A student used the procedure described in this module to determine the empirical

Question

A student used the procedure described in this module to determine the empirical formula of an oxides of titanium (Ti). A crucible and crucible cover were heated to a constant mass of 34.749 g. The mass of the crucible, crucible cover and a sample of TI was 36.383 g. Following the reaction of the Ti metal with O2, the crucible, the crucible cover, and contents were heated to a constant mass of 37.472 g.

1) determine the mass of the oxide of Ti.

2) Determine the mass of Ti in the compound

3) Determine the mass of O in the compound

4) calculate the number of moles of Ti in teh compound using equation 1.

5) Calculate the moles of O in the compound, using Equation 1.

Explanation / Answer

Empirical formulas show the simplest ratio of atoms in a compound. In this sort of problem, you make use of the fact that a mole of any kind of atom is always the same number - the effect being the empirical formula is related to the simplest MOLE ratio in a compound.

1) First thing you need is how many grams of each element.

Mg is easy - 6.031g

O can be found given the total amount of compound and amount of magnesium

10.000 g - 6.031 g = 3.969 g

Since we're trying to find MOLE ratios, convert each mass to moles

Mg = 6.031g / (1 mol / 24.305 g) = 0.2481 moles

O = 3.969 g / (1 mol / 15.999 g) = 0.2481 moles

To find the simplest ratio divide each mole value by the SMALLEST mole value

Mg = 0.2481 / 0.2481 = 1

O = 0.2481 / 0.2481 = 1

In this case it's a lot obvious since the values are the same, but this process of 1) divide by molar mass and 2) divide by the smallest value with help you in the future for any of this sort of problem

2)

a) mass of oxide = total mass of oxide/crucible/cover at the end - empty crucible/cover at the beginning

mass of oxide = 37.472 g - 34.749 g = 2.723 g

b) mass of Ti = total mass of titanium/crucible/cover at the end - empty crucible/cover at the beginning

mass of Ti = 36.383 g - 34.749 g = 1.634 g

c) mass of oxygen would be the change in mass as the reaction occured

mass of O = total mass of titanium/crucible/cover at the end - Ti/crucible/cover before reaction

mass of Ti = 37.472 g - 36.383 g = 1.089 g

d) mole Ti = 1.634 g (1 mol/ 47.90 g) = 0.03411 moles

e) mole O = 1.089 g / (1 mol / 16.00 g) = 0.6806 moles

You arent asked to find it, but finding the empirical formula wwould be as easy as dividing each mole value by 0.03411, giving a 1:2 Ti:O ratio, or a formula of TiO2

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