A binary compound created by reaction of aluminum and an unknown element E contains 52.99% Al and 47.01% E by mass. If the formula of the compound is Al2E3, calculate the atomic mass of E.

Respuesta :

Answer:

Atomic mass of E is 15.95 g/mol

Explanation:

To solve this we must begin with this knowledge:

1 mol of Al₂E₃ contains 2 moles of Al and 3 moles of E

The percentages shows the mass of elements in 100 g of salt, so we must think this rule of three.

2 moles of Al = 2 mol . 26.98 g/mol → 53.96 g (the mass of Al, in 1 mol of Al₂E₃)

If 52.99 g of Al, are contained in 100 g of salt

53.96 g of Al, would be contained in (53.96 . 100) / 52.99 = 101.8 g of salt.

This is the molar mass of the salt, 101.8 g/mol. Therefore, we can think another rule of three.

In 100 g of salt we have 47.01 g of E

In 101.8 g of salt we would have (101.8 g .47.01g) / 100g = 47.87 g of E

This number,  47.87 g represents the amount of 3 moles of E, so if we want to know the molar mass, we should do this g/mol

47.87 g / 3mol = 15.95 g/mol

We can make a verification of this. Imagine that we have the molar mass ( 101.8 g/mol ) and the percent of composition so we have to find out the molecular formula.

We divide the percent between the molar mass of each element

52.99 g /26.98 = 1.96

47.01 g / 15.95 = 2.95

Now we divide between the less value of both

1.96 / 1.96  = 1

2.95 / 1.96 = 1.50

As we must have integer numbers in the formula  we multiply x2, so the formula will be correctly Al₂E₃