A bromoalkane contains 34.9% carbon and 6.60% hydrogen by mass. The rest of the mass is made up by bromine. What is the empirical formula of this molecule?
A bromoalkane is a molecule which contains bromine, carbon and hydrogen (and nothing else) and has no carbon-carbon double bonds. An empirical formula gives the simplest ratio of atoms in a molecule. Here's how I'd go about tackling this question:1) Work out the % bromine by mass: 100 - 34.9 - 6.60 = 58.5%, so this is the % mass as we are told the rest of the mass comes from bromine.2) Work out the number of moles of C, H and Br: moles of C = mass/Ar = 34.9/12 = 2.91 (we can use "% mass" as "mass" because we are simply working out a ratio of the atoms relative to each other). Similarly, moles of H = 6.60/1 = 6.60. Moles of Br = 58.5/79.9 = 0.732.3) Work out the whole number ratio of the C:H:Br. You can do this by dividing all the moles by the value for the smallest numeber of moles (i.e. 0.732 moles of bromine). So 2.91 moles of C becomes 3.98 (because 2.91/0.732 = 3.98); 6.60 moles of H becomes 9.02; 0.732 moles of Br becomes 1. As these values are all very close to whole numbers we can simply round them to the nearest, so the ratio of C:H:Br is 4:9:14) You can conclude that the empirical formula of the bromoalkane is C4H9Br
