C, 8, 43 g of O, 16, 11 g of H

Step 3. Convert masses to moles

You have to convert the molecular masses of each element into moles. To do this, you need to divide the molecular masses by the atomic masses of each respective element.

• Look up the atomic masses of each element in the periodic table of elements. They are usually located in the lower part of the square of each element.

• Example:

  • 75.46 g C * (1 mol / 12.0107 g) = 6.28 mol of C
  • 8.43 g O * (1 mol / 15.9994 g) = 0.33 mol of O
  • 16.11 g H * (1 mol / 1.00794) = 15.98 mol of H.

  

  Step 4. Divide the moles by the smallest molar amount of each element

  You must divide the number of moles for each separate element by the smallest molar amount of all elements in the compound. Thus, the simplest molar ratios can be found.

  • Example: the smallest molar quantity is oxygen with 0.53 mol.

    • 6.28 mol / 0.33 mol = 11.83
    • 0.33 mol / 0.33 mol = 1
    • 15.98 mol / 0.33 mol = 30.15

    

    Step 5. Round off the molar ratios

    These numbers will become the subscripts of the empirical formula, so you should round to the nearest whole number. Once you have found these numbers, you can write the empirical formula.

    • Example: the empirical formula would be C.₁₂OH₃₀
      • 11, 83 = 12
      • 1 = 1
      • 30, 15 = 30

      Part 2 of 3: Finding the Molecular Formulas

      

      Step 1. Calculate the number of moles of the gas

      You can determine the number of moles based on the pressure, volume and temperature provided by the experimental data. The number of moles can be calculated using the following formula: n = PV / RT

      • In this formula, it is the number of moles, P. is the pressure, V. is the volume, T. is the temperature in Kelvin and R. is the gas constant.
      • This formula is based on a concept known as the ideal gas law.
      • Example: n = PV / RT = (0, 984 atm * 1 L) / (0, 08206 L atm mol^-1 K.^-1 * 318.15 K) = 0.0377 mol

      

      Step 2. Calculate the molecular weight of the gas

      This can be done by dividing the grams of gas present by the moles of gas in the compound.

      Example: 14.42 g / 0.0377 mol = 382.49 g / mol

      

      Step 3. Add the atomic weights

      Add all the separate weights of the atoms to find the overall weight of the empirical formula.

      Example: (12, 0107 g * 12) + (15, 9994 g * 1) + (1, 00794 g * 30) = 144, 1284 + 15, 9994 + 30, 2382 = 190, 366 g

      

      Step 4. Divide the molecular weight by the empirical formula weight

      In doing so, you can determine how many times the empirical weight is repeated within the compound used in the experiment. This is important, so that you know how many times the empirical formula repeats itself in the molecular formula.

      Example: 382, 49/190, 366 = 2, 009

      

      Step 5. Write the final molecular formula

      Multiply the subscripts of the empirical formula by the number of times the empirical weight is in the molecular weight. This will give you the final molecular formula.

      Example: C.₁₂OH₃₀ * 2 = C₂₄OR₂H.₆₀

      Part 3 of 3: Further Example Problem

      

      Step 1. Review the data

      Find the molecular formula of a compound containing 57.14% nitrogen, 2.16% hydrogen, 12.52% carbon and 28.18% oxygen. At 82.5 C (355.65 K) and pressure of 0.722 atm, 10.91 g of this compound has a volume of 2 L.

      

      Step 2. Change the mass percentages to masses

      This gives you 57.24g of N, 2.16g of H, 12.52g of C and 28.18g of O.

      

      Step 3. Convert the masses to moles

      You must multiply the grams of nitrogen, carbon, oxygen and hydrogen by their respective atomic masses per mole of each element. In other words, you divide the masses of each element in the experiment by the atomic weight of each element.

      • 57.25 g N * (1 mol / 14.00674 g) = 4.09 mol N
      • 2.16 g H * (1 mol / 1.00794 g) = 2.14 mol H.
      • 12.52 g C * (1 mol / 12.0107 g) = 1.04 mol C.
      • 28.18 g O * (1 mol / 15.9994 g) = 1.76 mol O

      

      Step 4. For each element divide the moles by the smallest molar amount

      The smallest molar amount in this example is carbon with 1.04 moles. The amount of moles of each element in the compound must, therefore, be divided by 1.04.

      • 4, 09 / 1, 04 = 3, 93
      • 2, 14 / 1, 04 = 2, 06
      • 1, 04 / 1, 04 = 1, 0
      • 1, 74 / 1, 04 = 1, 67

      

      Step 5. Round off the molar ratios

      To write the empirical formula for this compound, you need to round the molar ratios to the nearest whole number. Enter these integers in the formula next to their respective elements.

      • 3, 93 = 4
      • 2, 06 = 2
      • 1, 0 = 1
      • 1, 67 = 2
      • The resulting empirical formula is N₄H.₂CO₂

      

      Step 6. Calculate the number of moles of the gas

      Following the ideal gas law, n = PV / RT, multiply the pressure (0.722 atm) by the volume (2 L). Divide this product by the product of the ideal gas constant (0.08206 L atm mol^-1 K.^-1) and the temperature in Kelvin (355, 65 K).

      (0, 722 atm * 2 L) / (0, 08206 L atm mol^-1 K.^-1 * 355.65) = 1.444 / 29.18 = 0.05 mol

      

      Step 7. Calculate the molecular weight of the gas

      Divide the number of grams of the compound present in the experiment (10.91 g) by the number of moles of that compound in the experiment (mol of 0.05).

      10.91 / 0.05 = 218.2 g / mol

      

      Step 8. Add the atomic weights

      To find the weight that corresponds to the empirical formula of this particular compound, you need to add the atomic weight of nitrogen four times (14, 00674 + 14, 00674 + 14, 00674 + 14, 00674), the atomic weight of hydrogen twice (1, 00794 + 1, 00794), the atomic weight of carbon once (12, 0107) and the atomic weight of oxygen twice (15, 9994 + 15, 9994) - this gives you a total weight of 102, 05 g.

      

      Step 9. Divide the molecular weight by the empirical formula weight

      This will tell you how many molecules of N₄H.₂CO₂ are present in the sample.

      • 218, 2 / 102, 05 = 2, 13
      • This means that approximately 2 molecules of N are present₄H.₂CO₂.

      

      Step 10. Write the final molecular formula

      The final molecular formula would be twice as large as the original empirical formula since two molecules are present. Therefore, it would be N₈H.₄C.₂OR₄.