Question

Q. 1 A sample of an unknown compound was analyzed and was found to contain 0.0894 moles of gold and 0.0447 moles of sulfur. What is the formula of the compound? What is the name of the compound?

Q.2   A sample of an unknown compound was analyzed and was found to contain 1.49 grams of iron and 6.37 grams of bromine. What is the formula of the compound? What is the name of the compound?

Q.3   A sample of an unknown compound was analyzed and was found to contain 0.351 grams of water and 1.27 grams of CoCl2. What is the formula of the compound (written as a hydrate)?

Answer 1

1. Let the compound be Au_xS

If we try to mole ratio of atoms, we have

moles of Au / Moles of S =x/1          .......(1)

Data given: moles of Gold = 0.0894 moles

               moles of sulfur =0.0447

substituting the above values in equation (1)

                 0.0894/0.0447 =x

     therefore x= 2

Hence the formula of the compound is Au2s

Name of the compound =Gold sulfide

2. Data given : Mass of iron =1.49 grams

                       mass of bromine = 6.37 grams

Lets us convert mass to moles

         Molecular weight of iron =56 g/mole

        Molecular weight of Bromine= 80 g/mole

   moles of iron =1.49/55.845 = 0.0266 moles

moles of Bromine = 6.37/79.9 =0.0797 moles

Let us calculate the moles ratio of bromine to iron

moles of bromine/moles of iron =0.0797/0.0266 =2.97 which is close to 3

since moles ratio fo bromine to iron is 3 . We will have 3 moles of bromine for 1 mole of iron

Moelcular formula: FeBr3

Name of the compound = Iron bromide

3. Data given : mass of water =0.351 grams

                     mass of CoCl2 = 1.27 grams

          Molecular weight of water =18 g/mole

         Molecular weight of CoCL2 =129.839 g/mole

Moles of water = 0.351/18 = 0.0195

Moles of CoCL2 = 1.27/129.839 =0.00978134

let us take the ratio of moles of water to CoCL2

moles of water/moles of CoCL2 = 0.0195/0.00978134 =1.99 which is approximately 2

Hence molecular formulea: CoCL₂.2H₂O