Question

Two elements \(\mathrm{A}\) and \(\mathrm{B}\) combine to form compound \(\mathrm{X}\) and \(\mathrm{Y}\). 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 \(\mathrm{X}, 4 \mathrm{~g}\) of \(\mathrm{A}\) 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

a) 16 b) 24 c) 28 d) 32 Answer: (c) 28.

Step-by-step solution

Set up a proportion with the given ratio.

The problem states that the mass of B combined in compounds X and Y is in the ratio 3:7. So, we can set up a proportion as follows: $$(mass\ of\ B\ in\ X):(mass\ of\ B\ in\ Y) = 3:7$$

Plug in the known values.

In compound X, 4 g of A combines with 12 g of B. Thus, we can plug the known values into the proportion: $$(12\ g):(mass\ of\ B\ in\ Y) = 3:7$$

Cross-multiply and solve for the mass of B in compound Y.

To solve for the mass of B in compound Y, we can cross-multiply and get: $$(12\ g) \times 7 = 3 \times (mass\ of\ B\ in\ Y)$$ Which can be simplified as: $$84\ g = 3 \times (mass\ of\ B\ in\ Y)$$ Now, we can divide both sides by 3 in order to isolate the mass of B in compound Y: $$mass\ of\ B\ in\ Y= \frac{84}{3}= 28 \ \mathrm{g}$$ Therefore, in compound Y, 8 g of A will combine with 28 g of B. The correct answer is (c) 28.

Key concepts

Stoichiometry

Understanding stoichiometry is crucial for any student diving into the world of chemistry. It's basically the accounting system of chemistry. It involves calculating the quantities of reactants and products involved in chemical reactions. This concept lays the foundation for predicting the outcomes of reactions and balancing chemical equations.

Let's take a moment to link this to your textbook exercise. You noticed that for compound X, 4 grams of element A combine with 12 grams of element B. Stoichiometry allows us to use this ratio to predict how much of element B would react with a different amount of element A to form compound Y. By following the stoichiometric ratios given (3:7), you're essentially using the principles of stoichiometry to find the missing mass of element B.

In queries like these, stoichiometry not only helps us preserve the law of conservation of mass but also ensures that chemical equations are balanced, mirroring actual laboratory conditions.

Chemical Proportion

Chemical proportion refers to the fixed ratio of elements that compose a compound. This concept is based on the Law of Definite Proportions, which states that a chemical compound always contains its component elements in fixed ratio and does not depend on its source and method of preparation. Therefore, understanding chemical proportion is key in predicting the outcomes of chemical reactions.

In the exercise, you're told that elements A and B combine in a fixed mass ratio to form two different compounds, demonstrating chemical proportion in action. To solve the problem, you utilise that fixed ratio (3:7) between the two compounds. By setting up a proportion based on these ratios, you're applying the idea that elements combine in definite proportions, a tenet essential for this calculation and for chemistry as a whole.

Mole Concept

The mole concept is a bridge between the microscopic world of atoms and the macroscopic world we live in. It provides a way to count particles, like atoms, molecules, and ions, by grouping them into a quantity known as a 'mole', which is about 6.022 x 1023 entities, also known as Avogadro's number.

In your textbook solution, the mole concept isn’t explicitly used, but it's implied. It can be used here if we were to consider the atomic or molecular masses and calculate the amount of substance in moles. If the compounds in question were molecular, you could translate the given masses into moles, which would involve converting grams into moles using the molar mass. While this isn't necessary for solving the provided problem, recognizing how the mole concept intertwines with stoichiometry and chemical proportions is essential for a deeper understanding of chemistry.