Find the empirical and molecular formulas for a compound...

find the empirical and molecular formulas for a compound that gives the following mass percentages on analysis: 71.65% CI , 24.27% C , 4.07% H the known molar mass is to be 98.96 g/mol

Transcript

00:01 According to the question, for three elements, chlorine, carbon and hydrogen, percentage by mass is given and we have to find out the empirical and molecular formula.

00:12 So first of all, we will draw the table to find out empirical formula.

00:16 So we have elements, then percentage by mass, then the mass, if the substance is 100 gram, then the molar mass we have to find out, then the relative number.

00:30 So, this is the table which we have to fill relative number of atoms.

00:35 Then the simplest atomic number ratio.

00:39 The simplest atomic ratio which we have to find out for all these three elements.

00:44 And last, simplest whole number ratio.

00:47 So let's complete this table, the simplest whole number ratio.

00:52 For chlorine, the percentage abundance, percentage by mass given to us is 71 .65.

01:02 24 .27 for carbon and 4 .07 for hydrogen.

01:08 And if mass in gram, it would be 71 .65, 24 .27 and 4 .07.

01:15 Molar mass for chlorine is 35 .5, 12 for carbon and 1 for hydrogen.

01:21 Relative number of atom can be found out by dividing mass with molar mass and it comes out to be 2 .02 for chlorine.

01:30 2 .0 .2 for hydrogen.

01:31 2 .0 .2 .2.

01:31 2 for carbon and 4 .07 for hydrogen.

01:36 Now, simplest atomic ratio would be dividing the related number of atom with the smallest relative number of atom, which is 2 .02 in this case...

Question

find the empirical and molecular formulas for a compound that gives the following mass percentages on analysis: 71.65% CI , 24.27% C , 4.07% H the known molar mass is to be 98.96 g/mol

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