Problem 57 - Chemistry: An Atoms-Focused Approach

Question

Phosphorus burns in pure oxygen with a brilliant white light. The product of combustion is \(43.64 \%\) phosphorus and \(56.36 \%\) oxygen. a. What is the empirical formula of this compound? b. The molar mass of the compound is \(284 \mathrm{g} / \mathrm{mol}\). What is its molecular formula?

Step-by-Step Solution

Answer

Question: Using the given information, determine the empirical and molecular formulas of a compound containing 43.64% phosphorus and 56.36% oxygen, with a molar mass of 284 g/mol. Answer: The empirical formula of the compound is P2O5, and the molecular formula of the compound is P4O10.

Step 1: Calculate moles of each element

Given that the compound is \(43.64\%\) phosphorus and \(56.36\%\) oxygen, let's assume we have \(100 \mathrm{g}\) of the compound. Then, we'll have \(43.64 \mathrm{g}\) of phosphorus and \(56.36 \mathrm{g}\) of oxygen. To find the moles of each element, we'll divide the mass by their molar mass. Molar mass of phosphorus (P) = 30.97 \mathrm{g/mol}$ Molar mass of oxygen (O) = 16.00 \mathrm{g/mol}$ Moles of phosphorus = \(\frac{43.64}{30.97}\) Moles of oxygen = \(\frac{56.36}{16.00}\)

Step 2: Find the smallest mole value and divide each moles by the smallest value

Divide the moles of phosphorus and oxygen by the smallest mole value to find the ratio: Moles of phosphorus: \(\frac{43.64}{30.97} = 1.41\) Moles of oxygen: \(\frac{56.36}{16.00} = 3.5225\) Smallest mole value: \(1.41\) Ratio of moles of P: \(\frac{1.41}{1.41} = 1\) Ratio of moles of O: \(\frac{3.5225}{1.41} = 2.5 \approx 2\) (round to the nearest whole number)

Step 3: Write the empirical formula

Combining the ratio of moles of phosphorus and oxygen, we can write the empirical formula: Empirical formula: \(P_1O_2\), or simply \(P_2O_5\)

Step 4: Calculate the molecular formula

Given the molar mass of the compound (\(284 \mathrm{g/mol}\)), we need to find the molar mass of our empirical formula: Molar mass of \(P_2O_5 = 2 \times 30.97 + 5 \times 16 = 141.94 \mathrm{g/mol}\) Divide the molar mass of the compound by the molar mass of the empirical formula: Ratio = \(\frac{284}{141.94} = 2\) Multiply the empirical formula by this ratio to get the molecular formula: \(P_2O_5 \times 2 = P_4O_{10}\)

Step 5: State the answer for both parts

a) The empirical formula of the compound is \(P_2O_5\). b) The molecular formula of the compound is \(P_4O_{10}\).

Key Concepts

Chemical Composition

Chemical composition refers to the identity and quantity of the elements that make up a compound. In a chemical reaction, it's essential to know the composition to determine the empirical and molecular formulas. For example, if a compound consists of phosphorus and oxygen, its chemical composition tells us the percentage of each in the compound.

• The empirical formula is a simplified fraction of the actual number of atoms that are involved in forming a compound.
• The chemical composition will provide insights into finding this empirical formula by reflecting the simplest whole-number ratio of the elements.

Understanding the percentages of each element can help determine how many moles of each are present, leading to the empirical formula. Once we understand the chemical composition, it's easier to predict how these elements interact in further reactions, like combustion.

Combustion Reaction

A combustion reaction occurs when a substance reacts with oxygen to produce heat and light, often seen as burning. In the case of phosphorus and oxygen, combustion is particularly bright due to the intense reaction releasing energy. Combustion reactions are typically represented in form of general equations where a compound, usually containing hydrogen, carbon, or phosphorus, combines with oxygen, resulting in oxides such as carbon dioxide or phosphorus pentoxide in the case of phosphorus.

• These reactions are critical for determining the end products and net chemical changes, such as the transformation into specific oxides.
• Analyzing combustion helps us understand how different elements in a compound react uniquely with oxygen, which is essential in determining the resulting formulas like empirical and molecular formulas in chemical analysis.

Moles and Molar Mass

Moles and molar mass are foundational concepts in chemistry that allow us to quantify elements and compounds. A "mole" is a unit that describes the amount of a substance, containing Avogadro's number of particles, approximately \(6.022 \times 10^{23}\). The molar mass, measured in grams per mole, is the mass of one mole of a substance, corresponding to the atomic or molecular weight in atomic mass units (amu).

• To determine the amount of moles, divide the mass of an element by its molar mass. For example, phosphorus's molar mass is approximately \(30.97 \mathrm{g/mol}\) while oxygen's is \(16.00 \mathrm{g/mol}\).
• This relationship aids in finding the empirical formula by offering a baseline to calculate the simplest ratio of elements present.

Understanding the molecular and empirical formulas relies heavily on these calculations to ensure chemical reactions are accurately balanced and represented.

Phosphorus and Oxygen Compounds

Phosphorus and oxygen compounds, like phosphorus pentoxide, are known for their significant roles in various chemical processes. When phosphorus combusts in oxygen, it typically forms phosphorus(III) and (V) oxides, depending on the reaction conditions. Phosphorus pentoxide ( P_4O_{10} ) is a common form generated from such reactions, especially when analyzing substances containing phosphorus and oxygen.

• Phosphorus compounds are often used in industry for manufacturing elements and as dehydrating agents due to the compound's reactivity with water.
• The empirical formula for phosphorus compounds reflects the simplest ratio, such as P_2O_5 , derived from an empirical analysis.

Understanding these compounds' chemical behavior and interaction is crucial in fields ranging from material sciences to pharmaceuticals, highlighting the importance of mastering concepts like compound formulas in chemistry.