Fundamentals of Chemistry: Measurement, Matter, and Nomenclature

Density and Temperature

Quantitative skills also extend to physical properties like density. Density is an intensive property that relates the mass of a substance to the volume it occupies. The mathematical relationship is expressed as: $\text{Density} = \frac{\text{mass}}{\text{volume}}$ For instance, if you have a liquid with a density of $2.70\text{ g/mL}$ and a volume of $36.0\text{ mL}$, you can calculate the mass by multiplying the two: $36.0\text{ mL} \cdot 2.70\text{ g/mL} = 97.2\text{ g}$.

Additionally, you must be comfortable with temperature conversion. The exam may require you to calculate a temperature change (interval) between Fahrenheit and Celsius scales. Note that a change in temperature is different from a specific temperature reading; a change of $16.0$ degrees Fahrenheit must be converted to its equivalent Celsius interval.

Classification of Matter

Matter can be classified based on its composition. One key concept is the homogeneous mixture. In a homogeneous mixture, the components are distributed uniformly throughout the entire sample, meaning the composition is the same no matter where you take a sample.

A classic example is sugar water. Unlike a heterogeneous mixture—such as a rock collection, chocolate chip cookies, or oil and vinegar dressing, where you can clearly see the different parts—a homogeneous mixture appears as a single phase.

Chemical Nomenclature and Formulas

Perhaps the most challenging section of the midterm involves naming and writing formulas for chemical substances. This requires distinguishing between different types of bonding and following specific naming conventions.

Ionic Compounds and Naming

Ionic compounds are typically formed between a metal and a non-metal. For example, $\text{Na}_3\text{P}$ is an ionic compound. When naming these, you must be careful with several factors:

1. Polyatomic Ions: You must recognize specific groups of atoms that stay together. For example, the formula for rubidium sulfite is $\text{Rb}_2\text{SO}_3$. Be careful not to confuse "sulfite" ($\text{SO}_3^{2-}$) with "sulfate" ($\text{SO}_4^{2-}$).
2. The Stock System: Transition metals can have multiple possible charges. To clarify which one is being used, we use Roman numerals. For example, $\text{Cu}(\text{ClO}_3)_2$ is named copper(II) chlorate.
3. Acids: Naming oxyacids (acids containing oxygen) follows specific patterns based on the polyatomic ion. For instance, chlorous acid is written as $\text{HClO}_2$.

Worked Example: Finding a Molecular Formula

A common high-level question involves using empirical formulas and molar mass to find a molecular formula.

Problem: A compound has a molecular weight of $146\text{ g/mol}$. A sample contains $0.321\text{ g}$ of Carbon, $0.044\text{ g}$ of Hydrogen, and $0.285\text{ g}$ of Sulfur. What is the molecular formula?

Step 1: Find the Empirical Formula Convert mass to moles: $\text{Moles C} = \frac{0.321\text{ g}}{12.01\text{ g/mol}} \approx 0.027\text{ mol}$ $\text{Moles H} = \frac{0.044\text{ g}}{1.01\text{ g/mol}} \approx 0.044\text{ mol}$ $\text{Moles S} = \frac{0.285\text{ g}}{32.0\text{ g/mol}} \approx 0.0089\text{ mol}$

Divide by the smallest number of moles ($0.0089$): $\text{C: } \frac{0.027}{0.0089} \approx 3, \quad \text{H: } \frac{0.044}{0.0089} \approx 5, \quad \text{S: } \frac{0.0089}{0.0089} = 1$ The empirical formula is $\text{C}_3\text{H}_5\text{S}$.

Step 2: Find the X-factor Calculate the empirical formula mass: $\text{Mass} = (3 \cdot 12.01) + (5 \cdot 1.01) + 32.0 = 73\text{ g/mol}$ Find the multiplier ($X$): $X = \frac{\text{Given Molar Mass}}{\text{Empirical Mass}} = \frac{146}{73} = 2$ Multiply the empirical formula by $2$ to get the molecular formula: $\text{C}_6\text{H}_{10}\text{S}_2$.

Final Study Tip: For stoichiometry and limiting reactant problems, always balance your chemical equation first! A mistake in balancing will lead to incorrect mole ratios and an incorrect final answer.