What is the empirical formula of the oxide?

Question

when 1.010g Zinc vapor is burned in air, 1.257g of the oxide was produced.

Answer 1

mass of zinc in zinc oxide = 1.010g

mass of oxide in zinc oxide = 1.257g - 1.010g = 0.247g

number of mole of zinc = 1.010 / 65

                                        = 0.0155 mol

number of mole of oxide = 0.247 / 16

                                           = 0.0154 mol

to find the empirical formula, we have to get the simplest ratio of the two elements

0.0155 / 0.0154 = 1

0.0154 / 0.0154 = 1

therefore the empirical formula of the oxide is ZnO.

besides this, there is another way to find out the empirical formula...

the ion of zinc is Zn2+

and the ion of oxygen is O2-

cancel out the positive and negative charge and we get ZnO...

Answer 2

First you will want to determine how many moles of Zinc you have. Divide your mass of Zinc by the atomic mass and you get 1.01/65.407 = 0.01544 moles of Zinc

We know the total mass is 1.257g so there must be 1.257g minus 1.01g of Oxygen. This gives 0.247 grams of Oxygen. Now we find the number of moles of Oxygen. 0.247/16 = .015437 or approximately equal to the number of moles of Zinc. The difference is within the number of significant digits.

So we have One mole of Oxygen and One mole of Zinc. Therefore the formula for Zinc Oxide would be ZnO.