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anonymous
 4 years ago
A 0.4791g sample was analyzed and was found to contain the following masses of elements: Carbon, 0.1929g; Hydrogen, 0.01079g; Oxygen, 0.08566g; Chlorine, 0.1898g. Determine the empirical formula of the compound.
This is the first question of the assignment, and if anyone could help me solve this and explain the process to me that would be helpful too
anonymous
 4 years ago
A 0.4791g sample was analyzed and was found to contain the following masses of elements: Carbon, 0.1929g; Hydrogen, 0.01079g; Oxygen, 0.08566g; Chlorine, 0.1898g. Determine the empirical formula of the compound. This is the first question of the assignment, and if anyone could help me solve this and explain the process to me that would be helpful too

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Rogue
 4 years ago
Best ResponseYou've already chosen the best response.1I don't have the time to write a full solution, but here are the basic steps you need to follow: 1. Divide the masses of each of the elements in the sample by their molar masses to get the moles of each elements in the sample. 2. Compare the ratio of each element to one another to find the formula of the molecule.

Rogue
 4 years ago
Best ResponseYou've already chosen the best response.1Here is a better guide with examples: http://dl.clackamas.cc.or.us/ch10404/empirica.htm

Xishem
 4 years ago
Best ResponseYou've already chosen the best response.2To give support to Rogue, here is the work done:\[0.1929g(C)*\frac{1mol(C)}{12.01g(C)}=0.01606mol(C)\]\[0.01079g(H)*\frac{1mol(H)}{1.008g(H)}=0.01070mol(H)\]\[0.08566g(O)*\frac{1mol(O)}{16.00g(O)}=0.005354mol(O)\]\[0.1898g(Cl)*\frac{1mol(Cl)}{35.45g(Cl)}=0.005354mol(Cl)\]Now express each element as a ratio to an element with the fewest number of moles, and round to the nearest whole number: \[mols(C)=\frac{0.01606}{0.005354}=3\]\[mols(H)=\frac{0.01070}{0.005354}=2\]\[mols(O)=\frac{0.005354}{0.005354}=1\]\[mols(Cl)=\frac{0.005354}{0.005354}=1\]Based on that ratio of moles, what do you think the final empirical formula is?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thank you both for the answers :) both extremely helpful
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