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.
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
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?
I had the same question as Tarun had and your response was very helpful.
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
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.
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!!
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!
what is a mol
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.
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?
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!
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?
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
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.
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!
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!
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!
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
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
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.
Thank you so much for your help. The information you provided helped me tons. I appreciate it greatly.
You are most welcome
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.
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?
yes thankx a lot! that makes things a lot clearer!
Hi Mrs. Berger, the knowledge you have shared with us (me and my homework buddy) took care of our chemistry needs. Thanks a lot!!!
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.
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.
Thanks, Michelle. I am glad I could help. Come back and see me again if you have any more questions
What is the empirical formula for acetone, C2H4O2?
What is the empirical formula for acetone, C2H4O2? thanks!
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.
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
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?
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!
Yes, you are correct. Glad I could help!
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!
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 =]
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.
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. =)
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…