if 1.3860 g of mg ribbon combusts to form 2.1252 g of magnesium oxide product, calculate the experimental empirical formula from this data.

Explanation:

MgO

Explanation:

The following data were obtained from the question:

mass of crucible and cover + magnesium metal = 33.741 g

mass of crucible and cover = 33.5 g

mass of crucible and cover + magnesium oxide = 33.899 g

Next, we shall determine the mass of magnesium metal. This can be obtained as follow:

mass of crucible and cover + magnesium metal = 33.741 g

mass of crucible and cover = 33.5 g

Mass of magnesium metal =..?

Mass of magnesium metal = (mass of crucible and cover + magnesium metal) – (mass of crucible and cover)

Mass of magnesium metal = 33.741 – 33.5

Mass of magnesium metal = 0.241g

Next, we shall determine the mass of magnesium oxide. This can be obtained as follow:

mass of crucible and cover + magnesium oxide = 33.899 g

mass of crucible and cover = 33.5 g

Mass of magnesium oxide =?

Mass of magnesium oxide = (mass of crucible and cover + magnesium oxide) – (mass of crucible and cover)

Mass of magnesium oxide = 33.899 –. 33.5

Mass of magnesium oxide = 0.399g

Next, we shall determine the mass of oxygen. This can be obtained as follow:

Mass of magnesium oxide = 0.399g

Mass of magnesium metal = 0.241g

Mass of oxygen =..?

Mass of oxygen = (Mass of magnesium oxide) – (Mass of magnesium metal)

Mass of oxygen = 0.399 – 0241

Mass of oxygen = 0.158g

Now, we can obtain the empirical formula for the magnesium oxide as follow:

Mg = 0.241g

O = 0.158g

Divide by their molar mass

Mg = 0.241 / 24 = 0.01

O = 0.158 / 16 = 0.0099

Divide by the smallest

Mg = 0.01 / 0.0099 = 1

O = 0.0099 / 0.0099 = 1

Therefore, the empirical formula for the magnesium oxide is MgO

i have know clue i guess aluminum

i think lol