Problem 142 - Chemistry and Chemical Reactivity

Question

In a reaction, 2.04 g of vanadium combined with 1.93 g of sulfur to give a pure compound. What is the empirical formula of the product?

Step-by-Step Solution

Answer

The empirical formula is V2S3.

Step 1: Convert Masses to Moles

To find the empirical formula, we first need to convert the given masses of vanadium (V) and sulfur (S) into moles. The atomic mass of vanadium is approximately 50.94 g/mol, and for sulfur, it is approximately 32.06 g/mol.For vanadium: \[\text{Moles of vanadium} = \frac{2.04 \text{ g}}{50.94 \text{ g/mol}} \approx 0.04006 \text{ mol}\]For sulfur: \[\text{Moles of sulfur} = \frac{1.93 \text{ g}}{32.06 \text{ g/mol}} \approx 0.06019 \text{ mol}\]

Step 2: Determine the Simplest Mole Ratio

The simplest whole number ratio of the moles gives us the empirical formula. Divide each quantity in moles by the smallest number of moles calculated:For vanadium:\[\frac{0.04006}{0.04006} = 1\]For sulfur:\[\frac{0.06019}{0.04006} \approx 1.503\]The closest whole number ratio is needed, so we multiply both by 2 to avoid fractions, giving us vanadium as 2 and sulfur as 3.

Step 3: Write the Empirical Formula

The whole number ratio between vanadium and sulfur is 2:3. Therefore, the empirical formula of the compound is V2S3.

Key Concepts

Vanadium Sulfide

In chemistry, vanadium sulfide is an important compound that results from the reaction between vanadium and sulfur elements. Understanding its empirical formula can reveal much about its composition and chemical characteristics.
Vanadium sulfide is a compound formed through the combination of vanadium and sulfur, whose reaction yields different ratios based on experimental conditions. In the context of the empirical formula, the focus is on determining the simplest whole number ratio of atoms. This gives insight into the relative amount of each element in the compound. The empirical formula for vanadium sulfide shown in the example above is V₂S₃, which means in any given sample of this compound, there are always 2 atoms of vanadium for every 3 atoms of sulfur. This ratio is crucial for understanding the basic molecular structure needed for further studies in chemistry, such as crystallography and reaction mechanisms.
It's important to remember that the empirical formula is not always indicative of other properties like molecular formula, which could differ depending on the actual number of atoms in a definitive molecule.

Mole Calculation

To find the empirical formula of a compound, we start with mole calculations. Moles are a standard unit of measurement in chemistry that helps translate between mass and number of particles.
In the context of the example provided, we were given masses of vanadium and sulfur. To convert these masses to moles, we use the molar mass, which is the mass of one mole of an element in grams. For vanadium, the molar mass is approximately 50.94 g/mol, and for sulfur, it's 32.06 g/mol. By dividing the mass of the elements by their respective molar masses, you obtain the number of moles:

• Moles of vanadium = \( \frac{2.04 \text{ g}}{50.94 \text{ g/mol}} \approx 0.04006 \text{ mol} \)
• Moles of sulfur = \( \frac{1.93 \text{ g}}{32.06 \text{ g/mol}} \approx 0.06019 \text{ mol} \)

This conversion is critical as it allows chemists to compare quantities of substances at the atomic level, making further calculations and reactions understandable and manageable.

Simplest Mole Ratio

The concept of the simplest mole ratio is essential in determining the empirical formula. It reflects the smallest whole number ratio of moles of each element in a compound.
In our example of vanadium sulfide, we calculated the moles of vanadium and sulfur. The next step involves determining the simplest ratio by dividing each element's mole quantity by the smallest mole value obtained:

• Vanadium's ratio = \( \frac{0.04006}{0.04006} = 1 \)
• Sulfur's ratio = \( \frac{0.06019}{0.04006} \approx 1.503 \)

As the result isn't a whole number, it's necessary to adjust to the nearest whole number by multiplying all numbers by the same integer to eradicate fractions. Here, multiplying by 2 gives a 2:3 ratio for vanadium and sulfur, respectively. This simplification step is crucial for converting precise pounds or grams into practical, understandable chemical formulas.

Stoichiometry

Stoichiometry is the study of the quantitative relationships or ratios between reactants and products in a chemical reaction. It's an essential part of figuring out empirical formulas.
Stoichiometry allows chemists to use these relationships to predict outcomes of chemical reactions, including the amounts of products formed. By knowing empirical formulas, one can understand the stoichiometric relationships in reactions.
When determining empirical formulas as in the vanadium sulfide example, stoichiometry ensures that the chemical equations reflect the balanced proportions necessary for reactions. This provides the foundation for calculating reactants or products in grams, liters, or moles, depending on the context. These calculations are vital for chemical manufacturing, pharmaceuticals, and any scientific research involving chemical analysis. Recognizing these foundational balance equations helps in accurately conducting experiments and generating reproducible results.