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A compound contains equal masses of the elements \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\). If the atomic masses of \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) are 20,40 and 60 , respectively, the empirical formula of the compound is (a) \(\mathrm{A}_{3} \mathrm{~B}_{2} \mathrm{C}\) (b) \(\mathrm{AB}_{2} \mathrm{C}_{3}\) (c) \(\mathrm{ABC}\) (d) \(\mathrm{A}_{6} \mathrm{~B}_{3} \mathrm{C}_{2}\)

Short Answer

Expert verified
The empirical formula of the compound is \(\mathrm{A}_{6} \mathrm{~B}_{3} \mathrm{C}_{2}\), option (d).

Step by step solution

01

- Calculate the mole ratio

Assume that the mass of each element in the compound is 1 unit for simplicity. Calculate the number of moles of each element by dividing the mass by its atomic mass. For element A, the number of moles is 1 g / 20 g/mol = 0.05 mol. For B, it's 1 g / 40 g/mol = 0.025 mol. For C, it's 1 g / 60 g/mol = 0.0167 mol.
02

- Determine the simplest whole number ratio

Find the simplest whole number ratio of the moles of each element. To do this, divide each mole value by the smallest number of moles calculated in Step 1. This gives the ratio for A: 0.05 mol / 0.0167 mol = 3, for B: 0.025 mol / 0.0167 mol = 1.5, and for C: 0.0167 mol / 0.0167 mol = 1.
03

- Convert ratios to integers

Since we are looking for the empirical formula, which is represented by whole numbers, we need to multiply each ratio by a factor that converts all the ratios to the smallest possible whole numbers. Multiplying the ratio for B by 2 gives 3, which also leads to multiplying the ratio for A by 2 to get a whole number, giving 6. Thus, the whole number ratios are A: 6, B: 3, and C: 2.
04

- Write the empirical formula

Using the whole number mole ratios, the empirical formula of the compound is A6B3C2. This formula corresponds to the option (d) A6B3C2.

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Key Concepts

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

Stoichiometry
Stoichiometry involves the calculation of reactants and products in chemical reactions. It is a section of chemistry that focuses on the quantities of substances consumed and produced in reactions. Effective stoichiometry calculations depend on the balanced chemical equation, as it provides the ratios in which reactants combine and products form. These ratios are necessary to predict yields of reactions and to ensure that experiments are planned with the appropriate amounts of materials.
To understand stoichiometry, one should be familiar with the concept of the mole, which is a standard unit of measurement for amount of substance in chemistry. Stoichiometry is not just about balancing equations; it's about understanding the relationships between different components in a reaction. By mastering stoichiometry, students can comprehend how to conserve mass in reactions, as matter is neither created nor destroyed in chemical processes, aligning with the law of conservation of mass. When working with stoichiometric calculations, always check units and ensure that they are consistent across the substances involved.
Mole Concept
The mole concept is a fundamental aspect in chemistry that allows chemists to count atoms, molecules, ions, and other entities by weighing them. One mole is equivalent to Avogadro's number (\(6.022 \times 10^{23}\)) of particles. The mass of one mole of a substance in grams corresponds to its atomic or molecular mass in atomic mass units (u).
In the context of the empirical formula calculation, we use the mole concept to determine the ratio of atoms in a compound. By comparing the moles of each element present, we establish the simplest whole number ratio of elements in the compound. This method ensures that we can reliably determine formulas and compare different substances on an equal basis. Remember that the accuracy of the mole calculations directly impacts the determination of the empirical formula, so precise measurement and conversion are crucial.

Importance of the Mole

The mole allows chemists to:
  • Translate microscopic measurements into macroscopic quantities.
  • Perform calculations involving the mass and volume of substances.
  • Determine the composition of compounds and their molecular formulas.
It serves as a bridge between the atomic scale and the scale of everyday chemical amounts.
Chemical Formula Determination
Determining the chemical formula of a compound is a multi-step process that involves empirical and molecular formulas. The empirical formula represents the simplest integer ratio of the elements in the compound, while the molecular formula is the actual number of atoms of each element in a molecule of the compound.
To find the empirical formula, one must:
  1. Obtain the mass or percentage composition of each element.
  2. Convert these masses to moles using the atomic masses.
  3. Calculate the simplest mole ratio of the elements.
  4. Adjust these ratios to whole numbers if necessary, because empirical formulas are based on whole number ratios.
In the exercise provided, we used masses and atomic weights to determine the ratio of elements in moles, then found the simplest whole number ratio to write the empirical formula. Students should pay special attention to rounding and adjustment in the ratio part, as slight variations can lead to incorrect empirical formulas. Additionally, understanding that empirical formulas may not reflect the true number of atoms in molecules but rather the smallest ratio is key. This understanding leads to the determination of molecular formulas, when molecular masses are known, as a later step in chemical formula determination.

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Most popular questions from this chapter

A certain vitamin extracted from plant sources has carbon and hydrogen in \(8: 1\) mass ratio. The percentage of oxygen is nearly 7.3. The compound gave no test for nitrogen or sulphur or any other element. What should be the empirical formula of the compound? (a) \(\mathrm{C}_{30} \mathrm{H}_{45} \mathrm{O}_{2}\) (b) \(\mathrm{C}_{15} \mathrm{H}_{23} \mathrm{O}\) (c) \(\mathrm{C}_{29} \mathrm{H}_{45} \mathrm{O}_{3}\) (d) \(\mathrm{C}_{10} \mathrm{H}_{15} \mathrm{O}\)

Out of \(1.0 \mathrm{~g}\) dioxygen, \(1.0 \mathrm{~g}\) atomic oxygen and \(1.0 \mathrm{~g}\) ozone, the maximum number of oxygen atoms are contained in (a) \(1.0 \mathrm{~g}\) of atomic oxygen (b) \(1.0 \mathrm{~g}\) of ozone (c) \(1.0 \mathrm{~g}\) of oxygen gas (d) All contain the same number of atoms

Ethanol is the substance commonly called alcohol. The density of liquid alcohol is \(0.8 \mathrm{~g} / \mathrm{ml}\) at \(293 \mathrm{~K}\). If \(1.2\) moles of ethanol is needed for a particular experiment, what volume of ethanol should be measured out? (a) \(55.2 \mathrm{ml}\) (b) \(57.5 \mathrm{ml}\) (c) \(69 \mathrm{ml}\) (d) \(47.9 \mathrm{ml}\)

A quantity of \(4.35 \mathrm{~g}\) of a sample of pyrolusite ore, when heated with conc. HCl, gave chlorine. The chlorine, when passed through potassium iodide solution, liberated \(6.35 \mathrm{~g}\) of iodine. The percentage of pure \(\mathrm{MnO}_{2}\) in the pyrolusite ore is \((\mathrm{Mn}=55, \mathrm{I}=127)\) (a) 40 (b) 50 (c) 60 (d) 70

An amount of 1 mole of calcium cyanamide and 1 mole of water are allowed to react. The number of moles of ammonia produced is (a) \(3.0\) (b) \(2.0\) (c) \(1.0\) (d) \(0.67\)

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