CHM1045L Test 1

An unknown sample weighed 2.129g. It contains CoCl * 6H20 and NaCl and was heated to drive off all of the water. The solid residue that remained weighed 1.712g. Calculate the percentage of CoCl2 * 6H20 in the unknown sample.

gH20 lost = gSample - gResidue

gH20 lost -----> gHydrate

% of CoCl2 * 6H20 = g CoCl2 * 6H20 / gSample x 100.

How can you tell if a compound that gives off water when heated is a true hydrate?

When residue dissolves completely in water.

Are are hygroscopic substances deliquescent?

Not all hygroscopic substances deliquescent. They can both absorb water but deliquescents dissolve completely while hygroscopic just get clumpy.

What practical use can be made of hygroscopic compounds?

Can be used as drying agents.

A dry agent called Drierite, contains anhydrous CaSo4, a white compound. It is hygroscopic and picks up water. It serves as a drying agent. However, when it gets too wet, it no longer serves as drying agent. Therefore, some anhydrous cobalt (II) chloride is mixed with the calcium sulfate. When the calcium sulfate is dry so that it can function as a drying agent, Drierite is blue. However, Drierite turns violet or red when it can no longer serve as a useful drying agent. Suggest an explanation of these color changes and explain why we can tell if Drierite in a bottle can be useful as a drying agent or not by merely looking at its color.

The change in color can be useful. It can tell us when substance has absorbed moisture vs when its dry.

Vcylinder

πr²h

Vsphere

4/3πr^3

Determining volume via water displacement

Record initial volume of water in cylinder.

Slowly slide solid into cylinder.

Record new volume of water in cylinder with solid.

Vsolid = Difference between final volume and initial volume.

Determining volume via mathematical formula

Use a caliper to record necessary dimensions for each solid in mm. (Convert to cm)

Plug in dimensions to mathematical formula.

If the measured volume of the cylinder was lower than the true volume, would the calculated density be higher or lower than the true value?

The calculated density would be higher than the true value due to its inverse relationship.

Density increases as volume decreases.

Suppose a student performed everything perfectly except the balance was no zeroed, but weighed 0.100 g too high throughout the experiment. Would the calculated value for the density of the unknown liquid be correct?

Density would be equal to the true value because the difference method that cancels out the error.

We would still have true value of mass so we would have true value of density.

If the same balance was used as before (0.100 too high) was used to determine the mass of the metal cylinder, would the calculated value be equal to, higher, or lower than the true value?

Density would be higher than the true value because the mass value would be skewed up 0.100 error.

Volume would stay constant.

Direct relationship between mass and density would cause both values to be higher.

Error would not cancel out because difference method was not used.

Why does the Spectronic 20 only detect Cu²⁺?

The only species in the solution that absorbs at 620 nm, which corresponds to the visible portion of electromagnetic waves, is Cu²⁺ — the other species do not absorb at this wavelength of light (and tin is precipitated).

How were the liquid brass samples made? (The ones you put into the Spectronic 20)

Brass was dissolved in concentrated Nitric Acid (HNO3) to produce Lead, Copper, and Zinc ions in solution as well as hydrated tin (IV) oxide precipitate (SnO₂).

Why is the reaction between the brass sample and HNO₃ done under the lab hood?

To prevent the toxic brown NO₂ gas from entering the lab.

What is the cloudy stuff in the final copper solution? Why must it be filtered out and removed? How do you remove it?

It is tin (IV) oxide. It must be removed or else an incorrect high reading will be obtained for the absorbance reading due to the scatter of the light waves. It is removed by using special filter paper.

If you weigh out 0.5g of brass instead of 0.7g, how is it going to affect absorbance?

Measured absorbance will be lower that what it should be.

When the brass sample was dissolved in 8 M nitric acid in a flask, suppose some of the mixture was spilled onto the table top. Would the percent of copper experimentally be higher, lower or equal to the actual percent copper in the brass sample?

Less copper particles, less absorption consequently calculating lower molarity.

If 0.521g of brass was dissolved in 8 M nitric acid, diluted to 50.0mL volumetric flask with distilled water, filtered, and found to have a Cu 2+ molarity of 0.0931 mole/liter, then what is the percent copper in the brass sample?

0.05000L x 0.0931 mole/liter x 63.5g/1 mole = 0.296g Cu

% Cu = 0.296g/0.521g x 100 = 58.4%

Suppose the Cu 2+ determination described in question 3 was done by using a 50.0 mL volumetric flask. If the student thought a 25.0mL volumetric flask used used instead of 50.0mL volumetric flask, what would be the true percentage actually reported by the student?

0.02500L x 0.0931 mol/L x 63.5/1 mol Cu = 0.148g Cu

% Cu = 0.148/0.521 x 100 = 28.4

Should the straight line always go through the origin of a standard curve in which absorbance values are being plotted as a function of molarity? Why?

A = KM

Yes, M = O, A = O, hence the line passes through origin.

If a solution of 0.750g of CuSo4 * 5H20 in 80.0mL of solution has an absorbance of 0.100, what is the absorbance of a solution of 1.200g of CuSo4 * 5H20 in 50.0mL of solution? Show your set-up and the results of your calculation.

Solution 1:

0.750g of CuSo4 * 5H20

80.0mL

Absorbance of 0.100

Solution 2:

1.200g of CuSo4 * 5H20

50.0mL

Absorbance ??

S1: 0.750g / 80.0mL = 0.009375 g/mL

S2: 1.200g / 50.0mL = 0.024g/mL

Absorbance 2 = 0.100 = 0.024g/mL / 0.009375g/mL = 0.256

Multiply each side by 0.100 = 0.256