Potassium chromate is isomorphous to potassium sulphate \(\left(\mathrm{K}_{2} \mathrm{SO}_{4}\right)\) and it is found to have \(26.78 \%\) Cr. Calculate the at.wt. of \(\mathrm{Cr}\) if at.wt. of potassium is \(39.10\).

Short Answer

Expert verified
The atomic weight of chromium (Cr) is calculated to be approximately 52.00 g/mol.

Step by step solution

01

Understand the Concept of Isomorphous Substances

Isomorphous substances are those which have the same crystal structure but may contain different elements. Since potassium chromate (\(\mathrm{K}_2\mathrm{CrO}_4\)) is isomorphous to potassium sulphate (\(\mathrm{K}_2\mathrm{SO}_4\)), they have similar formulas. Therefore, in the formula for potassium chromate, there are two atoms of potassium (K).
02

Calculate the Molar Mass Contribution of Potassium

The atomic weight of potassium (K) is given as 39.10. Since there are two potassium atoms in potassium chromate, their combined contribution to the molar mass of potassium chromate is \(2 \times 39.10 = 78.20\) g/mol.
03

Determine the Fractional Contribution of Chromium

Given that chromium (Cr) accounts for 26.78% of the weight of potassium chromate, we can express this as a fraction by dividing by 100, resulting in \(0.2678\).
04

Calculate the Molar Mass of Chromium

The molar mass of potassium chromate can be determined by adding the molar mass of all the atoms in the compound. To find the atomic weight of chromium (Cr), we can use the percentage and the molar mass contribution of potassium. Let's assume the molar mass of potassium chromate as 'M'. The equation can be written as: \(0.2678 \times M = \text{molar mass contribution of Cr}\). We can subtract the molar mass contribution of potassium \((2 \times 39.10)\) from 'M' to get the molar mass of Cr.
05

Solve for the Atomic Weight of Chromium

Let the molar mass of potassium chromate (\(\mathrm{K}_2\mathrm{CrO}_4\)) be 'M'. The molar mass contribution of Cr is \(0.2678 \times M\) and the remainder is from K and O atoms. The molar mass of oxygen (O) is known and constant (16.00 g/mol), and there are four oxygen atoms in potassium chromate. Now we can write the equation \(M = 2 \times 39.10 + 4 \times 16.00 + 0.2678 \times M\). Solving this equation will give us the atomic weight of Cr.
06

Rearrange and Simplify the Equation

Rearrange the equation \(M = 78.20 + 64.00 + 0.2678 \times M\) to isolate M on one side. This can be done by subtracting \(0.2678 \times M\) from both sides of the equation, yielding \((1 - 0.2678) \times M = 78.20 + 64.00\). The left side simplifies to \(0.7322 \times M\) and the right side adds up to \(142.20\).
07

Solve for 'M'

Now we have \(0.7322 \times M = 142.20\). Dividing both sides by 0.7322, we get \(M = \frac{142.20}{0.7322}\), which is the molar mass of potassium chromate.
08

Calculate the Atomic Weight of Chromium

After calculating the molar mass of potassium chromate 'M', plug this value back into the equation \(0.2678 \times M = \text{molar mass contribution of Cr}\) to find the atomic weight of chromium.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Calculating Atomic Weight
The atomic weight, or atomic mass, of an element is a crucial concept in chemistry, detailing the average mass of atoms of an element, measured in atomic mass units (u). The atomic weight of an element can be derived from its isotopic composition and the relative atomic masses of its isotopes. In exercises involving isomorphous substances, like potassium chromate and potassium sulfate, understanding this concept is essential.

When given the percentage of an element in a compound and knowing that the compound is isomorphous with another with a known molecular formula, one can deduce the atomic weight of the element in question. The calculation involves determining the molar mass of the isomorphous compound and using the percentage to isolate the contribution of the given element to this molar mass, ultimately revealing the atomic weight of the element.
Percentage Composition in Compounds
Percentage composition is pivotal in understanding the makeup of a compound. It reflects the percent by mass of each element present in the compound. The calculation demands the molar mass of the entire compound and the molar mass of the individual elements within it. You can determine the percentage composition of an element by dividing the total mass of the element in one mole of the compound by the molar mass of the compound, then multiplying by 100%.

In practice, this provides vital information regarding the proportion of elements, which can be used to derive empirical and molecular formulas of the compound. Additionally, for isomorphous substances, the percentage composition can be applied to infer the atomic weight of elements when one isomorph has unknown parameters and the other is already characterized.
Molar Mass Calculation
The molar mass of a compound is the mass of one mole of its molecules, reported in grams per mole (g/mol). It's calculated by summing the atomic weights of each atom contained within the molecule, multiplying each by the number of times the atom appears in the formula. Knowledge of the molar mass is indispensable when converting between the mass of a substance and the number of moles or number of particles.For example, in the problem involving potassium chromate, the molar mass quantifies the grams of potassium chromate equivalent to one mole. This calculation is crucial to deduce the atomic weight of a constituent element, such as chromium, from its percentage composition in the isomorphous compound. Understanding molar mass calculations translates directly to substantial insights into the stoichiometry of reactions and the quantification of chemical compounds involved.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free