Understanding Empirical Formula, Molecular Formula, and Molar Mass.?
Q. I am looking for a site that would explain in an understandable detailed method how to work through the homework I was given. I do not want to just post my questions begging for answers so please do not post answers to the questions. I would like to understand the process of solving them so I can pass the test as well. For Reference at what I need help with here are a few problems I was given... --- Determine the empirical formula for compounds that have the following percent composition- 28.4 Cu,% 71.6% Br --- Determine the empirical formula for compounds that have the following mass composition. 0.537 g Cu, 0.321 g F --- Determine the molecular formula for compounds with the following empirical formulas and actual molecular… [cont.]
Asked by ioncann0ns - Tue Jan 20 17:17:55 2009 - - 1 Answers - 0 Comments
A. 1. The percentages are equal to the amount of grams because if you add 28.4% + 71.6%= 100%. This allows you use 28.4gCu for your given and 71.6gBr for your other given. Now you have to use your givens and convert them to moles. 28.4gCu x 1 mol Cu/63.6 g Cu = 0.447 (I got the ratio 1 mol Cu/63.6 g Cu because if you look at the periodic table the mass numbers listed are equal to 1 mol- this applies to each element) Same thing for this one: 71.6gBr x 1 mol Br/79.9 = 0.896 Now you take the smallest of those two numbers- 0.447 - and divide each answer you got by that number: 0.447/0.447= 1 ... 0.896/0.447= about 2 (round it. *if you get a number like 1.5, you have to multiply each number you got by 2) These numbers tell you the number of atoms… [cont.]
Answered by anonymous - Tue Jan 20 17:52:18 2009
Q. I am looking for a site that would explain in an understandable detailed method how to work through the homework I was given. I do not want to just post my questions begging for answers so please do not post answers to the questions. I would like to understand the process of solving them so I can pass the test as well. For Reference at what I need help with here are a few problems I was given... --- Determine the empirical formula for compounds that have the following percent composition- 28.4 Cu,% 71.6% Br --- Determine the empirical formula for compounds that have the following mass composition. 0.537 g Cu, 0.321 g F --- Determine the molecular formula for compounds with the following empirical formulas and actual molecular… [cont.]
Asked by ioncann0ns - Tue Jan 20 17:17:55 2009 - - 1 Answers - 0 Comments
A. 1. The percentages are equal to the amount of grams because if you add 28.4% + 71.6%= 100%. This allows you use 28.4gCu for your given and 71.6gBr for your other given. Now you have to use your givens and convert them to moles. 28.4gCu x 1 mol Cu/63.6 g Cu = 0.447 (I got the ratio 1 mol Cu/63.6 g Cu because if you look at the periodic table the mass numbers listed are equal to 1 mol- this applies to each element) Same thing for this one: 71.6gBr x 1 mol Br/79.9 = 0.896 Now you take the smallest of those two numbers- 0.447 - and divide each answer you got by that number: 0.447/0.447= 1 ... 0.896/0.447= about 2 (round it. *if you get a number like 1.5, you have to multiply each number you got by 2) These numbers tell you the number of atoms… [cont.]
Answered by anonymous - Tue Jan 20 17:52:18 2009
How do I get the actual molecular weight ?
Q. I am having a lot of trouble with this. I have an unknown sample and am given the weight in a 10g sample. C 5.453 g H 0.915 g O 3.632 h The liquid density is .901 g/mL It says to use the analysis to calculate the empirical formula of the unknown sample. divide the experimental molecular weight by the emprical formula weight to determine the molecular formula. Round the ratio down to the nearest whole number. calculate the actual molecular weight from this formula. I did the work for the emprical formula... mol C= (5.453 g)times( 1 mol C)/ 12 g C = .454 mol H= (.915 g)times (1 mol H)/ 1.00794 g H= .9077 mol O= (3.632g)times (1 mol O)/15.9994 g 0)= .22700 I divided all of the numbers by the smallest amount .22700. C=2 H=3.998 0= 1 i… [cont.]
Asked by ap - Sat Jan 24 12:38:30 2009 - - 1 Answers - 0 Comments
A. Now you need to work out the formula weight og the empirical formula. (44.05 g/mol) Divide your experimental molecular weight by the empirical formula weight. This tells you how many times the empirical formula fits into the molecula formula Multiplying the empirical formula by this number gives you the molecular formula ::: EDIT ::: I don't know what experiment you did, but since it is an experimental molecular mass then it must have been what you were trying to determine from the results of your experiment.
Answered by Lexi R - Sat Jan 24 19:58:39 2009
Q. I am having a lot of trouble with this. I have an unknown sample and am given the weight in a 10g sample. C 5.453 g H 0.915 g O 3.632 h The liquid density is .901 g/mL It says to use the analysis to calculate the empirical formula of the unknown sample. divide the experimental molecular weight by the emprical formula weight to determine the molecular formula. Round the ratio down to the nearest whole number. calculate the actual molecular weight from this formula. I did the work for the emprical formula... mol C= (5.453 g)times( 1 mol C)/ 12 g C = .454 mol H= (.915 g)times (1 mol H)/ 1.00794 g H= .9077 mol O= (3.632g)times (1 mol O)/15.9994 g 0)= .22700 I divided all of the numbers by the smallest amount .22700. C=2 H=3.998 0= 1 i… [cont.]
Asked by ap - Sat Jan 24 12:38:30 2009 - - 1 Answers - 0 Comments
A. Now you need to work out the formula weight og the empirical formula. (44.05 g/mol) Divide your experimental molecular weight by the empirical formula weight. This tells you how many times the empirical formula fits into the molecula formula Multiplying the empirical formula by this number gives you the molecular formula ::: EDIT ::: I don't know what experiment you did, but since it is an experimental molecular mass then it must have been what you were trying to determine from the results of your experiment.
Answered by Lexi R - Sat Jan 24 19:58:39 2009
What is the molecular formula of the compound?
Q. The simplest ratio of the elements in a compound containing C, H and Cl is 1:2:1 The actual molecule is twice as large as the empirical formula indicates
Asked by Genie2299 - Tue Feb 24 23:38:03 2009 - - 2 Answers - 0 Comments
A. C2H4Cl2
Answered by ricky - Tue Feb 24 23:59:33 2009
Q. The simplest ratio of the elements in a compound containing C, H and Cl is 1:2:1 The actual molecule is twice as large as the empirical formula indicates
Asked by Genie2299 - Tue Feb 24 23:38:03 2009 - - 2 Answers - 0 Comments
A. C2H4Cl2
Answered by ricky - Tue Feb 24 23:59:33 2009
What is the molecular formula of ..?
Q. A sample of NiCl*H2O is found experimentally to contain 0.0116 moles of H2O, 0.0055 moles of Ni, and 0.00121 moles of Cl. What is the molecular formula of NiCl*H2O? How can I calculate it? In the actual question, it says Ni(subscript x)Cl(subscript y) * zH2O, but i dont really think those are important. And does anyone know why they are there or why the z is infront of the water molecule and not a subscript?
Asked by Jamie - Sat Oct 13 20:09:20 2007 - - 4 Answers - 0 Comments
A. The good news is that everything is already given in moles and you don't need to convert from grams to moles first. Select the smallest number of moles given (0.00121 mol Cl) This has to be the lowest subscript. To find the other subscripts you will determine the lowest whole-number ratio of the moles. This sounds hard, but all you do is divide each number of moles given by the lowest number of moles (0.00121 mol Cl) 0.0116 mol / 0.00121 mol = 9.56 0.0055 mol / 0.00121 mol = 4.55 0.00121 mol / 0.00121 mol = 1 Since you cannot have a subscript with a decimal, multiply each answer by 2 to remove the fractional denominator (0.5 = 1/2) 9.56 x 2 = 19 4.55 x 2 = 9 1 x 2 = 2 Ni9Cl2*19 H2O
Answered by physandchemteach - Sat Oct 13 20:22:53 2007
Q. A sample of NiCl*H2O is found experimentally to contain 0.0116 moles of H2O, 0.0055 moles of Ni, and 0.00121 moles of Cl. What is the molecular formula of NiCl*H2O? How can I calculate it? In the actual question, it says Ni(subscript x)Cl(subscript y) * zH2O, but i dont really think those are important. And does anyone know why they are there or why the z is infront of the water molecule and not a subscript?
Asked by Jamie - Sat Oct 13 20:09:20 2007 - - 4 Answers - 0 Comments
A. The good news is that everything is already given in moles and you don't need to convert from grams to moles first. Select the smallest number of moles given (0.00121 mol Cl) This has to be the lowest subscript. To find the other subscripts you will determine the lowest whole-number ratio of the moles. This sounds hard, but all you do is divide each number of moles given by the lowest number of moles (0.00121 mol Cl) 0.0116 mol / 0.00121 mol = 9.56 0.0055 mol / 0.00121 mol = 4.55 0.00121 mol / 0.00121 mol = 1 Since you cannot have a subscript with a decimal, multiply each answer by 2 to remove the fractional denominator (0.5 = 1/2) 9.56 x 2 = 19 4.55 x 2 = 9 1 x 2 = 2 Ni9Cl2*19 H2O
Answered by physandchemteach - Sat Oct 13 20:22:53 2007
Molecule of water from a peptide bond?
Q. Hi! I'm struggling with the high-school biology question below, which involves peptide bonds between amino acids: "When two monomers, such as amino acids, join together, a molecule of water is produced. Use two amino acid molecules to explain where this water comes from." I realise that, given that the peptide bond between amino acids is between the carboxyl group (COOH) and amino group (H2N), so obviously the water comes from somewhere... but could someone please give me an *actual* example, hopefully with a molecular formula or something? Thanks in advance.
Asked by Andrew M - Thu Jan 24 23:36:37 2008 - - 2 Answers - 0 Comments
A. RCOOH + H2N-R' --> RCONHR' + H2O Now I will write the same equation with small oh and h to show you where the water comes from- RCOoh + hHNR' --> RCONHR' + h2o (RCOOH is the acid part of an amino acid and RNH2 is the amine part of an amino acid.)
Answered by skipper - Thu Jan 24 23:44:14 2008
Q. Hi! I'm struggling with the high-school biology question below, which involves peptide bonds between amino acids: "When two monomers, such as amino acids, join together, a molecule of water is produced. Use two amino acid molecules to explain where this water comes from." I realise that, given that the peptide bond between amino acids is between the carboxyl group (COOH) and amino group (H2N), so obviously the water comes from somewhere... but could someone please give me an *actual* example, hopefully with a molecular formula or something? Thanks in advance.
Asked by Andrew M - Thu Jan 24 23:36:37 2008 - - 2 Answers - 0 Comments
A. RCOOH + H2N-R' --> RCONHR' + H2O Now I will write the same equation with small oh and h to show you where the water comes from- RCOoh + hHNR' --> RCONHR' + h2o (RCOOH is the acid part of an amino acid and RNH2 is the amine part of an amino acid.)
Answered by skipper - Thu Jan 24 23:44:14 2008
Chemistry help!! Need help with chemistry homework!?
Q. 1. What is the molality of a solution of 90 grams of glucose (C6H12O) dissolved in 800 grams of water ? A. .01 B. .4 C. .625 D. 1.6 2. How many grams of chlorine (Cl2 would you have to add to 1 kg of water to increase the boiling point by 1.56oC? A. 200 B. 213 C. 230 3. One quality of metals is that they are ___, which means that they can be hammered into thin sheets. A. ductile B. malleable C. bendable 4. What is the molecular mass of carbon dioxide,CO2?___ Round your answer to the nearest tenth. 5. What do you call an ion that carries a positive charge? ___. 6. A process that needs to take in energy to get started is called a/an ___ process. A. endothermic B. exothermic 7.… [cont.]
Asked by Jessica` M - Tue Oct 7 16:11:18 2008 - - 1 Answers - 0 Comments
A. Being lazy?
Answered by Mskitty - Tue Oct 7 16:19:43 2008
Q. 1. What is the molality of a solution of 90 grams of glucose (C6H12O) dissolved in 800 grams of water ? A. .01 B. .4 C. .625 D. 1.6 2. How many grams of chlorine (Cl2 would you have to add to 1 kg of water to increase the boiling point by 1.56oC? A. 200 B. 213 C. 230 3. One quality of metals is that they are ___, which means that they can be hammered into thin sheets. A. ductile B. malleable C. bendable 4. What is the molecular mass of carbon dioxide,CO2?___ Round your answer to the nearest tenth. 5. What do you call an ion that carries a positive charge? ___. 6. A process that needs to take in energy to get started is called a/an ___ process. A. endothermic B. exothermic 7.… [cont.]
Asked by Jessica` M - Tue Oct 7 16:11:18 2008 - - 1 Answers - 0 Comments
A. Being lazy?
Answered by Mskitty - Tue Oct 7 16:19:43 2008
Help me with this chemistry problem please...?
Q. A 4.99 gram sample of a compound contains 1.52 grams of nitrogen atoms and 3.47 grams of oxygen atoms. The molar mass of the compound is between 90.0 g and 95.0 g. Determine the empirical and molecular formulas. Also, calculate the actual molar mass of this compound.
Asked by Farid M - Mon Apr 27 19:42:02 2009 - - 1 Answers - 0 Comments
A. 1.52 grams of N 3.47 grams of O Divide the given grams of each element by its molar mass. The gram units cancel to give you mols. mols of N = 1.52/14 = .109 mols of O = 3.47/16 = .217 Divide the mols of N by the mols of O to find out the molar ratio. .109/.217 = .5 This is the ratio of Nitrogen to Oxygen in the compound. This means that for every 1 Nitrogen you have 2 Oxygen, for every 2 Nitrogen you have 4 Oxygen, etc. This also gives you the empirical formula; NO2, which has a molar mass of 46 g/mol. But the problem tells you that the sample weighs between 90-95 grams. So simply multiply the empirical formula by a factor of 2, this gives: N2O4, and has a molar mass of 92 grams/mol and this is the "actual" molar mass of the compound.… [cont.]
Answered by DAP1MP13 - Mon Apr 27 20:05:19 2009
Q. A 4.99 gram sample of a compound contains 1.52 grams of nitrogen atoms and 3.47 grams of oxygen atoms. The molar mass of the compound is between 90.0 g and 95.0 g. Determine the empirical and molecular formulas. Also, calculate the actual molar mass of this compound.
Asked by Farid M - Mon Apr 27 19:42:02 2009 - - 1 Answers - 0 Comments
A. 1.52 grams of N 3.47 grams of O Divide the given grams of each element by its molar mass. The gram units cancel to give you mols. mols of N = 1.52/14 = .109 mols of O = 3.47/16 = .217 Divide the mols of N by the mols of O to find out the molar ratio. .109/.217 = .5 This is the ratio of Nitrogen to Oxygen in the compound. This means that for every 1 Nitrogen you have 2 Oxygen, for every 2 Nitrogen you have 4 Oxygen, etc. This also gives you the empirical formula; NO2, which has a molar mass of 46 g/mol. But the problem tells you that the sample weighs between 90-95 grams. So simply multiply the empirical formula by a factor of 2, this gives: N2O4, and has a molar mass of 92 grams/mol and this is the "actual" molar mass of the compound.… [cont.]
Answered by DAP1MP13 - Mon Apr 27 20:05:19 2009
BIOLOGY HELP! please?(;?
Q. please help! help is GREATLY appreciated!!! 1. what term is used to describe reactions where two monomers are joined together with the loss of H2O [water] 2. based on the molecular formula of glycerol, which class of macromolecules does it belong in? 3.what type of reaction is it when you join glycerol and unsatuated fatty acid? 3a. this is an example of what type of reaction? 4. how does an actual trigyceride differ from models? 5. what does glycine and alanin have in common? how are they different? 6. what do you have to remove in order to join alanine and glycine? 6a. what type of reaction does it demonstrate? please, if you know any of these, help! thank you very much! <3ya.
Asked by Jennifer - Sat Sep 22 16:15:18 2007 - - 1 Answers - 0 Comments
A. 1. dehydration 2. it's a sugar alcohol 3. you get triglyceride, which is a lipid. this is also dehydration. 4. models are more neat 5. look it up on google 6. see 5. 7. see 6.
Answered by ATP-Boy - Sat Sep 22 16:59:53 2007
Q. please help! help is GREATLY appreciated!!! 1. what term is used to describe reactions where two monomers are joined together with the loss of H2O [water] 2. based on the molecular formula of glycerol, which class of macromolecules does it belong in? 3.what type of reaction is it when you join glycerol and unsatuated fatty acid? 3a. this is an example of what type of reaction? 4. how does an actual trigyceride differ from models? 5. what does glycine and alanin have in common? how are they different? 6. what do you have to remove in order to join alanine and glycine? 6a. what type of reaction does it demonstrate? please, if you know any of these, help! thank you very much! <3ya.
Asked by Jennifer - Sat Sep 22 16:15:18 2007 - - 1 Answers - 0 Comments
A. 1. dehydration 2. it's a sugar alcohol 3. you get triglyceride, which is a lipid. this is also dehydration. 4. models are more neat 5. look it up on google 6. see 5. 7. see 6.
Answered by ATP-Boy - Sat Sep 22 16:59:53 2007
Combustion analysis help please?
Q. A molecule being considered as a drug candidate (CxHyOz) consists of only carbon, hydrogen, and oxygen. The molar mass of this substance was determined to be 334 g/mol in a freezing point depression experiment. Freezing point analysis of a pure substance provides an accurate, but not exceptionally precise, molar mass. It s not more than 5% off. A combustion analysis was attempted to elucidate the molecular formula of this substance. 2.253 grams of the drug candidate was ignited and 6.035 g of carbon dioxide and 1.481 g of water vapor were detected. (Hint: see Sample Problem 3.7 on page 101 in your textbook if you need help!) a. Determine the empirical formula of the drug candidate. b. Determine the molecular formula of the drug candidate. [cont.]
Asked by :sigh: - Wed Oct 28 23:30:35 2009 - - 1 Answers - 0 Comments
A. a) C5H6O b) C20H24O4 c) 328 g/mole d) 2% error e) 99.94% *** the key to solving this and other combustion analysis problems are the following steps... 1) think about the reaction. Write an unbalanced equation 2) convert mass CO2 to mass C and mass H2O to mass H 3) determine mass O in the compound 4) convert mass C, H, and O to moles C, H, and O 5) simplify mole ratio to get empirical formula 6) compare molar mass of emperical formula to molecular formula solution... *** 1 *** the reaction looks like this right? __CxHyOz + __O2 ---> __CO2 + __ H2O where does the C in CO2 come from? how about the H in H2O? both only come from CxHyOz right? O2 has no C and no H. so if you calculate the mass of C in 6.035g CO2 and H in 1.481g H2O and… [cont.]
Answered by m w - Thu Oct 29 11:17:46 2009
Q. A molecule being considered as a drug candidate (CxHyOz) consists of only carbon, hydrogen, and oxygen. The molar mass of this substance was determined to be 334 g/mol in a freezing point depression experiment. Freezing point analysis of a pure substance provides an accurate, but not exceptionally precise, molar mass. It s not more than 5% off. A combustion analysis was attempted to elucidate the molecular formula of this substance. 2.253 grams of the drug candidate was ignited and 6.035 g of carbon dioxide and 1.481 g of water vapor were detected. (Hint: see Sample Problem 3.7 on page 101 in your textbook if you need help!) a. Determine the empirical formula of the drug candidate. b. Determine the molecular formula of the drug candidate. [cont.]
Asked by :sigh: - Wed Oct 28 23:30:35 2009 - - 1 Answers - 0 Comments
A. a) C5H6O b) C20H24O4 c) 328 g/mole d) 2% error e) 99.94% *** the key to solving this and other combustion analysis problems are the following steps... 1) think about the reaction. Write an unbalanced equation 2) convert mass CO2 to mass C and mass H2O to mass H 3) determine mass O in the compound 4) convert mass C, H, and O to moles C, H, and O 5) simplify mole ratio to get empirical formula 6) compare molar mass of emperical formula to molecular formula solution... *** 1 *** the reaction looks like this right? __CxHyOz + __O2 ---> __CO2 + __ H2O where does the C in CO2 come from? how about the H in H2O? both only come from CxHyOz right? O2 has no C and no H. so if you calculate the mass of C in 6.035g CO2 and H in 1.481g H2O and… [cont.]
Answered by m w - Thu Oct 29 11:17:46 2009
chemistry q's?
Q. 1.Consider the unbalanced reaction: H2 + Br2 > HBr If the theoretical yield is 78.2 grams HBr, and the percent yield is 45.0%, what is the actual yield? 2. For the following reaction, calculate the number of moles of Cu that can be produced from 30.1 grams of Cu2SO4 with excess Na. 2 Na + Cu2SO4 > Na2SO4 + Cu 3. The reaction of 1.45 moles of Cl2 with excess F2 produced 66.9 grams of ClF3. What percentage yield of ClF3 was obtained? The balanced equation is as follows: Cl2 + 3F2 > 2ClF3 4. A compound that contains carbon, hydrogen and oxygen is found to have an empirical formula of C3H4O. If the molar mass of this compound is 224 g/mol, what is its molecular formula? 5.Consider the following balanced equation: PBr3 + 3H2O > H3PO3 +… [cont.]
Asked by janga - Thu May 10 19:26:37 2007 - - 1 Answers - 0 Comments
A. Question 1: Plug the numbers into this equation Percent yield = actual yield / theoretical yield .45 = actual / 78.2 actual yield = 35.19 grams Question 2: The words "excess Na" means that Cu2So4 is the limiting reagent. Do stoichiometry to get the number of moles of Cu2SO4. 30.1 g CuSO4 * (1 mole CuSO4 / 223.17 grams) You get .135 moles of CuSO4. Question 3: Find the theoretical yield (what you should have got if the experiment was perfect with no mistakes). To find this, you do stoichiometry: 1.45 moles Cl2 * (2 moles ClF3 / 1 mole Cl2) * (92.45 g / 1 mole ClF3) You get that 268 grams is the theoretical yield. Then, plug it into the equation: Percent yield = actual yield / theoretical yield Percent yield = 66.9 g / 268 =… [cont.]
Answered by C_F15H - Thu May 10 19:47:29 2007
Q. 1.Consider the unbalanced reaction: H2 + Br2 > HBr If the theoretical yield is 78.2 grams HBr, and the percent yield is 45.0%, what is the actual yield? 2. For the following reaction, calculate the number of moles of Cu that can be produced from 30.1 grams of Cu2SO4 with excess Na. 2 Na + Cu2SO4 > Na2SO4 + Cu 3. The reaction of 1.45 moles of Cl2 with excess F2 produced 66.9 grams of ClF3. What percentage yield of ClF3 was obtained? The balanced equation is as follows: Cl2 + 3F2 > 2ClF3 4. A compound that contains carbon, hydrogen and oxygen is found to have an empirical formula of C3H4O. If the molar mass of this compound is 224 g/mol, what is its molecular formula? 5.Consider the following balanced equation: PBr3 + 3H2O > H3PO3 +… [cont.]
Asked by janga - Thu May 10 19:26:37 2007 - - 1 Answers - 0 Comments
A. Question 1: Plug the numbers into this equation Percent yield = actual yield / theoretical yield .45 = actual / 78.2 actual yield = 35.19 grams Question 2: The words "excess Na" means that Cu2So4 is the limiting reagent. Do stoichiometry to get the number of moles of Cu2SO4. 30.1 g CuSO4 * (1 mole CuSO4 / 223.17 grams) You get .135 moles of CuSO4. Question 3: Find the theoretical yield (what you should have got if the experiment was perfect with no mistakes). To find this, you do stoichiometry: 1.45 moles Cl2 * (2 moles ClF3 / 1 mole Cl2) * (92.45 g / 1 mole ClF3) You get that 268 grams is the theoretical yield. Then, plug it into the equation: Percent yield = actual yield / theoretical yield Percent yield = 66.9 g / 268 =… [cont.]
Answered by C_F15H - Thu May 10 19:47:29 2007
chemistry plz HELP!!!!!?
Q. Problems. 1. Determine the empirical formula and the molecular formula of a compound containing 40.0%C,6.7%H and 53.3% O. The molecular weight of the compounds is 60.02 g. 2. Calculate the percentage composition of a compound containing 27.93g Fe,24.05gS,and 48.00 g O. 3.How many grams of silver is present in 3.45g Ag2S? 4.The reaction 2Al+3MnO--->Al2O2+3Mn proceeds until the limiting reagent is all consumed. A mixture containing 110g Al and 200g MnO was heated to initiate the reaction. Which substance is the excess reagent and by how much? 5. Calculate the theoretical yield, potassium chloride produced from the reaction of 2.56g of K and 3.85 g Cl2 if an actual experimental produces 3.81g KCL. Calculate the… [cont.]
Asked by Bryan G - Sun Jul 27 06:41:58 2008 - - 2 Answers - 0 Comments
A. 1. Empirical formula of a compound is the lowest whole ratio of each atom in a molecule. The molecular formula is the exact number of each atom that is present in a molecule. You can work out an empirical formula from the mass of each atom in a sample. Firstly, assume that you have 100 g of sample. So in 100 g The mass ratio of Carbon : Hydrogen : Oxygen is 40.0 g : 6.7 g : 53.3 g You need to convert this to moles in order to get the moles ratio of elemnts in the sample. To convert to moles divide each mass by the atomic mass of the atom because moles = mass / molecular mass So the mole ratio of C : H : O is (40.0/12.0) : (6.7 / 1.008) : (53.3/16.0) = 3.33 : 6.64 : 3.33 Now, to get the lowest whole ratio divide each number by the lowest… [cont.]
Answered by Lexi R - Mon Jul 28 03:17:22 2008
Q. Problems. 1. Determine the empirical formula and the molecular formula of a compound containing 40.0%C,6.7%H and 53.3% O. The molecular weight of the compounds is 60.02 g. 2. Calculate the percentage composition of a compound containing 27.93g Fe,24.05gS,and 48.00 g O. 3.How many grams of silver is present in 3.45g Ag2S?
Asked by Bryan G - Sun Jul 27 06:41:58 2008 - - 2 Answers - 0 Comments
A. 1. Empirical formula of a compound is the lowest whole ratio of each atom in a molecule. The molecular formula is the exact number of each atom that is present in a molecule. You can work out an empirical formula from the mass of each atom in a sample. Firstly, assume that you have 100 g of sample. So in 100 g The mass ratio of Carbon : Hydrogen : Oxygen is 40.0 g : 6.7 g : 53.3 g You need to convert this to moles in order to get the moles ratio of elemnts in the sample. To convert to moles divide each mass by the atomic mass of the atom because moles = mass / molecular mass So the mole ratio of C : H : O is (40.0/12.0) : (6.7 / 1.008) : (53.3/16.0) = 3.33 : 6.64 : 3.33 Now, to get the lowest whole ratio divide each number by the lowest… [cont.]
Answered by Lexi R - Mon Jul 28 03:17:22 2008
Chemistry Review Packet?
Q. Maybe this seems like a lot, but it's out of 200 questions. Some of them might actually not be that hard, but it is late and I'm sick so I'm not thinking too well. So I appreciate any help! 1) The scale for expressing atomic masses is based on the mass of ___. 2) How many atoms are in a sample of an element whose masses in grams is numerically equal to the atomic mass? 3) The number of atoms of an element equal to the number of atoms in exactly 12.0 grams of carbon-12 is called a ___. 4)What is the molecular formula of a compound whoe empirical formula is CH and whose molecular mass is approximately 84.0 g/mol? 5)What is a mass-mass problem? 6) When dealing with theoretical yield, what are the expected and actual yields? 7)What is… [cont.]
Asked by Alexa - Tue May 19 02:05:46 2009 - - 2 Answers - 0 Comments
A. 1. I'm not sure about this one actually. The unit used, amu, is the mass of a hydrogen atom, so it might be hydrogen. It is actually DEFINED though exactly by using the mass of an atom of carbon-12. That is the answer that popped into my head when I read it. 2. The atomic mass expressed in grams is the molar mass, so there would be one mole of atoms present. This is 6.022 x 10^23 atoms. 3. mole 4. I think you have a mistake in that one. You would rarely have an empirical formula of CH, and there is no solution. C6H6 would give you 78 g/mol and C7H7 would give you 91 g/mol 5. a problem where you have a given mass and you must use a balanced reaction equation to calculate the corresponding mass of another species. For example,… [cont.]
Answered by jacee - Tue May 19 02:29:15 2009
Q. Maybe this seems like a lot, but it's out of 200 questions. Some of them might actually not be that hard, but it is late and I'm sick so I'm not thinking too well. So I appreciate any help! 1) The scale for expressing atomic masses is based on the mass of ___. 2) How many atoms are in a sample of an element whose masses in grams is numerically equal to the atomic mass? 3) The number of atoms of an element equal to the number of atoms in exactly 12.0 grams of carbon-12 is called a ___. 4)What is the molecular formula of a compound whoe empirical formula is CH and whose molecular mass is approximately 84.0 g/mol? 5)What is a mass-mass problem? 6) When dealing with theoretical yield, what are the expected and actual yields? 7)What is… [cont.]
Asked by Alexa - Tue May 19 02:05:46 2009 - - 2 Answers - 0 Comments
A. 1. I'm not sure about this one actually. The unit used, amu, is the mass of a hydrogen atom, so it might be hydrogen. It is actually DEFINED though exactly by using the mass of an atom of carbon-12. That is the answer that popped into my head when I read it. 2. The atomic mass expressed in grams is the molar mass, so there would be one mole of atoms present. This is 6.022 x 10^23 atoms. 3. mole 4. I think you have a mistake in that one. You would rarely have an empirical formula of CH, and there is no solution. C6H6 would give you 78 g/mol and C7H7 would give you 91 g/mol 5. a problem where you have a given mass and you must use a balanced reaction equation to calculate the corresponding mass of another species. For example,… [cont.]
Answered by jacee - Tue May 19 02:29:15 2009
Plz help with any of these problems...?
Q. Hydrogen gas and bromine gas react to form hydrogen bromide gas. a.Write a balanced chemical equation for this reaction. b.How many grams of hydrogen bromide gas can be produced from 3.2 g of hydrogen gas and 9.5 g of bromine gas? c.How many grams of which reactant is left unreacted? d. What volume of HBr, measured at STP, is produced in When ammonia gas, oxygen gas and methane gas (CH4) are combined, the products are hydrogen cyanide gas and water. a.Write a balance chemical equation for this reaction. b.Calculate the mass of each product produced when 225 g of oxygen gas is reacted with an excess of the other two reactants. c.If the actual yield of the experiment in b) is 105 g of HCN, calculate the percent yield. Cinnamic acid… [cont.]
Asked by MiCk - Sun Aug 24 00:40:14 2008 - - 1 Answers - 0 Comments
A. H2 + Br2 >> 2 HBr Moles H2 = 3.2 / 2.016 g/mol =1.59 Moles Br2 = 9.5 / 159.808 g/mol = 0.0594 ( limiting reactant) we get 2 x 0.0594 moles of HBr => 0.119 moles Mass HBr = 0.119 mol x 80.912 g/mol =9.63 g Moles H2 needed = 1.59 - 0.0594 = 1.53 => 1.53 mol x 2.016 g/mol = 3.08 g V of HBr = 22.4 x 0.119 = 2.67 g 3NH3 + 2CH4 +3 O2 >>2 HCN +6 H2O Moles O2 = 225 g / 32 g/mol = 7.0 moles HCN = 7.0 x 2 / 3 = 4.7 Mass HCN = 4.7 mol x 27 g/mol =127 g moles H2O = 7.0 x 6 / 3 = 14 Mass water = 14 x 18 = 252 g % yield = 105 x 100 / 127 = 82.7 Moles C = 73.0 / 12.011 = 6.08 moles H = 5.4 / 1.008 = 5.4 moles O = 21.6 / 16 = 1.35 6.08/ 1.35 =4.5 => C 5.4 / 1.35 = 4 =>H 1.35 / 1.35 = 1 => O we multiply by 2 to get whole numbers C9H8O2 is the… [cont.]
Answered by Dr.A - Sun Aug 24 09:04:28 2008
Q. Hydrogen gas and bromine gas react to form hydrogen bromide gas. a.Write a balanced chemical equation for this reaction. b.How many grams of hydrogen bromide gas can be produced from 3.2 g of hydrogen gas and 9.5 g of bromine gas? c.How many grams of which reactant is left unreacted? d. What volume of HBr, measured at STP, is produced in When ammonia gas, oxygen gas and methane gas (CH4) are combined, the products are hydrogen cyanide gas and water. a.Write a balance chemical equation for this reaction. b.Calculate the mass of each product produced when 225 g of oxygen gas is reacted with an excess of the other two reactants. c.If the actual yield of the experiment in b) is 105 g of HCN, calculate the percent yield. Cinnamic acid… [cont.]
Asked by MiCk - Sun Aug 24 00:40:14 2008 - - 1 Answers - 0 Comments
A. H2 + Br2 >> 2 HBr Moles H2 = 3.2 / 2.016 g/mol =1.59 Moles Br2 = 9.5 / 159.808 g/mol = 0.0594 ( limiting reactant) we get 2 x 0.0594 moles of HBr => 0.119 moles Mass HBr = 0.119 mol x 80.912 g/mol =9.63 g Moles H2 needed = 1.59 - 0.0594 = 1.53 => 1.53 mol x 2.016 g/mol = 3.08 g V of HBr = 22.4 x 0.119 = 2.67 g 3NH3 + 2CH4 +3 O2 >>2 HCN +6 H2O Moles O2 = 225 g / 32 g/mol = 7.0 moles HCN = 7.0 x 2 / 3 = 4.7 Mass HCN = 4.7 mol x 27 g/mol =127 g moles H2O = 7.0 x 6 / 3 = 14 Mass water = 14 x 18 = 252 g % yield = 105 x 100 / 127 = 82.7 Moles C = 73.0 / 12.011 = 6.08 moles H = 5.4 / 1.008 = 5.4 moles O = 21.6 / 16 = 1.35 6.08/ 1.35 =4.5 => C 5.4 / 1.35 = 4 =>H 1.35 / 1.35 = 1 => O we multiply by 2 to get whole numbers C9H8O2 is the… [cont.]
Answered by Dr.A - Sun Aug 24 09:04:28 2008
Chemistry Help.....need Major help=)?
Q. okay..so here are the questions. Thank You!! =D 1: A Compound's percent composition by mass in 49.5% carbon, 5.20% hydrogen, 28.8% nitrogen and 16.5% oxygen. a:Determine the empirical formula of the Compound. b:Find the molecular formula if the molar mass of the compound is 194.2g/mole 2:Given a Container of Solid ZnCI2, describe how you would make 200ml of a 0.15mol/L solution ZnCI2? 3: If 226.1 g tin(II) Chloride is the actual yield obtained in a reaction for which the theoretical yield is 288.0g, what is the percent yield?
Asked by Coco009 - Thu Jun 11 21:43:49 2009 - - 2 Answers - 0 Comments
Q. okay..so here are the questions. Thank You!! =D 1: A Compound's percent composition by mass in 49.5% carbon, 5.20% hydrogen, 28.8% nitrogen and 16.5% oxygen. a:Determine the empirical formula of the Compound. b:Find the molecular formula if the molar mass of the compound is 194.2g/mole 2:Given a Container of Solid ZnCI2, describe how you would make 200ml of a 0.15mol/L solution ZnCI2? 3: If 226.1 g tin(II) Chloride is the actual yield obtained in a reaction for which the theoretical yield is 288.0g, what is the percent yield?
Asked by Coco009 - Thu Jun 11 21:43:49 2009 - - 2 Answers - 0 Comments
chemistryhelp plzzzzzzzzzzzzzzzzzzzz i have formula just need to know how to do the problems?
Q. mass of flask +robber cap+liquid-97.39g mass of flask+robber cup-97.13g Temperature of water bath at the last trace of the unknown liquid -100degres celcius mass of vapor condensed-0.26g actual volume of flask markd 125 ml --146 ml barometric pressure from lab barometer---1.00atm 1. using Boyles and charles law combined,calculate the volume of the vapor at STP. my teacher said we have to use P1V1 over T1=P2V2 over T2 can us how me how to solve this I don t know how to plug in the numbers to solve 2.calculate the molecular mass of the gas: molecular mass of unknown over mass of condensed vapor=22.4/mole over volume of gas in liters at STP 3.How should i calculate my percentage error? Actual mass- measured mass over actual mass X 100 4.Why… [cont.]
Asked by Ivan C - Sun Mar 16 16:06:13 2008 - - 1 Answers - 0 Comments
Q. mass of flask +robber cap+liquid-97.39g mass of flask+robber cup-97.13g Temperature of water bath at the last trace of the unknown liquid -100degres celcius mass of vapor condensed-0.26g actual volume of flask markd 125 ml --146 ml barometric pressure from lab barometer---1.00atm 1. using Boyles and charles law combined,calculate the volume of the vapor at STP. my teacher said we have to use P1V1 over T1=P2V2 over T2 can us how me how to solve this I don t know how to plug in the numbers to solve 2.calculate the molecular mass of the gas: molecular mass of unknown over mass of condensed vapor=22.4/mole over volume of gas in liters at STP 3.How should i calculate my percentage error? Actual mass- measured mass over actual mass X 100 4.Why… [cont.]
Asked by Ivan C - Sun Mar 16 16:06:13 2008 - - 1 Answers - 0 Comments
Need Help With Chemistry!?
Q. Why is the actual yield of a reaction often less than the theoretical yield? (Select all that apply.) 1.Side reactions may occur generating other products as well. 2.The reaction may not go to completion. 3.Reactants may be impure. 4.The products may not be completely dry. 5.Some products may be lost in the process of extraction or purification. 6.The reaction's equilibrium position may not lie completely on the side of the products. Terephthalic acid is an important chemical used in the manufacture of polyesters and plasticizers. It contains only C, H, and O. Combustion of 17.83 mg terephthalic acid produces 37.78 mg CO2 and 5.81 mg H2O. The molar mass of terephthalic acid is 166 g/mol. Calculate the empirical and molecular formulas for… [cont.]
Asked by kanu978 - Wed Jul 18 23:55:32 2007 - - 3 Answers - 0 Comments
A. Fist question: all apply Second question- 13.78 mg of CO2 contains (12.011/44.011)x13.78 mg C 3.7607 mg C 5.81 mg H2O contains (2.016/18.016)x5.81 mg H 0.6501 mg H 17.83 mg terephthalic acid - 3.7607 - 0.6501 = 13.42 mg O 3.7607/12.011=0.3131 millimoles C; 0.6501/1.008=0.6449 millimoles H; 13.42/16=0.83875 millimoles O; divide all by 0.3131 C=1.00, H=2.06, O=2.68 O looks like 2 2/3; get a whole number by multiplying by 3 C=1.00 x 3 = 3.0 H=2.06x3 = 6.2 (call it 6.0) O=2.68x3 = 8.0 Empirical formula C3H6O8 Molecular formula is C3H6O8 (MW=166 g/mole)
Answered by skipper - Thu Jul 19 00:21:01 2007
Q. Why is the actual yield of a reaction often less than the theoretical yield? (Select all that apply.) 1.Side reactions may occur generating other products as well. 2.The reaction may not go to completion. 3.Reactants may be impure. 4.The products may not be completely dry. 5.Some products may be lost in the process of extraction or purification. 6.The reaction's equilibrium position may not lie completely on the side of the products. Terephthalic acid is an important chemical used in the manufacture of polyesters and plasticizers. It contains only C, H, and O. Combustion of 17.83 mg terephthalic acid produces 37.78 mg CO2 and 5.81 mg H2O. The molar mass of terephthalic acid is 166 g/mol. Calculate the empirical and molecular formulas for… [cont.]
Asked by kanu978 - Wed Jul 18 23:55:32 2007 - - 3 Answers - 0 Comments
A. Fist question: all apply Second question- 13.78 mg of CO2 contains (12.011/44.011)x13.78 mg C 3.7607 mg C 5.81 mg H2O contains (2.016/18.016)x5.81 mg H 0.6501 mg H 17.83 mg terephthalic acid - 3.7607 - 0.6501 = 13.42 mg O 3.7607/12.011=0.3131 millimoles C; 0.6501/1.008=0.6449 millimoles H; 13.42/16=0.83875 millimoles O; divide all by 0.3131 C=1.00, H=2.06, O=2.68 O looks like 2 2/3; get a whole number by multiplying by 3 C=1.00 x 3 = 3.0 H=2.06x3 = 6.2 (call it 6.0) O=2.68x3 = 8.0 Empirical formula C3H6O8 Molecular formula is C3H6O8 (MW=166 g/mole)
Answered by skipper - Thu Jul 19 00:21:01 2007
Calculate the pH for a 0.365M solution of acetylsalycilic acid. The Ka and structure can be found in table.?
Q. Pleas include: a) A chemical equation showing this acid ionization equilibrium in solution the molecular acid and water will be reactants. You may use HC9H8O4 for the acid version, and C9H8O4^- as its conjugate base. b) A generic Ka expression (listing the acid formula, and ionized species formula and the actual Ka value) c) An ICE table showing your logic/calculation of the hydrogen ion concentration d)your final answer for pH.
Asked by qwer - Fri Jul 4 21:38:18 2008 - - 1 Answers - 0 Comments
Q. Pleas include: a) A chemical equation showing this acid ionization equilibrium in solution the molecular acid and water will be reactants. You may use HC9H8O4 for the acid version, and C9H8O4^- as its conjugate base. b) A generic Ka expression (listing the acid formula, and ionized species formula and the actual Ka value) c) An ICE table showing your logic/calculation of the hydrogen ion concentration d)your final answer for pH.
Asked by qwer - Fri Jul 4 21:38:18 2008 - - 1 Answers - 0 Comments
Chem. Lab Help please please?
Q. I am so lost I dont understand this. 1. Record and calculate the following about the copper sulfate pentahydrate: (a) Mass of CuSO4*5H2O (g): (b) Molecular weight of CuSO4*5H2O: (c) Moles of CuSO4*5H2O: (d) Moles of Cu in the copper sulfate salt: 2. Calculate the following for the oxygen: (a) The initial mass of copper in the moles of copper from the copper sulfate salt (MW of Cu = 63.55 g/mole) (g): (b) The of copper oxide obtained (g): (c) The gain in mass, equal to the added mass of oxygen (g): (d) The number of moles of oxygen in the copper oxide, given a molecular weight of oxygen (O) equal to 16.00 g/mole: 3. Calculate the molar ratio of copper to oxygen from: (moles copper) / (moles of… [cont.]
Asked by lovintsh - Wed Feb 18 13:12:35 2009 - - 1 Answers - 0 Comments
A. Your question is to long for me to help you answer each one but I hope I can help you get started on your own. You should have recorded the mass of the copper sulfate when you did the experiment. From there you need to perform the necessary calculations to get the mass into moles and whatever else you need to do. Molecular masses can be calculated using the periodic table. You should also have recorded the mass of the copper oxide. Again perform necessary calculations. If you are completely confused see your teacher/professor for help. PS If there are certain parts you are having trouble with you should try asking question only about that part. You might get more people to answer you if you are more specific.
Answered by Timmy D - Wed Feb 18 14:01:53 2009
Q. I am so lost I dont understand this. 1. Record and calculate the following about the copper sulfate pentahydrate: (a) Mass of CuSO4*5H2O (g): (b) Molecular weight of CuSO4*5H2O: (c) Moles of CuSO4*5H2O: (d) Moles of Cu in the copper sulfate salt: 2. Calculate the following for the oxygen: (a) The initial mass of copper in the moles of copper from the copper sulfate salt (MW of Cu = 63.55 g/mole) (g): (b) The of copper oxide obtained (g): (c) The gain in mass, equal to the added mass of oxygen (g): (d) The number of moles of oxygen in the copper oxide, given a molecular weight of oxygen (O) equal to 16.00 g/mole: 3. Calculate the molar ratio of copper to oxygen from: (moles copper) / (moles of… [cont.]
Asked by lovintsh - Wed Feb 18 13:12:35 2009 - - 1 Answers - 0 Comments
A. Your question is to long for me to help you answer each one but I hope I can help you get started on your own. You should have recorded the mass of the copper sulfate when you did the experiment. From there you need to perform the necessary calculations to get the mass into moles and whatever else you need to do. Molecular masses can be calculated using the periodic table. You should also have recorded the mass of the copper oxide. Again perform necessary calculations. If you are completely confused see your teacher/professor for help. PS If there are certain parts you are having trouble with you should try asking question only about that part. You might get more people to answer you if you are more specific.
Answered by Timmy D - Wed Feb 18 14:01:53 2009
chemistry question?
Q. 1.Carbon dioxide called a greenhouse gas because bacterial degradation of fertilizers in a greenhouse environment produce large quantities of carbon dioxide. Is this True? 2. The molecular weight is ALWAYS a whole-number multiple of the empirical formula weight. Is this true? 3. The mass of a single atom of an element (in amu) is numerically EQUAL to the mass in grams of 1 mole of that element. Is this True? 4. A great deal of the carbon dioxide produced by the combustion of fossil fuels is absorbed into the oceans. Is this True? 5. The quantity of product that is calculated to form when all of the limiting reagent reacts is called the actual yield. Is this True?
Asked by bob - Fri Feb 29 11:58:10 2008 - - 1 Answers - 0 Comments
A. 1) False, most CO2 generation is due to combustion of hydrocarbons in fossil fuels in internal combustion engines and industry. 2) True 3) False - the difference is huge, there are 6.022 *10^23 atoms/mole 4) False - some is, but not - "a great deal" 5) True
Answered by Dr Dave P - Fri Feb 29 12:08:31 2008
Q. 1.Carbon dioxide called a greenhouse gas because bacterial degradation of fertilizers in a greenhouse environment produce large quantities of carbon dioxide. Is this True? 2. The molecular weight is ALWAYS a whole-number multiple of the empirical formula weight. Is this true? 3. The mass of a single atom of an element (in amu) is numerically EQUAL to the mass in grams of 1 mole of that element. Is this True? 4. A great deal of the carbon dioxide produced by the combustion of fossil fuels is absorbed into the oceans. Is this True? 5. The quantity of product that is calculated to form when all of the limiting reagent reacts is called the actual yield. Is this True?
Asked by bob - Fri Feb 29 11:58:10 2008 - - 1 Answers - 0 Comments
A. 1) False, most CO2 generation is due to combustion of hydrocarbons in fossil fuels in internal combustion engines and industry. 2) True 3) False - the difference is huge, there are 6.022 *10^23 atoms/mole 4) False - some is, but not - "a great deal" 5) True
Answered by Dr Dave P - Fri Feb 29 12:08:31 2008
Chemistry help please?
Q. Mass beaker=47.420g Mass of beaker and copper sulfate pentahydrate=49.420g test tube= 18.200g test tube with the solid= 18.981g test tube after cooked on burner= 18.837g 1. Record and calculate the following about the copper sulfate pentahydrate: (a) Mass of CuSO *5H O (g): (b) Molecular weight of CuSO *5H O: (c) Moles of CuSO *5H O: (d) Moles of Cu in the copper sulfate salt: 2. Calculate the following for the oxygen: (a) The initial mass of copper in the moles of copper from the copper sulfate salt (MW of Cu = 63.55 g/mole) (g): (b) The mass of copper oxide obtained (g): (c) The gain in mass, equal to the added mass of oxygen (g): (d) The number of moles of oxygen in the copper oxide, given a molecular weight of oxygen (O)… [cont.]
Asked by vitamin C - Fri Oct 3 01:54:09 2008 - - 1 Answers - 0 Comments
A. This is a but too much stuff for anyone to do, you really should ask your parents or teacher if you dont know, getting some1 here to do all of that for you, as well as explaining it, is really useless to you. Good luck.
Answered by Jazza - Fri Oct 3 02:02:32 2008
Q. Mass beaker=47.420g Mass of beaker and copper sulfate pentahydrate=49.420g test tube= 18.200g test tube with the solid= 18.981g test tube after cooked on burner= 18.837g 1. Record and calculate the following about the copper sulfate pentahydrate: (a) Mass of CuSO *5H O (g): (b) Molecular weight of CuSO *5H O: (c) Moles of CuSO *5H O: (d) Moles of Cu in the copper sulfate salt: 2. Calculate the following for the oxygen: (a) The initial mass of copper in the moles of copper from the copper sulfate salt (MW of Cu = 63.55 g/mole) (g): (b) The mass of copper oxide obtained (g): (c) The gain in mass, equal to the added mass of oxygen (g): (d) The number of moles of oxygen in the copper oxide, given a molecular weight of oxygen (O)… [cont.]
Asked by vitamin C - Fri Oct 3 01:54:09 2008 - - 1 Answers - 0 Comments
A. This is a but too much stuff for anyone to do, you really should ask your parents or teacher if you dont know, getting some1 here to do all of that for you, as well as explaining it, is really useless to you. Good luck.
Answered by Jazza - Fri Oct 3 02:02:32 2008
From Yahoo Answer Search: 'actual molecular formula'
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