Limiting Reactants M19

Semi Yoon
Posts: 60
Joined: Fri Sep 28, 2018 12:27 am

Limiting Reactants M19

A stimulant in coffee and tea is caffeine, a substance of molar mass 194 g mol 1. When 0.376 g of caffeine was burned, 0.682 g of carbon dioxide, 0.174 g of water, and 0.110 g of nitrogen were formed. How would you find the empirical and molecular formulas of caffeine, and write the equation for its combustion?

Kobe_Wright
Posts: 83
Joined: Fri Sep 28, 2018 12:16 am

Re: Limiting Reactants M19

Well you would have to get the info that the composition of caffeeing is c(8)H(10)N(4)O(2), then you could construct a chemical equation and balance it, then using the known molecular mass, and the weight of the products we could use dimensional analysis and come up with the empirical, then molecular formulas.

Posts: 73
Joined: Fri Sep 28, 2018 12:15 am

Re: Limiting Reactants M19

I was also wondering about this question. What would you do with the grams given for the molecules like h20 and co2? I am used to the problem giving the mass of just the individual elements. I tried to solve for the individual masses by saying there are 2 mol of o in 1 mol co2 and 1 mol of c in 1 mol of co2 etc but I did not get the answer. Anyone know what to do?

jessicahe4Elavelle
Posts: 61
Joined: Fri Sep 28, 2018 12:27 am
Been upvoted: 1 time

Re: Limiting Reactants M19

How do we get the info that the composition of caffeine is c(8)H(10)N(4)O(2)? I thought that that was what we were trying to find.

Bingcui Guo
Posts: 30
Joined: Fri Sep 28, 2018 12:19 am

Re: Limiting Reactants M19

First, use the product and the mass percentage to calculate the mass of C. H. N in the caffeine. And subtract these mass from the 0.376g to calculate the mass of O in the caffeine. Second, use the mass you get from the products to calculate the moles of C.H.N.O. Third, find the simplest proportion of these four elements. Fourth, use the molar mass to find the molecular formula. Hope this would help