Aluminum reacts with oxygen to yield aluminum oxide. if 5.0 g of al reacts with 4.45 g of o2, what is the empirical formula of aluminum oxide? (type your answer using the format co2 for co2.)

Let us assume that the given values of each element completely react with each other. To solve for the empirical formula, first we have to calculate for the number of moles of each element.

Molar mass Al = 27 g / mol

Molar mass Oxygen = 16 g / mol

Therefore the number of moles each is:

moles Al = 5 / 27 = 0.185 mol

moles O = 4.45 / 16 = 0.278 mol

Next we divide each moles with the smallest mol value, in this case 0.185 mol, therefore divide the two with 0.185 mol:

moles Al = 0.185 / 0.185 = 1

moles O = 0.278/0.185 = 1.5

The empirical formula must be in whole numbers, however moles O is still in decimal. Therefore multiply the two by 2:

moles Al = 1 * 2 = 2

moles O = 1.5 * 2 = 3

Therefore the empirical formula is:

Al2O3

i'm pretty sure it is letter c.) car