Moles, Percent Composition, and Empirical Formulas Notes

The concept of molar mass serves as a fundamental conversion factor between the mass of a substance and the number of moles. This is a review of material typically covered in early introductory chemistry (Chapter 3).
Review Question 1: What is the mass in grams of $3.04\,mol$ of Nitrogen?
- To solve this, one must determine if the problem refers to atomic nitrogen ( $N$ ) or nitrogen gas ( $N_2$ ).
- Using atomic nitrogen ( $14.01\,g/mol$ ): $3.04\,mol \times 14.01\,g/mol = 42.59\,g$
- Using diatomic nitrogen ( $28.02\,g/mol$ ): $3.04\,mol \times 28.02\,g/mol = 85.18\,g$
New Application Question 2: What is the mass in grams of $3.04\,mol$ of ammonia ( $NH_3$ )?
- Step 1: Calculate molar mass of $NH_3$ . $14.01\,g/mol\text{ (N)} + (3 \times 1.01\,g/mol\text{ (H)}) = 17.04\,g/mol$ .
- Step 2: Convert moles to grams: $3.04\,mol \times 17.04\,g/mol = 51.80\,g$
New Application Question 3: Calculate the mass of $0.257\,mol$ of Calcium Nitrate ( $Ca(NO_3)_2$ ).
- Step 1: Calculate molar mass of $Ca(NO_3)_2$ . $40.08\,g/mol\text{ (Ca)} + (2 \times 14.01\,g/mol\text{ (N)}) + (6 \times 16.00\,g/mol\text{ (O)}) = 164.10\,g/mol$ .
- Step 2: Convert moles to grams: $0.257\,mol \times 164.10\,g/mol = 42.17\,g$

Definition: Percent composition refers to the percentage by mass of each individual element contained within a chemical compound.
Invariance Principle: The percentage of an element in a specific compound is constant and does not change based on the size or origin of the sample.
Percent composition is defined as a ratio:
- \text{% element in compound} = \frac{\text{mass of element in 1 mol of compound}}{\text{molar mass of compound}} \times 100

Step 1: Identify the chemical formula of the compound.
Step 2: Find the molar mass of the compound and the individual atomic masses of the elements within it (using the periodic table).
Step 3: Calculate the mass percentage of each individual element by dividing its total mass contribution in one mole by the compound's total molar mass.
Step 4: Verification Check. Ensure that the sum of the percentages for all elements in the compound equals $100\%$ .
Example Problem: Calculate the percentage composition of Sodium Nitrate ( $NaNO_3$ ).
- Determine molar mass: $Na\, (22.99) + N\, (14.01) + 3 \times O\, (16.00) = 85.00\,g/mol$ .
- Percentage $Na$ : $\frac{22.99}{85.00} \times 100 = 27.05\%$
- Percentage $N$ : $\frac{14.01}{85.00} \times 100 = 16.48\%$
- Percentage $O$ : $\frac{48.00}{85.00} \times 100 = 56.47\%$

The empirical formula represents the simplest whole-number ratio of atoms in a substance.
Conversion Method: If the percent composition is given, the researcher can assume a sample mass of exactly $100\,g$ . Under this assumption, the numerical value of the percentage translates directly into an equivalent mass in grams.
- Example: In Copper (I) Sulfide ( $Cu_2S$ ), if the substance is $79.85\%\,Cu$ and $20.15\%\,S$ , a $100\,g$ sample contains exactly $79.85\,g\,Cu$ and $20.15\,g\,S$ .
Converting Grams to Moles:
- For Copper ( $Cu$ ): $79.85\,g \rightarrow 1.256\,mol\,Cu$
- For Sulfur ( $S$ ): $20.15\,g \rightarrow 0.6283\,mol\,S$
Determining the Molar Ratio:
- Divide each calculated mole value by the smallest mole value in the set to find the relative ratio.
- Calculation: $\frac{1.256\,mol\,Cu}{0.6283\,mol\,S} = 2/1$
- Conclusion: There are $2\,moles$ of $Cu$ for every $1\,mole$ of $S$ . Therefore, the empirical formula is $Cu_2S$ .

Empirical Formula: This is the smallest possible whole-number ratio of the atoms present in a compound.
Molecular Formula: This is the actual chemical formula representing the real number of atoms in a molecular compound.
Example Case (Ethene):
- Structural representation: Two carbon atoms double-bonded to each other, with each carbon bonded to two hydrogen atoms.
- Molecular Formula: $C_2H_4$
- Empirical Formula: $CH_2$ (Simplified from a $2:4$ ratio to a $1:2$ ratio).

The molecular formula is always a whole-number multiple ( $X$ ) of the empirical formula.
Fundamental Relationships:
- $X \times (\text{empirical formula}) = \text{molecular formula}$
- $X \times (\text{empirical formula mass}) = \text{molecular formula mass}$
To solve for $X$ , use the ratio of the experimentally determined molecular mass to the empirical formula mass:
- $X = \frac{\text{Molecular mass}}{\text{Empirical mass}}$
Example Problem: Find the molecular formula for a compound with the following data:
- Empirical formula: $P_2O_5$
- Experimental mass (molecular mass): $283.89\,g/mol$
- Step 1: Calculate the empirical mass of $P_2O_5$ . $(2 \times 30.97) + (5 \times 16.00) = 141.94\,g/mol$ .
- Step 2: Solve for $X$ . $X = \frac{283.89\,g/mol}{141.94\,g/mol} \approx 2$ .
- Step 3: Determine molecular formula: $2 \times (P_2O_5) = P_4O_{10}$ .