1.261 grams sample of pure caffeine contains 0.624g C, 0.065g H, 0.364g N and 0.208g O.

a) What is the empirical formula of caffeine?

b) The molecular weight of caffeine is 194 g.mol^-1, what is the molecular formula of caffeine?

Question

1.261 grams sample of pure caffeine contains 0.624g C, 0.065g H, 0.364g N and 0.208g O.

a) What is the empirical formula of caffeine?

b) The molecular weight of caffeine is 194 g.mol^-1, what is the molecular formula of caffeine?

Vocaloid · Answer

calculate the percent C, H, N, and O based on the sample.

starting with 0.624g C out of 1.261g total sample, this would be (0.624/1.261)*100% = 49.5% carbon. repeat this process with the other elements.

for part A) assume you have 100g caffeine. starting with C, we found that caffeine has 49.5% C by mass. so a 100g sample would have 49.5g C. convert this to moles by dividing by the molar mass of carbon. repeat with the other elements.

now, look at how many moles of each element you have. whichever number of moles is the smallest, divide all the molar quantities by the smallest number. in the end you should end up with whole number ratios of each element.*

that should give you the number of moles of each element in the empirical formula. write the formula based on those numbers.

Vocaloid · Answer

for part B) use the empirical formula you determined in part A. calculate the molecular weight of the empirical formula by summing the masses of its elements.

divide the molecular weight (given as 194g/mol) by the molecular weight of the empirical formula from the previous step. you should get a whole number. that tells you how much to "scale" up the empirical formula by to get the molecular formula. ***for example*** if you get 2, you need to double all the subscripts in the empirical formula.

Vocaloid · Answer

* regarding this part: if you get ratios like 1.5 or 0.333, etc. you can try multiplying all the numbers by some whole number to eliminate the decimals. for example, if you get 1.5, 0.5, etc. as your numbers, you can try multiplying everything by 2 to convert everything to whole numbers.

joancrawford · Answer

thank you so much for your helpp

Vocaloid · Answer

for part A) you forgot the nitrogen. the results for C, O, and H are correct.

for part B) you'll need to add in the nitrogen into the empirical formula and repeat the calculations

joancrawford · Answer

@vocaloid wrote:

for part A) you forgot the nitrogen. the results for C, O, and H are correct. for part B) you'll need to add in the nitrogen into the empirical formula and repeat the calculations

so can an formula be 4 different things for example C4H5O2N4?

Vocaloid · Answer

Yes, a formula can have any number of elements.