Vanadium forms four different oxides in which the percent by...

Key Concepts

Stoichiometry of Metal Oxides

In metal oxides, stoichiometry is applied to link the mass percentages of metal and oxygen to their molar amounts, allowing for the determination of the ratio between metal and oxygen atoms. This ratio is fundamental in establishing the correct chemical formula of the oxide.

Atomic Mass and Mole Concept

The atomic mass of an element serves as a conversion factor between the mass of the element and the number of moles it contains. By using the atomic mass, one can convert a given percentage of mass into moles, which is critical for calculating the ratios in a compound.

Oxidation States and Chemical Nomenclature

Oxidation states indicate the degree of oxidation of an atom within a compound, which can vary in metal oxides. Understanding oxidation states is important for both deducing the correct formula (by balancing the charges) and assigning systematic names to the compounds.

Percent Composition

Percent composition is the mass percentage of each element in a compound. It is used to determine the relative amounts of each element by mass and is essential when reverse-engineering a compound’s formula from experimental data.

Empirical Formula Determination

Determining an empirical formula involves converting the percent composition of each element into moles by dividing by the atomic mass, then finding the simplest whole-number ratio of these moles. This process reveals the simplest representation of the compound’s composition.

Transcript

00:01 All right, guys, we're doing problem 138 of chapter 3 of chemistry and molecular approach.

00:06 Nadium forms four different oxides in which the percent by mass of vanadium is respectively 76 percent, 68 percent, 56 percent.

00:15 Terminum molecular formula and name of each compound.

00:20 So let's do this.

00:21 So 0 .76 is going to equal 51x, which is going to represent the total mass of our vanadium, over 51x, plus 16 y.

00:32 Just going to represent the mass of our vanadium compound, which x and y being the respective number of atoms of vanadium and oxygen.

00:39 So we're going to do this.

00:40 We're going to bring this over, 76 .6 .6 times 51x plus 61x plus 16 .5 .15 times 51x plus 16 plus 1 .16 plus 12 .16 .16.

01:20 Is going to equal 51x, 12 .16y is going to equal 12 .24.

01:38 And since 12 .16 and 12 .24, they're very practically the same.

01:45 And y is approximately going to be equal to x.

01:48 So that means that our compound is going to be vo.

01:55 And that means that our oxygen tends to have a minus two charge so vanadium is going to be plus two so this is going to be vanadium 2 oxide so that's part a now part b we're going with 68 % and i'll just skip a step so we're going to start out with 0 .68 times 51x plus 16y is going to equal 51x.

03:03 So 34 .68 is x plus 10.

03:17 So 34 .68 is x plus 10 .88.

03:23 That's equal to 51x.

03:30 8.

03:30 So that's going to be 10 .88y is equal to 16 .32x.

03:43 So that means y is equal to 1 .5x.

03:48 And let's just change out the whole numbers.

03:50 Multiply both sides by 2.

03:52 So that means 2y is going to be equal to 3x.

03:56 And this is a very simple equation to figure out.

04:00 So 2y is going to be equal to 3x.

04:03 So that means that, so what's the least common multiple of two and three? three.