Molecular formula for caffeine?
Q. I know that the molecular formula will turn out to be C8H10N4O2 because I've looked it up. But how would you determine the moelcular formula knowing the empirical fomula (C4H5N2O) and that the molar mass of caffeine is 195 g/mol?
Asked by osnap.itstom - Mon Apr 7 00:57:57 2008 - - 2 Answers - 0 Comments
A. Divide the molar mass of coffiene by the molar mass of the empirical formula and multiply the subscripts of the empirical formula by that numbber
Answered by Sly_Salmon - Mon Apr 7 01:01:40 2008
Q. I know that the molecular formula will turn out to be C8H10N4O2 because I've looked it up. But how would you determine the moelcular formula knowing the empirical fomula (C4H5N2O) and that the molar mass of caffeine is 195 g/mol?
Asked by osnap.itstom - Mon Apr 7 00:57:57 2008 - - 2 Answers - 0 Comments
A. Divide the molar mass of coffiene by the molar mass of the empirical formula and multiply the subscripts of the empirical formula by that numbber
Answered by Sly_Salmon - Mon Apr 7 01:01:40 2008
What is the molecular weight, density, boiling point, structure, and formula for Caffeine Salicylate?
Q. What is the molecular weight, density, boiling point, structure, and formula for Caffeine Salicylate?
Asked by Eileen B - Thu Oct 5 10:28:58 2006 - - 1 Answers - 0 Comments
A. It's actually caffeine sodium salicylate. It's the salt that forms when you combine caffeine with sodium salicylate. C8 H10 N4 O2 . C7 H6 O3 . Na (you can figure out the molecular weight from this). You can't post structures on here... Google the registry number (8002-85-5). You should get what you need then...
Answered by The ~Muffin~ Man - Thu Oct 5 12:58:31 2006
Q. What is the molecular weight, density, boiling point, structure, and formula for Caffeine Salicylate?
Asked by Eileen B - Thu Oct 5 10:28:58 2006 - - 1 Answers - 0 Comments
A. It's actually caffeine sodium salicylate. It's the salt that forms when you combine caffeine with sodium salicylate. C8 H10 N4 O2 . C7 H6 O3 . Na (you can figure out the molecular weight from this). You can't post structures on here... Google the registry number (8002-85-5). You should get what you need then...
Answered by The ~Muffin~ Man - Thu Oct 5 12:58:31 2006
The molar mass of caffeine is 194.1g. Is the molecular formula C4H5N2O or C8H10N4O2?
Q. The molar mass of caffeine is 194.1g. Is the molecular formula C4H5N2O or C8H10N4O2?
Asked by Keepsby - Thu Feb 19 18:35:44 2009 - - 2 Answers - 0 Comments
A. C4H5N2O = (4x12.0) + (5x1.0) + (2x14.0) + (2x16.0) = 113 C8H10N4O2 = (8x12.0) + (10x1.0) + (4x14.0) + (2x16.0) = 194 Second option is correct.
Answered by RobA - Thu Feb 19 18:47:27 2009
Q. The molar mass of caffeine is 194.1g. Is the molecular formula C4H5N2O or C8H10N4O2?
Asked by Keepsby - Thu Feb 19 18:35:44 2009 - - 2 Answers - 0 Comments
A. C4H5N2O = (4x12.0) + (5x1.0) + (2x14.0) + (2x16.0) = 113 C8H10N4O2 = (8x12.0) + (10x1.0) + (4x14.0) + (2x16.0) = 194 Second option is correct.
Answered by RobA - Thu Feb 19 18:47:27 2009
The empirical and molecular formula...?
Q. Determine the empirical and molecular formula a)caffeine, a stimulant found in coffee, contains 49.5 percent C, 5.15 percent H, 28.9 percent N, and 16.5 percent O by mass; molar mass about 195g b)Modosodium glotamate, a flavor enhancer in certain foods contains 35.51 percent C, 4.77 percent H, 37.85 percent O, 8.29 percent N and 13.60 percent Na and has a molar mass of 169g Thank you for your time 10 points
Asked by pianoman3434 - Sat Aug 16 20:16:28 2008 - - 1 Answers - 0 Comments
A. a) Assume that we have 100 g of caffeine. Based on the percentages given, we have 49.5 g C, 5.15 g H, 28.9 g N, and 16.5 g O. We need to calculate moles of each. C: 49.5 g / 12.0 g/mole = 4.13 moles C H: 5.15 g / 1.01 g/mole = 5.10 moles H N: 28.9 g / 14.0 g/mole = 2.06 moles N O: 16.5 g / 16.0 g/mole = 1.03 moles O Dividing by the smallest (1.03) we get C - 4 H - 5 N - 2 O - 1 The empirical formula is C4H5N2O. Its formula weight is (4 x 12) + (5 x 1) + (2 x 14) + (1 x 16) = 97. If the molar mass is 195, that's equal to 2 empirical formula units (2 x 97 = 194). So the molecular formula is 2 x C4H5N2O = C8H10N4O2. b) Work this exactly like the one above. You should get a molecular formula of C5H8O4NNa.
Answered by HPV - Sat Aug 16 20:35:17 2008
Q. Determine the empirical and molecular formula a)caffeine, a stimulant found in coffee, contains 49.5 percent C, 5.15 percent H, 28.9 percent N, and 16.5 percent O by mass; molar mass about 195g b)Modosodium glotamate, a flavor enhancer in certain foods contains 35.51 percent C, 4.77 percent H, 37.85 percent O, 8.29 percent N and 13.60 percent Na and has a molar mass of 169g Thank you for your time 10 points
Asked by pianoman3434 - Sat Aug 16 20:16:28 2008 - - 1 Answers - 0 Comments
A. a) Assume that we have 100 g of caffeine. Based on the percentages given, we have 49.5 g C, 5.15 g H, 28.9 g N, and 16.5 g O. We need to calculate moles of each. C: 49.5 g / 12.0 g/mole = 4.13 moles C H: 5.15 g / 1.01 g/mole = 5.10 moles H N: 28.9 g / 14.0 g/mole = 2.06 moles N O: 16.5 g / 16.0 g/mole = 1.03 moles O Dividing by the smallest (1.03) we get C - 4 H - 5 N - 2 O - 1 The empirical formula is C4H5N2O. Its formula weight is (4 x 12) + (5 x 1) + (2 x 14) + (1 x 16) = 97. If the molar mass is 195, that's equal to 2 empirical formula units (2 x 97 = 194). So the molecular formula is 2 x C4H5N2O = C8H10N4O2. b) Work this exactly like the one above. You should get a molecular formula of C5H8O4NNa.
Answered by HPV - Sat Aug 16 20:35:17 2008
please help me..The stimulant in coffee and tea is caffeine, a substance of molar mass 194g/mol.?
Q. when 0.376g of caffeine was burned, 0.682g carbon dioxide, 0.174g of water and 0.110g of nitrogen were formed. determine teh empirical and molecular formulas of caffeine.. im realy confuse.
Asked by Rubieanne - Sun Apr 22 02:23:14 2007 - - 1 Answers - 0 Comments
A. Let Caffeine be CxHyOzNs where x,y,z and s have to be determined. Moles of Caffeine = 0.376/194 Moles of N2 = 0.11/28 Thus 0.376*s/(194*2) = 0.11/28 So, s = 4 Now, Moles of CO2 released = 0.376x/194 = 0.682/44 So, x = 8 Moles of H2O = 0.376*y/388 = 0.174/18 So, y = 10 Mass of O = 194-(4*14+8*12+10) = 32 g So, z = 2 Molecular formula is C8H10O2N4 and empirical formula is C4H5ON2
Answered by ag_iitkgp - Sun Apr 22 02:50:18 2007
Q. when 0.376g of caffeine was burned, 0.682g carbon dioxide, 0.174g of water and 0.110g of nitrogen were formed. determine teh empirical and molecular formulas of caffeine.. im realy confuse.
Asked by Rubieanne - Sun Apr 22 02:23:14 2007 - - 1 Answers - 0 Comments
A. Let Caffeine be CxHyOzNs where x,y,z and s have to be determined. Moles of Caffeine = 0.376/194 Moles of N2 = 0.11/28 Thus 0.376*s/(194*2) = 0.11/28 So, s = 4 Now, Moles of CO2 released = 0.376x/194 = 0.682/44 So, x = 8 Moles of H2O = 0.376*y/388 = 0.174/18 So, y = 10 Mass of O = 194-(4*14+8*12+10) = 32 g So, z = 2 Molecular formula is C8H10O2N4 and empirical formula is C4H5ON2
Answered by ag_iitkgp - Sun Apr 22 02:50:18 2007
CH4 is methane, there is no name but that it is a methyl group with an unpaired electron.?
Q. CH3- is a partial moleculewhich is usually part of an aromatic hydrocarbon or also found in the molecular formula for chocolate and caffeine.
Asked by Scott - Tue Jul 7 13:56:16 2009 - - 1 Answers - 0 Comments
A. It is not quite clear what your question is. Here are some suggestions. Methane, CH4, is a compound that is common in natural gas. Methyl radical can refer to .CH3, a short-lived free radical that is thought to occur as an unstable intermediate structure in reaction mechanisms such as the free-radical chlorination of methane. Methyl radical is also used in an older nomenclature to describe a CH3 bonded as a branch to some chain or ring such as you mention. Nowadays, this is more commonly referred to as a methyl group to distinguish it from .CH3.
Answered by Buck - Tue Jul 7 14:14:18 2009
Q. CH3- is a partial moleculewhich is usually part of an aromatic hydrocarbon or also found in the molecular formula for chocolate and caffeine.
Asked by Scott - Tue Jul 7 13:56:16 2009 - - 1 Answers - 0 Comments
A. It is not quite clear what your question is. Here are some suggestions. Methane, CH4, is a compound that is common in natural gas. Methyl radical can refer to .CH3, a short-lived free radical that is thought to occur as an unstable intermediate structure in reaction mechanisms such as the free-radical chlorination of methane. Methyl radical is also used in an older nomenclature to describe a CH3 bonded as a branch to some chain or ring such as you mention. Nowadays, this is more commonly referred to as a methyl group to distinguish it from .CH3.
Answered by Buck - Tue Jul 7 14:14:18 2009
Can anyone help me with these Chemistry questions?
Q. ...PLEASE!!! I have been working on them for 2 hours now, and I just don't understand them! What is the empirical formula of C6H12? A compound is analyzed and is found to contain 5.657 g C; 0.3165 g H; and 5.566 g Cl. What is the empirical formula? A 2.573 g oxide of iron conatins 2.000 g iron. What is its empirical formula? Caffeine is a compound containing carbon, hydrogen, nitrogen and oxygen. The percent composition of carbon is 49.47% C, 5.191% H, 28.86% N, and 16.48% oxygen. The molar mass of caffeine is approximately 194 g. What is the molecular formula of caffeine? A compound is 25.94% nitrogen and 74.06% oxygen. What is the empirical formula of the compound?
Asked by ~*PrInCeSs*~ - Sun Feb 24 12:54:02 2008 - - 1 Answers - 0 Comments
A. For question 2 it is just the least amount of C and H which can maintain the right stoichiometric equivalents for the molecule or CH2 For question 2 what you want to do first is find out how many moles of each element there are C= 5.657g /(1 mole C/12.0107g)= 0.471 moles C H= 0.3165g/(1 mole H/1.00794g)= 0.316 moles H Cl= 5.566g/(1 mole Cl/35.453g)= 0.157 moles Cl Now which one is limiting? Cl is, divide each one by # moles of Cl So C is 0.471 moles/0.157 =3 H is 0.316/0.157=2 Cl is 0.157/0.157=1 So your empirical formula is C3H2Cl Now for question 3 We know 2.000 grams is iron so 0.573 must be Oxygen. Know that we know the grams repeat the steps above. For question 4 Multiply each % by the molecular weight (194g) … [cont.]
Answered by Zachary P - Sun Feb 24 13:21:17 2008
Q. ...PLEASE!!! I have been working on them for 2 hours now, and I just don't understand them! What is the empirical formula of C6H12? A compound is analyzed and is found to contain 5.657 g C; 0.3165 g H; and 5.566 g Cl. What is the empirical formula? A 2.573 g oxide of iron conatins 2.000 g iron. What is its empirical formula? Caffeine is a compound containing carbon, hydrogen, nitrogen and oxygen. The percent composition of carbon is 49.47% C, 5.191% H, 28.86% N, and 16.48% oxygen. The molar mass of caffeine is approximately 194 g. What is the molecular formula of caffeine? A compound is 25.94% nitrogen and 74.06% oxygen. What is the empirical formula of the compound?
Asked by ~*PrInCeSs*~ - Sun Feb 24 12:54:02 2008 - - 1 Answers - 0 Comments
A. For question 2 it is just the least amount of C and H which can maintain the right stoichiometric equivalents for the molecule or CH2 For question 2 what you want to do first is find out how many moles of each element there are C= 5.657g /(1 mole C/12.0107g)= 0.471 moles C H= 0.3165g/(1 mole H/1.00794g)= 0.316 moles H Cl= 5.566g/(1 mole Cl/35.453g)= 0.157 moles Cl Now which one is limiting? Cl is, divide each one by # moles of Cl So C is 0.471 moles/0.157 =3 H is 0.316/0.157=2 Cl is 0.157/0.157=1 So your empirical formula is C3H2Cl Now for question 3 We know 2.000 grams is iron so 0.573 must be Oxygen. Know that we know the grams repeat the steps above. For question 4 Multiply each % by the molecular weight (194g) … [cont.]
Answered by Zachary P - Sun Feb 24 13:21:17 2008
Determine the empirical and molecular formulas of each of the following substances.?
Q. Determine the empirical and molecular formulas of each of the following substances. A) Caffeine, a stimulant found in coffee, contains 49.5% C, 5.15% H, 28.9% N, and 16.5% O by mass and has a molar mass of 195g/mole. B) Monosodium glutamate (MSG), a flavor enhancer in certain foods, contains 35.51% C, 4.77% H, 37.85% O, 8.29% N, and 13.60% Na, and has a molar mass of 169g/mole . Please work it out so I will be able to know what to do! Thanks.
Asked by Big Dan - Tue Sep 22 14:55:51 2009 - - 2 Answers - 0 Comments
A. in 100 g of compound we have Carbon49.5g or4.125moles we also have 5.15g of hydrogen or 5.15moles of hydrogen there is 28.9g of nitrogen or 2.064285714moles of nitrogen for oxygen we have 16.5g or 1.03125moles of Oxygen in summary we have C4.125 H5.15 N2.064285714 O1.03125 dividing by the smallest we get C4 H4.99 N2.00 O1.00 the ratio is C:H:N:O=4:5:2:1 so empirical formula is C4H5N2O Empirical FW = 97 molecular weight is 195 or 2 * empirical FW molecular formula is C8H10N4O2 ~~~ in 100 g of compound we have 35.51gof carbon or 2.95916667molesof carbon we also have 4.77g of hydrogen or 4.77moles of hydrogenthe is 37.85g of oxygen or 2.365625moles of oxygento finish there is8.29g of Nitrogen or 0.592142857moles and 13.6g of… [cont.]
Answered by unknown - Tue Sep 22 15:26:27 2009
Q. Determine the empirical and molecular formulas of each of the following substances. A) Caffeine, a stimulant found in coffee, contains 49.5% C, 5.15% H, 28.9% N, and 16.5% O by mass and has a molar mass of 195g/mole. B) Monosodium glutamate (MSG), a flavor enhancer in certain foods, contains 35.51% C, 4.77% H, 37.85% O, 8.29% N, and 13.60% Na, and has a molar mass of 169g/mole . Please work it out so I will be able to know what to do! Thanks.
Asked by Big Dan - Tue Sep 22 14:55:51 2009 - - 2 Answers - 0 Comments
A. in 100 g of compound we have Carbon49.5g or4.125moles we also have 5.15g of hydrogen or 5.15moles of hydrogen there is 28.9g of nitrogen or 2.064285714moles of nitrogen for oxygen we have 16.5g or 1.03125moles of Oxygen in summary we have C4.125 H5.15 N2.064285714 O1.03125 dividing by the smallest we get C4 H4.99 N2.00 O1.00 the ratio is C:H:N:O=4:5:2:1 so empirical formula is C4H5N2O Empirical FW = 97 molecular weight is 195 or 2 * empirical FW molecular formula is C8H10N4O2 ~~~ in 100 g of compound we have 35.51gof carbon or 2.95916667molesof carbon we also have 4.77g of hydrogen or 4.77moles of hydrogenthe is 37.85g of oxygen or 2.365625moles of oxygento finish there is8.29g of Nitrogen or 0.592142857moles and 13.6g of… [cont.]
Answered by unknown - Tue Sep 22 15:26:27 2009
I need help with my chemistry please.?
Q. Here is my problem I need help with. Caffeine, a stimulant found in coffee and soda pop, has the following mass percent composition: C, 49.48%; H, 5.19%; N, 28.85%; O, 16.48%. The molar mass of the caffeine is 194.19g/mol. Find the molecular formula of caffeine.
Asked by mjhappy17 - Sun Nov 1 13:22:29 2009 - - 6 Answers - 0 Comments
A. First establish that your going to measure using 1 mole So now you have 194.19g caffeine Carbon is 49.48% of the mass so you have (49.48%)*(194.19)=96.0852 g carbon to see how many carbons you have you do (mass/Molar mass) (96.0852/12.0107) = 10 So you have 10 carbons Do it for each one: Hydrogen: (5.19%)*(194.19) = 10.0785 g (mass/molar mass) (10.0785/1.00794) = 10 Nitrogen: (28.85%)(194.19) = 56.0238 g (56.0238/14.00674) = 4 Oxygen: (16.48%)(194.19) = 32.0025g (32.0025/15.9994) = 2 Final formula: C8 H10 N4 O2
Answered by sambkup - Sun Nov 1 13:44:10 2009
Q. Here is my problem I need help with. Caffeine, a stimulant found in coffee and soda pop, has the following mass percent composition: C, 49.48%; H, 5.19%; N, 28.85%; O, 16.48%. The molar mass of the caffeine is 194.19g/mol. Find the molecular formula of caffeine.
Asked by mjhappy17 - Sun Nov 1 13:22:29 2009 - - 6 Answers - 0 Comments
A. First establish that your going to measure using 1 mole So now you have 194.19g caffeine Carbon is 49.48% of the mass so you have (49.48%)*(194.19)=96.0852 g carbon to see how many carbons you have you do (mass/Molar mass) (96.0852/12.0107) = 10 So you have 10 carbons Do it for each one: Hydrogen: (5.19%)*(194.19) = 10.0785 g (mass/molar mass) (10.0785/1.00794) = 10 Nitrogen: (28.85%)(194.19) = 56.0238 g (56.0238/14.00674) = 4 Oxygen: (16.48%)(194.19) = 32.0025g (32.0025/15.9994) = 2 Final formula: C8 H10 N4 O2
Answered by sambkup - Sun Nov 1 13:44:10 2009
Help with Stochiometry question please!?
Q. Caffeine: C8H10N4O2 Consider the molecular formula for caffeine above. Using only whole numbers with no decimal point, 4 mol of caffeine has ___ mol of carbon, ___ mol of nitrogen, ___mol of hydrogen, and ___ mol of oxygen. how would you do this? help!
Asked by Jill - Wed Jan 14 21:20:41 2009 - - 2 Answers - 0 Comments
A. The formula gives you the numbers of each element for one mol of caffeine, so just multiply each of those numbers by 4. 4 x 8 = 32 mol carbon 4 x 4 = 16 mol nitrogen 10 x 4 = 40 mol hydrogen 2 x 4 = 8 mol oxygen
Answered by margarita - Wed Jan 14 21:25:24 2009
Q. Caffeine: C8H10N4O2 Consider the molecular formula for caffeine above. Using only whole numbers with no decimal point, 4 mol of caffeine has ___ mol of carbon, ___ mol of nitrogen, ___mol of hydrogen, and ___ mol of oxygen. how would you do this? help!
Asked by Jill - Wed Jan 14 21:20:41 2009 - - 2 Answers - 0 Comments
A. The formula gives you the numbers of each element for one mol of caffeine, so just multiply each of those numbers by 4. 4 x 8 = 32 mol carbon 4 x 4 = 16 mol nitrogen 10 x 4 = 40 mol hydrogen 2 x 4 = 8 mol oxygen
Answered by margarita - Wed Jan 14 21:25:24 2009
Chemistry Question, Please Help!?
Q. Caffeine, a stimulant found in coffee and tea, contains 49.5% carbon, 5.15% hydrogen, 28.9% nitrogen and 16.5% oxygen by mass. What is the empirical formula of caffeine? If its molar mass is about 195g, what is its molecular formula?
Asked by Susan - Tue Sep 22 17:45:19 2009 - - 1 Answers - 0 Comments
A. Moles C = 49.5 / 12.011 = 4.12 moles H = 5.15 / 1.008= 5.11 moles N = 28.9 / 14.0067 =2.06 Moles O = 16.5 / 15.9994 = 1.03 divide by the smallest number C 4 H 5 N2 O ( empirical formula = MM = 97 g/mol) 195 / 97 = 2 multiply by 2 C8 H 10 N4 O2 is the molecular formula
Answered by Andrea - Tue Sep 22 21:50:45 2009
Q. Caffeine, a stimulant found in coffee and tea, contains 49.5% carbon, 5.15% hydrogen, 28.9% nitrogen and 16.5% oxygen by mass. What is the empirical formula of caffeine? If its molar mass is about 195g, what is its molecular formula?
Asked by Susan - Tue Sep 22 17:45:19 2009 - - 1 Answers - 0 Comments
A. Moles C = 49.5 / 12.011 = 4.12 moles H = 5.15 / 1.008= 5.11 moles N = 28.9 / 14.0067 =2.06 Moles O = 16.5 / 15.9994 = 1.03 divide by the smallest number C 4 H 5 N2 O ( empirical formula = MM = 97 g/mol) 195 / 97 = 2 multiply by 2 C8 H 10 N4 O2 is the molecular formula
Answered by Andrea - Tue Sep 22 21:50:45 2009
pleaseee help..this has to b turned in soon..take a look!!?
Q. do what makes sense to u..thank u An element consists of 1.40% of an isotope with mass 203.973 amu, 24.10% of an isotope with mass 205.9745 amu, 22.10% of an isotope with mass 206.9759 amu, and 52.40% of an isotope with mass 207.9766 amu. Calculate the average atomic mass. 1 amu Identify the element (Give the symbol). 2 2. 0/2 points All Submissions Notes Question: ZumChem8 3.E.042. Question part Points Submissions 1 2 0/1 0/1 5/50 0/50 Total 0/2 What number of Fe atoms and what amount (moles) of Fe atoms are in 450.0 g of iron? 1 atoms 2 moles 3. /3 points Notes Question: ZumChem8 3.E.047. Question part Points Submissions 1 2 3 0/1 0/1 0/1 0/50 0/50 0/50 Total 0/3 Calculate the… [cont.]
Asked by ThatGuy88 - Thu Sep 24 02:51:04 2009 - - 1 Answers - 1 Comments
A. If you don't even try it yourself how can you hope to learn something? I will gladly explain things and help you out if you try to solve the questions yourself Since you explained yourself in your message I will give you a hand. Question 1 calculate the average atomic mass (1.40 * 203.973g/mol + 24.10 * 205.9745 g/mol + 22.10 * 206.9759 g/mol + 52.40 * 207.9766 g/mol)/100 =207.216889 g/mol Identify the element (Give the symbol). Atomic mass of 207.2 points us to lead and the symbol is Pb (I know this by looking it up in the table of mendeljev) Question 2 The amount of atoms in 450.0 g Iron The molar mass of iron is 55.847 g/mol so 450.0g Fe / 55.847g/mol =8.05772915 moles Fe now if you multiply moles with the number of Avogadro you… [cont.]
Answered by Ruben J - Thu Sep 24 04:51:38 2009
Q. do what makes sense to u..thank u An element consists of 1.40% of an isotope with mass 203.973 amu, 24.10% of an isotope with mass 205.9745 amu, 22.10% of an isotope with mass 206.9759 amu, and 52.40% of an isotope with mass 207.9766 amu. Calculate the average atomic mass. 1 amu Identify the element (Give the symbol). 2 2. 0/2 points All Submissions Notes Question: ZumChem8 3.E.042. Question part Points Submissions 1 2 0/1 0/1 5/50 0/50 Total 0/2 What number of Fe atoms and what amount (moles) of Fe atoms are in 450.0 g of iron? 1 atoms 2 moles 3. /3 points Notes Question: ZumChem8 3.E.047. Question part Points Submissions 1 2 3 0/1 0/1 0/1 0/50 0/50 0/50 Total 0/3 Calculate the… [cont.]
Asked by ThatGuy88 - Thu Sep 24 02:51:04 2009 - - 1 Answers - 1 Comments
A. If you don't even try it yourself how can you hope to learn something? I will gladly explain things and help you out if you try to solve the questions yourself Since you explained yourself in your message I will give you a hand. Question 1 calculate the average atomic mass (1.40 * 203.973g/mol + 24.10 * 205.9745 g/mol + 22.10 * 206.9759 g/mol + 52.40 * 207.9766 g/mol)/100 =207.216889 g/mol Identify the element (Give the symbol). Atomic mass of 207.2 points us to lead and the symbol is Pb (I know this by looking it up in the table of mendeljev) Question 2 The amount of atoms in 450.0 g Iron The molar mass of iron is 55.847 g/mol so 450.0g Fe / 55.847g/mol =8.05772915 moles Fe now if you multiply moles with the number of Avogadro you… [cont.]
Answered by Ruben J - Thu Sep 24 04:51:38 2009
Chemistry help..please?
Q. Current Score: 0/33 Due: Friday, September 25, 2009 05:00 pm cdt view Last Response View Saved Work About this Assignment An element consists of 1.40% of an isotope with mass 203.973 amu, 24.10% of an isotope with mass 205.9745 amu, 22.10% of an isotope with mass 206.9759 amu, and 52.40% of an isotope with mass 207.9766 amu. Calculate the average atomic mass. 1 amu Identify the element (Give the symbol). 2 2. 0/2 points All Submissions Notes Question: ZumChem8 3.E.042. Question part Points Submissions 1 2 0/1 0/1 5/50 0/50 Total 0/2 What number of Fe atoms and what amount (moles) of Fe atoms are in 450.0 g of iron? 1 atoms 2 moles 3. /3 points Notes Question: ZumChem8 3.E.047. Question part… [cont.]
Asked by ThatGuy88 - Thu Sep 24 02:11:13 2009 - - 1 Answers - 0 Comments
A. I wish i had assignments like this, so easy...
Answered by AJ - Thu Sep 24 02:49:49 2009
Q. Current Score: 0/33 Due: Friday, September 25, 2009 05:00 pm cdt view Last Response View Saved Work About this Assignment An element consists of 1.40% of an isotope with mass 203.973 amu, 24.10% of an isotope with mass 205.9745 amu, 22.10% of an isotope with mass 206.9759 amu, and 52.40% of an isotope with mass 207.9766 amu. Calculate the average atomic mass. 1 amu Identify the element (Give the symbol). 2 2. 0/2 points All Submissions Notes Question: ZumChem8 3.E.042. Question part Points Submissions 1 2 0/1 0/1 5/50 0/50 Total 0/2 What number of Fe atoms and what amount (moles) of Fe atoms are in 450.0 g of iron? 1 atoms 2 moles 3. /3 points Notes Question: ZumChem8 3.E.047. Question part… [cont.]
Asked by ThatGuy88 - Thu Sep 24 02:11:13 2009 - - 1 Answers - 0 Comments
A. I wish i had assignments like this, so easy...
Answered by AJ - Thu Sep 24 02:49:49 2009
From Yahoo Answer Search: 'Molecular formula of caffeine'
Sat Nov 21 15:06:47 2009 [ refresh local cache ]
