A compound used to generate O2 gas in the laboratory has mass percentage composition 31.91% K, and 28.93% Cl, the remainder being oxygen. What is the empirical formula of the compound?1. KClO2. KClO23. KClO34. K3ClO45. K2ClO3

Answer:KClO3

Explanation:

Start with the number of grams of each element, given in the problem.

If percentages are given, assume that the total mass is 100 grams so that

the mass of each element = the percent given.

Convert the mass of each element to moles using the molar mass from the periodic table.

Divide each mole value by the smallest number of moles calculated.

Round to the nearest whole number. This is the mole ratio of the elements and is

represented by subscripts in the empirical formula.

If the number is too far to round (x. 1 ~ x. 9), then multiply each solution by the same

factor to get the lowest whole number multiple.

e. g. If one solution is 1.5, then multiply each solution in the problem by 2 to get 3.

e. g. If one solution is 1.25, then multiply each solution in the problem by 4 to get 5.

Once the empirical formula is found, the molecular formula for a compound can be determined if the molar mass of the compound is known. Simply calculate the mass of the empirical formula and divide the molar mass of the compound by the mass of the empirical formula to find the ratio between the molecular formula and the empirical formula. Multiply all the atoms (subscripts) by this ratio to find the molecular formula

To find percentage of oxygen 100 - (31.91 + 28.93) = 39.16

Divide through by their molar masses

K = 31.91/39=0.8182

Cl = 28.93/35.5=0.8149

O = 39.16/16 = 2.4475

Divide through by the lowest value and round up to nearest whole number

K = 0.8182/0.8149=1

Cl=0.8149/0.8149=1

O=2.4475/0.8149=3

Empirical Formula = KClO3

1. KClO2

Explanation:

We calculate the empirical formula using the percentage compositions. We divide the percentage compositions by the relative atomic masses. The relative atomic mass of potassium is 39, the relative atomic mass of chlorine is 35.5 while the atomic mass of oxygen is 16.

Now to get the percentage composition of oxygen, we simply subtract the addition of chlorine and potassium. 100 - (31.91 + 28.93) = 39.16%

We then calculate the ratio as follows:

Potassium = 31.91/39 = 0.818

Chlorine = 28.93/35.5 = 0.815

Oxygen = 39.16/16 = 2.4475

Approximating the ratios, potassium has 1, chlorine has 1 and oxygen has 2.

The empirical formula is thus KClO2