Empirical Formula of a Compound PURPOSE: Determine the empirical formula of magn

Question

Empirical Formula of a Compound PURPOSE: Determine the empirical formula of magnesium oxide formed when LABORATORY EQUIPMENT REAGENTS Magnesium metal Distilled water Clay trangle Tripod Meker Burner Balance Safety goggles THEORY: According to the Law of Definite Proportions, different samples of a pure compound always contain the same elements in the same proportions by mass. After a compound, a quantitative analysis is performed to deterrnine the pe each element. chemist has d by qualitative analysis which elements are present in a percentage by mass of The simplest or empirical formula of a compound is the smallest whole ratio of atoms present in a compound. The empirical formula may be the same as the molecular (true) formula. For example, both the empirical and molecular formula for carbon dioxide is CO2. On the other hand, the molecular formula for glucose is C,H120s, while its empirical formula is CH2O. Therefore, the molecular weight of a compound must be known to determine its molecular formula from its empirical formula. To find the empirical formula of a compound, it is essential to know the number of grams of each element in the compound. In this experiment, a known mass of magnesium metal is reacted with oxygen in the air and the mass of the magnesium oxide is determined. The mass of oxygen combined with the magnesium can be calculated as follows:

Explanation / Answer

Let me generate an experimental data in the provided in the question.

Mass of crucible + cover = 100 g

Mass of crucible + cover + magnesium = 101 g              (suppose ~1 g Mg has been taken)

So, mass of magnesium = (101-100) g = 1 g

Similarly,

Mass of crucible + cover = 100 g

Mass of crucible + cover + magnesium oxide = 101.666 g              

So, mass of magnesium oxide = (101.666-100) g = 1.666 g

Calculations:

Now, mass of only oxygen = (mass of magnesium oxide – mass of magnesium)

                                          = (1.666-1) g = 0.666 g

Number of moles magnesium present = 1 g/ atomic weight of magnesium

                                                             = 1 g/ 24 g = 0.0416

Similarly,

Number of moles oxygen present = 0.666 g/ atomic weight of oxygen

                                                             = 0.666 g/ 16 g = 0.0416

Remarks:

So, the empirical formula of magnesium oxide contains 1:1 magnesium and oxygen atoms, and it is “MgO”.

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