Equivalent Weight of Magnesium
Experiment # 16
What we are doing today:
• We are going to verify the atomic mass of Magnesium.
• We are going to determine the # of moles of
hydrogen displaced during a reaction.
• This will be accomplished using the Ideal Gas Law.
•Once we calculate the # of moles of Hydrogen produced
we can calculate the # of moles of Mg that reacted.
•We will verify the atomic mass of magnesium by
comparing the grams of Mg used to the number of moles
of hydrogen displaced.
Idea Gas Law
PV=NRT
In order to use the Ideal gas law the
following information must be true:
P
pressure in atmospheres
V
volume in liters
N
number of moles
R
gas constant = 0.0821 L atm /mol K
T
temperature in Kelvin
Mg + HCl → H2 + MgCl2
Procedure:
• Obtain and scrape with steel wool a 5 cm piece of Mg.
do not touch the Mg until it has been weighed.
• Using a mg balance accurately determine the mass of the
Mg.
• Coil the Mg around a pencil
• Add 10 of 6M HCl to your eudiometer tube.
•
Slowly fill the tube with water.
• Do NOT MIX THE WATER AND THE
ACID.
Procedures Continued
• Obtain a 1000 mL. beaker and fill it with water.
• Carefully lower the coil into the tube holding the
string.
• Use your thumb to close the tube and then invert it.
• Place your tube in your 1000 mL. beaker and
clamp it in place.
• Once your eudiometer tube is secured measure the
amount of gas produced.
• Also measure the water column measurement in mm.
• Record the barometric pressure
Use my data as trial # 1 and you do two trials
#1
Mg mass = 0.040g
#2
Volume of H 2 collected = 47.0 mL from eudiometer tube
#4
Barometric pressure: 759 mm of Hg.
#5
Water temp: 20 OC
#6
Height of water = top of water to the gas line:
34.8 cm = 348 mm
K = 273 + OC = 293 K
Pressure of water column: must be in mm of Hg
so we must convert from mm of H2O to mm of Hg
#6
348 mm of H20 is the height of my water column
#7
348 mm of H 2 O 1mm of Hg
 25.58 mm of Hg
1
13.6 mm of H 2 O
#8
Water vapor pressure (P H2O) see chart on pg
172 of lab book. My water temp was 20OC
and my pressure is 17.5 mm of Hg
Pressure Dry H2 = PAtm - Pwater Vapor – P water column
= 759 mm of Hg – 17.5mm of Hg - 25.58 mm of Hg
= 715.92 mm of Hg
To use the ideal gas law pressure must be in atms.
715.92mm of Hg
1atm
 .942atms
1
760mm of Hg
#9
.942 atms is the pressure of the dry H2
Now that we have pressure we can solve
for moles of H2 produced :
PV  NRT
PV
N
RT
.942atms 0.047L 

0.0821L atms/ mol K 293K 
 1.84X10 3 mol H 2 produced
# 13 on chart
g Mg
0.040g

 21.85 g/mole
-3
mole H 2 1.83X10 mole
21.85 g/mole was the mass of Mg that I determined.
The actual mass of Mg is 24.3 g/mole
SHOW ALL WORK FOR FULL CREDIT