Chapter 1-6 Review: Empirical Formulas, Combustion Analysis, and Molarity
Empirical formula from percent composition
The empirical formula is the simplest whole-number ratio of atoms or moles of elements in a compound. It represents the most reduced stoichiometric ratio. It can be determined using only the percent composition (mass percent) of the elements. The molecular formula, on the other hand, represents the actual number of atoms of each element in a molecule and may be an integer multiple of the empirical formula (e.g., a compound with empirical formula CH2O could have molecular formulas CH2O, C2H4O2, C3H6O3, etc.).
What information you need to determine the empirical formula:
The percent composition (mass percent) of each element present in the compound. This information is typically obtained through elemental analysis techniques.
Optional, but required for the molecular formula: The molar mass of the compound is needed to determine the molecular formula multiplier. Without it, only the empirical formula can be found.
Key idea: The core principle is to convert the given percentages to relative masses, then to the number of moles for each element, and finally express these moles as the simplest whole-number ratio. Converting to moles allows us to compare the relative number of particles of each element, which is what chemical formulas represent.
From percent to empirical formula: step-by-step procedure (percent
grams
moles
empirical formula)
Step 1: Assume a convenient total mass for the sample. The standard and most convenient practice is to assume 100 g for the sample. This assumption simplifies calculation because the percent values directly translate to grams:
mass of element i: \text{m}i = \text{percent}i \times \frac{100\,\text{g}}{100} = \text{percent}_i\ \text{g} For example, if a compound is 40% carbon, in a 100 g sample, there would be 40 g of carbon.
Step 2: Convert each mass to moles using the atomic/molar masses of the elements (found on the periodic table).
This conversion gives the number of moles (ni) for each element: \text{n}i = \frac{\text{m}i}{\text{M}i} where \text{M}i is the atomic molar mass of element i. For a multi-element compound, it is crucial to use the atomic mass of each individual element (e.g., MC for carbon, M_H for hydrogen).
Step 3: Form the mole ratio by dividing all calculated ni values by the smallest n{\text{min}} among the elements.
The ratio for each element i is: \text{ratio}i = \frac{\text{n}i}{\text{n}_{\text{min}}} This step normalizes the mole values, establishing a relative ratio where the smallest element's subscript becomes 1. You'll typically obtain decimal or fractional subscripts (e.g., 1.3333, 0.6667, 0.25).
Step 4: Clear fractions to obtain whole-number subscripts. This involves multiplying all ratios by the smallest possible integer multiplier, k, such that all subscripts become integers. Common fractional approximations and their multipliers:
If a ratio is approximately 1.3333 (≈ 4/3) or 0.6667 (≈ 2/3), multiply all ratios by 3 to clear the fraction.
If a ratio is approximately 0.25 (≈ 1/4) or 0.75 (≈ 3/4), multiply all ratios by 4.
If a ratio ends in 0.5 (≈ 1/2), multiply all ratios by 2.
If a ratio ends in other simple fractions (e.g., 0.2, 0.4, 0.6, 0.8 which correspond to 1/5, 2/5, 3/5, 4/5), multiply accordingly (by 5, etc.). It is important to remember to multiply all ratios by the same integer k to maintain the correct overall ratio.
Step 5: The resulting whole-number subscripts give the empirical formula. For example, an outcome such as 7 moles of Carbon, 9 moles of Hydrogen, and 7 moles of Oxygen, after simplifying and clearing fractions, would give C7H9O7.
Step 6: (If molar mass is known) Compute the empirical formula mass:
The empirical formula mass, M{\text{emp}}, is the sum of the atomic masses of all atoms in the empirical formula: \text{M}{\text{emp}} = \sumi \text{a}i \text{M}i where \text{a}i are the empirical subscripts determined in Step 5.
Compare this to the given molar mass of the compound, M_{\text{mol}}. The ratio, n (the molecular formula multiplier), should be a small positive integer:
\text{n} = \frac{\text{M}{\text{mol}}}{\text{M}{\text{emp}}} This value of n tells you how many empirical formula units are present in one molecular formula unit.
The molecular formula is then obtained by multiplying all subscripts in the empirical formula by n: \text{Molecular formula} = \text{n} \times \text{empirical formula}.
Important note: If you only know the percentages and do not know the molar mass, you can determine the empirical formula but cannot uniquely determine the molecular formula because there could be multiple compounds (e.g., CH2O, C2H4O2) with the same empirical formula.
Worked ideas and common in-class routes
Route commonly discussed in class: Start with a 100 g sample, convert the masses of C, H, O (and any other elements) to moles, find the lowest mole ratio by dividing by the smallest mole value, clear any fractions to get whole numbers, and deduce the empirical formula. If the resulting empirical molar mass is a factor of the unknown molar mass, multiply the empirical formula accordingly by that factor to obtain the molecular formula. This is the most robust and widely applicable method.
If you obtain fractional subscripts (e.g., 1.3333, which is 4/3), multiplying all subscripts by 3 yields whole numbers (e.g., 4 after scaling, if the original value was 1.3333). Similarly for other common fractions (0.5 (1/2) factor of 2, 0.25 (1/4) factor of 4, etc.).
A second route mentioned (alternative but less generally applicable): This method might use a known mass basis for a specific element to anchor a direct path to the molecular formula. For example, if you know the exact mass of carbon in a specific mass of the molecular compound, you could deduce molar relationships directly to the molecular formula. However, this path is correct in principle but depends on having the known molar mass of the molecule from the outset, and it isn't as universally reliable or straightforward as the standard percent → grams → moles route, which can always yield at least the empirical formula.
The instructor emphasized that different valid routes can lead to the same molecular formula as long as each step is clearly described, justified, and mathematically sound. Understanding the underlying chemical principles is more important than memorizing a single approach.
Example context discussed in lecture
An example scenario mentioned involved a fractional subscript like 1.3333, which clearly demonstrates the need for multiplying by 3 to give whole numbers (e.g., convert 1.33 moles to 4 units in the whole-number ratio).
Another common example mentioned an empirical formula like C7H9O7 (as a result of the step-by-step process) and highlighted the idea that the empirical formula mass would typically be smaller than the molecular mass by a whole-number factor (n = 1, 2, 3…). This 'n' value scales the empirical formula to the molecular formula.
In one slide, a hypothetical molecular mass of 130 g/mol was used to illustrate how to determine the multiplier n between the empirical formula mass and the molecular mass, demonstrating how M{mol} is divided by M{emp} to find this integer factor.
Combustion analysis: concept, data, and calculation flow
Purpose: Combustion analysis is a crucial technique used primarily to determine the elemental composition (specifically carbon, hydrogen, and sometimes oxygen, nitrogen, and sulfur) of organic compounds. It is particularly useful when the compound's structure is unknown but its empirical formula is sought.
Key idea: An unknown organic sample is completely burned (combusted) in an atmosphere of excess oxygen. All the carbon in the sample is converted to carbon dioxide (CO2) and all the hydrogen to water (H2O). Any oxygen in the original sample cannot be directly determined from the CO2 and H2O masses because additional oxygen is supplied by the excess O_2 gas. Therefore, you first deduce the masses of C and H from the products, and then determine the oxygen content of the original sample by mass balance.
General reaction picture (conceptual):
CxHyOz + \text{excess } O2 \rightarrow xCO2 + \frac{y}{2}H2O + \text{unreacted } O_2
The chemical equation demonstrates how each carbon atom in the sample forms one CO2 molecule, and each pair of hydrogen atoms forms one H2O molecule. Excess oxygen ensures complete combustion.
Practical workflow described in class:
Precisely weigh a sample of the unknown organic compound.
Place the sample in a combustion apparatus and burn it completely in a stream of excess oxygen at high temperatures.
The gaseous products (CO2 and H2O) are then swept by the gas stream through absorption traps. H2O is typically absorbed first (e.g., by magnesium perchlorate), followed by CO2 (e.g., by soda lime). The increase in mass of these traps directly corresponds to the mass of H2O and CO2 produced.
Use these measured masses of CO2 and H2O to compute the grams (and subsequently moles) of C and H that were present in the original sample.
Subtract the calculated masses of C and H from the original total sample mass to obtain the mass of oxygen (if any) in the original sample. This is the mass balance step: \text{mass of O} = \text{mass of sample} - \text{mass of C} - \text{mass of H}.
Important caveat: As oxygen atoms in CO2 and H2O can originate from either the sample or the O2 gas reacting with it, only the carbon and hydrogen content can be directly and unambiguously determined from the CO2 and H_2O product masses. The oxygen in the sample must be found by difference.
Equations often used in combustion analysis:
From CO2 mass to carbon mass: If you measure the mass of CO2 produced, m{\text{CO2}}, then the carbon mass (mC) is derived from the molecular weight ratio:
mC = m{\text{CO2}} \cdot \frac{MC}{M{\text{CO2}}} = m{\text{CO2}} \cdot \frac{12.01\,\text{g/mol}}{44.01\,\text{g/mol}} Each mole of CO2 contains one mole of C.
From H2O mass to hydrogen mass: If you measure the mass of H2O produced, m{\text{H2O}}, then the total hydrogen mass (mH) is calculated considering there are two hydrogen atoms per water molecule:
mH = m{\text{H2O}} \cdot \frac{2 \cdot MH}{M{\text{H2O}}} = m{\text{H2O}} \cdot \frac{2.016\,\text{g/mol}}{18.015\,\text{g/mol}} Each mole of H2O contains two moles of H.
Oxygen mass in the original sample: If the original sample mass is m{\text{sample}}, then the mass of oxygen (mO) in the original sample is determined by subtracting the masses of C and H from the total sample mass:
mO = m{\text{sample}} - mC - mH
Practical notes on data handling:
Due to the nature of stoichiometric calculations, even small calculation errors, especially in early steps (like measuring product masses), can significantly propagate and throw off the final empirical/molecular conclusions. It is crucial to carry enough significant figures through intermediate calculations.
If a result yields non-physical fractional subscripts (e.g., 0.1 or 0.9 for an integer) or inconsistent totals, it's important to flag it. In an academic setting, it is often advisable to acknowledge and justify the suspected error to potentially gain partial credit rather than presenting a clearly erroneous answer without comment.
It is common to present the steps clearly and note where a slight deviation from perfectly whole-number ratios might have occurred; instructors often reward partial credit for correctly arranged setup and clearly outlined steps, even if the final answer has minor rounding issues.
Chemical notation reminders (brief):
States in reactions are written as (g) for gas, (l) for liquid, (s) for solid, and (aq) for aqueous solution.
Delta (\Delta) typically denotes heat transfer, often shown above the reaction arrow to indicate heating.
The chemical formulas on the left side of the arrow represent reagents (reactants), and those on the right side represent products; the arrow shows the direction of the reaction.
A reaction like combustion typically produces CO2 and H2O when a hydrocarbon (a compound with C, H, and possibly O) reacts with O_2.
Introduction to molarity and related concepts
Molarity (M): Molarity is one of the most common units used to express the concentration of a chemical solute in a solution. It is defined as the number of moles of solute per liter of the entire solution:
\text{M} = \frac{\text{n}{\text{solute}}}{\text{V}{\text{solution}}}
Where n{\text{solute}} is the number of moles of the dissolved substance (solute) and V{\text{solution}} is the total volume of the solution in liters. It is crucial that the volume is for the entire solution, not just the solvent.
Example in lab practice (as described):
To prepare a 1 M NaCl solution, one would precisely weigh 1 mole of solid NaCl (approximately 58.44 g/mol for NaCl). This precisely measured amount is then quantitatively transferred into a volumetric flask. To achieve exactly 1.00 L of solution, water (the solvent) is added to the flask until the bottom of the meniscus precisely aligns with the 1.00 L mark. This yields a 1 M NaCl solution, meaning there is 1 mole of NaCl dissolved in 1 liter of solution total.
Important nuance: The volume used in the denominator for molarity is the volume of the solution, which includes the volume occupied by the dissolved solute, not merely the initial volume of the solvent. Dissolving a solute can affect the total volume, so it's about making the solution up to a specified total volume, typically using a volumetric flask.
Related concept: Molality (m): Molality is another measure of concentration, defined as the number of moles of solute per kilogram of solvent. It is given by:
\text{m} = \frac{\text{n}{\text{solute}}}{\text{m}{\text{solvent}}} with the solvent's mass expressed in kilograms ($\text{kg}$). Typically, molality is often used for colligative properties.
Distinction from molarity:
Molarity depends on the volume of the solution, which can change significantly with temperature (as liquids expand or contract) and with the addition of different solutes. This makes molarity temperature-dependent.
Molality depends on the mass of the solvent (not volume) and is therefore relatively constant with changes in temperature. Mass is a temperature-independent property, making molality a more suitable unit for studies where temperature variations are significant or for highly concentrated solutions.
Practical takeaway from the class discussion:
The lab workflow for preparing solutions emphasizes the importance of precise measurement using appropriate glassware (like volumetric flasks) and a clear understanding of what is being counted (solute vs. solvent vs. total solution) and the exact definition of the concentration variable being used (e.g., molarity vs. molality).
Connections to previous topics and real-world relevance
Empirical formulas are foundational not just for determining molecular formulas but also for understanding the fundamental composition of all chemical compounds. They are crucial in materials science (e.g., in characterizing polymers, ceramics, and alloys), biochemistry (e.g., determining the elemental ratios in macromolecules like sugars), and stoichiometry, providing the basis for quantitative chemical reactions.
Combustion analysis links directly to principles of mass balance and conservation of mass, stoichiometry (mole ratios in reactions), and elemental analysis. It is a cornerstone technique in analytical chemistry, used for quality control in pharmaceuticals, characterizing new organic compounds, environmental testing (e.g., analyzing pollutants), and even forensic science (e.g., identifying unknown substances and residues).
The concepts of molarity and molality underpin countless chemical calculations in solution chemistry, essential for tasks like preparing reagents for experiments, understanding reaction rates in solutions, and drug dosages in pharmacology. They are vital in biochemistry (e.g., buffer preparation, cell culture media) and chemical engineering (e.g., process design, industrial scaling of reactions).
Understanding how to handle fractional subscripts, deal with potential calculation errors, and the awarding of partial credit emphasizes the broader message of good problem-solving practices, clear communication of steps and assumptions, and physical reasoning. These are critical skills applicable across all quantitative scientific and engineering disciplines.
Quick reference: key formulas (LaTeX)
Percent to grams (assuming 100 g sample): \text{m}i = \%i \quad\text{g}
Moles from mass: \text{n}i = \frac{\text{m}i}{\text{M}_i}
Empirical formula mass: \text{M}{\text{emp}} = \sumi \text{a}i \text{M}i
Molecular formula multiplier: \text{M}{\text{mol}} = \text{n} \cdot \text{M}{\text{emp}}, \quad \text{n} = \frac{\text{M}{\text{mol}}}{\text{M}{\text{emp}}}
Molarity: \text{M} = \frac{\text{n}{\text{solute}}}{\text{V}{\text{solution}}} (Volume in Liters)
Combustion: from CO2 to C mass: \text{m}C = \text{m}{\text{CO2}} \cdot \frac{\text{M}C}{\text{M}{\text{CO2}}} = \text{m}{\text{CO2}} \cdot \frac{12.01}{44.01}
Combustion: from H2O to H mass: \text{m}H = \text{m}{\text{H2O}} \cdot \frac{2 \cdot MH}{M{\text{H2O}}} = \text{m}{\text{H2O}} \cdot \frac{2.016}{18.015}
Oxygen in sample (by mass difference): \text{m}O = \text{m}{\text{sample}} - \text{m}C - \text{m}H