A sample of a compound contains 7.89 g potassium, 2.42 g carbon, and 9.69 g oxygen. Determine the empirical and molecular formulas of this compound, which has a molar mass of 198.22 g/mol.

The empirical formula of the compound =

The molecular formula of the compound =

Explanation:

Mass of potassium = 7.89 g

Molar mass of potassium = 39 g/mol

Moles of potassium =

Mass of carbon  = 2.42 g

Molar mass of carbon = 12 g/mol

Moles of carbon =

Mass of oxygen = 9.69 g

Molar mass of oxygen = 16 g/mol

Moles of oxygen =

For the empirical mass of the compound, divide the lowest moles of an element with all the moles of the elements:

Potassium :

Carbon :

The empirical formula of the compound =

Empirical mass of the compound= (39 g/mol+12 g/mol+3 × 16 g/mol) = 99 g/mol

Molar mass of the compound = 198.2 g/mol

Let the molecular formula of the compound be :

So, molecular mas of the compound :

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.