Some elements can form several oxides. Element X forms several different oxides,

Question

Some elements can form several oxides. Element X forms several different oxides, data for FOUR of them are listed below. In each the amount of X that reacts with oxygen stays constant, yet the amount of oxygen varies: Using this data determine the empirical formulas for the oxides A, B and C, the empirical form for one of the four oxides is already given in the table. Be sure to show your work. When doing this problem, use element X as your reference. It was discovered that the element X forms a fifth oxide (compound D) that has a mass percent of X = 36.8%. What is the empirical formula for compound D? SHOW YOUR WORKJLOGIC!

Explanation / Answer

The law of multiple proportions is to be used. Take XO as the reference compound. Let the molar mass of X be x; the molar mass of O is 16.

The compound given has the empirical formula XO which means

Moles of X: Moles of O = 1:1

Mass of X in XO = 1.000 g.

Mass of O in XO = 1.140 g.

Moles of X in XO = (1.000/x) mole.

Moles of O in XO = (1.140/16) mole

Given

[(1.000/x)/(1.140/16)] = 1/1

===> 1.000*16/1.140x = 1/1

===> x = 1.000*16/1.140 = 14.035

The molar mass of X is 14.035.

To find out the empirical formula of the three compounds, we will first find out the moles of X and the moles of O present; then calculate the molar ratio and round of the molar ratio to the simplest integer values. The simplest ratio will give us the number of atoms of X and O in the compounds and hence the empirical formula of the compounds.

Moles of X = mass of X/molar mass of X

Moles of O = mass of O/molar mass of O

Ratio of moles of X:moles of O

Simplest ratio of moles of X:moles of O

Emperical formula of compound

(1.000 g)/(14.035 g/mol) = 0.07125 mol

(0.571 g)/(16 g/mol) = 0.03568 mol

0.07125:0.03568 = 1.0:0.513 (take X as reference)

2:1

X2O

(1.000 g)/(14.035 g/mol) = 0.07125 mol

(2.280 g)/(16 g/mol) = 0.1425 mol

0.07125:0.1425 = 1:2

1:2

XO2

(1.000 g)/(14.035 g/mol) = 0.07125 mol

(4.570 g)/(16 g/mol) = 0.2856 mol

0.07125:02856 = 1:4.008

1:4

XO4

Moles of X = mass of X/molar mass of X

Moles of O = mass of O/molar mass of O

Ratio of moles of X:moles of O

Simplest ratio of moles of X:moles of O

Emperical formula of compound

(1.000 g)/(14.035 g/mol) = 0.07125 mol

(0.571 g)/(16 g/mol) = 0.03568 mol

0.07125:0.03568 = 1.0:0.513 (take X as reference)

2:1

X2O

(1.000 g)/(14.035 g/mol) = 0.07125 mol

(2.280 g)/(16 g/mol) = 0.1425 mol

0.07125:0.1425 = 1:2

1:2

XO2

(1.000 g)/(14.035 g/mol) = 0.07125 mol

(4.570 g)/(16 g/mol) = 0.2856 mol

0.07125:02856 = 1:4.008

1:4

XO4

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