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In a reaction \(2.04 \mathrm{g}\) of vanadium combined with \(1.93 \mathrm{g}\) of sulfur to give a pure compound. What is the empirical formula of the product?

Short Answer

Expert verified
The empirical formula is \( \text{V}_2\text{S}_3 \).

Step by step solution

01

Convert Mass to Moles

First, we need to convert the given masses of vanadium and sulfur into moles using their respective molar masses. The molar mass of vanadium (V) is approximately 50.94 g/mol, and the molar mass of sulfur (S) is approximately 32.07 g/mol.For vanadium: \[ \text{Moles of V} = \frac{2.04 \text{ g}}{50.94 \text{ g/mol}} \approx 0.0400 \text{ mol} \]For sulfur:\[ \text{Moles of S} = \frac{1.93 \text{ g}}{32.07 \text{ g/mol}} \approx 0.0602 \text{ mol} \]
02

Determine the Mole Ratio

Calculate the mole ratio of vanadium to sulfur by dividing the moles of each by the smallest number of moles calculated.The smaller number of moles is that of vanadium (0.0400 mol).\[ \text{Ratio of V} = \frac{0.0400}{0.0400} = 1 \]\[ \text{Ratio of S} = \frac{0.0602}{0.0400} \approx 1.505 \]Round the ratio to the nearest whole number to get the simplest ratio of atoms.
03

Adjust for the Whole Number Ratio

The ratio 1.505 for sulfur is close to 1.5, suggesting a 2:3 ratio when simplified. Therefore, we can multiply each ratio by 2 to eliminate the fraction. Vanadium Ratio: 1 × 2 = 2. Sulfur Ratio: 1.5 × 2 = 3.
04

Write the Empirical Formula

Using the whole number ratio, write the empirical formula of the compound. Vanadium and sulfur have the ratio of 2:3, respectively.Therefore, the empirical formula is \( \text{V}_2\text{S}_3 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mole Conversion
Mole conversion is an essential process in chemistry that allows us to relate the mass of a substance to the number of particles, specifically atoms or molecules, by using the concept of moles. The mole is a fundamental unit in chemistry that represents approximately 6.022 x 10\(^ {23} \) entities, whether they be atoms, molecules, ions, etc. This number is known as Avogadro's Number.

To convert mass to moles, we use the formula:
  • Moles = \( \frac{\text{mass}}{\text{molar mass}} \)
In the given exercise, we converted 2.04 grams of vanadium and 1.93 grams of sulfur to moles by dividing their masses by their respective molar masses (50.94 g/mol for vanadium and 32.07 g/mol for sulfur). This helped us find that there are 0.0400 moles of vanadium and 0.0602 moles of sulfur, establishing a foundation for further calculations.
Mole Ratio
The mole ratio is fundamental in determining the simplest relationship between moles of different elements in a compound. After converting mass to moles, the next step is to establish how these moles compare to each other.

To find the mole ratio, we:
  • Identify the smallest quantity of moles among the elements.
  • Divide each element's mole quantity by this smallest number.
In our example, the smallest number of moles belongs to vanadium (0.0400 moles). By dividing the amount of each element by 0.0400, we derived a mole ratio of 1:1.505. However, empirical formulas need whole numbers, prompting us to round the sulfur ratio to 1.5 and then double all values to get a clear whole number ratio of 2:3.
Empirical Formula Calculation
An empirical formula represents the simplest whole-number ratio of all elements in a compound. Calculating the empirical formula involves a step-by-step simplification of the molar relationships into the smallest whole numbers.

Once mole ratios are determined, as seen in our example where the ratio 1:1.505 became 2:3 after multplication by two, writing the empirical formula requires aligning these ratios with the chemical symbols of the elements involved. Vanadium and sulfur, with a final ratio of 2:3, direct us to write the empirical formula as \( \text{V}_2\text{S}_3 \). This formula shows the simplest ratio of vanadium and sulfur atoms within the compound.

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Most popular questions from this chapter

Stibnite, \(\mathrm{Sb}_{2} \mathrm{S}_{3},\) is a dark gray mineral from which antimony metal is obtained. What is the weight percent of antimony in the sulfide? If you have \(1.00 \mathrm{kg}\) of an ore that contains \(10.6 \%\) antimony, what mass of \(\mathrm{Sb}_{2} \mathrm{S}_{3}\) (in grams) is in the ore?

Which of the following are correct formulas for ionic compounds? For those that are not, give the correct formula. (a) \(\mathrm{Ca}_{2} \mathrm{O}\) (c) \(\mathrm{Fe}_{2} \mathrm{O}_{5}\) (b) \(\mathrm{SrBr}_{2}\) (d) \(\mathrm{Li}_{2} \mathrm{O}\)

Transition metals can combine with carbon monoxide (CO) to form compounds such as \(\mathrm{Fe}(\mathrm{CO})_{5}\) (Study Question 3.82 ). Assume that you combine 0.125 g of nickel with CO and isolate \(0.364 \mathrm{g}\) of \(\mathrm{Ni}(\mathrm{CO})_{x} .\) What is the value of \(x ?\)

Which of the following pairs of elements are likely to form ionic compounds when allowed to react with each other? Write appropriate formulas for the ionic compounds you expect to form, and give the name of each. (a) chlorine and bromine (b) phosphorus and bromine (c) lithium and sulfur (d) indium and oxygen (e) sodium and argon (f) sulfur and bromine (g) calcium and fluorine

Calculate the mass percent of each element in the following compounds: (a) \(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{N}_{2} \mathrm{O}_{2},\) caffeine (b) \(\mathrm{C}_{10} \mathrm{H}_{20} \mathrm{O},\) menthol (c) \(\mathrm{CoCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\)

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