1) You are given an unknown containing KClO 3 . Following the procedure in your

ID: 1050365 • Letter: 1

Question

1) You are given an unknown containing KClO3.
Following the procedure in your manual, you obtain the following fictional data
Note! For this problem use the given (allbeit incorrect) vapor pressure of water

3) Know....
How accurately you will measure the volumes and weights?
How accurately you will be able to measure molar volume, % KClO3
What is the function of MnO2, the leveling tank
When do you use the Triple Beam Balance and the Analytical Balance and why.
Then think how would your results be affected IF...:
You started with a wet test tube?
You made an error in one or more of your weighings?
You made an error of 1 cm in using the leveling tank?

Mass of empty test tube 54.5679 g Mass of test tube + unknown 54.8523 g Mass of test tube + unknown + MnO2 54.9786 g Mass of erlenmeyer flask with residual water 308.81 g Mass of erlenmeyer flask filled with water 364.23 g Atmospheric pressure 754.9 mmHg Temperature of water 38.5 °C Partial pressure of water at 38.5 37.94 mmHg Mass of test tube + Residue after heating 54.9032 g Density of water .996 g/mL

Explanation / Answer

A) Molar volume of oxygen

Pick up the relevant data from the table to complete this part. Also, we shall include data that we can compute from the given data.

Mass of Erlenmeyer flask with residual water (g)

308.81

Mass of Erlenmeyer flask filled with water (g)

364.23

Mass of water in the flask (g) = b – a

55.42

Atmospheric Pressure (mmHg)

754.9

Temperature of water (C)

38.5

Partial pressure of water at 38.5C (mmHg)

37.94

Partial pressure of dry oxygen at 38.5C (mmHg) = d – f

716.96

Temperature of water (K) = e + 273

311.5

Volume of water in flask = volume of displaced water = c/(Density of water at 38.5C) = (55.42 g)/(0.996 g mL-1) = 55.6426 mL (I have kept a few guard digits extra) = (55.6426 mL)*(1 L/1000 mL) = 0.0556426 L

We know that 760 mmHg = 1 atm.

Therefore, partial pressure of dry oxygen gas = (716.96 mmHg)*(1 atm/760 mmHg) = 0.9434 atm (I have kept a few guard digits extra).

Volume of oxygen collected = volume of water displaced = 0.0556426 L.

STP is defined as 1 atm pressure and 273 K temperature. Let V L be the volume of the gas at STP. Therefore, we can use the gas law to write

(0.9434 atm)*(0.0556426 L)/(311.5 K) = (1 atm)*(V L)/(273 K)

=====> V = (0.9434)*(0.0556426)*(273)/(311.5)(1) = 0.0460053

The volume of oxygen gas collected is 0.0460053 0.046 L at STP (ans)

We can calculate the moles of oxygen gas collected by employing the equation of state n = PV/RT where P = pressure of the gas; V = volume of the gas; T = temperature of the gas and R = gas constant = 0.082 L-atm/mol.K.

Therefore, n = (1 atm)*(0.046 L)/(0.082 L-atm/mol.K).(273 K) = 2.0548*10-3 2.055*10-3 mole (ans).

Molar mass of oxygen gas produced (O2) = 32 g/mol.

Therefore, mass of oxygen gas produced = (2.055*10-3 mole)*(32 g/1 mole) = 0.06576 g 0.066 g (ans).

B) Percent KClO3 IN UNKNOWN

We shall pick out and calculate data that are essential for solving this part of the problem.

Mass of empty test tube (g)

54.5679

Mass of empty test tube + unknown (g)

54.8523

Mass of unknown used (g) = b – a

0.2844

Write down the decomposition of KClO3 reaction as below:

2 KClO3 ---------> 2 KCl + 3 O2

There is a 3:2 molar ratio between O2 produced and KClO3 used. We have obtained in part (A) above that 2.055*10-3 mole O2 gas was produced.

Therefore, moles KClO3 used = (2.055*10-3 mole O2)*(2 mole KClO3/3 mole O2) = 1.37*10-3 mole (ans).

Molar mass of KClO3 = 122.55 g/mol.

Therefore, mass of KClO3 used = (1.37*10-3 mole)*(122.55 g/1 mole) = 0.167893 g 0.1679 g (ans).

Percent of KClO3 in the unknown sample = (0.1679 g/0.2844 g)*100 = 59.0365 59.04 (ans).