Composition of Substances and Solutions

Formula Mass and the Mole Concept

The formula mass of a substance is defined as the sum of the average atomic masses of all the atoms represented in the chemical formula of that substance.
For covalent substances, which exist as discrete molecules, the formula mass is often referred to specifically as the molecular mass or molecular weight.
For ionic compounds, which exist as a crystal lattice or array rather than individual molecules, the formula mass is the sum of the atomic masses of the ions in the empirical formula.
Example: Chloroform ( $CHCl_3$ ) has an average molecular mass of $119.37\,amu$ . This is calculated by adding the mass of one carbon atom ( $12.01\,amu$ ), one hydrogen atom ( $1.008\,amu$ ), and three chlorine atoms ( $3 \times 35.45\,amu$ ).
Example: Aspirin ( $C_9H_8O_4$ ) has an average molecular mass of $180.15\,amu$ .
Example: Sodium Chloride ( $NaCl$ ) exists in a $1:1$ ratio of sodium and chloride ions. Its formula mass is $58.44\,amu$ .

The Mole Concept and Avogadro's Number

The mole is the standard unit used by scientists to bridge the gap between microscopic atoms and macroscopic weighable amounts.
Previous Definition: The number of entities (atoms) in exactly $12\,g$ of Carbon-12 ( $^{12}C$ ).
Current Definition: Exactly $6.02214076 \times 10^{23}$ entities. This constant is known as Avogadro's Number or Avogadro's constant ( $N_A$ ).
The atomic weight listed for an element on the periodic table represents the mass of one mole of atoms of that element (using typical Earth isotopic ratios) in grams.

Molar Mass of Elements and Compounds

Molar mass is defined as the mass in grams of one mole of a substance ( $g/mol$ ).
For any element, its molar mass is numerically equivalent to its atomic mass in atomic mass units ( $amu$ ). - Carbon ( $C$ ): $12.01\,amu$ atomic mass; $12.01\,g/mol$ molar mass. - Hydrogen ( $H$ ): $1.008\,amu$ atomic mass; $1.008\,g/mol$ molar mass. - Oxygen ( $O$ ): $16.00\,amu$ atomic mass; $16.00\,g/mol$ molar mass. - Sodium ( $Na$ ): $22.99\,amu$ atomic mass; $22.99\,g/mol$ molar mass. - Chlorine ( $Cl$ ): $35.45\,amu$ atomic mass; $35.45\,g/mol$ molar mass.
Calculating Molar Mass for a Compound: - Identify the number of moles of each element in one mole of the compound. - Multiply the number of moles of each element by its respective molar mass. - Sum the resulting masses. - Example: Methane ( $CH_4$ ) contains $1\,mol$ of $C$ and $4\,mol$ of $H$ . - Mass = $(1 \times 12.011\,g) + (4 \times 1.008\,g) = 16.043\,g/mol$ .

Mass Percent Composition

Percent composition is the percentage by mass of each element in a compound.
For a known formula, it can be derived from formula mass and atomic masses: - \text{Mass % of Element} = \frac{\text{Mass of element in 1 mol of compound}}{\text{Molar mass of compound}} \times 100
Example: Ammonia ( $NH_3$ ) with a formula mass of $17.03\,amu$ . - $\% N = \frac{14.01\,amu}{17.03\,amu} \times 100 = 82.27\%$ - $\% H = \frac{3 \times 1.008\,amu}{17.03\,amu} \times 100 = 17.76\%$
Application: Percent composition acts as a conversion factor between the mass of a specific element and the total mass of the compound. For instance, in mining, knowing the mass percentage helps determine if an ore is worth refining (e.g., Iron in Pyrite, $FeS_2$ ). - Calculation for Pyrite: $Fe = 55.84\,g/mol$ , $FeS_2 = 119.98\,g/mol$ . Mass \% Fe = $\frac{55.84}{119.98} = 46.5\%$ .

Determining Empirical and Molecular Formulas

Empirical Formula Determination from Mass: 1. Convert the mass of each element into moles using molar masses. 2. Divide each molar amount by the smallest number of moles obtained. 3. If results are not whole numbers, multiply all subscripts by the same small integer (e.g., $2, 3, \text{ or } 4$ ) to achieve the smallest whole-number ratio.
Empirical Formula Determination from Percent Composition: 1. Assume a $100\,g$ sample (this converts percentages directly to grams). 2. Follow the steps for mass-to-empirical formula conversion.
Molecular Formula Determination: - The molecular formula is a whole-number multiple ( $n$ ) of the empirical formula. - Calculate the ratio: $n = \frac{\text{Molecular or Molar Mass}}{\text{Empirical Formula Mass}}$ . - Multiply subscripts of the empirical formula by $n$ . - Example: Empirical formula $CH_2O$ (mass $30\,amu$ ) and molecular mass $180\,amu$ . - $n = \frac{180}{30} = 6$ . - Molecular formula = $(CH_2O)_6 = C_6H_{12}O_6$ .

Elemental and Combustion Analysis

Historically, elemental analysis involved decomposing substances into pure elements to determine composition. Issues included time consumption, incomplete yields, and missing minor impurities (e.g., Amethyst is $>99.9\% \, SiO_2$ but contains $20\,ppm$ iron).
Combustion Analysis: - Primarily used for organic compounds containing carbon and hydrogen. - The sample is accurately weighed and burned in excess oxygen. - Combustion products are collected and weighed: - All Carbon ( $C$ ) becomes $CO_2$ . - All Hydrogen ( $H$ ) becomes $H_2O$ . - All Sulfur ( $S$ ) becomes $SO_2$ . - Nitrogen ( $N$ ) becomes $N_2$ (after converting oxides). - Oxygen Determination: Oxygen mass is determined by difference. $\text{Mass O} = \text{Total sample mass} - (\text{Mass C} + \text{Mass H} + \text{Mass N} + \text{Mass S})$ .
Comprehensive Example provided: $15.45\,mg$ unknown substance yields $16.83\,mg\,CO_2$ , $8.09\,mg\,H_2O$ , $8.17\,mg\,SO_2$ , and $1.79\,mg\,N_2$ . - Calculates to: $29.73\% \, C$ , $5.86\% \, H$ , $26.47\% \, S$ , $11.59\% \, N$ , and $26.35\% \, O$ (by difference). - Resulting Empirical Formula: $C_3H_7NSO_2$ .

Molarity and Solutions

Solutions are homogeneous mixtures with uniform composition.
Components: 1. Solvent: Component with significantly higher concentration. 2. Solute: Component present at lower concentration. 3. Aqueous Solution: A solution where water is the solvent.
Molarity ( $M$ ): The number of moles of solute in exactly $1\,L$ of solution. - Formula: $M = \frac{\text{mol solute}}{L \text{ solution}}$ . - Example: soft drink containing $0.133\,mol$ sucrose in $355\,mL$ . - $M = \frac{0.133\,mol}{0.355\,L} = 0.375\,M$ .

Dilution of Solutions

Dilution is the process of lowering the concentration of a solution by adding more solvent.
The amount of solute ( $n$ ) does not change during dilution ( $n_1 = n_2$ ).
Dilution Equation: $M_1 L_1 = M_2 L_2$ or more generally $C_1 V_1 = C_2 V_2$ .
Example: Diluting $0.850\,L$ of a $5.00\,M$ $Cu(NO_3)_2$ solution to $1.80\,L$ . - $C_2 = \frac{5.00\,M \times 0.850\,L}{1.80\,L} = 2.36\,M$ .

Alternative Concentration Units

Mass Percentage: $\frac{\text{mass of component}}{\text{mass of solution}} \times 100$ . - Used for liquid mixtures (e.g., bleach is $7.4\%\,NaOCl$ ) and low-abundance solids in ores.
Volume Percentage: $\frac{\text{volume of solute}}{\text{volume of solution}} \times 100$ . - Typically used for liquid-liquid solutions.
Mass-Volume Percentage: $\frac{\text{mass solute}}{\text{volume solution}} \times 100$ . - Common in medical settings (e.g., physiological saline is $0.9\%\,m/v$ or $0.9\,g$ solute per $100\,mL$ solution). Glucose in blood is measured in $mg/dL$ .
Parts per Million (ppm) and Parts per Billion (ppb): - Used for very low solute concentrations. - $ppm = \frac{\text{mass solute}}{\text{mass solution}} \times 10^6$ - $ppb = \frac{\text{mass solute}}{\text{mass solution}} \times 10^9$

Questions & Discussion

Capsaicin and Scoville Scale Case Study: - Capsaicin formula: $C_{18}H_{27}NO_3$ ; Molecular weight: $305.425\,g/mol$ . - A scorpion pepper weighing $10\,g$ with $1\%\,capsaicin$ ( $0.1\,g$ ) has $1 \times 10^6\,SHU$ . - Calculation shows that approximately $2 \times 10^{14}$ molecules of capsaicin are required to produce $1\,SHU$ .
Sample Exam Topics Include: - Nomenclature of ionic and molecular compounds (e.g., Sodium oxide, Cobalt(III) sulfate, $K_2Cr_2O_7$ , $P_4O_6$ ). - Calculating numbers of atoms in a specific mass of glucose. - Finding potential molar mass of hemoglobin based on iron percentage ( $0.34\% \, Fe$ ). - Identifying unidentified metals ( $M$ ) in compounds ( $M_3(PO_3)_2$ ) using mass percentage. - Calculating the final concentration after evaporation of a solution (e.g., $225.5\,mL$ of $0.09988\,M\,Na_2CO_3$ reduced to $45.00\,mL$ ). - Calculating the mass of solid NaOH needed for a weight-percent solution given density. - Converting concentration from $ppb$ to Molarity for contaminants like mercury in water.