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Chapter 3: Problem 67

Potassium manganate is a dark green, crystalline substance whose composition is \(39.6 \% \mathrm{~K}, 27.9 \% \mathrm{Mn}\), and \(32.5 \% \mathrm{O}\), by mass. What is its empirical formula?

Short Answer

Expert verified
The empirical formula for potassium manganate is \( \text{K}_2\text{MnO}_4 \).

Step by step solution

01

Convert Percentages to Grams

Assume you have 100 grams of potassium manganate. This means you have 39.6 grams of Potassium (K), 27.9 grams of Manganese (Mn), and 32.5 grams of Oxygen (O).
02

Convert Grams to Moles

Use the molar mass of each element to convert grams to moles. The molar mass is approximately 39.10 g/mol for K, 54.94 g/mol for Mn, and 16.00 g/mol for O. Thus:\[\text{Moles of K} = \frac{39.6}{39.10} \approx 1.013 \\text{Moles of Mn} = \frac{27.9}{54.94} \approx 0.508 \\text{Moles of O} = \frac{32.5}{16.00} = 2.031\]
03

Determine the Simplest Whole Number Ratio

To find the simplest ratio, divide each number of moles by the smallest number calculated in Step 2, which is 0.508:\[\frac{1.013}{0.508} \approx 2, \\frac{0.508}{0.508} = 1, \\frac{2.031}{0.508} \approx 4\]
04

Write the Empirical Formula

Using the whole number ratios from Step 3, the empirical formula of potassium manganate is derived as \( \text{K}_2\text{MnO}_4 \).

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Key Concepts

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

Understanding Percent Composition
The term "percent composition" refers to the percentage by mass of each element in a compound. It helps in determining the proportion of each element in a given substance. The percent composition can be crucial, especially when identifying the empirical formula of a compound. Here鈥檚 how it works in practice:
  • To find the percent composition, one must assume a specific mass of the substance. Often, assuming a 100-gram sample simplifies calculations.
  • With 100 grams, the percentage of each element directly translates to its mass. For instance, in potassium manganate, we find 39.6 grams of Potassium.
  • This straightforward conversion from percentage to mass aids in further calculations, such as determining moles.
By understanding percent composition, you lay the groundwork to efficiently proceed with mole calculations, bridging the gap between mass and chemical formulas.
Basics of Mole Calculation
Mole calculations are a fundamental concept in chemistry that allow you to convert between the mass of a substance and the number of particles that substance contains. This process is essential for determining the empirical formula.To convert grams to moles:
  • Use the molar mass of each element, often found on the periodic table. For Potassium (K), Manganese (Mn), and Oxygen (O), their molar masses are approximately 39.10 g/mol, 54.94 g/mol, and 16.00 g/mol, respectively.
  • Divide the mass (in grams) by the molar mass of the element. This formula, \( ext{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \), gives the amount of moles present in your sample.
For potassium manganate, converting mass to moles forms the backbone of finding the simplest whole number ratio and, ultimately, the empirical formula. These calculations connect the reality of laboratory measurements with fundamental chemical quantities.
Exploring Potassium Manganate
Potassium manganate is an inorganic compound notable for its unique properties and uses. This substance is often recognized by its dark green crystalline appearance, which can be deceptive; it transforms to permanganate, a vibrant violet, in certain reactions.Key facts about potassium manganate:
  • Its chemical structure comprises potassium (K), manganese (Mn), and oxygen (O), aligning to the empirical formula \(\text{K}_2\text{MnO}_4\).
  • The formula shows that in each unit of potassium manganate, there are two potassium atoms, one manganese atom, and four oxygen atoms.
  • Known for its strong oxidizing properties, potassium manganate is used in various chemical processes, including the synthesis of other manganese compounds.
Understanding the empirical formula highlights the exact composition and guides chemists in predicting how potassium manganate behaves in reactions. The information can be pivotal for tasks in both research and industrial applications.

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Most popular questions from this chapter

Phosphorus oxychloride is the starting compound for preparing substances used as flame retardants for plastics. An \(8.53\) -mg sample of phosphorus oxychloride contains \(1.72 \mathrm{mg}\) of phosphorus. What is the mass percentage of phosphorus in the compound?

Why is the mass spectrum of a molecule much more complicated than that of an atom?

A titanium ore contains rutile \(\left(\mathrm{TiO}_{2}\right)\) plus some iron oxide and silica. When it is heated with carbon in the presence of chlorine, titanium tetrachloride, \(\mathrm{TiCl}_{4}\), is formed. $$ \mathrm{TiO}_{2}(s)+\mathrm{C}(s)+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{TiCl}_{4}(g)+\mathrm{CO}_{2}(g) $$ Titanium tetrachloride, a liquid, can be distilled from the mixture. If \(35.4 \mathrm{~g}\) of titanium tetrachloride is recovered from \(17.4 \mathrm{~g}\) of crude ore, what is the mass percentage of \(\mathrm{TiO}_{2}\) in the ore (assuming all \(\mathrm{TiO}_{2}\) reacts)?

Find the formula masses of the following substances to three significant figures. a. sulfuric acid, \(\mathrm{K}_{2} \mathrm{SO}_{4}\) b. phosphorus pentachloride, \(\mathrm{PCl}_{5}\) c. potassium sulfite, \(\mathrm{K}_{2} \mathrm{SO}_{3}\) d. calcium hydroxide, \(\mathrm{Ca}(\mathrm{OH})_{2}\)

You perform combustion analysis on \(255 \mathrm{mg}\) of a molecule that contains only \(\mathrm{C}, \mathrm{H}\), and \(\mathrm{O}\), and you find that \(561 \mathrm{mg}\) \(\mathrm{CO}_{2}\) is produced, along with \(306 \mathrm{mg} \mathrm{H}_{2} \mathrm{O}\). a. If the molecule contains only \(\mathrm{C}, \mathrm{H}\), and \(\mathrm{O}\), what is the empirical formula? b. If the molar mass of the compound is \(180 \mathrm{~g} / \mathrm{mol}\), what is the molecular formula of the compound?
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