Magnesium + oxygen
Empirical Formula
Determination
• 1. Determine the mass of magnesium
used. (Subtract B – A)
•
19.06 g – 18.82 g = 0.24 g
• 2. Determine the number of moles of
magnesium used.
• Hint:
•
Answer #1:
0.24 g x 1 mole Mg = 0.0099 mole
•
24.305
• 3. Determine the mass of magnesium
oxide formed. (Subtract C – A)
19.21 – 18.82 =
0. 39 g
•
• 4. Determine the mass of oxygen that
combined with the magnesium.
(Subtract Answer #3 – Answer #1)
0.39 – 0.24 =
0.15 g
• 5. Calculate the number of moles of
oxygen atoms that were used.
• Hint:
•
• Answer #4: 0.15 g x 1 mole O = 0.0093 moles
•
15.9994
• 6. Calculate the ratio between moles of magnesium used
and moles of oxygen used. Express this ratio in simplest
whole number form. After dividing, round each to the
nearest whole number.
•
• Divide:
moles Mg =0.0099 mole=
moles Mg 0.0099 mole
1
moles=
moles O
=0.0093
________
moles Mg 0.0099 mole
1:1 ratio
1
0.947
Ratio of Mg:O is 1:1
•
• 7. Write the empirical formula for
magnesium oxide. (Lowest whole number
MgO
ratio of atoms) ____________
Empirical formula is the lowest whole
number ratio of atoms.
According to the charges of Mg and O what should
the formula be?
Magnesium oxide
+
1+
2+
MgO

3+
4-
0
3- 2-
O2-
Mg2+
4+
1-
If moles of Mg is lower than moles of O it would seem that the
group could have the following errors:
1. The crucible was still wet when it was weighed. The water would
have evaporated affecting the masses recorded later.
2. The magnesium strip may have not been cleaned well or had a coating
of oxide already on the strip. This would make the original mass of
Mg incorrect due to the added O.
All error examples represent a ratio less than 1:1.
If moles of Mg were higher than moles of O
the following errors may have occurred:
1. The sample was not heated long enough. This would result in not
all of the Mg completely reacting leading to less oxygen in the MgO
combustion product.
2. Some white powder was spilled or stuck to the tongs. This would
show the final mass lower than normal and since oxygen was calculated
by subtracting Mg from final mass the O would have been less.
3. The sample lost some mass when smoke escaped.
All error examples represent a ratio greater than 1:1.