A sample of an oxide of vanadium weighing 4.589 g was heated with hydrogen gas to form water an

step 1 Okay, so I'm going to write this out kind of in generic terms. We've got a vanadium oxide, and w

A sample of an oxide of vanadium weighing 4.589 g was...

A sample of an oxide of vanadium weighing $4.589 \mathrm{~g}$ was heated with hydrogen gas to form water and another oxide of vanadium weighing $3.782 \mathrm{~g} .$ The second oxide was treated further with hydrogen until only $2.573 \mathrm{~g}$ of vanadium metal remained. (a) What are the simplest formulas of the two oxides? (b) What is the total mass of water formed in the successive reactions?

A sample of an oxide of vanadium weighing $4.589 \mathrm{~g}$ was heated with hydrogen gas to form water and another oxide of vanadium weighing $3.782 \mathrm{~g} .$ The second oxide was treated further with hydrogen until only $2.573 \mathrm{~g}$ of vanadium metal remained.
(a) What are the simplest formulas of the two oxides?
(b) What is the total mass of water formed in the successive reactions?

Key Concepts

Stoichiometry

Stoichiometry involves calculating the quantitative relationships between reactants and products in a chemical reaction. It relies on the mole concept, balanced equations, and proportional reasoning to determine how much of each substance is involved, ensuring that mass and atoms are conserved in the reaction.

Redox Reactions

Redox (reduction-oxidation) reactions involve the transfer of electrons between species, with one substance being oxidized and another being reduced. Understanding redox processes is key when a reducing agent, such as hydrogen, converts an oxide to a lower oxidation state or the elemental metal, while another species gains oxygen or water is formed.

Empirical Formula Determination

Empirical formula determination is the process of finding the simplest whole number ratio of atoms in a compound based on experimental data. This method uses mass measurements and mole calculations to deduce the formula from the relative amounts of different elements present.

Hydrogen Reduction of Metal Oxides

Hydrogen reduction of metal oxides is a specific type of redox reaction where hydrogen gas is used as the reducing agent to convert a metal oxide into a lower oxide or even the elemental metal while producing water. This process highlights the interplay between oxidation states and the conservation of mass.

Conservation of Mass

The conservation of mass principle asserts that mass is neither created nor destroyed in a chemical reaction. This fundamental concept is used to track changes in mass during chemical transformations, such as the loss of oxygen in metal oxides during reduction, ensuring that all mass inputs equal mass outputs.

Transcript

00:01 Okay, so i'm going to write this out kind of in generic terms.

00:06 We've got a vanadium oxide, and we're going to react it with hydrogen.

00:13 And some of that oxygen is going to come off and make water, and we're going to be left with a new oxide.

00:22 So i left the x there because none of the vanadium is leaving, but the number for z will be less than y.

00:29 And then we'll take that oxide, vx, o, z, and react it with some more hydrogen.

00:39 Again, to make some more water.

00:42 And now we're just left with vanadium.

00:45 Okay.

00:45 So the amount of vanadium that we're left with, which is 2 .573 grams, that's got to be the same amount of vanadium that was in these compounds.

00:58 Okay? whatever we started with is what we ended up with.

01:01 So when we go to do some simple formulas, we'll know how many grams of vanadium we have.

01:08 For this first one, they told us we started with 4 .589 grams.

01:16 And the second one here we had 3 .782 grams.

01:22 So if i know the mass of vanadium in there, i should be able to subtract and find the mass of the oxygen in there.

01:29 So let's go ahead and start with our vxoz.

01:35 Okay.

01:37 So i know in here i have 2 .573 grams of vanadium.

01:44 Okay.

01:45 And i know that i have 3 .782 minus 2 .573.

01:55 So that's going to give me 1 .209 grams.

02:00 And that's our mass of oxygen, okay? because the whole thing has to be, has to add up to the total, and it's just vanadium and oxygen.

02:07 So now we're just going to go ahead and change our grams to moles and get our simplest ratio.

02:17 So vanadium is 50 .94, and oxygen is 16.

02:27 Okay, so we're going to get 0 .05, 0 .05 moles of vanadium here.

02:34 And we're going to get 0 .07, 6 moles of o.

02:42 So now to get a simple whole number ratio, we'll divide them both by the smallest number.

02:49 So we'll get one.

02:55 And for this one, we're going to get 1 .5.

03:00 So i'm going to have to go ahead and multiply this by two, and this by two, so we can get some whole numbers.

03:07 So we can see this is going to be 2 to 3.

03:11 So this compound right here was v2 -03.

03:16 So we found one of our simplest formulas.

03:19 Now let's do the other one...

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