The combustion of a 0.4255 g sample of a compound containing carbon, hydrogen, oxygen and bromine produces 0.4961 g CO2 and 0.1777 g H2O. A second sample of the compound (0.1523 g) produces 0.1894 g AgBr in a separate bromine analysis. The empirical formula for the compound is:.

The empirical formula is C11HO3Br

Explanation: 1ST SAMPLE;

First, we need to get the number of moles;

Molar mass of C=12, O=16, H=1, Ag =107.9, Br=79.9

0.4961g CO2 x (1 mol CO2) / (44.0g CO2) = 0.0113 mol of Co2

Since one mole of CO2 is made up of 1 mole of C and 2 moles of O, and we have 0.0113 moles of CO2 in our sample, then we know we have 0.0113 moles of C in the sample. Now, we need to know the mass of C we have in the sample:

(0.0113 mol C) (12.011g C/1 mol C) = 0.136g C

Now we follow the same pattern above and do it for the hydrogen:

0.1777g H2O x (1 mol H2O) / (18g H2O) = 0.01 mol

Now, for the mass of H:

(0.01 mol H) (1g H/1 mol H) = 0.01g H

But this is for 1 mole of H. Whereas, H has 2 moles from the question, therefore it's equal to 0.01 x 2 = 0.02g H

Since we combusted 0.4255g of the sample, the missing mass will be from the bromine and oxygen since we have gotten masses of Carbon and Hydrogen.

Therefore missing mass = 0.4255 - (0.136+0.02) = 0.2695 of bromine and oxygen

2ND SAMPLE:

First, we need to get the number of moles;

0.1894 g AgBr x (1 mol AgBr) / (107.9 + 79.9g AgBr) = 0.001 mol of AgBr

Since AgBr is made up of 1 mole of Ag and 1 mole of Br each,

We can say that there's 0.001 mole of Br in this second sample.

Now looking at the first and second samples, lets set up a proportion to know the number of moles of Br in the first sample.

If 0.1523g of the second sample produced 0.001 mol of Br, therefore 0.4255g in the first sample will produce: (0.4255g x 0.001)/0.1523 = 0.0028mol of Br

Therefore the mass of Br in the first sample is: (0.0028 mol C) (79.9g Br/1 mol C) = 0.22372g Br

From the first sample, we saw that the sum of Br and Oxygen equals 0.2695

Therefore,the mass of oxygen is: 0.2695 - 0.22372 = 0.04578g of oxygen

Therefore to find number of moles of oxygen;

(0.04578g O x (1 mol O) / (16.0g O) = 0.0029 mol of oxygen

Overall, we have; C=0.0113 moles ;H=0.001 moles; Br= 0.0028moles and O= 0.0029

The smallest is 0.001. So to simplify this for the empirical formula, we divide each by 0.001 to get approximately C= 11, H=1, Br=3, O=3

Therefore the empirical formula is C11HO3Br

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