A29.05 gram sample of cobalt is heated in the presence of excess oxygen. a metal oxide is formed with a mass of 40.88 g. determine the empirical formula of the metal oxide.

The empirical formula of the oxide is Co₂O₃.

Step 1. Calculate the mass of oxygen

Your reaction is

 Cobalt + oxygen ⟶ cobalt oxide

29.05 g +    x g    ⟶     40.88 g

According to the Law of Conservation of Mass, the total mass of the reactants must equal the total mass of the products. Thus,

29.05 g + x g ⟶ 40.88 g

x = 40.88 – 29.05 = 11.83

Step 2. Calculate the moles of each element

The empirical formula is the simplest whole-number ratio of atoms in a compound.

The ratio of atoms is the same as the ratio of moles.

So, our job is to calculate the molar ratio of Co to O.

Moles of Co = 29.05 g Co × (1 mol Co /(58.93 g Co) = 0.492 96 mol Co

Moles of O = 11.83 g O × (1 mol O/16.00 g O) = 0.739 38 mol O

Step 3. Calculate the molar ratio of the elements

Divide each number by the smaller number of moles

Co:O = 0.429 26:0.739 38 = 1:1.4999

Step 4. Multiply each number by a factor that makes the ratio close to whole numbers

Multiply by 2. Then

Co:O = 2:2.998 ≈ 2:3

Step 5: Write the empirical formula

EF = Co₂O₃

assuming that your periodic table has eighteen columns, you should highlight   the four elements in the third column from the left of the table.

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the have energy higher than those in the orbital while lower than those of the orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the orbital is therefore filled right after orbitals and before orbitals.  

potassium has electron configuration and calcium . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the and should be highlighted as well.

i think the answer is terrae but not 100 percent sure. ▓▬▬ §