Hello, everyone. This is my first time posting a question on chemistry community.
I am having difficulties with the sampling question #9.
The compound contains only C, H, and O and was experimentally found to have a molar mass of 110±10 g/mol . When a 1.000 g sample of caproic acid is burned in excess oxygen, 2.275 g CO2 and 0.929 g H2O are collected. Determine the empirical formula and molecular formula of caproic acid. Insert subscripts as necessary.
I have tried calculating the coefficients but I got lost. I would really appreciate some guidelines. Thank you!
Sampling #9
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Re: Sampling #9
Ok so you gotta find how much mass each element is. The easiest elements to find first is C and H. Calculate the molar mass of C and divide it to the molar mass of CO2. From there, multiply the ratio you got for C to the true mass of CO2 (in this case, 2.275g). After that, apply the same method with that of H with H20. When you get the masses of both C and H, you have O left. As previously mentioned, there was 1.000g of caproic acid burned, so you add the masses of both C and H and subtract that value from 1.000g. The answer from that will be your mass for O. From there, find the moles of all elements, divide those moles by the smallest mole among the elements, and hopefully you should get whole numbers in your calculation indicating the subscripts for each element. I hope this helps.
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Re: Sampling #9
IreneSeo3F wrote:Hello, everyone. This is my first time posting a question on chemistry community.
I am having difficulties with the sampling question #9.
The compound contains only C, H, and O and was experimentally found to have a molar mass of 110±10 g/mol . When a 1.000 g sample of caproic acid is burned in excess oxygen, 2.275 g CO2 and 0.929 g H2O are collected. Determine the empirical formula and molecular formula of caproic acid. Insert subscripts as necessary.
I have tried calculating the coefficients but I got lost. I would really appreciate some guidelines. Thank you!
You will need to calculate the grams of carbon and hydrogen in the reactant by isolating them from the co2 and h2o using molar mass and dimensional analysis conversions. After this you will subtract the total carbon and hydrogen from the total mass to get the total oxygen mass and go through the traditional empirical formula conversion to solve for the empirical formulas and molecular formulas. Hope this helps!
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Re: Sampling #9
Hi. I was really lost too until someone explained it to me.
So looking at the problem, the given information is that we have 1.000g of caproic acid. We have 2.275g of CO2 and 0.929g of H2O.
Step one would be to find the carbon in CO2. In order to do this, you must take the molar mass of carbon (12.01 g/mol) and divide it by the mass of CO2 (44.01). You can then multiply by 100 to get a percentage but then round to the nearest whole number. Once you have the percentage, multiply that percentage and the 2.275g of CO2. (You do this because every 1g of CO2 is that much percent carbon). The second step would be to repeat the process for hydrogen. In order to find oxygen, the formula is as follows: oxygen = 1g - (g of C + g of H) . Then once you get that result, you would follow the process from before (get a percent and multiply).
Next, we would find what percentage of the sample is each element. For C, we would take the grams of C and divide by the sample size (in this case 1g). Since it is one, we would get the same number. Then, you would multiply by 100% in order to make it a percentage. You follow this process for H and O as well.
Then, we would convert from grams to moles. In order to do this, we take the percentage you have gotten and divide it by the molar mass. (For example, if you got 62.08% C then you would write 62.08/12.01). Repeat the process for all of the letters (C, H, and O).
Once you have gotten all of your moles, pick which one is the smallest. This will become the limiting reactant which means that everything else gets divided by the smallest one. When you divide, round up to the nearest whole number (only if you get a .999999 or a .0000001 or something like that). This should give you your empirical formula.
Once you have the empirical formula, find the mass of that compound. Multiply the molar mass by the number of moles you have. (example: if my empirical formula contains C3, I would do 12.01 g/mol times 3 mol. ) Then take the 110 and divide by the mass of your compound. Since the problem says +/- 10 , you also have some wiggle room so find the number it is the closest to. Once you have that number, multiple the empirical formula by that number and that should give you your molecular formula.
Keep in mind: the formulas should have whole numbers.
So looking at the problem, the given information is that we have 1.000g of caproic acid. We have 2.275g of CO2 and 0.929g of H2O.
Step one would be to find the carbon in CO2. In order to do this, you must take the molar mass of carbon (12.01 g/mol) and divide it by the mass of CO2 (44.01). You can then multiply by 100 to get a percentage but then round to the nearest whole number. Once you have the percentage, multiply that percentage and the 2.275g of CO2. (You do this because every 1g of CO2 is that much percent carbon). The second step would be to repeat the process for hydrogen. In order to find oxygen, the formula is as follows: oxygen = 1g - (g of C + g of H) . Then once you get that result, you would follow the process from before (get a percent and multiply).
Next, we would find what percentage of the sample is each element. For C, we would take the grams of C and divide by the sample size (in this case 1g). Since it is one, we would get the same number. Then, you would multiply by 100% in order to make it a percentage. You follow this process for H and O as well.
Then, we would convert from grams to moles. In order to do this, we take the percentage you have gotten and divide it by the molar mass. (For example, if you got 62.08% C then you would write 62.08/12.01). Repeat the process for all of the letters (C, H, and O).
Once you have gotten all of your moles, pick which one is the smallest. This will become the limiting reactant which means that everything else gets divided by the smallest one. When you divide, round up to the nearest whole number (only if you get a .999999 or a .0000001 or something like that). This should give you your empirical formula.
Once you have the empirical formula, find the mass of that compound. Multiply the molar mass by the number of moles you have. (example: if my empirical formula contains C3, I would do 12.01 g/mol times 3 mol. ) Then take the 110 and divide by the mass of your compound. Since the problem says +/- 10 , you also have some wiggle room so find the number it is the closest to. Once you have that number, multiple the empirical formula by that number and that should give you your molecular formula.
Keep in mind: the formulas should have whole numbers.
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Re: Sampling #9
Hi! I had trouble with this one as well at first.
1) identify molar mass of CO2 and the given mass in grams. then find the mass of carbon alone. (grams/molar mass of CO2 x molar mass of carbon) This answer tells you how many grams of carbon are in one gram of the compound. Therefore, you can find the percent of carbon in 100 grams by moving decimal to the right twice.
2) next, identify molar mass of H20 and the grams given. There are two grams of hydrogen in H20. Therefore to find the mass of an individual hydrogen in one gram of the final compound= (grams/molar mass of H20 x 2). Find the percentage by moving decimal to the right twice.
3) Find the percentage of oxygen based on the other two percentages. (100- (% of C + % of H) = % of O)
4) Divide each of the 3 percentages by molar mass. (12 for C, 1 for H, 16 for O). Identify the smallest result. Divide all three by the smallest.
5) Round slightly to nearest whole number and you should get the subscripts for the empirical formula.
6) Find molar mass of EF, and divide the actual molar mass (it's given) by the molar mass of the EF to find a factor.
7) Finally, multiply the subscripts of the EF by the factor, in order to get the molecular formula.
1) identify molar mass of CO2 and the given mass in grams. then find the mass of carbon alone. (grams/molar mass of CO2 x molar mass of carbon) This answer tells you how many grams of carbon are in one gram of the compound. Therefore, you can find the percent of carbon in 100 grams by moving decimal to the right twice.
2) next, identify molar mass of H20 and the grams given. There are two grams of hydrogen in H20. Therefore to find the mass of an individual hydrogen in one gram of the final compound= (grams/molar mass of H20 x 2). Find the percentage by moving decimal to the right twice.
3) Find the percentage of oxygen based on the other two percentages. (100- (% of C + % of H) = % of O)
4) Divide each of the 3 percentages by molar mass. (12 for C, 1 for H, 16 for O). Identify the smallest result. Divide all three by the smallest.
5) Round slightly to nearest whole number and you should get the subscripts for the empirical formula.
6) Find molar mass of EF, and divide the actual molar mass (it's given) by the molar mass of the EF to find a factor.
7) Finally, multiply the subscripts of the EF by the factor, in order to get the molecular formula.
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Re: Sampling #9
Rich Luong 1B wrote:Ok so you gotta find how much mass each element is. The easiest elements to find first is C and H. Calculate the molar mass of C and divide it to the molar mass of CO2. From there, multiply the ratio you got for C to the true mass of CO2 (in this case, 2.275g). After that, apply the same method with that of H with H20. When you get the masses of both C and H, you have O left. As previously mentioned, there was 1.000g of caproic acid burned, so you add the masses of both C and H and subtract that value from 1.000g. The answer from that will be your mass for O. From there, find the moles of all elements, divide those moles by the smallest mole among the elements, and hopefully you should get whole numbers in your calculation indicating the subscripts for each element. I hope this helps.
This was super helpful as I was super stuck on how to approach this problem, thank you!
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Re: Sampling #9
Nathaly Cruz 2D wrote:Hi. I was really lost too until someone explained it to me.
So looking at the problem, the given information is that we have 1.000g of caproic acid. We have 2.275g of CO2 and 0.929g of H2O.
Step one would be to find the carbon in CO2. In order to do this, you must take the molar mass of carbon (12.01 g/mol) and divide it by the mass of CO2 (44.01). You can then multiply by 100 to get a percentage but then round to the nearest whole number. Once you have the percentage, multiply that percentage and the 2.275g of CO2. (You do this because every 1g of CO2 is that much percent carbon). The second step would be to repeat the process for hydrogen. In order to find oxygen, the formula is as follows: oxygen = 1g - (g of C + g of H) . Then once you get that result, you would follow the process from before (get a percent and multiply).
Next, we would find what percentage of the sample is each element. For C, we would take the grams of C and divide by the sample size (in this case 1g). Since it is one, we would get the same number. Then, you would multiply by 100% in order to make it a percentage. You follow this process for H and O as well.
Then, we would convert from grams to moles. In order to do this, we take the percentage you have gotten and divide it by the molar mass. (For example, if you got 62.08% C then you would write 62.08/12.01). Repeat the process for all of the letters (C, H, and O).
Once you have gotten all of your moles, pick which one is the smallest. This will become the limiting reactant which means that everything else gets divided by the smallest one. When you divide, round up to the nearest whole number (only if you get a .999999 or a .0000001 or something like that). This should give you your empirical formula.
Once you have the empirical formula, find the mass of that compound. Multiply the molar mass by the number of moles you have. (example: if my empirical formula contains C3, I would do 12.01 g/mol times 3 mol. ) Then take the 110 and divide by the mass of your compound. Since the problem says +/- 10 , you also have some wiggle room so find the number it is the closest to. Once you have that number, multiple the empirical formula by that number and that should give you your molecular formula.
Keep in mind: the formulas should have whole numbers.
thank you for this explanation! I was having a lot of difficulty trying to solve this but your explanation helped me understand what I needed to fix!
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Re: Sampling #9
I would start off first with the amount of grams you would get from the combustion formula for both Carbon and Hydrogen. First use the grams of CO2 and work backwords towards C to find the grams of C. Then use the grams of H2O and work backwords to find the grams of H. To get grams of O, take the total of the solution grams given (1.000) and subtract by the total of both grams of C and H to get O. To get the empirical, convert all grams to moles, then divide all numbers by the lowest number in order to get the correct ratio. After getting the correct ration divide the total mass given by the empirical mass to know how many times to multiply in order to get the molecular formula!
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