## HW M19

SGonzales_3L
Posts: 55
Joined: Thu Jul 11, 2019 12:17 am

### HW M19

I was a bit confused on how to begin calculating the empirical and molecular formulas for problem #M19, about the combustion of caffeine. We are given the information that there is 0.376 g of caffeine initially and, when burned, the products are 0.682g CO2, 0.174 g H2O, and 0.110 g N2. We also know that the molar mass of caffeine is 194 g/mol.

I was able to calculate that there is a total of 0.966 g of product, which means that by the law of conservation of mass, there should be 0.590 g of O2 on the reactants side. I'm stumped on where to go from here, though. How can I calculate the empirical formulas given only this information?

Cynthia Gong 1L
Posts: 51
Joined: Fri Aug 09, 2019 12:15 am

### Re: HW M19

For all combustion problems, you would use molar ratios to solve for the empirical formula. For instance, in this problem you are given that there is 0.682 g of CO2 produced. The molecular weight of CO2 is around 44g, so you divide 0.682/44 to calculate the number of moles of CO2 that is produced in this reaction, which gives you an answer of 0.0155. Then you would use mole ratios. Since there is one mole of carbon in one mole of CO2, there is a 1:1 ratio and therefore 0.0155 moles of carbon is produced in this combustion reaction. Using this same process for the other elements, you will end up with the number moles for each individual element and then divide each number of moles by the smallest amount of moles to come up with the empirical formula.

Maeve Miller 1A
Posts: 51
Joined: Fri Aug 09, 2019 12:16 am

### Re: HW M19

In order to calculate the empirical formula (and eventually the molecular formula), you will need to divide the mass of each element by is molar mass (in g*mol^-1) to attain the number of moles; then, divide each number by the smallest to get the molecular ratio. If the ratio is not in integers, you must multiply in order to get WHOLE numbers. This will give you the empirical formula. If they give you the molar mass of the compound you can divide this number by the molar mass of the empirical formula and then multiply the empirical ratio by the number you get. Hope this helps :)