What is the empirical formula of the compound? What is the molecular formula of the compound also?
Q. A compound that contains only nitrogen and oxygen is 30.4% N by mass; the molar mass of the compound is 92 g/mol. What is the empirical formula of the compound? What is the molecular formula of the compound also?
Asked by John - Thu Jan 22 01:01:38 2009 - - 2 Answers - 0 Comments
A. so you know that u have 30.4 % nitrogen by mass, you can write that as: 30.4 g N then u know you have 69.6 % O left but you can write that as 69.6 g O ( 100 - 30.4 = 69.4) ok so now you need to calculate the empirical formula: you can do that by taking each respective element and finding the value for one mole of that element for example: 1) 30.4gN x ( 1 mol N / 14.06g N) and you get: about 2.16 mol of N (since the grams cancel) 2) 69.6g O x (1 mol O / 16.0g O ) and you get: 4.35 mol O. now you found the two elements for one mol, so you take the smallest one and divide all of the values by it ex: 1) 2.16 / 2.16 = 1 2) 4.35 /2.16 = about 2 ok so now yuo know that the empirical formula is equal to (N1)(O2)… [cont.]
Answered by dts670 - Sun Jan 25 11:38:26 2009
Q. A compound that contains only nitrogen and oxygen is 30.4% N by mass; the molar mass of the compound is 92 g/mol. What is the empirical formula of the compound? What is the molecular formula of the compound also?
Asked by John - Thu Jan 22 01:01:38 2009 - - 2 Answers - 0 Comments
A. so you know that u have 30.4 % nitrogen by mass, you can write that as: 30.4 g N then u know you have 69.6 % O left but you can write that as 69.6 g O ( 100 - 30.4 = 69.4) ok so now you need to calculate the empirical formula: you can do that by taking each respective element and finding the value for one mole of that element for example: 1) 30.4gN x ( 1 mol N / 14.06g N) and you get: about 2.16 mol of N (since the grams cancel) 2) 69.6g O x (1 mol O / 16.0g O ) and you get: 4.35 mol O. now you found the two elements for one mol, so you take the smallest one and divide all of the values by it ex: 1) 2.16 / 2.16 = 1 2) 4.35 /2.16 = about 2 ok so now yuo know that the empirical formula is equal to (N1)(O2)… [cont.]
Answered by dts670 - Sun Jan 25 11:38:26 2009
What is the empirical formula from the given amount of moles?
Q. What is the empirical formula of a substance that contains 0.250mol of carbon, 0.500mol of hydrogen, and 0.250mol of oxygen?
Asked by Kelsey Z - Tue Nov 18 19:29:58 2008 - - 2 Answers - 0 Comments
A. Divide by the smallest number of moles they give you. So, in this case, the smallest number is 0.250 mol 0.250 mol C/0.250 mol = 1 0.500 mol H/0.250 mol = 2 0.250 mol O/0.250 mol = 1 Thus, there's 1 carbon, 2 hydrogen, and 1 oxygen Therefore, the empirical formula is: CH2O
Answered by unknown - Tue Nov 18 19:34:21 2008
Q. What is the empirical formula of a substance that contains 0.250mol of carbon, 0.500mol of hydrogen, and 0.250mol of oxygen?
Asked by Kelsey Z - Tue Nov 18 19:29:58 2008 - - 2 Answers - 0 Comments
A. Divide by the smallest number of moles they give you. So, in this case, the smallest number is 0.250 mol 0.250 mol C/0.250 mol = 1 0.500 mol H/0.250 mol = 2 0.250 mol O/0.250 mol = 1 Thus, there's 1 carbon, 2 hydrogen, and 1 oxygen Therefore, the empirical formula is: CH2O
Answered by unknown - Tue Nov 18 19:34:21 2008
What is the empirical formula of the compound?
Q. When a 0.860 g sample of an organic compound containing C, H, and O was burned completely in oxygen, 1.64 g of CO2 and 1.01 g of H2O were produced. What is the empirical formula of the compound?
Asked by Belynda N - Thu Feb 19 23:06:18 2009 - - 2 Answers - 0 Comments
A. 1.64 g CO2 x (12.011 / 44.01) = 0.4476 g C 1.01 g H2O x (2.0158 / 18.0152) = 0.1130 g H 0.860 g sample - 0.4476 g C - 0.1130 g H = 0.2994 g O 0.4476 g C / 12.011 = 0.0373 mol C atoms 0.1130 g H / 1.0079 = 0.1121 mol H atoms 0.2994 g O / 15.9994 = 0.0187 mol O atoms C: 0.0373 / 0.0187 = 2.0 H: 0.1121 / 0.0187 = 6.0 O: 0.0187 / 0.0187 = 1.0 C2H6O
Answered by skipper - Thu Feb 19 23:18:22 2009
Q. When a 0.860 g sample of an organic compound containing C, H, and O was burned completely in oxygen, 1.64 g of CO2 and 1.01 g of H2O were produced. What is the empirical formula of the compound?
Asked by Belynda N - Thu Feb 19 23:06:18 2009 - - 2 Answers - 0 Comments
A. 1.64 g CO2 x (12.011 / 44.01) = 0.4476 g C 1.01 g H2O x (2.0158 / 18.0152) = 0.1130 g H 0.860 g sample - 0.4476 g C - 0.1130 g H = 0.2994 g O 0.4476 g C / 12.011 = 0.0373 mol C atoms 0.1130 g H / 1.0079 = 0.1121 mol H atoms 0.2994 g O / 15.9994 = 0.0187 mol O atoms C: 0.0373 / 0.0187 = 2.0 H: 0.1121 / 0.0187 = 6.0 O: 0.0187 / 0.0187 = 1.0 C2H6O
Answered by skipper - Thu Feb 19 23:18:22 2009
How do you solve the empirical formula and how many hydrogen atoms are present?
Q. A molecule that contains only carbon and hydrogen is composed of 74.83% carbon by mass. a) What is the empirical formula? b) How many hydrogen atoms are present in 25.8 g of this substance?
Asked by bryon_barker - Mon Nov 5 19:59:49 2007 - - 1 Answers - 0 Comments
A. a) What is the empirical formula? go for moles 74.83 grams carbon @ 12.01g/mol = 6.28 moles carbon 25.17 grams hydrogen @ 1g/mol = 25.17 moles of hydrogen that is a 1:4 ratio of moles answer ch4 --- b) How many hydrogen atoms are present in 25.8 g of this substance go for moles of CH4: 25.8 grams @ 16.04 grams /mol = 1.608 mole of CH4 times 4 = 6.434 moles of H in 25,8 g of CH4 --- go for atoms using avagadro's # 6.434 moles X (6.022 e23 atoms / mol) = aqnswer= 3.87 e24 atoms of hydrogen
Answered by Steve O - Mon Nov 5 22:03:51 2007
Q. A molecule that contains only carbon and hydrogen is composed of 74.83% carbon by mass. a) What is the empirical formula? b) How many hydrogen atoms are present in 25.8 g of this substance?
Asked by bryon_barker - Mon Nov 5 19:59:49 2007 - - 1 Answers - 0 Comments
A. a) What is the empirical formula? go for moles 74.83 grams carbon @ 12.01g/mol = 6.28 moles carbon 25.17 grams hydrogen @ 1g/mol = 25.17 moles of hydrogen that is a 1:4 ratio of moles answer ch4 --- b) How many hydrogen atoms are present in 25.8 g of this substance go for moles of CH4: 25.8 grams @ 16.04 grams /mol = 1.608 mole of CH4 times 4 = 6.434 moles of H in 25,8 g of CH4 --- go for atoms using avagadro's # 6.434 moles X (6.022 e23 atoms / mol) = aqnswer= 3.87 e24 atoms of hydrogen
Answered by Steve O - Mon Nov 5 22:03:51 2007
What is the empirical formula of the material?
Q. You have found a sample of high-temperature superconductor in your lab. A mass analysis of a 10.000 g sample of this material produced the following results: Compound/Mass: Y2O3 1.715 g BaO 4.659 g CuO 3.626 g What is the empirical formula of the material?
Asked by JohnD - Thu Apr 16 11:14:10 2009 - - 1 Answers - 0 Comments
A. ___Partial moles ___YxBayCuzOw x moles Y = 1.715g * (2 mole Y/mole Y2O3)/MW Y2O3 g/mole y moles Ba = 4.659g * (1 mole Ba/mole BaO)/MW BaO g/mole z moles Cu = 3.626g * (1 mole Cu/mole CuO)/MW CuO g/mole w moles O = [10.000g - moles Y*atwt Y - moles Ba*atwt Ba - mole Cu*atwtCu] / 16g/mole Make xyzw integers by dividing by the smallest (probably Y) Plug and SOLVE Basic mathematics is a prerequisite to chemistry I just try to help you with the methodology of solving the problem.
Answered by SciMann - Fri Apr 17 12:20:50 2009
Q. You have found a sample of high-temperature superconductor in your lab. A mass analysis of a 10.000 g sample of this material produced the following results: Compound/Mass: Y2O3 1.715 g BaO 4.659 g CuO 3.626 g What is the empirical formula of the material?
Asked by JohnD - Thu Apr 16 11:14:10 2009 - - 1 Answers - 0 Comments
A. ___Partial moles ___YxBayCuzOw x moles Y = 1.715g * (2 mole Y/mole Y2O3)/MW Y2O3 g/mole y moles Ba = 4.659g * (1 mole Ba/mole BaO)/MW BaO g/mole z moles Cu = 3.626g * (1 mole Cu/mole CuO)/MW CuO g/mole w moles O = [10.000g - moles Y*atwt Y - moles Ba*atwt Ba - mole Cu*atwtCu] / 16g/mole Make xyzw integers by dividing by the smallest (probably Y) Plug and SOLVE Basic mathematics is a prerequisite to chemistry I just try to help you with the methodology of solving the problem.
Answered by SciMann - Fri Apr 17 12:20:50 2009
What is the empirical formula for the compound p4o6?
Q. What is the empirical formula for the compound p4o6?
Asked by Ivan - Wed Apr 22 07:25:22 2009 - - 2 Answers - 0 Comments
A. It's quite similar to factorizing in maths apart from the fact that there are no brackets here, firstly to find the empirical formula for the compound p406, you would have yo determine but the largest multiple that fits in both 4 & 6, in this case it is 2. Meaning you would divide both of these numbers by 2, giving you an answer of p2o3. So the empirical formula for the compound p4o6 = p2o3
Answered by unknown - Wed Apr 22 09:09:08 2009
Q. What is the empirical formula for the compound p4o6?
Asked by Ivan - Wed Apr 22 07:25:22 2009 - - 2 Answers - 0 Comments
A. It's quite similar to factorizing in maths apart from the fact that there are no brackets here, firstly to find the empirical formula for the compound p406, you would have yo determine but the largest multiple that fits in both 4 & 6, in this case it is 2. Meaning you would divide both of these numbers by 2, giving you an answer of p2o3. So the empirical formula for the compound p4o6 = p2o3
Answered by unknown - Wed Apr 22 09:09:08 2009
What is the empirical formula and mole ratio of the sample?
Q. Elemental mercury was first discovered when a mercury oxide was decomposed with heat, forming mercury metal and oxygen gas. When a 0.667g sample of mercury oxide is heated, 0.618g of mercury metal remains. A. What is the mole ratio of mercury to oxygen in the sample? B. What is the empirical formula of the mercury oxide? (Just need to verify my answers before I hand in, thanks.)
Asked by Sciguy - Thu Feb 12 14:41:19 2009 - - 3 Answers - 0 Comments
A. (A) Mercury oxide mass = 0.667g Mecury mass = 0.618 g oxygen mass = 0.667 - 0.618 = 0.049 Mercury moles = 0.618 g / 200.59 g/mol = 0.00308 mol Hg atoms Oxygen moles = 0.049 g / 16 g/mol = 0.00306 mol O atoms Hg : O ratio = 1 : 1 (B) HgO
Answered by skipper - Thu Feb 12 14:53:30 2009
Q. Elemental mercury was first discovered when a mercury oxide was decomposed with heat, forming mercury metal and oxygen gas. When a 0.667g sample of mercury oxide is heated, 0.618g of mercury metal remains. A. What is the mole ratio of mercury to oxygen in the sample? B. What is the empirical formula of the mercury oxide? (Just need to verify my answers before I hand in, thanks.)
Asked by Sciguy - Thu Feb 12 14:41:19 2009 - - 3 Answers - 0 Comments
A. (A) Mercury oxide mass = 0.667g Mecury mass = 0.618 g oxygen mass = 0.667 - 0.618 = 0.049 Mercury moles = 0.618 g / 200.59 g/mol = 0.00308 mol Hg atoms Oxygen moles = 0.049 g / 16 g/mol = 0.00306 mol O atoms Hg : O ratio = 1 : 1 (B) HgO
Answered by skipper - Thu Feb 12 14:53:30 2009
What is the empirical formula of a hydrocarbon if complete combustion or 3.700 mg of the hydrocarbon produced?
Q. What is the empirical formula of a hydrocarbon if complete combustion or 3.700 mg of the hydrocarbon produced 12.507 mg of CO2 and 2.560 mg of H2O? Be sure to write C first in the formula. empirical formula = ? What is the molecular formula if the molar mass of the hydrocarbon is found to be about 40 molecular formula = ?
Asked by Derrick - Thu Aug 28 21:19:41 2008 - - 1 Answers - 0 Comments
A. There will be the same number of moles of C and H on both sides of the equation, so all the Carbon in your original hydrocarbon is now present in CO2, and all H is in the H2O. Hydrogen Calculate moles of H in H2O molar mass H2O = 16.00 + (2 x 1.008) = 18.016 g/mol molar mass H = 1.008 g/mol moles = mass / molar mass moles H2O = 0.002560 g / 18.016 g/mol = 1.421 x 10^-4 moles of H2O In ever H2O there are 2 moles of H therefore moles of H in H2O, and thus in your original hydrocarbon = 2 x 1.421 x 10^-4 = 2.842 x 10^-4 moles of H Carbon Calculate moles C in CO2 molar mass CO2 = (16.00 x 2) + 12.01 = 44.01 g/mol moles CO2 = mass / molar mass = 0.012507 g / 44.01 g/mol = 2.842 x 10^-4 moles of CO2 Every CO2 molecule has 1 Carbon, therefore… [cont.]
Answered by Lexi R - Thu Aug 28 22:38:04 2008
Q. What is the empirical formula of a hydrocarbon if complete combustion or 3.700 mg of the hydrocarbon produced 12.507 mg of CO2 and 2.560 mg of H2O? Be sure to write C first in the formula. empirical formula = ? What is the molecular formula if the molar mass of the hydrocarbon is found to be about 40 molecular formula = ?
Asked by Derrick - Thu Aug 28 21:19:41 2008 - - 1 Answers - 0 Comments
A. There will be the same number of moles of C and H on both sides of the equation, so all the Carbon in your original hydrocarbon is now present in CO2, and all H is in the H2O. Hydrogen Calculate moles of H in H2O molar mass H2O = 16.00 + (2 x 1.008) = 18.016 g/mol molar mass H = 1.008 g/mol moles = mass / molar mass moles H2O = 0.002560 g / 18.016 g/mol = 1.421 x 10^-4 moles of H2O In ever H2O there are 2 moles of H therefore moles of H in H2O, and thus in your original hydrocarbon = 2 x 1.421 x 10^-4 = 2.842 x 10^-4 moles of H Carbon Calculate moles C in CO2 molar mass CO2 = (16.00 x 2) + 12.01 = 44.01 g/mol moles CO2 = mass / molar mass = 0.012507 g / 44.01 g/mol = 2.842 x 10^-4 moles of CO2 Every CO2 molecule has 1 Carbon, therefore… [cont.]
Answered by Lexi R - Thu Aug 28 22:38:04 2008
How do i determine the empirical formula of copper chloride hydrate?
Q. We performed a lab and with the data i collected the formula came out to be Cu3Cl4, i know this is wrong but this is what i got from the data collected. The dehydrated sample was .76g, the hydrated sample was 1g. the mass of the copper was .43g and the mass of the chloride was .33g. and there were .013 moles of water. how would i find the hydration number for the empirical formula. my friend says to divide the amount of water by the amount of copper, but should i divide by the amount of total ions?
Asked by shish2kabob - Sun Nov 15 18:49:44 2009 - - 1 Answers - 0 Comments
A. Mass of water = 1.0 - 0.76 = 0.24 g = 0.013 moles, OK. 0.43 g Cu x (1 mole Cu / 63.5 g Cu) = 0.0068 moles Cu 0.33 g Cl x (1 mole Cl / 35.5 g Cl) = 0.0093 moles Cl. Dividing by the smallest (0.0068), we get a Cu / Cl ratio of 1/ 1.36 which is 3:4 so your Cu3Cl4 isn't unreasonable (but I've never heard of Cu3Cl4). However, I think that it's MUCH more likely that the ratio is 1 / 1.5 and the correct formula is CuCl2. That compound has a molar mass of 134.4. 0.76 g CuCl2 x (1 mole CuCl2 / 134.4 g CuCl2) = 0.0057 moles CuCl2 0.013 moles H2O / 0.0057 moles CuCl2 = 2.3 moles H2O / mole CuCl2. The likely formula is CuCl2 x 2H2O. That is a very common chemical found in most lab stockrooms.
Answered by HPV - Sun Nov 15 19:07:00 2009
Q. We performed a lab and with the data i collected the formula came out to be Cu3Cl4, i know this is wrong but this is what i got from the data collected. The dehydrated sample was .76g, the hydrated sample was 1g. the mass of the copper was .43g and the mass of the chloride was .33g. and there were .013 moles of water. how would i find the hydration number for the empirical formula. my friend says to divide the amount of water by the amount of copper, but should i divide by the amount of total ions?
Asked by shish2kabob - Sun Nov 15 18:49:44 2009 - - 1 Answers - 0 Comments
A. Mass of water = 1.0 - 0.76 = 0.24 g = 0.013 moles, OK. 0.43 g Cu x (1 mole Cu / 63.5 g Cu) = 0.0068 moles Cu 0.33 g Cl x (1 mole Cl / 35.5 g Cl) = 0.0093 moles Cl. Dividing by the smallest (0.0068), we get a Cu / Cl ratio of 1/ 1.36 which is 3:4 so your Cu3Cl4 isn't unreasonable (but I've never heard of Cu3Cl4). However, I think that it's MUCH more likely that the ratio is 1 / 1.5 and the correct formula is CuCl2. That compound has a molar mass of 134.4. 0.76 g CuCl2 x (1 mole CuCl2 / 134.4 g CuCl2) = 0.0057 moles CuCl2 0.013 moles H2O / 0.0057 moles CuCl2 = 2.3 moles H2O / mole CuCl2. The likely formula is CuCl2 x 2H2O. That is a very common chemical found in most lab stockrooms.
Answered by HPV - Sun Nov 15 19:07:00 2009
What is the empirical formula for the compound?
Q. A certain compound, is known to contain carbon and hydrogen. It might also contain oxygen. A 4.226 g sample of this compound was burned in pure oxygen gas. The measured amount of CO2 produced was 8.074 g, while that of H2O was 4.958 g. No other products were produced in the reaction. 2.202 grams of carbon were in the original sample and 0.5509 grams of hydrogen were in the original sample. What is the empirical formula?!! Please help. Ohh! I understand now! Thank you! :D
Asked by Kimi - Tue Nov 10 16:50:34 2009 - - 1 Answers - 0 Comments
A. Your sample definately contains Oxygen. The law of conservation of mass tells us that there must be the same mass in the products as there were in the reactants. The mass of C + mass H in you products = 2.202 g + 0.5509 g = 2.753 g But you started with 4.226 g of compound, so the difference in mass must be Oxygen mass O = 4.226 g - 2.202 - 0.5509 = 1.473 g Oxyygen in the compound Now work out the moles of C, H and O that were in the sample moles = mass / molar mass moles C = 2.202 g / 12.01 g/mol = 0.1833 moles C moles H = 0.5509 g / 1.008 g/mol = 0.5462 moles H moles O = 1.473 g / 16.00 g/mol = 0.0921 moles O Arrange the moles in a ratio format moles C : moles H : moles O = 0.1833 : 0.5462 : 0.0921 Divide each number by the… [cont.]
Answered by Lexi R - Tue Nov 10 19:17:38 2009
Q. A certain compound, is known to contain carbon and hydrogen. It might also contain oxygen. A 4.226 g sample of this compound was burned in pure oxygen gas. The measured amount of CO2 produced was 8.074 g, while that of H2O was 4.958 g. No other products were produced in the reaction. 2.202 grams of carbon were in the original sample and 0.5509 grams of hydrogen were in the original sample. What is the empirical formula?!! Please help. Ohh! I understand now! Thank you! :D
Asked by Kimi - Tue Nov 10 16:50:34 2009 - - 1 Answers - 0 Comments
A. Your sample definately contains Oxygen. The law of conservation of mass tells us that there must be the same mass in the products as there were in the reactants. The mass of C + mass H in you products = 2.202 g + 0.5509 g = 2.753 g But you started with 4.226 g of compound, so the difference in mass must be Oxygen mass O = 4.226 g - 2.202 - 0.5509 = 1.473 g Oxyygen in the compound Now work out the moles of C, H and O that were in the sample moles = mass / molar mass moles C = 2.202 g / 12.01 g/mol = 0.1833 moles C moles H = 0.5509 g / 1.008 g/mol = 0.5462 moles H moles O = 1.473 g / 16.00 g/mol = 0.0921 moles O Arrange the moles in a ratio format moles C : moles H : moles O = 0.1833 : 0.5462 : 0.0921 Divide each number by the… [cont.]
Answered by Lexi R - Tue Nov 10 19:17:38 2009
How to find the empirical formula of propane?
Q. The complete combustion of a sample of propane produced 2.641 grams of carbon dioxide and 1.442 grams of water as the only products. Find the empirical formula of propane. I understand how to do this if I was calculating based on the reactants, but I'm confused on this way.
Asked by kingjamesholdscourt23 - Sun Jan 4 00:44:04 2009 - - 1 Answers - 0 Comments
A. 2.641 g CO2 x (12.011 / 44.01) = 0.7208 g C 0.7208 g C / 12.011 g/mol = 0.0600 mol C atoms 1.442 g H2O x (2.0158 / 18.0152) = 0.1614 g H 0.1614 g H / 1.0079 g/mol = 0.1601 mol H atoms Divide both by the smallest- 0.0600 / 0.0600 = 1 C atom 0.1601 / 0.0600 = 2.67 H atom 2.67 H atom x 3 = 8 H atom 1 C atom x 3 = 3 C atom empirical formula = C3H8 (which is also the molecular formula)
Answered by skipper - Sun Jan 4 00:55:40 2009
Q. The complete combustion of a sample of propane produced 2.641 grams of carbon dioxide and 1.442 grams of water as the only products. Find the empirical formula of propane. I understand how to do this if I was calculating based on the reactants, but I'm confused on this way.
Asked by kingjamesholdscourt23 - Sun Jan 4 00:44:04 2009 - - 1 Answers - 0 Comments
A. 2.641 g CO2 x (12.011 / 44.01) = 0.7208 g C 0.7208 g C / 12.011 g/mol = 0.0600 mol C atoms 1.442 g H2O x (2.0158 / 18.0152) = 0.1614 g H 0.1614 g H / 1.0079 g/mol = 0.1601 mol H atoms Divide both by the smallest- 0.0600 / 0.0600 = 1 C atom 0.1601 / 0.0600 = 2.67 H atom 2.67 H atom x 3 = 8 H atom 1 C atom x 3 = 3 C atom empirical formula = C3H8 (which is also the molecular formula)
Answered by skipper - Sun Jan 4 00:55:40 2009
How to determine the empirical formula of the oxide?
Q. When 0.424 g of finely divided iron is butned, 0.606 g reddish brown oxide is obtained. Determine the empirical formula of the oxide. some tips to solve this kind problem on my own will get 10 points.
Asked by hetrp - Tue Sep 26 20:32:55 2006 - - 1 Answers - 1 Comments
A. OK. The 0.606 grams is Iron and Oxygen and if we subtract 0.424 from 0.606 we get the amount in grams that is oxygen, 0.182 grams. Now if we divide the 0.424 by the atomic mass of Iron to get 0.0076 and divide the 0.182 by the atomic mass of Oxygen gives us 0.011. 0.0076 and 0.011 seem like strange numbers but they are the key to finding the ratio of the atoms. Divide the smaller 0.0076 into itself and into 0.011. This gives you the values of 1 and 1.5. Ah! We are getting somewhere. If we multiple both of those numbers by 2 they are whole numbers of 2 and 3. So we can conclude that there are 2 Iron atoms for every 3 oxygen atoms and our empirical formula is Fe2O3. Make sense?
Answered by Alan Turing - Tue Sep 26 20:36:53 2006
Q. When 0.424 g of finely divided iron is butned, 0.606 g reddish brown oxide is obtained. Determine the empirical formula of the oxide. some tips to solve this kind problem on my own will get 10 points.
Asked by hetrp - Tue Sep 26 20:32:55 2006 - - 1 Answers - 1 Comments
A. OK. The 0.606 grams is Iron and Oxygen and if we subtract 0.424 from 0.606 we get the amount in grams that is oxygen, 0.182 grams. Now if we divide the 0.424 by the atomic mass of Iron to get 0.0076 and divide the 0.182 by the atomic mass of Oxygen gives us 0.011. 0.0076 and 0.011 seem like strange numbers but they are the key to finding the ratio of the atoms. Divide the smaller 0.0076 into itself and into 0.011. This gives you the values of 1 and 1.5. Ah! We are getting somewhere. If we multiple both of those numbers by 2 they are whole numbers of 2 and 3. So we can conclude that there are 2 Iron atoms for every 3 oxygen atoms and our empirical formula is Fe2O3. Make sense?
Answered by Alan Turing - Tue Sep 26 20:36:53 2006
How to find the empirical formula of this?
Q. Propane is a hydrocarbon. In a 26.80 gram sample of propane, 4.90g is hydrogen and the remainder is carbon. What is the empirical formula of propane?
Asked by myname_isalbert - Mon Mar 31 22:05:56 2008 - - 2 Answers - 0 Comments
A. You need to know the ratio of C and H by numbers of atoms (or moles) not weight. So, you could go through the work of calculating moles but it isn't nescessary, just divide the weight of each component by it's atomic weight: (26.80-4.90)/12.01 = 1.82 4.90/1.01 = 4.85 These numbers are arbitrary, but in the same ratio as the carbon-hydrogen in propane. We will divide both numbers by the smaller one to try to get them into whole numbers. 1.82/1.82 = 1.00 4.85/1.82 = 2.66 OK, we are getting close, we need to multiply both numbers by something that causes BOTH of them to be whole numbers. By multiplying by 3 you can get them to come out even: 1.00 x 3 = 3.00 2.66 x 3 = 7.98 Formula is C3H8, which matches the known formula of propane. (Not [cont.]
Answered by Flying Dragon - Mon Mar 31 22:25:10 2008
Q. Propane is a hydrocarbon. In a 26.80 gram sample of propane, 4.90g is hydrogen and the remainder is carbon. What is the empirical formula of propane?
Asked by myname_isalbert - Mon Mar 31 22:05:56 2008 - - 2 Answers - 0 Comments
A. You need to know the ratio of C and H by numbers of atoms (or moles) not weight. So, you could go through the work of calculating moles but it isn't nescessary, just divide the weight of each component by it's atomic weight: (26.80-4.90)/12.01 = 1.82 4.90/1.01 = 4.85 These numbers are arbitrary, but in the same ratio as the carbon-hydrogen in propane. We will divide both numbers by the smaller one to try to get them into whole numbers. 1.82/1.82 = 1.00 4.85/1.82 = 2.66 OK, we are getting close, we need to multiply both numbers by something that causes BOTH of them to be whole numbers. By multiplying by 3 you can get them to come out even: 1.00 x 3 = 3.00 2.66 x 3 = 7.98 Formula is C3H8, which matches the known formula of propane. (Not [cont.]
Answered by Flying Dragon - Mon Mar 31 22:25:10 2008
what is the empirical formula for nitroglycerin?
Q. I know the molecular formula is C3H6N3O9, but I don't know how to figure out the empirical formula.
Asked by frog - Tue Mar 13 04:06:59 2007 - - 3 Answers - 0 Comments
A. The empirical formula of any chemical compound is the smallest simple ratio of the relative number of each type of atom in the compound. C3H5N3O9 is both the molecular and the empirical formula for nitroglycerin. If C3H6N3O9 was the molecular formula (and it is not) then the empirical formula would be CH2NO3.
Answered by Richard - Tue Mar 13 05:03:14 2007
Q. I know the molecular formula is C3H6N3O9, but I don't know how to figure out the empirical formula.
Asked by frog - Tue Mar 13 04:06:59 2007 - - 3 Answers - 0 Comments
A. The empirical formula of any chemical compound is the smallest simple ratio of the relative number of each type of atom in the compound. C3H5N3O9 is both the molecular and the empirical formula for nitroglycerin. If C3H6N3O9 was the molecular formula (and it is not) then the empirical formula would be CH2NO3.
Answered by Richard - Tue Mar 13 05:03:14 2007
How can I put into words why the empirical formula for MgO will be the same?
Q. I made MgO by burning Mg metal and found the empirical formula. I need to explain why the empirical formula will not change even if the beginning mass of Mg is different. I understand it in my head that the ratio of Mg to O will still be the same, but I can't figure out how to explain why?
Asked by just chillin' - Sun Sep 21 21:11:29 2008 - - 1 Answers - 0 Comments
A. Mg ion has a charge of 2+ and O ion has a charge of 2- so the charges cancel out each other.
Answered by vietvan_89 - Sun Sep 21 21:29:14 2008
Q. I made MgO by burning Mg metal and found the empirical formula. I need to explain why the empirical formula will not change even if the beginning mass of Mg is different. I understand it in my head that the ratio of Mg to O will still be the same, but I can't figure out how to explain why?
Asked by just chillin' - Sun Sep 21 21:11:29 2008 - - 1 Answers - 0 Comments
A. Mg ion has a charge of 2+ and O ion has a charge of 2- so the charges cancel out each other.
Answered by vietvan_89 - Sun Sep 21 21:29:14 2008
What is the properly written empirical formula of a hydrocarbon that is 89.97% by mass carbon?
Q. What is the properly written empirical formula of a hydrocarbon that is 89.97% by mass carbon?
Asked by Love U - Fri Apr 10 14:12:25 2009 - - 2 Answers - 0 Comments
A. assume 100g 89.97 are C and 10.03 are H... moles C = 89.87 g x (1 mole /12.0g) = 7.5 moles H = 10.03 g x (1 mole / 1.0g) = 10 moles... simplify by dividing by the smallest.. 7.5 7.5 / 7.5 = 1 10 / 7.5 = 4/3 multiply both by 3 to get rid of the fraction and you get...3:4 C3H4
Answered by m w - Fri Apr 10 14:29:37 2009
Q. What is the properly written empirical formula of a hydrocarbon that is 89.97% by mass carbon?
Asked by Love U - Fri Apr 10 14:12:25 2009 - - 2 Answers - 0 Comments
A. assume 100g 89.97 are C and 10.03 are H... moles C = 89.87 g x (1 mole /12.0g) = 7.5 moles H = 10.03 g x (1 mole / 1.0g) = 10 moles... simplify by dividing by the smallest.. 7.5 7.5 / 7.5 = 1 10 / 7.5 = 4/3 multiply both by 3 to get rid of the fraction and you get...3:4 C3H4
Answered by m w - Fri Apr 10 14:29:37 2009
What is the empirical formula for 6.67 moles of carbon and 20 moles of hydrogen?
Q. There are 80grams of carbon and 20grams of hydrogen. i just need to know how to construct an empirical formula. Thanks
Asked by masjete - Tue Sep 16 01:33:59 2008 - - 3 Answers - 0 Comments
A. An empirical formula is the lowest whole number expression of the atomic ratios. Divide everything by the lowest number of moles, in this case 6.67. Carbon is therefore one mole, and hydrogen 2.99 moles, which would round up to 3. Your empirical formula is therefore CH3.
Answered by Dr. Buzz - Tue Sep 16 01:37:53 2008
Q. There are 80grams of carbon and 20grams of hydrogen. i just need to know how to construct an empirical formula. Thanks
Asked by masjete - Tue Sep 16 01:33:59 2008 - - 3 Answers - 0 Comments
A. An empirical formula is the lowest whole number expression of the atomic ratios. Divide everything by the lowest number of moles, in this case 6.67. Carbon is therefore one mole, and hydrogen 2.99 moles, which would round up to 3. Your empirical formula is therefore CH3.
Answered by Dr. Buzz - Tue Sep 16 01:37:53 2008
How wud the calculated value for the empirical formula of magnesium be affected?
Q. HOw wud your calculated value for the empirical formula of magnesium oxide have been affected if all the magnesium in the crucible had not burned?
Asked by jbcr123 - Thu Nov 6 23:53:16 2008 - - 1 Answers - 0 Comments
A. If all the magnesium was not burned, when you calculate the moles to get the empirical formula, the formula would be wrong. For example, if you have 0.20g of Mg and 0.08g of O (here I am exaggerating the amount of Mg not burned) You would try to find the formula like this: 0.20g of Mg X 1mol of Mg/24.31g of Mg = .008 moles of Mg 0.08g of O X 1 mol of O/16.00g of O = .005moles of O Since the moles of oxygen is smaller, you use that to divide. .008moles/0.005moles = 1.60 (Mg) approx. 2 .005moles/.005moles = 1 (O) So your formula would be Mg2O. But the formula is really supposed to be MgO from 2Mg + O2 --> 2MgO
Answered by Faye R - Fri Nov 7 00:19:05 2008
Q. HOw wud your calculated value for the empirical formula of magnesium oxide have been affected if all the magnesium in the crucible had not burned?
Asked by jbcr123 - Thu Nov 6 23:53:16 2008 - - 1 Answers - 0 Comments
A. If all the magnesium was not burned, when you calculate the moles to get the empirical formula, the formula would be wrong. For example, if you have 0.20g of Mg and 0.08g of O (here I am exaggerating the amount of Mg not burned) You would try to find the formula like this: 0.20g of Mg X 1mol of Mg/24.31g of Mg = .008 moles of Mg 0.08g of O X 1 mol of O/16.00g of O = .005moles of O Since the moles of oxygen is smaller, you use that to divide. .008moles/0.005moles = 1.60 (Mg) approx. 2 .005moles/.005moles = 1 (O) So your formula would be Mg2O. But the formula is really supposed to be MgO from 2Mg + O2 --> 2MgO
Answered by Faye R - Fri Nov 7 00:19:05 2008
What is the empirical formula of nicotine and what is its molecular formula?
Q. fw of nicotine is 162.2 g mol-1 The empirical formula (or simplest ratio between the elements) of nicotine which contains the elements C, H, and N, was determined by completely burning the substance and determining the amount of CO2, H2O produced. When 0.697 g of nicotine was burned, 1.841 g CO2 (fw 44.01) and 0.529 g H2O (fw 18.02) were produced.
Asked by ProgrammedDamned - Thu Aug 20 08:37:53 2009 - - 1 Answers - 0 Comments
A. Empirical is:C5H7N1; Molecular would be C10H14N2
Answered by anonymous - Thu Aug 20 08:53:25 2009
Q. fw of nicotine is 162.2 g mol-1 The empirical formula (or simplest ratio between the elements) of nicotine which contains the elements C, H, and N, was determined by completely burning the substance and determining the amount of CO2, H2O produced. When 0.697 g of nicotine was burned, 1.841 g CO2 (fw 44.01) and 0.529 g H2O (fw 18.02) were produced.
Asked by ProgrammedDamned - Thu Aug 20 08:37:53 2009 - - 1 Answers - 0 Comments
A. Empirical is:C5H7N1; Molecular would be C10H14N2
Answered by anonymous - Thu Aug 20 08:53:25 2009
what steps are used to convert an empirical formula into a molecular formula?
Q. Demonstrate, if possible, by converting the empirical formula NO2 into the molecular formula. The molar mass of this molecule is 92 g/mol.
Asked by phantom4muah - Thu May 28 16:09:50 2009 - - 1 Answers - 0 Comments
A. You have the molar mass of the compound = 92g/mol What is the molar mass of the empirical formula? NO2 = 14+32 = 46g/mol Number of time empirical formula is repeated to get molecular formula: Molar mass compound / molar mass empirical formula = 92/46 = 2 Formula of compound = (NO2)2 = N2O4 answer
Answered by Trevor H - Thu May 28 16:30:48 2009
Q. Demonstrate, if possible, by converting the empirical formula NO2 into the molecular formula. The molar mass of this molecule is 92 g/mol.
Asked by phantom4muah - Thu May 28 16:09:50 2009 - - 1 Answers - 0 Comments
A. You have the molar mass of the compound = 92g/mol What is the molar mass of the empirical formula? NO2 = 14+32 = 46g/mol Number of time empirical formula is repeated to get molecular formula: Molar mass compound / molar mass empirical formula = 92/46 = 2 Formula of compound = (NO2)2 = N2O4 answer
Answered by Trevor H - Thu May 28 16:30:48 2009
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There is simply no compelling reason why such protectionism is needed unless one wants to simply ignore all empirical evidence. ...
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AntonioLao
ue, 13 Oct 2009 16:20:03 GM
These are generated from the . empirical formula. 3646n /(n -2 ). In this case 3646 is simply a multiplicative factor derived from the experimental apparatus. Therefore, the key factor is really n /(n -2 ) and for n=3, 4, 5, 6, ...
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