Question

When \(10.00 \mathrm{~g}\) of phosphorus was reacted with oxygen, it produced \(17.77 \mathrm{~g}\) of a phosphorus oxide. This phosphorus oxide was found to have a molecular weight of approximately 220 in the vapor state. Determine its molecular formula.

Short answer

Step 1: Find the mass of oxygen in the phosphorus oxide: \[m_{O} = 17.77 \mathrm{~g} - 10.00 \mathrm{~g} = 7.77 \mathrm{~g}\] Step 2: Calculate moles of phosphorus and oxygen: \[n_{P} = \frac{10.00 \mathrm{~g}}{30.97 \mathrm{~g/mol}} = 0.323 \, mol\] \[n_{O} = \frac{7.77 \mathrm{~g}}{16.00 \mathrm{~g/mol}} = 0.486 \, mol\] Step 3: Find the empirical formula: \[\frac{n_{P}}{n_{min}} : \frac{n_{O}}{n_{min}} = \frac{0.323}{0.323} : \frac{0.486}{0.323} = 1 : 1.5 \approx 2 : 3\] The empirical formula is \(P_2O_3\). Step 4: Determine the molecular formula: \[n = \frac{220}{(2 \times 30.97) + (3 \times 16.00)} = \frac{220}{109.94} \approx 2\] So, the molecular formula is \(2 \times P_2O_3 = P_4O_6\).

Step-by-step solution

Find the mass of oxygen in the phosphorus oxide

Given that the mass of phosphorus reacted is 10.00 g and the mass of the phosphorous oxide produced is 17.77 g, we can find the mass of oxygen in the compound by finding the difference between the two masses: \[m_{O} = m_{compound} - m_{P}\]

Calculate moles of phosphorus and oxygen

First, we need to find the number of moles for both the phosphorus and the oxygen in the compound, using their respective molar masses (P: 30.97 g/mol, O: 16.00 g/mol): \[n_{P} = \frac{m_{P}}{M_{P}}\] \[n_{O} = \frac{m_{O}}{M_{O}}\]

Find the empirical formula

To find the empirical formula, we need to determine the ratio of moles of phosphorus and oxygen in the compound. Calculate the mole ratio by dividing both mole values by the smallest one: \[\frac{n_{P}}{n_{min}} : \frac{n_{O}}{n_{min}}\] This will give us the empirical formula.

Determine the molecular formula

Using the molecular weight given in the problem (approximately 220), and the molecular weight calculated from the empirical formula, we can deduce the molecular formula: \[Molecular \, Formula = n \times Empirical \, Formula\] where \(n = \frac{Molecular \, Weight}{Empirical \, Formula \, Weight}\) Now, let's calculate the molecular formula using the given information:

Key concepts

Empirical Formula

Understanding the empirical formula is essential for anyone studying chemistry. In simple terms, the empirical formula of a compound tells us the simplest whole number ratio of the elements within it. It doesn't indicate the actual number of atoms, but rather the basic proportion in which the atoms combine to form the compound. For example, the empirical formula of hydrogen peroxide (H₂O₂) is simply HO.

To find an empirical formula, you begin by determining the mass of each element in a sample of the compound. This information is then converted into moles since chemical reactions depend on the mole ratio of reactants and products rather than their mass ratio. Once the number of moles of each element is known, these values are divided by the smallest mole value among the elements present to get the simplest ratio.

In the exercise provided, one would calculate the moles of phosphorus (P) and oxygen (O) from their respective masses and molar masses, then derive a mole ratio. This ratio would then be simplified to whole numbers, giving the empirical formula of the phosphorus oxide.

Molecular Weight

Molecular weight, also known as molecular mass, is the total mass of all the atoms in a molecule expressed in atomic mass units (u) or grams per mole (g/mol). It reflects the mass of one molecule when compared to one-twelfth the mass of a carbon-12 atom, which is the standard for measuring atomic masses.

To calculate the molecular weight, you add up the atomic masses of each atom in the molecule as indicated by its chemical formula. For instance, the molecular weight of water (H₂O) is approximately 18 g/mol, coming from two hydrogen atoms (each approximately 1 g/mol) and one oxygen atom (approximately 16 g/mol).

In the context of our problem-solving process, once we have the empirical formula, we can calculate its weight, and then compare that to the actual molecular weight given. If the weight of the empirical formula is lesser than the molecular weight provided in the problem, it indicates that the molecule's true formula is a multiple of the empirical formula.

Stoichiometry

Stoichiometry is a section of chemistry that involves calculating the quantities of reactants and products during chemical reactions. It is based on the conservation of mass where the mass of the reactants equals the mass of the products. Stoichiometry uses the relationship between the amounts in moles to ensure that atoms are conserved in a reaction, enabling accurate predictions of product formation.

In practice, stoichiometry allows us to convert from mass to moles using the molar mass, and then use the mole ratio (derived from the chemical equation of the reaction or the empirical formula) to understand the proportions in which different reactants combine or the amount of product produced. For the exercise we are analyzing, stoichiometry is employed to find out the amount of oxygen that reacted with the phosphorus and then use the mole ratio to deduce the empirical formula of the resulting phosphorus oxide. This will subsequently lead to determining the actual molecular formula of the compound.