Question

Two elements \(\mathrm{A}\) and \(\mathrm{B}\) combine to form compound \(X\) and \(Y\). For the fix mass of A, masses of \(\mathrm{B}\) combined for the compounds \(\mathrm{A}\) and \(\mathrm{B}\) are in \(3: 7\) ratio. If in compound \(X, 4 g\) of \(A\) combines with \(12 \mathrm{~g} \mathrm{~B}\), then in compound \(\mathrm{Y}, 8 \mathrm{~g}\) of \(\mathrm{A}\) will combine with \(\ldots \ldots \mathrm{g}\) of \(\mathrm{B}\). (a) 24 (b) 56 (c) 28 (d) 8

Short answer

The mass of B that combines with 8 g of A in compound Y is 28 g.

Step-by-step solution

Understanding the Problem

The problem states that there are two compounds, X and Y, formed by the elements A and B. For compound X, it is given that 4 g of A combines with 12 g of B. It is also given that the masses of B that combine with a fixed mass of A are in a 3:7 ratio for compounds X and Y respectively. We need to find the mass of B that combines with 8 g of A to form compound Y.

Calculate the mass ratio of B to A in compound X

In compound X, for 4 g of A, we have 12 g of B. Therefore, the mass ratio of B to A in compound X is 12 g of B per 4 g of A, or \(\frac{12}{4} = 3\). This means for every gram of A, 3 grams of B are needed.

Use the ratio to compute the mass of B in compound Y

Knowing the mass ratio for compound X and the mass ratio of B in compounds X and Y (3:7), we can set up a proportion to find the amount of B in compound Y when combining with 8 g of A. The proportion is \(\frac{3}{4} = \frac{7}{x}\) where x is the mass of B for compound Y when A is 8 g. By solving the proportion we get \(x = \frac{7 \times 8}{4}\).

Solve for the mass of B in compound Y

After solving the equation \(x = \frac{7 \times 8}{4}\), we get \(x = 14\). Since the amount of B in Y should be double of what it is in X (as we used 8 g of A instead of 4 g), we multiply 14 by 2 to get the mass of B in compound Y. Therefore, \(x = 14 \times 2 = 28 \mathrm{g}\).

Key concepts

Law of Multiple Proportions

The Law of Multiple Proportions is a cornerstone of chemical understanding. It states that when two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in the ratio of small whole numbers.

In our example, elements A and B form two compounds, X and Y. The law is observed through the ratios of the masses of B that combine with the fixed mass of A. For X, we have a 3:1 ratio (12 g of B to 4 g of A), and for Y, a 7:1 ratio must exist by the law's definition.

This problem perfectly illustrates how the Law of Multiple Proportions guides us to determine the mass of element B that must combine with A to form a new compound, ensuring that the ratio of the masses of B (correlating with X and Y) adheres to the simplicity of whole numbers.

Mole Concept

The mole concept is an essential unit in chemistry that connects the microscopic world of atoms and molecules with the macroscopic world we can measure. One mole is defined as the amount of substance containing the same number of particles (atoms, molecules, ions, or others) as there are in 12 grams of carbon-12.

When working on stoichiometry problems, understanding the mole concept enables us to convert grams into moles and vice versa, using the molar mass of the substance. While the original problem did not explicitly ask for moles, understanding how to use the mole concept can provide deeper insight into the chemical relationships, facilitating more complex chemical calculations beyond the scope of the question.

Chemical Composition

Chemical composition refers to the identity and proportion of the elements that make up a chemical compound. Each compound has a unique composition that can be represented with a chemical formula. This formula reflects the types of atoms present and the ratio in which they combine.

In the example of compounds X and Y, we inferred the composition based on the ratio of the mass of B to the mass of A. Understanding chemical composition is crucial as it provides information for calculating the formula's molar mass and ultimately helps in preparing and analyzing chemical reactions.

Stoichiometric Calculations

Stoichiometric calculations are a set of quantitative relationships (ratios) derived from the coefficients of substances in balanced chemical equations. These calculations allow us to predict the amounts of reactants and products involved in chemical reactions.

In the case of our two compounds X and Y, we executed stoichiometric calculations to determine the mass of B in compound Y. By using the given mass ratio from compound X, we created a proportional relationship to find the unknown mass in compound Y. Mastery of stoichiometric calculations is crucial for anyone aiming to work with chemical reactions, as it lays the groundwork for predicting reaction yields and for scaling up reactions from the lab bench to industrial processes.