Mrs. Berger's Chemistry and Physics Blog

Of Mice and Men… or Moles and Einstein

Finding an Empirical Formula and Molecular Formula November 10, 2006

Filed under: Chemistry,Chemistry I — Mrs. Berger @ 12:18 pm

This week in Chem I, we have added to our chemistry knowlege by using our understanding of moles and molar mass to find percent composition, empirical formulas, and molecular formulas.  A brief review is below:

% Composition =  (Mass of element / Mass of Compound) X 100

If given % composition, you can find the empirical formula by:

Step 1:  Assume you have 100g.  This way, your percentages will be the number of grams.  Example:  40% of a compound is carbon.  40% of 100 g is 40 grams.  It’s a miracle!

 Step 2:  Change grams to moles.  Since empirical formulas are mole comparisons, we need to put our numbers in moles first.

 Step 3:  Compare the number of moles of each to the smallest number of moles by dividing by the smallest number.

Step 4:  Round your ratio to the nearest whole number as long as it is “close.”  For example, 1.99987 can be rounded to 2, but 1.3333 cannot be rounded to 1.  It is four-thirds, so we must multiply all ratios by 3 to rid ourselves of the fraction.

Step 5:  Write the formula using your ratio numbers as subscripts.

 To find the molecular formula, you must be given the actual molar mass of the compound.  Find the ratio of the actual mass to the empirical mass.  For example, if it is one, the empirical formula is equal to the molecular formula.  If it is 2, the subscripts must be multiplied by two.

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130 Responses to “Finding an Empirical Formula and Molecular Formula”

  1. Tarun Says:

    Hey,,i Have a questions
    how do u write a molecular formula for a given empirical formula?
    for eg. the empirical formula is CH3O2 so wats the molecular formula for this,,,email it to tarunkakumanu@yahoo.com ,,will be waiting for u reply
    thanks

  2. Mrs. Berger Says:

    To find the molecular formula, you would need to know the molecular mass of the actual compound. Using your example of CH3O2, this empirical formula would have a mass of 1carbon+3hydrogens+2oxygens = 47.04. If the molecular mass is this, then the empirical formula is the same as the empirical. If the actual molecular mass is 94, then this is twice the empirical formula, or C2H6O4.

    Does this help? How did you find my website? Was it helpful to you?

  3. Matthew Aubuef Says:

    I had the same question as Tarun had and your response was very helpful.

  4. Jo Says:

    wondered if you could help me? I am stuck with this question, don’t really understand moles.

    the only bit i understand is n= w/Mwt

    Q How many moles of Ca are combined with 2.68 moles of Br in CaBr2? How many g of Ca are combined with 2.68g of Br in the same compound?

    many thanks
    Jo

  5. Mrs. Berger Says:

    Jo,

    You really have two questions here. The formula for calcium bromide (CaBr2) tells you the ratio of moles of each element in the compound. This tells us that for every calcium atom, there are 2 atoms of barium. Because the “mole” is a number of particles, this also tells us that for every one mole of calcium, there are two moles of bromine. You can use this to set up a proportion.

    1 mol Ca : 2 mol Br = ___ mol Ca : 2.68 mol Br

    Moles can be compared in this way because a mole of anything is 6.02 x 10^23 whatevers. You cannot compare masses in this way because each element has a different mass. However, if you know the number of moles of something, you can compare it to the number of moles of the other thing.

    In your problem, if you find out how many moles of bromine are in 2.68g (n= w/Mwt), you can compare the number of moles of bromine to calcium using the ratio 1 mol Ca : 2 mol Br = _____ mol Ca : n mol Br. Then once you know the number of moles of calcium, convert the number of moles of calcium to grams of calcium using n= w/Mwt again, but this time for calcium.

  6. Megan Says:

    I have a question. It says : Copper reacts with sulfur to form copper (I) Sulfide. How many grams of copper (I) sulfide can be formed when 80.0 grams of copper reacts with 25.0 grams of sulfer?
    I appreciate your help!!

  7. Mrs. Berger Says:

    I will not divluge the answer to you, Megan, since I do not believe you are one of my students and I know your teacher would not appreciate it. However, so that you can learn, I will give you some hints. First, you need to write a balanced chemical reaction equation for this reaction. Sulfur, when found by itself in a chemical reaction, is not just S like Copper is Cu. It is found as a group of sulfur atoms, so you need to know this formula. Then, you must also write the formula for copper(I) sulfide. Remember the Roman numeral tells you the charge on coppper, and you should probably know the charge on sulfur. Write the formula so the two charges cancel out. Once you have the reaction equation written, you have to find out which element, copper or sulfur, is the limiting reactant. You can do this by finding out how many moles of each you are starting with. Then convert, for example, from moles of copper to moles of sulfur. If the number of moles of sulfur you calculate is more that what you have, then sulfur is the L.R. and you need to use its amount to calculate moles of copper(I) sulfide. If the number you calcuate is less than what you have, then copper is the L.R. so you need to use this amount to calculate the amount of product. Check out http://www.chem.tamu.edu/class/majors/tutorialnotefiles/limiting.htm for some example problems. Good Luck!

  8. rooster Says:

    what is a mol

  9. Mrs. Berger Says:

    A mol is a unit of measurement. Just like meters measures distance and liters measure volume, mols measure a number of particles. We might buy “a dozen” eggs at the store. We know a dozen is 12. Or, we might pick up a “ream” of paper, which we know should contain 500 sheets. A mol is also a certain number just like “dozen” or “ream”. 1 dozen = 12, 1 ream = 500, and 1 mol = 6.02×10^23 So, a mol of eggs would be 602000000000000000000000 eggs, or a mol of paper would be 6.02×10^23 sheets! That’s a lot. But usually we talk about mols in terms of atoms, so since atoms are really small, it doesn’t take much of a substance to equal one mol of atoms or molecules. The best part about mols is that 1 mol of atoms is equal to the atomic mass in grams and 1 mol of molecules is equal to the molecular mass in grams. We can this the “molar mass” which is not the mass of your tooth! (dumb chemistry teacher joke – sorry!) Hope that helps.

  10. Jordan Says:

    When doing mole conversions, and you are asked to find the number of molecules in a certain number of grams of an element (for example, 3.5 grams of N2), is finding the number of molecules the same as finding the number of atoms, since you are working with an element and not a compound?

  11. Mrs. Berger Says:

    I think you are on the right track. If you have 3.5 grams of nitrogen (N2 because it is diatomic), you would need to convert to moles. Then, we know that in one mole of somethings there are 6.02×10^23 somethings. Therefore in one mole of nitrogen molecules there are 6.02×10^23 nitrogen molecules. If they asked for the number of atoms, then you would have to consider that there are two moles of nitrogen atoms in one mole of nitrogen molecules. However, I think the first part is what you were looking for.

    I am glad you like my blog!

  12. Tara Says:

    the combustion of 7.02 grams of a compound that contains only C, H, N and O yields 12.07 grams of CO2 and 3.95 grams of H2O. another sample of the compound with the mass of 32.24 grams is found to contain 8.06 grams of O. what is the empirical formula of the compound?

  13. rasheed Says:

    I dont really understand how to appraoch this: I have to find the empirical formula of an organic compound containing 40% carbon, 6.7% hydrogen and 55.3% oxygen. Given that using a mass spectrometer the relative molecular mass of the compound is 180, suggest a molecular formula for the compound.

    much appreciated

  14. Angel Says:

    I have question I’m super stuck on & was hoping you could help me. Here it is:

    The molecular formula of a hydrocarbon is to be determined by analyzing its combustion products & investigating its colligative properties.

    The hydrocarbon burns completely, producing 7.2 grams of water & 7.2 liters of CO2 at standard conditions. What is the empirical formula of the hydrocarbon?

    Your help would be greatly appreciated. (: Thanks in advance.

  15. Mrs. Berger Says:

    I am assuming you want to know HOW to get the answer and not what the answer is, so I will go with that being your question! 🙂

    The empirical formula is the lowest whole-number ratio of each of the elements in the compound. Since it is a hydrocarbon, it contains some number of carbons and some number of hydrogens. If it is being burnt completely, it is reacting with oxygen to produce carbon dioxide and water. Therefore our reaction equation is:

    CxHy + O2 —> CO2 + H2O

    Looking at the products, the only place that hydrogen could’ve come from to form the water was our hydrocarbon. According to the law of conservation of matter, in any chemical reaction, atoms are neither created nor destroyed, just rearranged. Therefore, the amount of hydrogen we end with is the same amount we started with. If you can figure out what percent of your water is hydrogen and use this to find what mass of the 7.2 grams of water produced is hydrogen, you can figure out how many grams of hydrogen you started with. Since we are trying to find the mole ratio, then change this number of grams of hydrogen into moles of hydrogen and come back to that in a second.

    Secondly, the only place the carbon in CO2 could’ve come from is also from the hydrocarbon. For the same reasons above, you can find out the original number of moles of carbon you started with because it has to be the same as what is in the products. Since this is at “standard conditions” that is a REALLY big clue. At standard conditions, one mole of gas has a volume of 22.4 L. If you only have 7.2 L, then you can find how many moles of gas you have, obviously less than one. You know that in CO2, there is one mole of carbon for every one mole of CO2, so this gives you the number of moles of carbon.

    Once you have the number of moles of each element, you can find the ratio of them to each other by dividing them both by the smallest one. Obviously, one of them will be “1” because it was divided by itself but you can see how many moles of the other substance you have. For example, if you were to end up with .500 moles of C and 2.00 moles of H, you could say that this is a ratio of CH4. If you end up with a decimal that is not “near” a whole number, like .3333, multiply by the common denominator of this fraciton. For example, if you get .3333 moles of carbon and one mole of hydrogen, multiplying these both by 3 (because .3333 is like 1/3) gives us CH3.

    Try this, let me know if this was helpful, or what you got for your answer and I will tell you whether or not you got the right thing. Sorry this was so long, but I wanted to make sure to explain it well. Good luck!

    • nabila amira Says:

      how about this question?! it didnt give the mass of water.
      when 0.02 mole of hydrocarbon (X) was completely burnt in oxygen, 960 cm3 of carbon dioxide was formed at room temperature and pressure. the molecular mass of hydrocarbon (X) is 28.

      a)1) calculate the volume of carbon dioxide formed at room conditions from one mole of hydrocarbon X.
      2) determine the number of carbon atoms in one molecule of hydrocarbon X.
      b)give the molecular formula of hydrocarbon X.
      d)write the equation for the complete burning of hydrocarbon X in oxygen

      can you explain me on how to solve this question?! i really weak in this chapter and hope you can help me. your help would be greatly appreciated.

  16. Alex Dunn Says:

    hi, im a sophomore in sarasota florida, and i needed some help with mole and the empirical formula, and i googled it, found your site, and it really helped! thanks!

  17. Mrs. Berger Says:

    Thanks, Alex. I am glad my website helped! Please check back if you have any more dilemmas or let me know if there is something you would like to see here. Enjoy the sunshine!

  18. Carolyn Says:

    I have two questions that I am stumped on.

    1) How many grams of water are released when a 10.0-g sample of CaSO4•2H20 is heated?

    2) A hydrate of MgSO4 is analyzed for water content. A sample of the hydrate weighs 5.378g before heating and 2.628g after heating. What is the molecular formula of this hydrate?

    I would appreciate some help on these questions. Thank you so much!

    Carolyn

  19. Anna Says:

    i need help with a problem see my teacher gave us the answers but we have to find the work with a chart thing well my question is. What mass of glucose can be produced from a photosynthesis reaction that occurs using 10mol CO2

  20. Mrs. Berger Says:

    First for Carolyn…
    For your first question, if you can figure out what the total mass of your hydrate is by adding masses of Ca + S + four O+ 4H +two O, you are on your way. Then figure out what percent of that compound is just water. If you take this percentage times your mass of hydrate (10.0 g), you will find what part of that ten grams is just water. For the second question, you can start out by finding what percent of the total mass you lost in the same way. Each water molecule weighs about 18.02 g/mol, so you can estimate the number of waters from this.

    Now for Anna, I don’t know what “chart thing” you are talking about, but if you know your equation for photosynthesis, you can use stoichiometry to find the amount of product from a reactant.
    6CO2 + 6H2O + Energy –> C6H12O6 + 6O2

    So if you have 10 moles of CO2, you know the mole ratio of glucose to CO2 is 1:6 (6 moles of CO2 will produce one mole of glucose), so for 10 moles of CO2, you will produce 10/6 moles of glucose. If you add up the molar mass of glucose and take it multiplied by 10/6, you will find your mass of glucose produced.

  21. Carolyn Says:

    Thank you so much for your help. The information you provided helped me tons. I appreciate it greatly.

  22. Mrs. Berger Says:

    You are most welcome 🙂

  23. Chloe Says:

    hello, i found your site and it helped me a lot, but now im revising for a test, and i dont understand the difference when calculating the empirical formula and molecular formula of some compound. If you could explain it would help me lots.

  24. Mrs. Berger Says:

    Hi Chloe – the molecular formula is the ACTUAL formula for a compound whereas the empirical formula is the LOWEST RATIO of the elements. For example, the MOLECULAR formula for glucose is C6H12O6 but its empirical formula is CH2O. Some molecules, such as methane CH4, are already in the lowest ratio so the molecular formula and empirical formula are the same thing. Make sense?

  25. Chloe Says:

    yes thankx a lot! that makes things a lot clearer!

  26. Lois Says:

    Hi Mrs. Berger, the knowledge you have shared with us (me and my homework buddy) took care of our chemistry needs. Thanks a lot!!! 🙂

  27. Angee Says:

    hi there! I have problems in Chem to be answered so I hope you can help me.
    1. A 1.20g sample of a compound gave 2.92 g of CO2 and 1.22 g of H2O on combustion in Oxygen. The compound is known to contain only C, H and O. What is its simplest formula?

    2. Vinegar is 5% acetic acid, C2H4O2, by mass.
    a. How many grams of acetic acid are contained in 143.7 g vinegar?
    b. How many pounds of acetic acid are contained in 143.7 pounds vinegar?
    c. How many grams of sodium chloride, NaCl, are contained in 34.0 g of saline solution that is 5% NaCl by mass?

    How you can teach me how to solve those problems.
    I badly need them before tomorrow.
    Thank you very much. Hope you can reply immediately.
    God bless.

  28. Michelle Says:

    Hi Mrs. Berger, I stumbled upon your site when trying to find the true molecular mass given the empirical formula. I just want to say Thank you so much for all your help. You truly are a great teacher. The fact that you put so much detail and explained everything step by step to those who need help, make it seem much more clearer.

  29. John Says:

    What is the empirical formula for acetone, C2H4O2?

  30. Mike Says:

    What is the empirical formula for acetone, C2H4O2? thanks!

    • Mrs. Berger Says:

      Mike & John – I am not in the business of giving answers but if you read the previous comments, you will find that I explain that a molecular formula, like what you have for acetone, is the ACTUAL formula of a compound and the empirical formula is found by taking the lowest ratio of those elements. My response to Chloe explains this for glucose – by dividing C6H12O6 all by the common factor of 6, we get the empirical formula of CH2O. So, for acetone, C2H4O2, what factor do all your subscripts have in common, divide them all by that, and rewrite your formula in this lowest whole number ratio. If you post what you think it might be, I will tell you if you are right or not. 🙂

  31. nina Says:

    I totally dont understand the case of changing molecular to empirical formula, is there a simpler way u can explain this please. For example i have to Write the empirical formula corresponding to this molecular formulas, Al2Br6.

    Thank you

    • Mrs. Berger Says:

      Nina,
      In your molecular formula, you have two aluminum atoms and six bromine atoms. This is a 2:6 ratio. I am not going to do your problem but one similar so maybe you can get the idea. Take, for example, the formula P5O10 (five phosphorus and ten oxygens). This is a 5:10 ratio. Both of these numbers can be divided by the same factor – 5. If I divide 5 by 5, I get 1 and if I divide 10 by 5 I get 2. Therefore the REDUCED ratio of 5:10 is actually 1:2. So the EMPIRICAL formula would be PO2. Another example – molecular formula is C3H8. The ratio is 3:8. This cannot be reduced so the Empirical Formula is C3H8 – the same as the molecular formula. This is ok. The empirical formula is just the LOWEST WHOLE NUMBER RATIO of the elements in a compound. Did that make any sense?

  32. nina Says:

    ok so i would get AlBr3, since the ratio for Al2Br6 is 1:3. Thank you for taking the time to reply, ur explaination was very clear and helpful.
    thank you!

  33. Adrianna Says:

    I was reading your website and it helped me a great deal when trying for find the molecular formula while only having the empirical formula and the molecular mass… Thanks a ton!

  34. Adrianna Says:

    I actually have a question also…
    I am asked to find the molecular formula while being given the following information. How can i come to the final answer?

    formula mass= 42.08 amu
    85.64g C
    14.36g H

    I already converted the two masses above to mol’s and got the following:
    7.13 mol C
    14.22 mol H

    After finding those i think that the empirical formula is this:
    CH2

    If you could i help me i would really appreciate it =]

    • Mrs. Berger Says:

      Hope I am not too late to help, Adrianna. YOu are on the right track with the empirical formula. Since we know the formula mass to be 42.08 and if you take your empirical formulas mass to be Carbon + 2 Hydrogen = 14.03, you will find that 42.08 is about three times that amount. So, the molecular formula must be three times the empirical, 3(CH2) = C3H6.

  35. Michelle Says:

    Hi Mrs.Berger, it’s me again. Is it possible if you could teach me how to do Net Ionic equations? My teacher explained it to me, but I don’t fully understand the concepts. Thank you for taking your time to read my message. =)

    • Mrs. Berger Says:

      Hey there, MIchelle. I would love to help you with your net ionic equations. I was gone most of today because I had to take my son to the doctor. He is ok but I didn’t get to your question. During his nap tomorrow, I will try to get to your question and if not then, sometime this weekend. I will probably make a new post so just check the main site. Until then…

  36. Amy Says:

    More than three years later, I stumble upon this post while Google-ing “determining molecular formula,” and it helped me immensely! Technology is truly great. Thank you for sharing your encyclopedic knowledge of Chem, Mrs. Berger.

  37. Emily Says:

    thank you so much for having this here! I was googling for any type of explanation for all of this and you answered perfectly =D

  38. canesia Says:

    Hey there, I would like to ask
    A question about moles
    How to calculate the number of
    Moles in 2.78 dm^3 of S2 in STP?
    Whatabout RTP?

  39. canesia Says:

    hello there, i have a question here about moles and mass and tomic stuff
    how many moles does 2.78 dm^3 of S2 molecules are there in STP? RTP?

    • Mrs. Berger Says:

      I have had a couple people ask this same question. Must be working on some homework? First, I am sorry it took awhile to get back. I have been “out of the office” for awhile but just saw a few questions along these lines. Let me clarify the help I give. I will HELP you figure out a problem but WILL NOT do the problem for you. Do you have a specific question about this problem? If so, I can get you started.

  40. Lauren Says:

    Hello, i was wondering if you could help me please asap

    I am trying to find the molecular formula of compounds. all i have is the structural formula which is C6H14. how would i work it out???

    Thanks a lot

  41. Mrs. Berger Says:

    As far as I know, you just answered your own question. A structural formula is the molecule drawn out. I assume you are given a picture with six carbons chained with fourteen hydrogens bonded to it in some fashion. The actual picture is the structural formula. The molecular formula is what you gave me, C6H14. This reports the total number of each element in the molecule but gives no clue as to its structure. If your question is different from this, let me know, but I think you know what you are doing without knowing what you are doing 🙂 Make sense?

  42. Lana Says:

    Hello, I’m going to be a senior in high school and I am taking AP Chemistry which required me to complete a packet of summer work and I need some help with a free response question.

    The molecular formula of a hydrocarbon is to be determined by analyzing its combustion products and investigating its colligative properties.

    Anwers to A & B…
    a) The empirical formulaa of the hydrocarbon is C2H5.
    b) Mass in grams of the 02 required for the complete combustion of the sample is 416 grams.

    I got the first 2 parts, but the last two (c and d) are what I need help with.

    C) The hydrocarbon dissolves readily in CHCl3. The freezing point of a solution prepared by mixing 100. grams of CHCl3 and .600 grams of te hydrocarbon is -64.0 C. The molal freezing point depression constant of CHCl3 is 4.68 C/molal and its normal freezing point is -63.5 C. Calculate the molecular weight of the hydrocarbon.

    If you could help me in any way, that would be GREATLY APPRECIATED!!!!

    Lana

  43. Mrs. Berger Says:

    Lana,

    First, you said you needed help with parts C & D and I only see part C. Kind of confused there. Also, are A and B that you included your answers? Not sure in part B what 02 is. A little clarification would be helpful.

    Sorry, I have been on vacation this last week but if you can get back to me, I can see if I can be of some more help. Meanwhile, I will look over your part C and start thinking of how I can help.

    Thanks for checkin’ in!

    Mrs. Berger

  44. Sasha Says:

    Hello, I’m high school student taking enriched chemistry and I’m stuck on this chemistry practice question. And i have a test the next time I see my teacher, so I don’t have access to my teacher. So, i was wondering if you could help me with this problem.
    The question states the following:
    We have an isolated amino acid, so we decide to do a combustion analysis. We start with 0.0235 grams of our isolated amino acid, and 0.0327 grams of CO2 and 0.0156 grams of water vapor are produced, but we notice some soot remaining in the sample chamber. What might have happened? How many grams of oxygen gas were consumed in the combustion? what mass of the sample remains un-combusted?

    The answer of this problem isn’t important to me, but i want to know how to do the problem so that if I encounter a problem like this on the test, I’ll know what to do.

    If you could help me in any way, I’ll totally appreciate it!
    Thank You!!
    Sasha…

    • Mrs. Berger Says:

      First, from the looks of the question, it is testing you on a few areas of knowledge. First, you must know that an amino acid is a hydrocarbon that contains both a carboxylic acid and an amine group. Therefore, the amino acid has a formula containing CHON in some ratio. It also sounds like it is testing you on your knowledge of combustion reactions and the law of conservation of mass. However, I may have been out of my chemistry thinking routine for too long, but I believe there is some information that is needed that is missing. Since this is a combustion of an amino acid, there should not only be carbon dioxide and water produced but also some nitrogen gas? Otherwise, where did the nitrogen go? It cannot be destroyed in a chemical reaction. It also seems to me that we would need to know how much soot is there. There was probably some incomplete combustion and this is a byproduct of such a thing. If we knew the amount of soot, water vapor, carbon dioxide and nitrogen gas produced, we could decide the mass of oxygen that was consumed in the reaction by adding up the masses of the products and calculating the total mass of the reactants. Then I thought, well I guess you could figure out what percent of oxygen is in both the products given, figure the mass of oxygen produced and that would be the amount of oxygen reacted, but we don’t know the percent of the mass of amino acid that is oxygen, so it doesn’t seem to me that there is enough info to try this either. Plus, it doesn’t sound like all the amino acid reacted anyway, so this would limit us in this case as well. Do you have any more info given to you?

      I must say, however, thank you for asking a question ABOUT the question and not just saying, “here’s my homework question. please do it for me.” I appreciate that. I don’t know if I helped, and please feel free to share if you come up with anything else. I just don’t think there is enough info here to solve it but I could be wrong.

      • Sasha Says:

        Thank You Mrs. Berger! Atleast if a similar question appears on my test I’ll know how to deal with it.

        Thanks again

        Sasha

  45. Mrs. Berger Says:

    I don’t know that you are. I don’t currently have any students.

  46. Sasha Says:

    Hi Mrs. Berger its me again!
    I have another question;

    Element A forms A2+ ions, and element B forms B- ions. 0.2198 grams of A and 0.01809 moles of B form a stable ionic compound that is 74.5 % B, by mass. What are the identities of A and B?

    Hint: The formation of the ionic compound could be thought of as:

    A2+ + 2 B- → AB2

    In this question what i don’t understand is the identities part. I don’t know what identities of A and B are they asking for? Do you see what identities they are asking for?

    Thank You.
    Sasha

    • Mrs. Berger Says:

      Well, I am guessing they are asking for what the elements are. They want to know is A Magnesium or Calcium or whatever. Now the reason I suggest Magnesium and Calcium as possibilities is that they are Alkaline Earth Metals in group II. All these elements form+2 cations. Now, it could also be a transition metal like iron or copper, but I would start with the easy ones first. Plus, we know that about 1/4 of the mass of the compound is made up of the cation and the other 3/4 is made up of two of the anions. Therefore, we are probably looking at something a little lighter and transition metals are heavier because they are further to the right. Just a guess. Now, for the anion, you will want to find the elements that tend to form -1 anions. There is one group where this is true. Like I said before, 3/4 of the mass is two of this element which means each one of these B elements weights 1.5 times more than the A element. I will go ahead and see if I figure this out, as I haven’t thought all the way through this but if this helps you, see what you come up with and then reply and let me know what you come up with and I will see if it matches my answer. Good luck!

  47. Maria Says:

    im having a really hard time on this chemistry problem…
    The pseudorlic acid contains carbon, hyrdogen and oxygen. A chemist wanting to determine the molecular formula of pseudorlic acid A, burned 1 g of the compound in an elemental analyzer. The products of the combustion were 2.492 g of CO2 and .6495 g of H2O. What is the empirical formula of the compound?

    i keep getting C4H5O but its wrong. i dont know what im doing wrong?

    • Mrs. Berger Says:

      I am working on your problem. I have to get going, but will get back to this later today. I have had four posts in the last couple days and I haven’t been at my computer to respond. Check back soon!

      *******************THIS JUST IN***********************
      Ok, so now that I have been out for my daily exercise and then had some caffeine and a piece of chocolate (yes, I exercise so I can eat chocolate!) I have a clear head and think I can help you now. Your question is a good one, Maria! So let’s talk about elemental analysis first.

      To run an elemental analyzer, a compound is burnt in the presence of oxygen or nitrogen gas. In this case, since your products are only carbon dioxide and water and there are no nitrogen compounds, your pseudolaric acid must’ve been burnt in oxygen gas.

      Secondly, we must consider the law of conservation of mass. Think of the combustion (burning) reaction:

      CxHyOz + O2 –> CO2 + H2O

      We cannot create any new carbon, oxygen, or hydrogen that we don’t have in the beginning, or on the reactants side. We know that we used 1 g of the acid and it was burnt in excess oxygen. Therefore, the oxygen will have to be what we calculate after we figure out the carbon and hydrogen. Looking at the products, we see there is only one product with carbon, the CO2. Knowing that CO2 is 44.01 g/mol and that 12.01 g of that is carbon, we can calculate the percent of carbon in CO2. Using that percentage we can multiply it by the mass of CO2 produced to see what mass of carbon we have in the products and therefore the mass of carbon we started with in the reactant acid. Do the same for the water to find the hydrogen.

      Now, you will notice that the total mass of the products is more than the mass of the acid so that extra mass comes in the form of the oxygen used to combust the acid in the analyzer. Therefore, once you know the mass of carbon and the mass of hydrogen, you can subtract them from the 1 g of reactant acid to get the mass of oxygen in the acid.

      We are almost there. Just hang on a few more steps…

      So, recall the formula of a compound is not the mass ratio of the elements but the MOLE RATIO of the elements in a compound. So, figure out how many moles of each element you now have in your acid. To find the mole ratio, divide each of the number of moles by the smallest number of moles. That will give one element with a ratio of 1 (because the smallest moles divided by itself) and then ratios for the other two. If you do not get whole numbers, you will need to find a common number to multiply all ratios by to get a whole number. The answer I have figured out matches your answer for carbon but my numbers for hydrogen and oxygen are different. Try it again following these steps and let me know what you come up with or if you get stuck somewhere along the way.

      Good luck!

  48. Nisreen Says:

    I have a question regarding molecular formulas. To find the molecular formula i know you need the molar mass of the compound and the molar mass of the empirical formula where you divide the former by the latter and get a number, n. If say my number is 1.27 would i round it to 1 and multiply my subscripts from the empirical formula by 1, or would i multiply the subscripts my 1.27 and then round each subscript individually? or would i multiply each subscript then try and make the entire formula contain whole integer subscripts by multiplying each subscript by a common multiple? Thank you, much appreciated. I hope i was clear.

    • Mrs. Berger Says:

      Typically, the compound molar mass divided by the empirical formula molar mass should give a whole number since you can’t have a fraction of an element. Subscripts in formulas, as far as I know, aren’t decimal numbers but whole numbers as well. 1.27 is too far away from 1 to round it to that. I am not sure where 1.27 came from. Is this something you calculated or was this number given to you? Typically you would take the “n” value and multiply it by the subscripts in your empirical formula to get the molecular formula. However, I am not sure that is the case here. Sorry I can’t be of more help.

  49. Megan Says:

    Thanks SO much for putting up this site! The textbook I have for my advanced Chem class is very vague and doesn’t help me to grasp concepts quickly. After about a minute of looking through this site, I learned more than I would have in an hour of trying to decipher the examples in my book.

    Thanks!!
    Megan

    • Mrs. Berger Says:

      Thanks, Megan! You made my day. Since I am not teaching anymore, I try to keep tabs on this site in case there is a student out there that just needs that little extra help and if I can provide that, I feel like I am still connected. Thanks for your positive comments and please let me know if there are any other topics that you would like to see posts on in the future.

  50. Ariana Vazquez Says:

    Hi I was wondering if when finding empirical formulas for a compound like for example for Fe(SCN)3 that contains 19.0% water. Do you still find it by getting the % by mass of each element, then convert it to moles of each element, and last get the ratio? or does the % of water affect the way you find the empirical formula? or is that irrelevant to the question? the answer I got was FeSCN 1:1:1:1 ratio. Is that wrong? sorry I have a lot of questions. Thank you for your help!!!!

    • Mrs. Berger Says:

      Really quick help (and then if you still have more questions let me know):

      So it looks like the idea behind this problem is to find how many water molecules are attached to the iron (III) thiocyanate in a hydrate. A hydrate is an ionic compound with water molecules attached to its crystal structure. Their formulas are xxx*nH2O where xxx is the ionic compound, * is a dot, and n is the number of water molecules. Fe(SCN)3 cannot be reduced any further because the ratio is already 1:3:3:3. Because the iron is 1, this is in the lowest ratio possible. What you need to do is find out how many water molecules would make up 19% of the total compound. I will let you think about that, and then can give more guidance if you need. Just think, if water is 19% then the other 81% is the Fe(SCN)3. Solve for the unknown water mass and then figure out how many that would have to be so you would end up with a formula of Fe(SCN)3*nH2O.

  51. […] and chemistry students all over the world!  I have had a lot of activity lately on my post on Empirical and Molecular Formulas.  However, I know you won’t all be learning about this all year so please post a comment to […]

  52. Ariana Vazquez Says:

    Yes that really helped. I was assuming that Fe(SCN)3 = 100% but I didnt take into account the 19% of water. So I get that 19% of the the compound is water, and the remaining % is of Fe(SCN)3=80%. Thank you so much for your help!!! By the way the empirical formula I got is Fe(SCN)3*3H2O.
    You explain everything very well.
    I am truely thankful and this is a great website!!!
    -Ariana

  53. Chenille Says:

    Mrs. Berger, I was amazed on how you explain each question posted by different people, and I am a big fan. I admire you because you never seem tired of answering their questions. Right now, I’m working on my assignments in chemistry and i’m having a hard time answering some of them. Please if you could also help me I would highly appreciate it. Here are the last questions I’m having a hard time answering, if you could please help me out, and explain what i should do, I would be very grateful.
    j. Vinegar is 5.0% acetic acid, C2H4O2, by mass. (1) How many grams of acetic acid are contained in 24.0 g of vinegar? (2) The molecular weight of hemoglobin is about 65,000 g/mol. Hemoglobin contains 0.35% Fe by mass. How many iron atoms are in a hemoglobin molecule?
    k. Silver nitrate solution reacts with calcium chloride solution to form a solid (precipitate) at the bottom of the flask. Suppose we mix together a solution containing 12.6 g of silver nitrate and one containing 8.40 g of calcium chloride. (1) Write the balanced complete chemical and net ionit equation. (2) Identify the precipitate formed. (3) What mass of the precipitate is formed?
    l. Solid silver nitrate undergoes thermal decomposition to form silver metal, nitrogen dioxide, and oxygen. Write the chemical equation for this reaction. A 0.443-g sample of silver metal is obtained from the decomposition of a 0.784-g sample of silver nitrate. What is the percent yield of the reaction?
    m. What is the molarity of a solution that contains 555 g of phosphoric acid, in 3.00 L of solution?
    n. The density of an 18.0% solution of ammonium chloride solution is 1.05 g/mL. What mass of the salt does 275 mL of this solution contain?

  54. Chenille Says:

    Please help me, I don’t know what to do… it’s due tomorrow…

    • Chenille Says:

      for letter j number 1 I got 1.2 grams of acetic acid, and I’m not sure, so not sure of my answer

      • Mrs. Berger Says:

        Chenille,

        Your answer for j.1 is correct. The mass of the substance times the percentage by mass of a component gives you the mass of the component. To figure out j.2, you will do this same calculation. However, you need to first find the mass of one molecule of hemoglobin before you find .35% of that for the mass of iron in that one molecule. You know there are 6.02×10^23 molecules in one mole and you know the molar mass of hemoglobin in g/mol so you can calculate the mass of one mole. Let me know what you get and when I get home later I will check and let you know if you are right.

        SIDE NOTE: This posting that you have commented on is not really related to the questions you are asking. For the future, please start at the home page (chemberger.wordpress.com) and comment on the latest post as it asks for questions on other topics. Thanks!

        Continuing with your problem…
        Problem k is encompassing a couple different topics. First, writing chemical reactions. There are five basic types of chemical reactions – synthesis, decomposition, single replacement, double replacement, and combustion. We can rule out combustion in this case because nowhere in your problem does it say that anything is burned in oxygen or nitrogen. Synthesis is usually the combining of two singluar elements or small compounds into one lareger compound and decomposition is the opposite of this. Single replacement is one element replacing a similar element in another compound and double replacement is two, usually ionic, compounds that have the cations and anions changing places.

        In your case since you are reacting silver nitrate and calcium chloride, both ionic compounds, you have a double replacement reaction. The trick to writing equations for this are as follows:
        1. Write the correct formula for each of the reactants, paying attention to the charge on each ion. For example, silver nitrate would be written as AgNO3 since silver is a +1 cation and NO3 is a -1 anion and the charges on the ions must cancel. However, calcium chloride must be written as CaCl2 since Calcium is +2 and chloride is -1.
        2. The cations will change places in the products so now silver will be bonded with chloride and calcium with nitrate. These formulas must be written correctly based on their charges just as they were in step 1.
        3. Determine which compounds are soluble (aqueous) and which substance(s) are insoluble (solid). Any solid in the products from a reaction between two aqueous substances is a precipitate. This should answer k.2
        4. Balance your equation so that you have equal numbers of each element on both sides. Use coefficients in front of the compounds and the coefficient applies to that entire compound. For example if you have 2AgCl, then this means you have two silver atoms and two chlorine atoms. In other words, if you have 2 chlorine atoms and 5 oxygen atoms on the left side of the equation, you must also have two chlorine atoms and 5 oxygen atoms on the right side. This is the law of conservation of matter 🙂
        5. Finally, if you need to get to the net ionic equation, you need to write all aqueous ionic substances as separate ions, cancel out ions that are alike on the left and the right, make sure the coefficients are in lowest terms, and you should have the net ionic equation for the formation of the precipitate. For example:
        Pb(NO3)2 + 2NaCl –> PbCl2 + 2NaNO3
        Pb+2 + 2NO3 -1 + 2Na+ + 2Cl- –> PbCl2 (not written as ions because this substance is insoluble and is therefore the precipitate) + 2 Na+1 + 2 NO3 -1

        On each side there are two nitrate ions and two sodium ions so these cancel and you are left with
        Pb +2 + 2 Cl- –> PbCl2

        The charges are balanced and number of each element are balanced.

        Now, part k.3 is a limitiing reactant problem. I am going to assume you have some examples of this so now that you know what you are looking for, see if you can figure out where to go with that. Remember, a limiting reactant is a substance that is used up first in the reaction. For example, say we are making a cake and each cake requires one box of cake mix and 3 eggs. We have 30 boxes of cake mix and 12 eggs. can we make 30 cakes? No, because we only have enough eggs to make four cakes, therefore the eggs are the “limiting reactant.”

        For question L, calculate how much silver metal you would expect to produce. The percent yield is the percent of what you expected to get. Equation is as follows: percent yield = ( (what you actually got) / (what you expected to get) ) x 100

        For question M, molarity is #moles of solute / L of solution. You need to change grams of phosphoric acid to moles to do this calculation.
        For question N, you need to use the density to change from mL to grams. For example, 275 mL x 1.05 g/mL should give you the mass in grams.

        Hope this helps!

      • Chenille Says:

        Thank you very much for the steps. Here are my answers, I’m not sure abou them. For letter j #1 I got 1.2 grams of acetic acid, for letter j #2 I got 4 Fe atoms. For letter k#1, the balanced equation I got is 2AgNO3(aq) + CaCl2(aq) -> Ca(NO3)2(aq) + 2AgCl(s), for letter k #2 the precipitate is AgCl, for k #3 I got 2.66 grams of AgCl. For letter L, I got a percent yield of 88.96% for letter M my aswer is 1.89M and for letter N I got 51.98 grams of NH4CL

      • Mrs. Berger Says:

        All your answers sound good except I got something different for k.3. It could be my mistake but just double-check that the molar masses are correct. Also, if your teacher is picky about significant digits, you may want to check yoru answer for L and N. Good job!

  55. Chenille Says:

    Thank you very much Mrs. Berger. I hope you could help me again next time. :)) How about physics, can you answer questions regarding physics?

  56. Jaleesa Says:

    Mrs. Berger,

    I have a question regarding molecular formula. The question I am working says,
    “A hydrocarbon gas composed of 85.7% Carbon and 14.3% Hydrogen by mass is typically mixed with Oxygen for use in general anesthetic. If 1,560 mg of this gas at 122(degrees)F has a volume of 1*10^-3kL at 99.7038 kPa, what is the molecular formula of the hydrocarbon gas”?

    Can I go on and find the molecular formula from the given percentages of mass (85.7%C and 14.3%H) and disregard the other information? If not, how do I begin this problem?
    Thank you in advance.

    • Mrs. Berger Says:

      Hey Jaleesa! Hope I am not getting back to you too late but I was gone for Thanksgiving and just got back last night. Man, this problem is a great one! So, to answer your question about ignoring all that extra info, you can find the empirical formula with just the percentages. However, since you are asked to find the molecular formula, you do need all that other info. Sorry 😦 The good news is, I can help you!

      First, use those percentages to find the mole ratio of carbons to hydrogens. You will have an empirical formula from this. Go ahead and calculate the mass of this formula in grams per mole and save that answer somewhere.

      Then, when I look at the rest of the info you are given, it seems like you are given pressure, temperature, and volume. What pops in my head whenever I see these three values is the ideal gas equation, PV=nRT. If given P, V, and T, and we know R is a constant, we can solve for n, which is the number of moles. You are given the mass in mg for this sample of gas, so if you convert this to grams you could use the mass in grams divided by the number of moles to find the molar mass of the compound in grams per mole. Now if this matches the molar mass of your empirical formula, then we know it is also the molecular formula. However, chances are that the grams/mole calculation will be some multiple of your empirical formula. So, divide the grams per mole calculation by the empirical formula calculation in grams per mole and that should give you a number. So if you were to get 5, that would mean the molecular formula is five times the empirical. Does that make sense? Let me know what you get and I will let you know if you are right 🙂

  57. Dani Says:

    Hi,
    I was wondering if you could explain to me how to find the empirical formula for a hydrate?

    The problem that I’m working on is “A 1.628-g sample of a hydrate of magnesium iodide is heated until its mass is reduced to 1.072 g and all the water has been removed. What is the formula of the hydrate?”

    Thanks 🙂

    • Mrs. Berger Says:

      Hopefully this response will answer both Dani and Anna’s questions. First, a hydrate is an ionic crystal that that contains a certain ratio of water molecules within its crystal structure. The neat thing is that these water molecules aren’t just randomly thrown in but that they always combine in a certain ratio to the crystal. For example, cobalt (II) chloride can exist as an anhydrous (without water) crystal or as a hydrate with six water molecule contained within its crystal structure, cobalt (II) chloride hexahydrate. These two compounds, because of the addition of water molecules, appear very different. The anyhydrous version is a blue color but the hydrated version is red. Pretty cool!

      So, what does this have to do with molecular formulas? Well, normally when finding molecular formulas, we look at each element individually. We would find, for example, in cobalt (II) chloride the percent cobalt and the percent chloride. However, when we look at a hydrate, the water can be treated as one unit. Don’t split it up. Water is 18.02 g/mol and you just need to find the ratio of water molecules attached. It will always be a whole number you are looking for, just as the mole ratio for the other elements present should also be a mole ratio.

      Dani, for your problem, you will want to first find out how much water was lost, which you can get by subtracting the two masses. When you heat a hydrate, the water vaporizes and you are left with just the crystal. The difference in masses is the mass of water lost. The mass that is left after the hydrate has lost its water is just the mass of the ionic crystal. Magnesium Iodide you should be able to write the formula for since magnesium is in group 2 (charge of +2) and Iodide is in group 7 (-1 charge). Then find the molar mass of magnesium iodide by adding this all together. You now have the molar mass of water (18.02 g/mol) and the molar mass of your ionic crystal (which I am going to let you find).

      After you have found the molar masses, use these values to convert the actual mass of what you have to moles. For example, in this problem you have 0.556 grams of water. If water is 18.02 grams per mole, this comes out to 0.0308546 moles of water that were lost. Do the same thing with the anhydrous crystal – you have 1.072 grams left after the water is gone, use the molar mass of magnesium iodide to find the number of moles.

      Finally, once you have the number of moles of both the anhydrous crystal and the water, divide them both by the smallest number, which should be the anhydrous crystal. This should give you a ratio of anhydrous to water of 1:?. Let me know what you get and I can let you know if you are correct. 🙂 Also let me know if you have any more questions.

      Have a great day and thanks for looking me up!

  58. Anna T Says:

    I came across this site in my confusion, and it really helped me. You’ve put this into such a simple, understandable way that makes me feel a lot better about the material.

    I have a question though. Can you explain hydrated crystals in relation to these formulas? My teacher flew through that section and I can’t even get any more specific than that. Thanks for your help!

    • Mrs. Berger Says:

      Thanks, Anna. I totally geeked out when you said I come up before Wikipedia 🙂 Glad nobody was here to witness my dance of joy!

      Anyways, I answered a question from a girl named Dani that posted right before or after you did. Please read my response to her and see if that gets you headed in the right direction. If you still have questions, let me know but hopefully that helped.

  59. Anna T Says:

    also I hope you know- your site comes up first in a google search of “how to write an empirical formula from a molecular formula” even before wikipedia 🙂

  60. Paige Says:

    Hey, for a part of my Yr 12 chem assignment we are given the molecular mass and the colour of the flame of two alkanes, alkenes, or alcohols. We have to say what the compound is. My clues are.

    Unknown A- burns with a clean blue flame 86.2gmol-1 (i think this could be hexane?)

    Unknown B- burns with a clean blue flame 130.26gmol-1

    • Mrs. Berger Says:

      So what is your question? What have you done so far? I do not do your homework for you. I answer questions ABOUT homework. Give me something to work with and I will help you. Thanks!

  61. Chloe Says:

    I understand how to find a molecular formula, but one of our sample problems has a question with decomposition at all! I am totally stuck. its “7.36g of a compound has decomposed to give 6.93 g of ofygen. the only other element in the compound is hydrogen. The molar mass is 34.0g, what is the empirical formula?”
    please help! thank you!

    • Mrs. Berger Says:

      If the compound is only oxygen and hydrogen, and the mass of oxygen produced upon decomposition (where there would be no other reactants. it would just be like AX –> A + X) is 6.93 g, the the remaining mass (7.36 g – 6.93 g) would be your mass of hydrogen produced. According to the Law of Conservation of Mass, the mass of the reactants has to equal the mass of the products, so you could find the mole ratio of hydrogen to oxygen atoms. The lowest whole number ratio is the empirical formula.

  62. Ryan Says:

    Thank you so much for this site. It is the only site I found using Google that helped me prep for my college Inorganic Chemistry quiz.

    -Ryan
    Texas, USA

  63. tatyana Says:

    here is my question:
    Estradiol is a female sexual hormone that causes maturation and maintenance of the female reproductive system. Elemental analysis of estradiol gave the following mass percent composition: C, 79.37; H, 8.88; , O 11.75. The molar mass of estradiol is 272.37 .

    and the question is what is the molecular formula of estradiol?

    • Mrs. Berger Says:

      What is your question? This is the question you were given so I need to know what you have tried, what you think you should do, maybe where you are stuck, and then I can help you.

  64. Jesse Says:

    15.2g of an oxide of metal Q was formed when 10.64g of Q reacted with excess oxygen gas.Determine the molecular formula of the oxide of Q.
    (Q=56,O=16).help me with a clear explanation please.i did not get your explanation to Tarun ,Mrs.Berger.i will deeply appreciate your help.

  65. Scott Says:

    Hello, i am not in your class. In fact i don’t even know what state you live in but i live in New York. i have a question similar to ” The empirical formula of a compound is NO2. its molecular mass is 92g/mol. What is its molecular formula.” would i add N- 14.0067 and O2- 15,9994(2)= 46.0055 then divide that by 92g/mol so it looks like 92/46.0055. Then get 1.99 so round to 2. then multiple the 2 into NO2 and get N2O4 as the answer?

    thx for any help

    • Mrs. Berger Says:

      You don’t have to be in my class to ask a question, so thanks for checking in. I hope I can be of some help even if I didn’t get to your question right away. Sorry about that. I must also say thank you for asking a good question. You didn’t just give me your homework question but described how you would go about solving it so I could help YOU. Great! So I accept your challenge…

      Oh wait, I don’t have to do anything because you are just right! Good work 🙂

  66. Krissy D Says:

    hi i’m stuck on a question & i was wondering if you could help….
    Question: A chemist is given 1.08g of a compound labelled X, and is asked to determine the molecular and emperical formula. Analysis of X by mass spectrometry gives a relative molecular mass of 108. Elemental analysis shows that compound X contains carbon, hydrogen and one other element. Among the products obtained when X is burnt completely in oxygen are 1340cm3 of CO2 and 448cm3 of NO2. [Volumes of gases are measured at standard temperature and pressure]
    Determine the molecular formula of X !

    • Mrs. Berger Says:

      What is your question? This is the question you were given so I need to know what you have tried, what you think you should do, maybe where you are stuck, and then I can help you.

      • Krissy D Says:

        ok well what i did was to find the number of moles of X present and i found that to be 0.01 because
        No. of moles present = mass given/ mass of 1 mole
        = 1.08/108
        = 0.01
        then i tried writin an equation for the reaction && that is where i got stuck
        CxHyNz + O2 —-> CO2 + NO2 + some H2O
        is that right so far ? :s

      • Mrs. Berger Says:

        ok, i’m workin’ on it. be back soon 🙂

      • Mrs. Berger Says:

        Ok, you are right so far. This is a hard one because you are not given water but by process of elimination we can figure it out. This problem is a great one because it involves the ideal gas equation (PV=nRT), law of conservation of mass, stoichiometry, and empirical/molecular formulas. I can see why this was a bit confusing. I am going to give you some hints and then see where you come out. I know what your answer should be so if you think you got it, just let me know and I can tell you if you are right or not. Here goes…

        Your equation is set up correctly.
        Using the ideal gas equation and remembering that cm^3 is a mL, figure out the number of moles of carbon dioxide and nitrogen dioxide produced. In each gas, the Carbon or Nitrogen is present in a 1:2 ratio, so for every 1 mole of gas, there is one mole of Carbon or Nitrogen. For every 5 moles of gas there are 5 moles of Nitrogen or Carbon. For every 7.3 moles of gas… You get the pictures? So, knowing how many moles you have of each gas will give you the number of moles of each element present. Don’t worry about the oxygen because it was in excess in this case.
        Once you have the moles of carbon dioxide converted to moles of Carbon, and moles of nitrogen dioxide converted to moles of Nitrogen, using your stoichiometry skills, convert each of these moles (although write down the number of moles, you will need it later) to mass of Carbon or mass of Nitrogen.
        You know the mass of the entire compound and since mass is neither created nor destroyed, you can subtract the mass of carbon and nitrogen and you will be left with the mass of hydrogen. From here, I will leave you for now because it is a normal empirical formula problem from here out. Mass of each element … moles of each element… divide by the smallest number of moles… etc. Let me know what you come up with. 🙂

  67. Krissy D Says:

    ok well i used the ideal gas equation and found that CO2 has 0.06 mols while NO2 has 0.02 mols. but then i get confused. does this mean that CO2 contains 0.06 mols of carbon because of the ratio ? :s and if so do i jus convert to grams normally using the equation
    no. of moles present = mass given / mass of 1 mole

    • Mrs. Berger Says:

      Yes, for .06 moles of CO2 you have .06 moles of carbon and .12 moles of oxygen (1:2 ratio). Side note, don’t use such a rounded number when doing these problems. I actually got .05981507. Since we are dealing with such small numbers, sometimes more decimal places makes a difference. Anyways, take your moles of carbon and multiply that by the molar mass of carbon (12.01 g/mol) and that gives you the mass of carbon present. Same for nitrogen.

  68. Krissy D Says:

    kay i think i got it all figured out !
    Mass of carbon present = 0.71784
    mass of nitrogen present = 0.27910
    mass of hydrogen present = 0.08306
    therefore emperical formula = C3H4N

    and then molecular mass = 108
    C3H4N = (12×3)+4+14 =54
    108/54 =2
    therefore molecular mass is C3H4N multiplied by 2 = C6H8N2
    i hope thats correct 🙂

  69. huNy Says:

    i hv a question n i must say i m super stuckd in it…
    here it is..??

    how to find molecular mass of NA2S2O3 n NaPO3..??
    wud luv if u helpd me out..
    thAnk u..=)

    • Mrs. Berger Says:

      The molecular mass is the total mass of all the elements in the compound. Using your periodic table, you can see the mass of each individual element. This is called the atomic mass. You must also determine how many of each element is present. In the compound NaCl, there is one sodium atom and one chlorine atom per unit, so we would add the atomic mass of sodium and the atomic mass of chlorine to get the molecular mass of NaCl. In a compound like Mg(NO3)3, we could determine there is one magnesium atom, but then what do we do with all those other elements? The numbers down below the elements, called the subscripts, tell us how many of each element we have. Oxygen has a 3 next to it so we know there are three oxygens. However, Nitrogen and Oxygen are both in parentheses with a 3 down below that. Just like in math class when you have a number next to parentheses, that number gets distributed to the elements inside the parentheses. Therefore, instead of one nitrogen, there are three. Instead of three oxygens there are 3X3, or 9 oxygens. This is a bit harder of an example than what you have, but I wanted to give a more difficult example so hopefully yours seemed easier. Anyways, so if we add the mass of one magnesium, three nitrogens, and nine oxygens, we would have the molecular mass of magnesium nitrate. In your molecule Na2S2O3, figure out how many sodiums, sulfurs, and oxygens. Then add the mass of each element present for the number of times it is present.

  70. huNy Says:

    thAnxx alot….
    u saved my day….=)

  71. Alexander Says:

    This website, and its many comments, are fantastic!

    Thank you Mrs. Berger for your very thorough and easy to understand post and explanation from a Las Vegas High School student studying for his final exam. 😀

  72. Patrick Says:

    Mr. Berger,

    I have a question that im having some problems with.
    Here is the chemical equation:
    H2O+CO2—–> O2+C6H12O6+H2O
    Question one is What are the reactants in the equation?(give their moleculat formulas)
    Question Two is What are the products in the equation?(give their moleculat formulas)

    I think I know what they are, but im not 100%.

    Thanks,
    Patrick

    • Mrs. Berger Says:

      Reactants are what you start with and products are what are “produced” as a finished product. What do you think they are and I will tell you if you are right 🙂

  73. Emillie Says:

    Thanks for this post, it really helped me on my homework.

  74. inke-benzie Says:

    hi, i am tryng to make friends with moles but i got stucks when it comes to molality and molarity, how do this two are differ?

  75. Louisa Says:

    You website was of great assistance to me. Thank you so much!

  76. Crysh Says:

    How do I calculate the true molecular weight if it is not given in the problem?
    Is there a difference between molecular weight and true molecular weight?
    Thank you!

    • Mrs. Berger Says:

      As for molecular weight vs true molecular weight, I don’t know that there is a difference. I have never heard of it if there is. Truthfully, I did some googling and it sounds the same to me. As for calculating it, it depends on what other info you’re given in the problem. If you’re given the molecular formula, you can just add the masses of each of the elements (like H2O would be 2 hydrogen and one oxygen about 18.02 g/mol). However, I’m not sure what you’re asking because I need a little more context.

      • iamcrysh Says:

        actually, I was computing for the molecular formula, it was an assignment of mine and my teacher haven’t thought that exact lesson yet but she says we just have to follow the book and so I did, but when i was about to compute for the molecular ratio, i realized that we weren’t given the true molecular mass in the problem because there was only 2 elements and both was given as %(percent) of element.

      • iamcrysh Says:

        I read your previous answers from others people’s question, just to clear things…I think actual molar mass is the same as true molecular weight and I know how to compute for the molecular weight but not the true molecular weight. I’m just confused.

      • iamcrysh Says:

        the actual problem was this. deternmine the molecular formula of octane in gasoline containing 32.7% C and 67.3% H…..when I calculated for its empirical formula what I got was CH25. this is where I’m stuck, I’m about to get its molecular ration but I don’t have the true molecular weight….

        Side Note: sorry for fragment replies, I was just so confused because I can’t proceed to the next number. I just have this thing about not doing the next thing if I don’t resolve the first thing. Sorry if I bother too much.

      • Mrs. Berger Says:

        Two things: I did the problem as you have stated here. You CANNOT round the hydrogen to 25. The decimal for the ratio is too far away to round. If you read my previous explanations you will see that if it were 24.99779 or something, it is fine to round. However, it is more like 24.473377. This is like 24.5 and would need to be multiplied by 2. Then you would have C2H49. I am not sure I like this answer. Doesn’t feel right. Not really possible.

        However, the second thing, the problem states it is for OCTANE. Octane is always C8H18. So I am confused. Please be sure you have read the problem correctly and have your percentages right and such. Plus, in what you have given me, it asks for the molecular formula. You must have more info if the empirical and molecular formulas are not the same. So either we have missed something in reading it, I am not getting all the info for the question, or this is a really poorly written problem. In any case, as it is written. I am afraid I am not much help. Please let me know what you find out.

      • iamcrysh Says:

        first of all, Thank you so much!

        well, that question came from my book and that was all of it. anyways, my teacher already taught us how to solve it and as it turns out, the problem that was given in out book was incomplete and she also told us that we have to solve for the empirical formula before the molecular formula. But, nonetheless, thank you so much for having the time to answer my questions.

      • Mrs. Berger Says:

        Well good to know i hadn’t lost my touch 🙂

  77. Anthony Says:

    This still an updated/working site?

    • Mrs. Berger Says:

      I still answer questions when someone needs help, asks good questions, and isn’t just looking for someone to do their homework for them. I am also open to topic suggestions if there is a subject that I haven’t posted about and you need help with that.

  78. Suman Koorathota Says:

    Is there any possibilities of getting extra molecular weight as 77.Actually the molecular weight of my reaction product is 382.but i am getting molecular ion peak as 459.As of my knowledge we usually get [M+H] OR M+1 OR M+Na or M+Cl peaks, and as my knowledge the extra mass 77 matches with phenyl cation(m/z 77), but in my reaction i did not use any Pheny ring containing molecules. please help as soon as possible.

    • Mrs. Berger Says:

      I wanted you to know I received your question. However, I am afraid your question is not the same as this thread and I am afraid I cannot be of help to you at this time.

  79. Suman Koorathota Says:

    Ya, thank you for your reply. Actually i am doing my Ph.D in Synthetic Organic Chemistry, i have faced this problem while characterizing my molecule. I thought i may get help here…..anyway thank you for your reply..


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