Unit 1, Section 3: Chemical Formulas and Their Origins

Unit 1, Section 3: Chemical Formulas and Their Origins

Joseph Proust and the Law of Constant Composition

In the $1700$s, French chemist Joseph Proust established the Law of Constant Composition, also known as the Law of Definite Proportions. This law states that different samples of a chemical compound will always have the same composition of elements by mass. Proust deduced this by analyzing various samples of the same compound. For instance, any sample of water $(H_2O)$, regardless of its origin, will consistently be composed of approximately $11\%$ hydrogen and $89\%$ oxygen by mass. Similarly, a pure sample of sodium chloride $(NaCl)$ will always contain about $39\%$ sodium and $61\%$ chlorine by mass.

John Dalton's Atomic Theory

In the early $1800$s, British chemist John Dalton built upon the ideas of predecessors like Proust and Antoine Lavoisier to propose his atomic theory. This theory explained chemical behavior based on the existence of atoms and had four main statements:

1. All elements are composed of atoms. This was a revolutionary concept at the time, as the existence of atoms was not widely accepted.

2. Atoms cannot be created or destroyed in a chemical reaction. This statement aligns with the Law of Conservation of Mass and remains true today in the context of chemical reactions.

3. Atoms of one element are different from atoms of other elements, and all atoms of an element are identical to each other. The first part of this statement is correct: atoms of different elements (e.g., hydrogen and oxygen) are distinct. However, the second part, which claims all atoms of a given element are identical, is incorrect. As discussed in Unit 1, Section 2, isotopes exist, meaning an element can have different varieties of atoms with varying numbers of neutrons (e.g., the three isotopes of hydrogen). Thus, not all atoms of an element are identical in mass.

4. All compounds have a fixed ratio of atoms of elements. This statement strongly supports and explains Proust's Law of Constant Composition, indicating that compounds are formed when atoms combine in whole-number ratios.

Calculating Mass Percent Composition (Review)

We previously learned how to calculate the mass percent of elements in a compound. For example, consider caffeine with the formula $(C<em>8H</em>{10}N<em>4O</em>2)$. To find its molecular mass and mass percents:

• Atomic Masses (approximate): Carbon $(C)$ $=12$ amu, Hydrogen $(H)$ $=1$ amu, Nitrogen $(N)$ $=14$ amu, Oxygen $(O)$ $=16$ amu.

• Contribution from each element:

  • Carbon: $8 \text{ atoms} \times 12 \text{ amu/atom} = 96 \text{ amu}$

  • Hydrogen: $10 \text{ atoms} \times 1 \text{ amu/atom} = 10 \text{ amu}$

  • Nitrogen: $4 \text{ atoms} \times 14 \text{ amu/atom} = 56 \text{ amu}$

  • Oxygen: $2 \text{ atoms} \times 16 \text{ amu/atom} = 32 \text{ amu}$

• Molecular Mass of Caffeine: $96 + 10 + 56 + 32 = 194$ amu (approximately $194.0$ amu as given in the context).

• Mass Percent of each element:

  • Carbon: $(96 / 194) \times 100\% \approx 49.5\%$

  • Hydrogen: $(10 / 194) \times 100\% \approx 5.2\%$

  • Nitrogen: $(56 / 194) \times 100\% \approx 28.9\%$

  • Oxygen: $(32 / 194) \times 100\% \approx 16.5\%$

Determining the Empirical Formula from Percent Composition

The empirical formula is the simplest whole-number ratio of atoms of each element in a compound. It's like a chemical formula reduced to its lowest terms. We can determine the empirical formula by working backward from mass percent data.

Steps for determining empirical formula (using caffeine's calculated percents as an example):
Assume a $100$-gram sample of the compound, turning percentages into gram values:

• Carbon: $49.5 \text{ g}$ (49.5\%$)

• Hydrogen: 5.2 \text{ g}$($5.2\%$)

• Nitrogen: $28.9 \text{ g}$ (28.9\%$)

• Oxygen: 16.5 \text{ g}$($16.5\%$)

1. Convert grams to moles for each element: Divide each gram value by the element's atomic mass.

   • Carbon: $49.5 \text{ g} / 12.01 \text{ g/mol} \approx 4.122 \text{ mol}$

   • Hydrogen: $5.2 \text{ g} / 1.01 \text{ g/mol} \approx 5.149 \text{ mol}$

   • Nitrogen: $28.9 \text{ g} / 14.01 \text{ g/mol} \approx 2.063 \text{ mol}$

   • Oxygen: $16.5 \text{ g} / 16.00 \text{ g/mol} \approx 1.031 \text{ mol}$
     (Note: The provided transcript used slightly different atomic mass approximations, resulting in $\text{C} \approx 4.125 \text{ mol}, \text{H} \approx 5 \text{ mol}, \text{N} \approx 2 \text{ mol}, \text{O} \approx 1.031 \text{ mol}$ for simplicity in subsequent calculations.)

2. Divide all mole values by the smallest mole value: This normalizes the ratios.

   • Smallest mole value is for Oxygen: $1.031 \text{ mol}$.

   • Carbon: $4.122 / 1.031 \approx 4.00$

   • Hydrogen: $5.149 / 1.031 \approx 4.99$

   • Nitrogen: $2.063 / 1.031 \approx 2.00$

   • Oxygen: $1.031 / 1.031 = 1.00$

3. Round to the nearest whole number (if very close) to get subscripts: These numbers represent the ratio of atoms.

   • Carbon: $4$

   • Hydrogen: $5$

   • Nitrogen: $2$

   • Oxygen: $1$

   The empirical formula for caffeine derived from its percent composition is $(C<em>4H</em>5N<em>2O)$. This is different from the actual molecular formula, $(C</em>8H<em>{10}N</em>4O_2)$.

Examples of Empirical vs. Molecular Formulas:

• Glucose: Molecular formula $(C<em>6H</em>{12}O<em>6)$. Dividing by the common factor of $6$ gives the empirical formula $(CH</em>2O)$.

• Hydrazine: Molecular formula $(N<em>2H</em>4)$. Dividing by $2$ gives the empirical formula $(NH_2)$.

• Water: Molecular formula $(H<em>2O)$. Cannot be reduced further; so, its empirical formula is also $(H</em>2O)$.

• Tetraphosphorus Decoxide: Molecular formula $(P<em>4O</em>{10})$. Dividing by $2$ gives the empirical formula $(P<em>2O</em>5)$.

Determining the Molecular Formula from the Empirical Formula

To convert an empirical formula to a molecular formula, you need the molecular mass (or molar mass) of the compound.

Steps for determining molecular formula (using caffeine as an example):
Given:

• Empirical formula: (C4H5N2​O)

• Known molecular mass of caffeine: $194.0$ amu

1. Calculate the empirical formula mass (EFM): Sum the atomic masses for the atoms in the empirical formula.

   • Carbon: $4 \times 12 = 48$ amu

   • Hydrogen: $5 \times 1 = 5$ amu

   • Nitrogen: $2 \times 14 = 28$ amu

   • Oxygen: $1 \times 16 = 16$ amu

   • EFM = $48 + 5 + 28 + 16 = 97$ amu

2. Determine the ratio (n) between the molecular mass and the empirical formula mass:

   • $n = \text{Molecular Mass} / \text{Empirical Formula Mass}$

   • $n = 194.0 \text{ amu} / 97 \text{ amu} = 2$

3. Multiply the subscripts of the empirical formula by 'n' to get the molecular formula:

   • $(C<em>4H</em>5N<em>2O) \times 2 = C</em>{(4\times2)}H<em>{(5\times2)}N</em>{(2\times2)}O<em>{(1\times2)} = C</em>8H<em>{10}N</em>4O_2$
     This matches the known molecular formula of caffeine.

Example: Finding the Empirical Formula of a Nitrogen and Oxygen Compound

Let's find the empirical formula of a compound containing $25.93\%$ nitrogen and $74.07\%$ oxygen by mass.

1. Convert mass percents to grams (assuming a $100$-gram sample):

   • Nitrogen: $25.93 \text{ g}$

   • Oxygen: $74.07 \text{ g}$

2. Convert grams to moles for each element:

   • Nitrogen: $25.93 \text{ g} / 14.01 \text{ g/mol} \approx 1.851 \text{ mol}$

   • Oxygen: $74.07 \text{ g} / 16.00 \text{ g/mol} \approx 4.629 \text{ mol}$

3. Divide all mole values by the smallest mole value: ($1.851$ mol for nitrogen)

   • Nitrogen: $1.851 / 1.851 = 1.00$

   • Oxygen: $4.629 / 1.851 \approx 2.50$

4. Obtain whole-number ratios: Since we have $1$ and $2.5$, we cannot have a fractional atom (e.g., $O_{2.5})$. We must multiply both values by the smallest integer that converts the fraction to a whole number. For $2.5$, this integer is $2$ (since $2.5 \times 2 = 5$).

   • Multiply Nitrogen by $2$: $1.00 \times 2 = 2$

   • Multiply Oxygen by $2$: $2.50 \times 2 = 5$

   The empirical formula for this compound is (N2O5).

   Self-correction note: If you get a ratio like $1$ to $2.33$, this is $2 \frac{1}{3}$ or $\frac{7}{3}$. You would then multiply both numbers by $3$ to get whole numbers (e.g., $1 \times 3 = 3$ and $2.33 \times 3 \approx 7$). Always ensure subscripts in a chemical formula are whole numbers.