Question

1.3 x 10^-3 mole metal sample was used for the experiment

Data

Mass of crucible: 30.802 g

Mass of crucible with metal: 30.834 g

Mass of crucible and metal after heating: 30.848 g

1. Based on what you know about the unknown metal sample propose a balanced chemical equation for

a) The formation of the Group 1 metal oxide

b) the formation of the Group 2 metal oxide

2. Using the experimental data that you collected, determine the mass of the metal and oxygen in your final sample.

3. Using the experimental data that you collected, show your complete calculations for the empirical formula of the metal oxide that you have formed. (Be sure that all numbers are labeled with the correct units and your math is shown in a logical order.)

4. The identity of the metal can be further determined by calculating the molar mass of the metal. Show your complete calculations.   (Be sure that all numbers are labeled with the correct units and your math is shown in a logical order.)​

Answer 1

1)

Group 1 has the valency 1 whereas Group 2 elements have valency 2

Thus, the reaction with Oxygen will be

$\textrm{Group 1: }4M(s)+O_2(g)\rightarrow 2 M_2O$

$\textrm{Group 2: }2M(s)+O_2(g)\rightarrow 2 MO$

2)

Mass of metal = Mass of Crucible with Metal - Mass of Crucible = 30.834 g - 30.802 g = 0.032 g

Converting to mg, we have 0.032 g= 0.032 g * 1000mg/g = 32mg

Mass of Oxygen = Mass of heated Crucible - Mass of Crucible with Metal = 30.848 g - 30.834 g = 0.014 g

Converting to mg, we have 0.014 g= 0.014 g * 1000mg/g = 14mg

3)

milli Moles of Oxygen = Weight of Oxygen / Atomic Mass of Oxygen = 14 mg / (16 g/mol) = 0.875 mmol

milli Moles of Metal = 1.3 mmol

Thus, Ratio of M:O = 1.3:0.875 = 1.49:1 =3:2

Thus, the empirical formula based on calculation should be M₃O₂ but such an oxide is not possible. Thus there might have been experimental errors. What could have happened?

The ratio should have been 2:1 for Group 1 and 1:1 for group 1. The ratio we got is in between. For 1.3 mmol Group 1 element, the mmol of oxygen on complete reaction would have been 1.3/2 = 0.650 mmol. This is the maximum amount of oxygen for Group 1. But for Group 2 element, the maximum amount of oxygen would have been 1.3 mmol. But we have only 0.875 mmol. This means that the sample of Metal which was heated might not have reacted completely and only 0.875 mmol of the sample would have reacted with 0.875 mmol of Oxygen

Thus, the empirical formula should be MO for a Group 2 oxide.

4)

Mass of Metal = 32 mg

Moles of Metal = 1.3 mmol

Molecular Mass = Mass/ Moles = (32 mg)/ (1.3 mmol) = 24.61 mg/mmol = 24.61 g/mol

This is the approximately the molecular Mass of Magnesium which confirms that the molecular formula will be of the type MO and the metal is a group 2 element.