Lecture 4: Empirical formula

Empirical Formula Calculations

Learning Aims

• Calculate empirical formula using data given as a percentage or as a mass.
• Use the relative molecular mass of a compound to derive its molecular formula from the empirical formula.

Definition of Empirical Formula

• The simplest ratio of elements in a compound.
• Example:
  • $C<em>2H</em>6$ can be simplified to $CH_3$.
  • $N<em>2O</em>4$ can be simplified to $NO_2$.

Recap: Calculating Number of Moles

• Number of moles = Mass in grams / Molar mass in grams per mole.
• $number \ of \ moles = \frac{mass \ (g)}{molar \ mass \ (g/mol)}$

Question 1: Calculating Empirical Formula from Percentages

• A compound contains 75% carbon and 25% hydrogen. What is its empirical formula?
• Method:
  • Set up a table with element names (carbon and hydrogen).
  • Write down the amounts as if they are grams.
  • Write down the molar masses (relative atomic masses).
  • Calculate the moles using the formula: moles = mass / atomic mass.
  • Find the ratio by dividing each number of moles by the smallest number of moles.
  • The resulting ratio gives the empirical formula.
• Solution:
  • Moles of Carbon: $\frac{75}{12} = 6.25$
  • Moles of Hydrogen: $\frac{25}{1} = 25$
  • Divide by the smallest (6.25): Carbon: $\frac{6.25}{6.25} = 1$, Hydrogen: $\frac{25}{6.25} = 4$
  • Empirical formula: $CH_4$

Question 2: Calculating Empirical Formula with a Missing Percentage

• Fluorospar is made of calcium and fluorine. It is 51% calcium. Calculate the empirical formula.
• Find the percentage of fluorine by subtracting the percentage of calcium from 100%.
  • Percentage of Fluorine: $100 - 51 = 49 \%$
• Solution:
  • Moles of Calcium: $\frac{51}{40} = 1.275$
  • Moles of Fluorine: $\frac{49}{19} = 2.579$
  • Divide by the smallest (1.275): Calcium: $\frac{1.275}{1.275} = 1$, Fluorine: $\frac{2.579}{1.275} = 2.02 \approx 2$
  • Empirical formula: $CaF_2$

Question 3: Finding Empirical and Molecular Formula from Masses

• A compound contains 40 grams of carbon, 6.7 grams of hydrogen, and 53.5 grams of oxygen. It has a relative molecular mass of 60. Find both the empirical and molecular formula of the compound.
• Solution for Empirical Formula:
  • Moles of Carbon: $\frac{40}{12} = 3.33$
  • Moles of Hydrogen: $\frac{6.7}{1} = 6.7$
  • Moles of Oxygen: $\frac{53.5}{16} = 3.34$
  • Divide by the smallest (3.33): Carbon: $\frac{3.33}{3.33} = 1$, Hydrogen: $\frac{6.7}{3.33} = 2$, Oxygen: $\frac{3.34}{3.33} = 1$
  • Empirical formula: $CH_2O$
• Finding the Molecular Formula:
  • Start with the empirical formula: $CH_2O$.
  • Calculate the molecular mass of the empirical formula using atomic masses.
  • Molecular mass of $CH_2O = 12 + (2 \times 1) + 16 = 30$
  • The relative molecular mass from the question is 60.
  • Divide the given molecular mass by the mass of the empirical formula: $\frac{60}{30} = 2$
  • Multiply the empirical formula by 2 to get the molecular formula.
  • Molecular formula: $C<em>2H</em>4O_2$

Question 4: Higher Level Difficulty - Combustion Analysis

• This question involves combustion analysis, where a compound containing carbon, hydrogen, and oxygen is burned in excess oxygen.
• Assumptions:
  • All carbon is converted to carbon dioxide ($CO_2$).
  • Moles of carbon = moles of carbon dioxide
  • All hydrogen is converted to water ($H_2O$).
  • Moles of hydrogen = 2 x moles of water
• General Equation for Combustion:
  • $C<em>xH</em>yO<em>z + O</em>2 \rightarrow xCO<em>2 + \frac{y}{2} H</em>2O$
• Calculations:
  • Given: Compound X (0.43 grams) produces $CO<em>2$ and $H</em>2O$.
  • Calculate moles of $CO<em>2$ and $H</em>2O$ from their masses.
  • Moles of Carbon = Moles of $CO_2$
  • Moles of Hydrogen = 2 x Moles of $H_2O$
  • Calculate the mass of carbon and hydrogen in the original compound.
  • Determine the mass of oxygen in compound X by subtracting the mass of carbon and hydrogen from the total mass of X.
  • Calculate the moles of oxygen.
  • Find the mole ratios of carbon, hydrogen, and oxygen to determine the empirical formula.
  • Compare the molar mass of the empirical formula with the given molecular mass to determine the molecular formula.
• Example Values:
  • Moles of Carbon = 0.01
  • Moles of Hydrogen = 0.02
  • Mass of Carbon = 0.12 grams
  • Mass of Hydrogen = 0.02 grams
  • Mass of Oxygen = 0.43 - 0.12 - 0.02 = 0.29 grams
  • Moles of Oxygen = $\frac{0.29}{16} = 0.018125$
  • Empirical Formula : $C<em>3H</em>4O_3$
  • Molar mass of the empirical formula = 86.
  • Since the relative molecular mass of the compound is also 86, the empirical and molecular formulas are the same.
  • Final Answer for the molecular formula: The same as before

Summary

• Definition of empirical formula: simplest ratio of elements in a compound.
• Calculating empirical formula from percentage or mass data.
• Using molar mass to find the molecular formula from the empirical formula.

Additional Notes

• When dealing with combustion analysis, remember to balance the chemical equation correctly to accurately determine the moles of each element.
• Always double-check your calculations to avoid errors, especially when dividing by the smallest number of moles to find the ratio.
• Ensure that the units are consistent throughout the calculation.

Practice Questions

1. A compound contains 60% carbon, 5% hydrogen, and 35% nitrogen. What is its empirical formula?
2. A compound contains 52.17% carbon, 13.04% hydrogen, and 34