Animated Solution for Chemistry - Basic Concepts in Chemistry: 5 moles of $A{B}_2$ weight $125\times 10^{-3}$ kg and 10 moles of $A_2{B}_2$ weight $300\times 10^{-3}$ kg. The molar mass of $A(M_A)$ and molar mass of $B(M_B)$ in kg mol${}^{-1}$ are

The journey to mastering physical chemistry often begins with a deep understanding of the mole concept. It is the bridge between the microscopic world of atoms and the macroscopic world of grams and kilograms that we can actually measure in a laboratory.

In this problem, we are presented with a classic scenario: we have two different compounds made of the same two elements, $A$ and $B$. We are given the macroscopic data—the number of moles and the total mass of each sample—and we need to play detective to find the microscopic data, which are the individual molar masses of elements $A$ and $B$.

Imagine you are standing in a chemistry lab. On your workbench, you have two weighing scales.

On the first scale, you place a sample of compound $A{B}_2$. The label says there are exactly $5$ moles of this substance. The scale reads a mass of $125\times 10^{-3}\text{ kg}$.

On the second scale, you place a sample of a different compound, $A_2{B}_2$. This time, you have $10$ moles of the substance, and the scale reads $300\times 10^{-3}\text{ kg}$.

Our mission is to find the molar mass of element $A$, denoted as $M_A$, and the molar mass of element $B$, denoted as $M_B$.

To solve this mystery, we need our master key: the fundamental formula of the mole concept. The relationship between the given mass ($w$), the number of moles ($n$), and the molar mass ($M$) is beautifully simple:

This equation tells us that the total mass of a sample is simply the number of moles multiplied by the mass of a single mole.

But what is the molar mass $M$ for a compound? It is the sum of the molar masses of its constituent elements.

For our first compound, $A{B}_2$, the molar mass is the mass of one mole of $A$ plus the mass of two moles of $B$. Mathematically, we write this as:

Similarly, for our second compound, $A_2{B}_2$, the molar mass is:

Now, let's translate our physical lab setup into the language of algebra.

For the first sample ($A{B}_2$), we have $5$ moles weighing $125\times 10^{-3}\text{ kg}$. Plugging this into our master equation:

Let's simplify this immediately to make our lives easier. Dividing both sides by $5$, we get our first linear equation:

Now, let's look at the second sample ($A_2{B}_2$). We have $10$ moles weighing $300\times 10^{-3}\text{ kg}$. Again, using our master equation:

Dividing both sides by $10$, we get our second linear equation:

We have successfully transformed a chemistry problem into a simple system of two linear equations.

Notice how beautifully these equations are set up. Both equations contain the exact same term: $2M_B$. This is a massive hint! If we subtract equation (1) from equation (2), the $2M_B$ terms will perfectly annihilate each other. Let's do it:

We have found our first target! The molar mass of element $A$ is $5\times 10^{-3}{\text{ kg mol}}^{-1}$.

Now, finding $M_B$ is a walk in the park. We just substitute the value of $M_A$ back into our simplest equation, which is equation (1):

Subtracting $5\times 10^{-3}$ from both sides:

Dividing by $2$:

And there we have it! The molar mass of $A$ is $5\times 10^{-3}{\text{ kg mol}}^{-1}$ and the molar mass of $B$ is $10\times 10^{-3}{\text{ kg mol}}^{-1}$.

Problems like this remind us that chemistry is not just about memorizing reactions; it is about logical deduction. By carefully setting up our equations and trusting the algebra, we can uncover the hidden properties of the elements themselves. Keep practicing, and you will start seeing these patterns everywhere!