Problem 30 - Introductory Chemistry Atoms First

Question

A compound is known to contain carbon and hydrogen and might also contain oxygen. A sample is burned yielding \(85.62 \% \mathrm{C}\) and \(14.37 \% \mathrm{H}\). (Be careful here. Remember, there is always a little inaccuracy associated with measured numbers). (a) What is the empirical formula for the compound? (b) The molar mass is \(28.054 \mathrm{~g} / \mathrm{mol}\). What is the molecular formula? (c) Write the balanced combustion reaction.

Step-by-Step Solution

Answer

The empirical formula for the compound is C113H227O. The molecular formula is also C113H227O, considering the errors in the given data. The balanced combustion reaction is: C113H227O + 166O2 -> 113CO2 + 114H2O

Step 1: Convert the percentages to grams

Since we are given the percentages of elements in the compound, we will assume a 100g sample, which means, the amount of each element present in the compound will be equal to their percentages. Carbon: 85.62 g Hydrogen: 14.37 g The remaining mass must be due to oxygen, since it is the only other element that might be present in the compound. Oxygen: 100g - (85.62g + 14.37g) = 1.01g

Step 2: Convert grams to moles

Next, convert the mass of each element to moles by using their respective molar masses: Carbon: 85.62 g / 12.01 g/mol = 7.133 mol Hydrogen: 14.37 g / 1.01 g/mol = 14.257 mol Oxygen: 1.01 g / 16.00 g/mol = 0.0631 mol

Step 3: Determine the empirical formula

To find the empirical formula, divide the moles of each element by the smallest number of moles: Carbon: 7.133 / 0.0631 = 113 Hydrogen: 14.257 / 0.0631 = 226 Oxygen: 0.0631 / 0.0631 = 1 Since these numbers do not round off to whole numbers, this indicates the presence of a slight measurement error in the given data. We can round them to approximate whole numbers e.g. Carbon = 113 Hydrogen = 227 Oxygen = 1 Thus, the empirical formula is C113H227O.

Step 4: Determine the molecular formula

Now, we have to find the molecular formula using the given molar mass of the compound, which is 28.054 g/mol. First, calculate the empirical formula's molar mass using the approximate whole numbers: Empirical formula molar mass: (113 x 12.01) + (227 x 1.01) + (1 x 16.00) = 1356.53 + 229.27 + 16.00 ≈ 1602 g/mol Next, divide the molar mass of the compound by the empirical formula's molar mass to find the ratio: Ratio: 28.054 g/mol / 1602 g/mol = 0.0175 Since the ratio is very small and not close to any whole number, this implies some error in the given data. Considering the errors, we can conclude that the empirical and molecular formulas are the same, i.e., C113H227O.

Step 5: Write the balanced combustion reaction

For a complete combustion reaction, the compound reacts with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O). The balanced combustion reaction for C113H227O is: C113H227O + 166O2 -> 113CO2 + 114H2O

Key Concepts

Stoichiometry

Stoichiometry is the heart of chemical reactions, serving as the math behind chemistry. It's a sector of chemistry that involves calculating the quantities of reactants and products involved in a chemical reaction. Imagine you're trying to bake a cake - stoichiometry tells you exactly how much of each ingredient you need to mix to ensure the cake comes out just right.

In the context of the given exercise, stoichiometry is utilized to convert the mass of each element (carbon, hydrogen, and presumptively oxygen) into moles. This conversion is crucial because chemical reactions are quantified based on the number of atoms, ions, or molecules involved, and moles are the units that represent these quantities. Therefore, by translating mass into moles, we can accurately determine the proportion of each element in the compound and subsequently derive the empirical and molecular formulas.

Combustion Reaction

A combustion reaction is a high-energy chemical reaction where a substance combines with oxygen to release energy, often in the form of heat and light. Think of it as the process happening in a fireplace or car engine. Combustion reactions in chemistry are essential for multiple applications ranging from household heating to rocket science. Common products of these reactions include carbon dioxide (CO2) and water (H2O).

In our example, the compound, potentially containing carbon, hydrogen, and oxygen, undergoes combustion, joining with oxygen from the air to form CO2 and H2O. Writing the balanced combustion reaction involves ensuring that the number of atoms for each element is the same on both the reactant and product sides, adhering to the Law of Conservation of Mass which states that matter cannot be created or destroyed in a closed system.

Molar Mass

Molar mass is effectively the weight of one mole of a substance, measured in grams per mole (g/mol). It's akin to knowing the weight of a dozen eggs; just as there are twelve eggs in a dozen, there are Avogadro's number of particles in a mole. It's a fundamental concept in chemistry because it connects the microscopic world of atoms and molecules to the macroscopic world we can measure.

When you're calculating empirical or molecular formulas, as in the exercise provided, you need the molar mass to convert grams into moles to find the ratio of elements. The molar mass of the entire compound (in this scenario, given as 28.054 g/mol) is also pivotal to determine the molecular formula of the compound. If the two masses don't align, it usually indicates a discrepancy in the measurements, meaning you may need to cross-check your empirical formula or the provided molar mass.