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. Determine the empirical and molecular formulas of caffeine, and write the equation for its combustion.
I'm not sure how to solve this problem, I got an answer that was pretty close to the answer in the textbook but it wasn't right. I'm specifically confused if nitrogen gas is being released as a product in this combustion reaction, which is something I'm not too familiar with. If anyone could help me figure out how to solve this problem, it would be much appreciated!
M19 Textbook Problem
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Re: M19 Textbook Problem
Hi!
This question is different from the some of the others because in the process of combustion, components from the original elements in caffeine reacted with oxygen to make up molecules containing oxygen that was not in the original caffeine composition.
Therefore, our goal for this problem is to use the masses of the products to determine the mass of the components we know were definitely in the original sample of caffeine (C, H, and N) and then subtract the sum of these masses from the total mass of the caffeine sample (0.376g) to determine how much oxygen was used.
Caffeine + O2 --> CO2 + H20 + N2
Step 1: Convert the mass of each product to moles by dividing the total the mass of the reactant by its molar mass. (for example: 0.682g CO2 / 44.01g). The results should be 0.155mol CO2, 0.00966 mol H2O, 0.00393 mol N2
When nitrogen is released, it's typically in its diatomic form, or N2. In the following step, we will convert moles of N2 to simply moles of N to be able to determine the empirical formula. However for this problem you could have simply found the moles of N from this step as well and received 0.00785 mol N (as we will in the following step).
Step 2: Convert moles of CO2, H20, and N2 from the previous step to moles of C, H, and N. To do so, use molar ratios.
For example, for every 1 mol of H20 there are 2 mol of H, so for every 0.00966 mol H20 there must be 0.0193 mol H (or double as much). For N2, for every one mole of N2, there are 2 mol of N.
Your results should be 0.155 mol C, 0.0193 mol H, and 0.00786 mol N.
We exclude oxygen because part of the total mass of oxygen in the products belonged to the O2 used for the combustion reaction.
Step 3: Convert moles of C, H, and N to grams of each element by using their molar masses. You should receive 0.186g C, 0.0195 g H, and 0.110g N (which you could have kept from the original mass listed in the problem).
Step 4: Subtract the sum of these masses from the total mass of the caffeine sample to determine the mass of oxygen in the original sample. We know this last element to be oxygen because no other kinds of elements were present in the reaction.
0.376g caffeine - 0.186gC - 0.0195gH - 0.110gN = 0.605 g O
Convert the mass of O to moles. You should receive 0.00378 mol O.
Step 5: Now that you have the moles of each component in caffeine (0.155 mol C, 0.0193 mol H, 0.00786 mol N, and 0.00378 mol O), you can find the empirical formula as usual. First, divided the # of moles of each component by the smallest number of moles (in this case 0.00378 mol) to receive a ratio of each component relative to the other. Round each result to the closest whole number (you can do this b/c the all the results are within around .15 of a whole #). You should get 4 C's, 5 H's, 2 N's, and 1 O. Therefore, your empirical formula is C4H5N2O.
Step 6: To determine the molecular formula, find the total molar mass of the components in your empirical formula, which should be 97.09g/mol. Since the problem states that caffeine has a molar mass of 194g/mol, this is approximately double the molar mass you found from your empirical formula. Thus, double the amount of each component (multiply the empirical formula by 2). You should get C8H10N4O2 as your molecular formula.
This question is different from the some of the others because in the process of combustion, components from the original elements in caffeine reacted with oxygen to make up molecules containing oxygen that was not in the original caffeine composition.
Therefore, our goal for this problem is to use the masses of the products to determine the mass of the components we know were definitely in the original sample of caffeine (C, H, and N) and then subtract the sum of these masses from the total mass of the caffeine sample (0.376g) to determine how much oxygen was used.
Caffeine + O2 --> CO2 + H20 + N2
Step 1: Convert the mass of each product to moles by dividing the total the mass of the reactant by its molar mass. (for example: 0.682g CO2 / 44.01g). The results should be 0.155mol CO2, 0.00966 mol H2O, 0.00393 mol N2
When nitrogen is released, it's typically in its diatomic form, or N2. In the following step, we will convert moles of N2 to simply moles of N to be able to determine the empirical formula. However for this problem you could have simply found the moles of N from this step as well and received 0.00785 mol N (as we will in the following step).
Step 2: Convert moles of CO2, H20, and N2 from the previous step to moles of C, H, and N. To do so, use molar ratios.
For example, for every 1 mol of H20 there are 2 mol of H, so for every 0.00966 mol H20 there must be 0.0193 mol H (or double as much). For N2, for every one mole of N2, there are 2 mol of N.
Your results should be 0.155 mol C, 0.0193 mol H, and 0.00786 mol N.
We exclude oxygen because part of the total mass of oxygen in the products belonged to the O2 used for the combustion reaction.
Step 3: Convert moles of C, H, and N to grams of each element by using their molar masses. You should receive 0.186g C, 0.0195 g H, and 0.110g N (which you could have kept from the original mass listed in the problem).
Step 4: Subtract the sum of these masses from the total mass of the caffeine sample to determine the mass of oxygen in the original sample. We know this last element to be oxygen because no other kinds of elements were present in the reaction.
0.376g caffeine - 0.186gC - 0.0195gH - 0.110gN = 0.605 g O
Convert the mass of O to moles. You should receive 0.00378 mol O.
Step 5: Now that you have the moles of each component in caffeine (0.155 mol C, 0.0193 mol H, 0.00786 mol N, and 0.00378 mol O), you can find the empirical formula as usual. First, divided the # of moles of each component by the smallest number of moles (in this case 0.00378 mol) to receive a ratio of each component relative to the other. Round each result to the closest whole number (you can do this b/c the all the results are within around .15 of a whole #). You should get 4 C's, 5 H's, 2 N's, and 1 O. Therefore, your empirical formula is C4H5N2O.
Step 6: To determine the molecular formula, find the total molar mass of the components in your empirical formula, which should be 97.09g/mol. Since the problem states that caffeine has a molar mass of 194g/mol, this is approximately double the molar mass you found from your empirical formula. Thus, double the amount of each component (multiply the empirical formula by 2). You should get C8H10N4O2 as your molecular formula.
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Re: M19 Textbook Problem
Hi Ryan!
First, you have to convert all the masses to mols. (Divide grams by the molar mass of each molecule). You get .0155 mol CO2, .009656 mol H2O, and .003927 mol N2.
Then you find the mols for C, H, and N by multiplying the mols by the ratio of the element in each molecule. There is only 1 C in CO2, so you only need to multiply it by 1, but there are 2 H in H20, so you have to multiply it by 2. .0155 mol C, .01931 mol H, and .007852 mol N.
Now you have to find how much Oxygen is in the caffeine molecule. Convert the mols of C, H, and N to grams and subtract it by the total weight in grams of caffeine. .186171 g C, .019466 g H, and .109983 g N. When added, you get .31562 grams. Then, subtract that from .376. You will get .06038 g O.
Now convert O grams to mols. you get .003774 mol O.
With all the mols accounted for, you can find the empirical formula by dividing all the mols by the smallest mol. To get the molecular formula, you divide the molar mass of caffeine(194g/mol) by the molar mass of the empirical formula.
Good luck!
Kainath
First, you have to convert all the masses to mols. (Divide grams by the molar mass of each molecule). You get .0155 mol CO2, .009656 mol H2O, and .003927 mol N2.
Then you find the mols for C, H, and N by multiplying the mols by the ratio of the element in each molecule. There is only 1 C in CO2, so you only need to multiply it by 1, but there are 2 H in H20, so you have to multiply it by 2. .0155 mol C, .01931 mol H, and .007852 mol N.
Now you have to find how much Oxygen is in the caffeine molecule. Convert the mols of C, H, and N to grams and subtract it by the total weight in grams of caffeine. .186171 g C, .019466 g H, and .109983 g N. When added, you get .31562 grams. Then, subtract that from .376. You will get .06038 g O.
Now convert O grams to mols. you get .003774 mol O.
With all the mols accounted for, you can find the empirical formula by dividing all the mols by the smallest mol. To get the molecular formula, you divide the molar mass of caffeine(194g/mol) by the molar mass of the empirical formula.
Good luck!
Kainath
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Re: M19 Textbook Problem
Ok, thank you both so much! I was able to understand the problem after realizing that O is also an element in caffeine as well, that's the main part I was struggling with. Both of your step-by-step guides were very helpful.
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