Element A has a mass of 0.444 grams and reacts with Element B via a combination reaction. The mass of the product is 0.988 grams. The molar mass of Element A is 111 g/mol and the molar mass of Element B is 68 g/mol. What is the empirical formula of the product.

AB2

Explanation:

The following data were obtained from the question:

Mass of element A = 0.444 g

Molar mass of element A = 111 g/mol

Mass of the compound = 0.988 g

Molar mass of Element B = 68 g/mol

Next, we shall determine the mass of element B. This is shown below:

Mass of element B = Mass of the compound – Mass of Element A

Mass of element B = 0.988 – 0.444

Mass of element B = 0.544 g

Now, we can obtain the empirical formula for the compound formed as follow:

A = 0.444

B = 0.544

Divide by their molar mass.

A = 0.444/111 = 4×10¯³

B = 0.544/68 = 8×10¯³

Divide by the smallest

A = 4×10¯³ /4×10¯³ = 1

B = 8×10¯³/4×10¯³ = 2

Therefore, the empirical formula for the compound formed is AB2

< > according to newton's third law of motion, forces always act in equal but opposite pairs. another way of saying this is for every action, there is an equal but opposite reaction.< >

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8