Analytical techniques, such as combustion analysis or mass spectrometry, allow chemists to determine the relative amounts of elements in a compound. We use this data to find the empirical and molecular formulas of the substance.
Definitions
The simplest whole number ratio of atoms of each element in a compound.
The actual number of atoms of each element in a compound.
🔑 Key Principle
A molecular formula is always a simple whole-number multiple of the empirical formula. For example, the empirical formula of glucose is \( \text{CH}_2\text{O} \), while its molecular formula is \( \text{C}_6\text{H}_{12}\text{O}_6 \) (which is exactly \( 6 \times \text{CH}_2\text{O} \)).
How to Calculate Empirical Formula
When given the percentage composition or masses of elements in a compound, follow these four structured steps:
Assume we have a 100 g sample of the compound. Write down the mass of each element:
- Mass of Carbon (\( \text{C} \)) = \( 40.0 \text{ g} \)
- Mass of Hydrogen (\( \text{H} \)) = \( 6.7 \text{ g} \)
- Mass of Oxygen (\( \text{O} \)) = \( 53.3 \text{ g} \)
Step 1: Convert masses to moles using \( n = \frac{m}{A_r} \).
\[ n(\text{C}) = \frac{40.0}{12.0} = 3.33 \text{ mol} \]
\[ n(\text{H}) = \frac{6.7}{1.0} = 6.70 \text{ mol} \]
\[ n(\text{O}) = \frac{53.3}{16.0} = 3.33 \text{ mol} \]
Step 2: Divide all moles by the smallest value (\( 3.33 \)).
\[ \text{C} = \frac{3.33}{3.33} = 1.0 \]
\[ \text{H} = \frac{6.70}{3.33} = 2.01 \approx 2.0 \]
\[ \text{O} = \frac{3.33}{3.33} = 1.0 \]
The simplest whole number ratio is \( 1 : 2 : 1 \).
Therefore, the empirical formula is \( \text{CH}_2\text{O} \).
Sometimes the mole ratio will not resolve directly to whole numbers. Do not round numbers such as 1.5, 1.33, or 1.25. Instead, multiply the entire ratio to make them whole:
- If ratio ends in .5, multiply all values by 2 (e.g. \( 1 : 1.5 \to 2 : 3 \))
- If ratio ends in .33, multiply all values by 3 (e.g. \( 1 : 1.33 \to 3 : 4 \))
- If ratio ends in .25, multiply all values by 4 (e.g. \( 1 : 1.25 \to 4 : 5 \))
How to Calculate Molecular Formula
To find the molecular formula, you must be given the relative molecular mass (\( M_r \)) of the compound along with either its empirical formula or the data to calculate it.
- Calculate the relative formula mass of the empirical formula.
- Divide the molecular \( M_r \) by the empirical mass to find the multiplier.
- Multiply all numbers of atoms in the empirical formula by this multiplier.
Step 1: Calculate the empirical formula mass of \( \text{CH}_2 \).
\[ \text{Empirical mass} = A_r(\text{C}) + 2 \times A_r(\text{H}) = 12.0 + 2 \times 1.0 = 14.0 \]
Step 2: Divide the molecular mass by the empirical mass.
\[ \text{Multiplier} = \frac{\text{Molecular } M_r}{\text{Empirical mass}} = \frac{84.0}{14.0} = 6.0 \]
Step 3: Multiply the empirical formula by the multiplier.
\[ \text{Molecular Formula} = 6 \times (\text{CH}_2) = \text{C}_6\text{H}_{12} \]
The molecular formula of the compound is \( \text{C}_6\text{H}_{12} \).
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