On the bench in a professor's laboratory are six bottles...

On the bench in a professor's laboratory are six bottles containing the products of the syntheses of the six halides $\mathrm{CrCl}_3, \mathrm{HgI}_2, \mathrm{NF}_3, \mathrm{NI}_3, \mathrm{SnCl}_2$, and $\mathrm{SnCl}_4$. Unfortunately the labels have come loose. Examine the bottles, and reattach the labels correctly. Bottle A contains a white solid; bottle B contains an intensely red solid; bottle $\mathrm{C}$ contains a colorless gas; bottle $\mathrm{D}$ contains light violet flakes; bottle E contains a colorless liquid; and bottle F consists of broken glass-all that remains after an explosive attempted preparation of its would-be contents.

Key Concepts

Phase Behavior and Volatility

The state of matter—whether a substance is a gas, liquid, or solid at room temperature—and properties such as volatility are key considerations. They influence both the handling of chemicals and the methods used for their identification and analysis.

Chemical Stability and Explosiveness

Some molecular compounds are inherently unstable and can be highly sensitive to physical disturbances, leading to explosive behavior. Recognizing the signs of instability is crucial in handling substances safely and in interpreting unexpected experimental outcomes.

Oxidation States and Their Effects

Differences in oxidation states can dramatically affect the properties and reactivity of inorganic compounds. For example, compounds differing only in the oxidation state of a metal may have contrasting physical properties, which is essential for understanding and predicting chemical behavior.

Qualitative Analysis

This concept involves using observable physical attributes and simple tests to deduce the identity of a substance, particularly when instrumental or quantitative methods are not immediately available. It is fundamental in both laboratory and field settings for classifying unknown compounds.

Transition Metal Complexes

Certain compounds, especially those containing transition metals, exhibit distinctive colors due to d-d electronic transitions. The specific color and appearance of a compound can give important clues about its oxidation state, coordination number, and ligand field environment.

Physical Properties

Observations such as color, state (solid, liquid, gas), texture, and visual appearance are essential for identifying substances in inorganic chemistry. The physical characteristics help to distinguish between compounds with similar compositions but different structures or coordination environments.

Transcript

00:01 This problem isn't that hard to do, but there's a ton of information.

00:05 So we're given this information.

00:07 You and a lab partner are asked to complete a lab titled fluorites of group 6b metals.

00:13 This lab is scheduled to extend over two lab periods.

00:18 The first period, which is completed by your partner, is devoted to carrying out compositional analysis.

00:24 The second day, you are going to determine melting points.

00:29 So you go to the lab and you find two unlabeled vials.

00:32 One contains a colorless liquid and one contains a green powder.

00:55 And you find the following notes in your lab partner's notebook.

01:00 Compound 1 is 47 .7 % chromium and 52 .3 % fluoride.

01:26 And these are both by mass.

01:32 Compound 2, put this down here, is 45 .7 % molybdenum and 54 .3 % fluoride.

02:05 For a and b, we're asked to determine the empirical formula for one and for b, two.

02:12 This is a and this is b.

02:14 We're going to find the empirical formula for each.

02:25 Okay, and empirical formula, the famous poem is percent to mass, so we're just going to turn this to mass.

02:32 These are going to be grams, and then mass to mold, divide by small, multiplied till whole.

02:37 So first order of business.

02:39 Let's go with a different color.

02:41 What am i at here? how about this one? so our first one, we're going to take 47 .7 grams of chromium.

02:54 I'm going to divide that by the molar mass of chromium, which is 51.

02:59 I'm going to go 52.

03:03 52 .00 grams per mole.

03:09 This will give me 0 .917.

03:16 Then i'm going to do the same thing for fluoride, 52 .3 grams, divided by 19 .0 .0.

03:26 Grams per mole and that's going to give me 2 .753.

03:31 I'm going to go to 2 .75 grams.

03:41 So then i can already see what this is going to be.

03:44 I'm going to take divide by small.

03:50 I should have done that a different color...

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