Just needed help with solving this problem:)
A stimulant in coffee and tea is caffeine, a substance of molar mass 194 g/mol. 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. Determine the empirical and molecular formulas of caffeine, and write the equation for its combustion.
I understand the gist of writing the chemical equation at the end, I'm just having trouble how I would get the individual masses of C, H, O, and N from what was given to get to the empirical formula.
Thanks!
Textbook Problem M.19
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Re: Textbook Problem M.19
Hello.
I like to write the chemical equation out like this:
(?) + O2 -> CO2 + H2O + N2
I just write caffeine as (?) because we don't know its molecular formula
You need to use the masses of CO2, H20, & N2 to find the moles of C, H, & N (using stoichiometry).
Use the masses of C, H, & N (from calculating their mole amounts), add them up and subtract them from the 0.376g of caffeine that was burned (that's the mass of O2). Find the amount of moles of O and find the empirical formula.
Then find the molar mass of the empirical formula.
Divide 194 g/mol (molar mass of caffeine) by the molar mass of the empirical formula.
Then multiply the number of each C, H, N, & O in the empirical formula by the quotient.
I like to write the chemical equation out like this:
(?) + O2 -> CO2 + H2O + N2
I just write caffeine as (?) because we don't know its molecular formula
You need to use the masses of CO2, H20, & N2 to find the moles of C, H, & N (using stoichiometry).
Use the masses of C, H, & N (from calculating their mole amounts), add them up and subtract them from the 0.376g of caffeine that was burned (that's the mass of O2). Find the amount of moles of O and find the empirical formula.
Then find the molar mass of the empirical formula.
Divide 194 g/mol (molar mass of caffeine) by the molar mass of the empirical formula.
Then multiply the number of each C, H, N, & O in the empirical formula by the quotient.
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Re: Textbook Problem M.19
So firstly, what I would do is I would convert the masses of carbon dioxide, water, and nitrogen to the moles of Carbon in carbon dioxide, Hydrogen in water, and simply Nitrogen.
To do this for carbon dioxide, you would essentially perform this calculation: 0.682gCO2 x (1molCO2/44.009gCO2) x (1molC/1molCO2).
For water, you would perform this calculation: 0.174gH2O x (1molH2O/18.015gH2O) x (2molH/1molH2O).
You would complete this calculation for each of the products, and once you determine the moles of Carbon, Hydrogen, and Nitrogen, then you will convert that to grams of Carbon, Hydrogen, and Nitrogen. Once you find the grams of Carbon, Hydrogen, and Nitrogen, then you will add all those masses up and subtract the mass of caffeine (0.376 g) from the combined mass of Carbon, Hydrogen, and Nitrogen. This will give you the mass of Oxygen. Then you will convert the mass of Oxygen to moles, and then proceed with the necessary steps to determine the empirical and molecular formulas.
Hope this helps!
To do this for carbon dioxide, you would essentially perform this calculation: 0.682gCO2 x (1molCO2/44.009gCO2) x (1molC/1molCO2).
For water, you would perform this calculation: 0.174gH2O x (1molH2O/18.015gH2O) x (2molH/1molH2O).
You would complete this calculation for each of the products, and once you determine the moles of Carbon, Hydrogen, and Nitrogen, then you will convert that to grams of Carbon, Hydrogen, and Nitrogen. Once you find the grams of Carbon, Hydrogen, and Nitrogen, then you will add all those masses up and subtract the mass of caffeine (0.376 g) from the combined mass of Carbon, Hydrogen, and Nitrogen. This will give you the mass of Oxygen. Then you will convert the mass of Oxygen to moles, and then proceed with the necessary steps to determine the empirical and molecular formulas.
Hope this helps!
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Re: Textbook Problem M.19
Christine Nguyen 1I wrote:Just needed help with solving this problem:)
A stimulant in coffee and tea is caffeine, a substance of molar mass 194 g/mol. 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. Determine the empirical and molecular formulas of caffeine, and write the equation for its combustion.
I understand the gist of writing the chemical equation at the end, I'm just having trouble how I would get the individual masses of C, H, O, and N from what was given to get to the empirical formula.
Thanks!
Hi! To find the individual masses you use the mass of carbon dioxide to find C, water to find H, and nitrogen to find N. Then you subtract each of these masses by the given 0.376g to find oxygen. Now you have the actual masses for each element and need to change that to moles in order to find the ratio between the elements. Hope this helps!
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Re: Textbook Problem M.19
The empirical formula is the lowest ratio of the molecule's composition using whole numbers, so just make sure that the numbers of each atom in the molecular formula you end up with can't be divided any further without having a fraction. For these kinds of problems where all of the masses for the products are provided, I first add up the mass of the products to find the missing masses on the reactant side (due to law of conservation of mass). Since the makeup of caffeine is the unknown, I just convert to moles the masses of the rest of the products and reactants that are known and from here fill in a skeleton equation of the reaction to see what must be filled in for C?H?N?O? for the equation to be balanced on both sides.
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Re: Textbook Problem M.19
Isabel_Eslabon_2G wrote:Hello.
I like to write the chemical equation out like this:
(?) + O2 -> CO2 + H2O + N2
I just write caffeine as (?) because we don't know its molecular formula
You need to use the masses of CO2, H20, & N2 to find the moles of C, H, & N (using stoichiometry).
Use the masses of C, H, & N (from calculating their mole amounts), add them up and subtract them from the 0.376g of caffeine that was burned (that's the mass of O2). Find the amount of moles of O and find the empirical formula.
Then find the molar mass of the empirical formula.
Divide 194 g/mol (molar mass of caffeine) by the molar mass of the empirical formula.
Then multiply the number of each C, H, N, & O in the empirical formula by the quotient.
Thank you!
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- Posts: 102
- Joined: Wed Sep 30, 2020 9:40 pm
Re: Textbook Problem M.19
SelenaDahabreh1K wrote:So firstly, what I would do is I would convert the masses of carbon dioxide, water, and nitrogen to the moles of Carbon in carbon dioxide, Hydrogen in water, and simply Nitrogen.
To do this for carbon dioxide, you would essentially perform this calculation: 0.682gCO2 x (1molCO2/44.009gCO2) x (1molC/1molCO2).
For water, you would perform this calculation: 0.174gH2O x (1molH2O/18.015gH2O) x (2molH/1molH2O).
You would complete this calculation for each of the products, and once you determine the moles of Carbon, Hydrogen, and Nitrogen, then you will convert that to grams of Carbon, Hydrogen, and Nitrogen. Once you find the grams of Carbon, Hydrogen, and Nitrogen, then you will add all those masses up and subtract the mass of caffeine (0.376 g) from the combined mass of Carbon, Hydrogen, and Nitrogen. This will give you the mass of Oxygen. Then you will convert the mass of Oxygen to moles, and then proceed with the necessary steps to determine the empirical and molecular formulas.
Hope this helps!
Thank you!
-
- Posts: 102
- Joined: Wed Sep 30, 2020 9:40 pm
Re: Textbook Problem M.19
Kimiya Aframian IB wrote:Christine Nguyen 1I wrote:Just needed help with solving this problem:)
A stimulant in coffee and tea is caffeine, a substance of molar mass 194 g/mol. 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. Determine the empirical and molecular formulas of caffeine, and write the equation for its combustion.
I understand the gist of writing the chemical equation at the end, I'm just having trouble how I would get the individual masses of C, H, O, and N from what was given to get to the empirical formula.
Thanks!
Hi! To find the individual masses you use the mass of carbon dioxide to find C, water to find H, and nitrogen to find N. Then you subtract each of these masses by the given 0.376g to find oxygen. Now you have the actual masses for each element and need to change that to moles in order to find the ratio between the elements. Hope this helps!
Thank you!
-
- Posts: 102
- Joined: Wed Sep 30, 2020 9:40 pm
Re: Textbook Problem M.19
Anna Yang 2A wrote:The empirical formula is the lowest ratio of the molecule's composition using whole numbers, so just make sure that the numbers of each atom in the molecular formula you end up with can't be divided any further without having a fraction. For these kinds of problems where all of the masses for the products are provided, I first add up the mass of the products to find the missing masses on the reactant side (due to law of conservation of mass). Since the makeup of caffeine is the unknown, I just convert to moles the masses of the rest of the products and reactants that are known and from here fill in a skeleton equation of the reaction to see what must be filled in for C?H?N?O? for the equation to be balanced on both sides.
Thank you!
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