Unit 3: Empirical and Molecular Formulae

From percentage composition to the true identity of a molecule.

3.17 Definitions of Empirical and Molecular Formulae

In chemistry, we use different types of formulae to represent compounds. The empirical formula is the simplest representation, while the molecular formula provides the complete picture for molecular compounds.

  • Empirical Formula: The simplest whole-number ratio of atoms of each element present in a compound.
  • Molecular Formula: The actual number of atoms of each element in one molecule of a substance. It is always a whole-number multiple of the empirical formula.
  • Unit Formula: Used for giant ionic structures, it represents the simplest whole-number ratio of ions in the crystal lattice.

For many compounds like water ($H_2O$) and carbon dioxide ($CO_2$), the empirical and molecular formulae are the same. However, for others, they differ. For example, ethane's molecular formula is $C_2H_6$, but its simplest ratio of C:H is 1:3, giving it an empirical formula of $CH_3$.

Solved Examples:
  1. What is the empirical formula for hydrogen peroxide, $H_2O_2$?
    Solution: The ratio of atoms is 2 H : 2 O. The simplest whole-number ratio is 1:1. The empirical formula is HO.
  2. The molecular formula for glucose is $C_6H_{12}O_6$. What is its empirical formula?
    Solution: The ratio is 6:12:6. Dividing by the greatest common divisor (6) gives a ratio of 1:2:1. The empirical formula is $CH_2O$.
  3. Determine the empirical formula for octane, $C_8H_{18}$.
    Solution: The ratio is 8:18. Dividing by the greatest common divisor (2) gives 4:9. The empirical formula is $C_4H_9$.
  4. What is the empirical formula for sodium peroxide, which has a unit formula of $Na_2O_2$?
    Solution: The ratio of ions is 2 Na : 2 O. The simplest ratio is 1:1. The empirical formula is NaO.
  5. A compound has an empirical formula of $CH_2$. What is a possible molecular formula?
    Solution: The molecular formula must be a multiple of $CH_2$. Possible answers include $C_2H_4$ (ethene), $C_3H_6$ (propene), etc.
  6. Why are the empirical and molecular formulas for methane ($CH_4$) the same?
    Solution: The ratio of atoms is 1 C : 4 H. This is already the simplest whole-number ratio, so it cannot be reduced further.
  7. What is the empirical formula for dinitrogen tetroxide, $N_2O_4$?
    Solution: The ratio is 2:4, which simplifies to 1:2. The empirical formula is $NO_2$.
  8. The unit formula for magnesium nitrate is $Mg(NO_3)_2$. What is its empirical formula?
    Solution: The formula unit contains 1 Mg, 2 N, and 6 O atoms. The ratio 1:2:6 is already in its simplest form. The empirical formula is $MgN_2O_6$.
  9. What is the empirical formula for acetic acid, $CH_3COOH$?
    Solution: First, sum the atoms: 2 C, 4 H, 2 O. The ratio is 2:4:2, which simplifies to 1:2:1. The empirical formula is $CH_2O$.
  10. If a compound's empirical formula is $P_2O_5$, can its molecular formula be $P_4O_{10}$?
    Solution: Yes. The molecular formula $P_4O_{10}$ is a whole-number multiple ($n=2$) of the empirical formula ($ (P_2O_5)_2 $).

3.18 Calculating Empirical Formulae

The empirical formula of a compound can be determined from its percentage composition by mass. The process involves converting masses to moles and then finding the simplest whole-number ratio of those moles.

Steps to Calculate Empirical Formula:
  1. If given percentages, assume a 100 g sample, so percentages convert directly to grams.
  2. Convert the mass of each element to moles by dividing by its molar mass (relative atomic mass).
  3. Divide the mole value of each element by the smallest mole value calculated in the previous step.
  4. If the results are not whole numbers, multiply all the ratios by the smallest integer that will make them whole numbers.
Solved Examples:
  1. A compound contains 85.8% Carbon and 14.2% Hydrogen. Find its empirical formula.
    Solution: Assume 100 g: 85.8 g C and 14.2 g H. Moles C = $85.8/12.0 = 7.15$ mol. Moles H = $14.2/1.0 = 14.2$ mol. Divide by smallest (7.15): C = $7.15/7.15 = 1$; H = $14.2/7.15 \approx 2$. Empirical formula is $CH_2$.
  2. Analysis of 3.36 g of iron oxide shows it contains 2.36 g of iron. What is its empirical formula?
    Solution: Mass of O = $3.36 - 2.36 = 1.00$ g. Moles Fe = $2.36/55.8 = 0.0423$ mol. Moles O = $1.00/16.0 = 0.0625$ mol. Divide by smallest (0.0423): Fe = 1; O = $0.0625/0.0423 \approx 1.5$. Multiply by 2 to get whole numbers: Fe₂O₃.
  3. A compound contains 22.02% C, 4.59% H, and 73.39% Br. Find its empirical formula.
    Solution: Moles C = $22.02/12.0 = 1.835$. Moles H = $4.59/1.0 = 4.59$. Moles Br = $73.39/79.9 = 0.9185$. Divide by smallest (0.9185): C = 2; H = 5; Br = 1. Empirical formula is $C_2H_5Br$.
  4. Find the empirical formula of an oxide of iron formed when 3.36 g of iron joins with 1.44 g of oxygen.
    Solution: Moles Fe = $3.36/55.8 \approx 0.0602$. Moles O = $1.44/16.0 = 0.09$. Divide by smallest (0.0602): Fe = 1; O = 1.5. Multiply by 2: Fe₂O₃.
  5. A sample contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen. Find its empirical formula.
    Solution: Moles C = $40.0/12.0=3.33$. Moles H = $6.7/1.0=6.7$. Moles O = $53.3/16.0=3.33$. Divide by smallest (3.33): C=1; H=2; O=1. Empirical formula is $CH_2O$.
  6. An ionic compound contains 21.2% N, 6.1% H, 24.3% S, and 48.4% O. Find its empirical formula.
    Solution: Moles N=1.51; H=6.1; S=0.758; O=3.025. Divide by smallest (0.758): N=2; H=8; S=1; O=4. Formula is $N_2H_8SO_4$. This is ammonium sulfate, $(NH_4)_2SO_4$.
  7. What is the empirical formula of a compound that is 52.2% C, 13.0% H, and 34.8% O?
    Solution: Moles C=4.35; H=13.0; O=2.175. Divide by smallest (2.175): C=2; H=6; O=1. Empirical formula is $C_2H_6O$.
  8. A 10.0 g sample of a hydrate of copper(II) sulfate ($CuSO_4 \cdot xH_2O$) is heated, and 6.4 g of anhydrous $CuSO_4$ remains. Find the value of x.
    Solution: Mass of H₂O = $10.0 - 6.4 = 3.6$ g. Moles CuSO₄ = $6.4/159.6 \approx 0.04$. Moles H₂O = $3.6/18.0 = 0.2$. Ratio of H₂O/CuSO₄ = $0.2/0.04 = 5$. The formula is $CuSO_4 \cdot 5H_2O$.
  9. A compound is 75% carbon and 25% hydrogen by mass. Determine its empirical formula.
    Solution: Moles C = $75/12.0 = 6.25$. Moles H = $25/1.0 = 25$. Divide by smallest (6.25): C=1; H=4. Empirical formula is $CH_4$.
  10. A 5.00 g sample of a tin oxide contains 3.93 g of tin. What is its empirical formula?
    Solution: Mass of O = $5.00 - 3.93 = 1.07$ g. Moles Sn = $3.93/118.7 \approx 0.0331$. Moles O = $1.07/16.0 \approx 0.0669$. Divide by smallest (0.0331): Sn=1; O=2. Empirical formula is $SnO_2$.

3.19 Calculating Molecular Formulae

While the empirical formula gives the simplest ratio, the molecular formula gives the true number of atoms in a molecule. To find the molecular formula, you need both the empirical formula and the compound's relative molecular mass ($M_r$).

Steps to Calculate Molecular Formula:
  1. Determine the empirical formula if it's not already known.
  2. Calculate the relative mass of the empirical formula (Empirical Formula Mass).
  3. Divide the given Relative Molecular Mass by the Empirical Formula Mass to find a whole-number multiple, 'n'. ($n = M_r$ / Empirical Formula Mass).
  4. Multiply the subscripts in the empirical formula by 'n' to get the molecular formula.
Solved Examples:
  1. A compound has an empirical formula of $CH_2$ and a relative molecular mass of 56. What is its molecular formula?
    Solution: Empirical formula mass of $CH_2$ is $12.0 + 2(1.0) = 14.0$. The multiple n = $56 / 14 = 4$. Molecular formula is $(CH_2)_4 = C_4H_8$.
  2. A compound contains C 62.08%, H 10.34%, and O 27.58%. Its $M_r$ is 58. Find its molecular formula.
    Solution: First, find the empirical formula: Moles C=5.17, H=10.34, O=1.72. Ratio gives $C_3H_6O$. Empirical mass is $3(12)+6(1)+16 = 58$. Since the empirical mass equals the molecular mass, n=1. The molecular formula is also $C_3H_6O$.
  3. A hydrocarbon contains 84.21% carbon and 15.79% hydrogen. Its $M_r$ is 114. Find its molecular formula.
    Solution: Moles C = $84.21/12 \approx 7.02$. Moles H = $15.79/1 \approx 15.79$. Ratio C:H is $1:2.25$. Multiply by 4 to get $C_4H_9$. Empirical mass = 57. Multiple n = $114/57=2$. Molecular formula = $C_8H_{18}$.
  4. Analysis of a 7.8 g hydrocarbon shows it contains 0.6 g of hydrogen. Its $M_r$ is 78. Find the molecular formula.
    Solution: Mass C = $7.8 - 0.6 = 7.2$ g. Moles C = $7.2/12=0.6$. Moles H = $0.6/1=0.6$. Ratio is 1:1, so empirical formula is CH. Empirical mass = 13. Multiple n = $78/13=6$. Molecular formula is $C_6H_6$.
  5. A compound with the empirical formula $P_2O_5$ has a molar mass of 283.9 g/mol. What is its molecular formula?
    Solution: Empirical mass = $2(31.0) + 5(16.0) = 142.0$. Multiple n = $283.9 / 142.0 \approx 2$. Molecular formula is $(P_2O_5)_2 = P_4O_{10}$.
  6. The empirical formula of a substance is CH. Its molar mass is 26.04 g/mol. What is its molecular formula?
    Solution: Empirical mass = $12.01 + 1.01 = 13.02$. Multiple n = $26.04/13.02 = 2$. Molecular formula is $(CH)_2 = C_2H_2$ (acetylene).
  7. A compound containing only carbon and hydrogen is 7.7% hydrogen by mass. The molar mass is 78 g/mol. What is the molecular formula?
    Solution: % Carbon = 92.3%. Empirical formula is CH. Empirical mass = 13. Multiple n = $78/13=6$. Molecular formula is $C_6H_6$ (benzene).
  8. A gas has an empirical formula of $NO_2$. Its density is 4.1 g/dm³ at STP. Find its molecular formula.
    Solution: Molar Mass = density $\times$ molar volume = $4.1 \, \text{g/dm}^3 \times 22.4 \, \text{dm}^3/\text{mol} \approx 92$ g/mol. Empirical mass of $NO_2$ = $14 + 2(16) = 46$. Multiple n = $92/46 = 2$. Molecular formula is $N_2O_4$.
  9. A compound's empirical formula is $C_3H_4O_3$ and its molar mass is 176 g/mol. What is its molecular formula?
    Solution: Empirical mass = $3(12)+4(1)+3(16) = 36+4+48 = 88$. Multiple n = $176/88 = 2$. Molecular formula is $C_6H_8O_6$ (ascorbic acid).
  10. The empirical formula for caffeine is $C_4H_5N_2O$. Its molecular mass is 194 g/mol. Find the molecular formula.
    Solution: Empirical mass = $4(12)+5(1)+2(14)+16 = 48+5+28+16 = 97$. Multiple n = $194/97 = 2$. Molecular formula is $C_8H_{10}N_4O_2$.

End of Topic Questions

Answer: The simplest whole-number ratio of atoms of each element in a compound.

Answer: Empirical mass is 13. Multiple is 78/13 = 6. Molecular formula is C₆H₆.

Answer: CH₂.

Answer: The relative molecular mass (or molar mass) of the compound.

Answer: CH₂O.

Answer: Yes, for example, ethene (C₂H₄) and propene (C₃H₆) both have the empirical formula CH₂.

Answer: Convert the percentage of each element to mass in grams (assuming a 100g sample).

Answer: Whole-number multiple.

Answer: Moles Hg=0.374, Moles Cl=0.704. Ratio is 1:1.88, which is close to 1:2. HgCl₂.

Answer: Multiply by 2.

Answer: C₂H₅.

Answer: Empirical mass = 29. Multiple n = 58/29 = 2. Molecular formula is C₄H₁₀.

Answer: Yes, if the simplest ratio is the actual ratio, as in H₂O or CO₂.

Answer: Moles Al=0.1, Moles O=0.15. Ratio is 1:1.5. Multiply by 2 gives Al₂O₃.

Answer: The simplest whole-number ratio of ions in a giant ionic structure.

Answer: Empirical mass = 17. Multiple n = 34/17 = 2. Molecular formula is H₂O₂.

Answer: Mass O=1.25g. Moles S=0.039, Moles O=0.078. Ratio is 1:2. SO₂.

Answer: Multiply all ratios by the smallest integer that will convert them to whole numbers.

Answer: The ratio is 4:8, which simplifies to 1:2. The empirical formula is CH₂.

Answer: It means the empirical formula and the molecular formula are the same.
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