Determining an Empirical Formula
Nick DiPreta
Jonny Masci
Chemistry
Due: 2/25/10
Pre-Lab Discussion:
Regardless of the size of a compound, the mole (g-atom) ratio of certain elements will always
be the same. These ratios are whole numbers and can be used to determine the subscripts in
the empirical formula of the compound. Thus, the amount of a certain compound will not
affect the ratio of the elements in the compound. In this lab, students will learn to determine
the mole ratio of two elements in a binary compound.
Purpose/Aim:
The purpose of this experiment is to use the relationships of the mass amount in grams, to find
out the ratio of magnesium to oxygen in magnesium oxide and to show that this is a definite
ratio.
Equipment/Materials: crucible, cover, iron ring, clay triangle, crucible tongs, scissors,
magnesium foil, Bunsen burner, striker, pencil, safety goggles, ruler, scissors
Safety
- Do not touch hot crucible with fingers
-Wear safety goggles
- Do not eat magnesium oxide
Procedure:
1. Clean a crucible and cover. Dry them by heating them in the hottest part of a burner
flame for 3 minutes. Allow them to cool. Measure the mass of just the crucible and
record this as a under “Observations and Data”
2. Cut a 35 cm length of magnesium ribbon into 1cm pieces. Place the pieces in the
crucible and measure the mass of the crucible and its contents.
3. Cover the crucible and place it in a clay triangle. Heat gently for 2 minutes. Using
crucible tongs, carefully tilt the cover to provide an opening for air to enter the crucible.
Heat the partially covered crucible strongly for 15-20 minutes.
4. Turn off the burner, cover the crucible and allow the contents to cool. When the
crucible is cool enough to touch, remove the cover and examine the contents. If any
unreacted magnesium remains, replace the cover at a slight tilt, and reheat the crucible
strongly for several minutes.
5. Put the cover all the way and allow to cool. After making sure that all the magnesium
has reacted, use a dropper pipet to add water to the crucible just to cover the contents.
Wash any material that may have spattered onto the inside of the cover into the
crucible.
6. Holding the burner in your hand, gently heat the contents of the uncovered crucible by
moving the burner slowly back and forth. Avoid spattering.
7. When all the liquid has boiled off, repeat steps 5 and 6.
8. When all the liquid has boiled off a second time, strongly heat the uncovered crucible
for 5 minutes.
9. Turn off the burner and all the crucible and contents to cool. Measure the combined
mass of the crucible and the contents.
Observations and Data:
Object
Empty Crucible
Crucible+Mg
Crucible+Mg Oxide
Mass in Grams
20.35 g
20.72 g
20.96 g
Calculations:
1. Find the mass of magnesium used: b-a
20.72g-20.35g=0.37g
2. Find the mass of oxygen that reacted: c-b
20.96-20.72=0.24g
3. Find the number of g-atoms of Mg used:
g-atoms Mg= mass of Mg in g/( 24g Mg/g atom Mg)
g-atoms Mg= 0.37g/24g=0.015 g-atm
4. Find the number of g-atoms of O that were used:
g-atoms O= Mass of O in g/16g 0 per g-atom
0.24g/16g= 0.015 g-atm
5. Find the ratio of g-atoms of Mg to g-atoms of O.
0.015 g-atoms Mg/ 0.015 g-atoms O= 1:1 ratio
Conclusions/Questions
1) Write the empirical formula of the oxide of magnesium based on your calculations from this
experiment.
The empirical formula is MgO due to the fact that the ratio of g-atoms of Mg to g-atoms
of O is 1:1.
2) What is the ratio of the mass in grams of magnesium used to the mass in grams of oxygen
that reacted? Relate this ratio to the law of definite proportions.
0.37g Mg/ 0.24 g O= 1.54:1 ratio.
This ratio is different in this question from question one. In question 1, there was a substantial
amount of rounding going from grams to moles (g-atm), which led to the 1:1 molar ratio. There
was less rounding and thus a more accurate (different) answer in question 2. The law of
definite proportions states that the elements in a pure compound combine in definite
proportion to each other. Only by making more precise measurements can one truly maintain
this law in their calculations.
3) Why is the ratio found in question 2 different from the ratio found in calculation 5 above?
Rounding issues when going from grams to g-atoms leads to slight differences between
ratios. Thus, different ratio calculations are obtained when different types of calculations are
done in questions 1 and 2.
4) In a chemical formula, explain the significance of subscripts In terms of atoms of molecules.
In terms of gram-atoms and molecules.
The subscript shows how many atoms or molecules there are in a specific compound of
a specific elements but they do not have an effect on moles or grams. The coefficient
determines the moles in a chemical formula.
5) The molecular formula for hydrogen peroxide is H2O2. What is its empirical formula?
The empirical formula for hydrogen peroxide is therefore, HO.
6) How is the chemical composition of carbon monoxide, CO, similar to that of carbon dioxide,
Co2? How is it different?
Both CO and CO2 contain carbon and oxygen elements. However, CO2 has one more
oxygen atom than does CO in each one of its molecules.
7) A sample of sulfur having a mass of 1.28 grams combines with oxygen to form a compound
with a mass of 3.20 g. What is the empirical formula of the compound?
3.20g-1.28g=1.92g
1.28g S/ 32g g-atm S= 0.04 mol S
1.92g 0/ 16g atm O= 0.12 mol 0
0.04 mol S/ 0.04= 1mol S
0.12 mol 0/ 0.04 mol S= 3 mol O
The empirical formula would be SO3, or sulfur trioxide.
Conclusion: The law of definite composition states that no matter the amount in grams of a
compound, the molar ratio of that compound always remains constant. The amount the
compound does not have any effect in any way on the molar ratios in the chemical formula, and
this can never be changed.