How to find empirical formula given percentages of Carbon, Hydrogen and Oxygen? I need to identify it.?
Q. I have a liquid unknown sample with Carbon= 70.5% Hydrogen= 13.8% Oxygen= 15.7%
Asked by Jay J - Sat Sep 26 23:23:20 2009 - - 2 Answers - 0 Comments
A. assume mass of compound 100g mass of carbon = 70.5g moles of carbon = 5.875 mass of = H13.8g moles of = H13.8 mass of O =15.7g moles of O= 0.98125 in summary moles of C :5.875 moles of H :13.8 moles of O :0.98125 dividing by the smallest we get C :5.987261146 H :14.06369427 O :1 empirical formula is C6H14O possibly hexanol
Answered by unknown - Sun Sep 27 00:14:30 2009
Q. I have a liquid unknown sample with Carbon= 70.5% Hydrogen= 13.8% Oxygen= 15.7%
Asked by Jay J - Sat Sep 26 23:23:20 2009 - - 2 Answers - 0 Comments
A. assume mass of compound 100g mass of carbon = 70.5g moles of carbon = 5.875 mass of = H13.8g moles of = H13.8 mass of O =15.7g moles of O= 0.98125 in summary moles of C :5.875 moles of H :13.8 moles of O :0.98125 dividing by the smallest we get C :5.987261146 H :14.06369427 O :1 empirical formula is C6H14O possibly hexanol
Answered by unknown - Sun Sep 27 00:14:30 2009
A gas containing only carbon and hydrogen has an empirical formula of CH2. The gas has a density of 1.65g/L at?
Q. 27 degrees C and 734torr. Determine the molar mass and molecular formula of the gas.
Asked by James - Tue Apr 21 17:45:10 2009 - - 1 Answers - 0 Comments
A. PV = nRT n = mass / mw substitute... PV = (mass / mw) RT rearrange... mw = (mass / V) x RT / P and since density = mass / V mw = density x RT/P and since T = 27C = 300K P = 734 torr x (1 atm / 760 torr) = 0.966 atm mw = (1.65 g/L) x (0.0821 Latm/moleK) x (300K) / (0.966 atm) mw = 42 g/mole finally, the unit mass of CH2 = 14 g / units 42 g / mole x (1 unit / 14 g) = 3 units / mole of gas... ie, the molecular formula = 3 x CH2 = C3H6 *** molar mass = 42 g/mole molecular formula = C3H6
Answered by m w - Tue Apr 21 17:57:28 2009
Q. 27 degrees C and 734torr. Determine the molar mass and molecular formula of the gas.
Asked by James - Tue Apr 21 17:45:10 2009 - - 1 Answers - 0 Comments
A. PV = nRT n = mass / mw substitute... PV = (mass / mw) RT rearrange... mw = (mass / V) x RT / P and since density = mass / V mw = density x RT/P and since T = 27C = 300K P = 734 torr x (1 atm / 760 torr) = 0.966 atm mw = (1.65 g/L) x (0.0821 Latm/moleK) x (300K) / (0.966 atm) mw = 42 g/mole finally, the unit mass of CH2 = 14 g / units 42 g / mole x (1 unit / 14 g) = 3 units / mole of gas... ie, the molecular formula = 3 x CH2 = C3H6 *** molar mass = 42 g/mole molecular formula = C3H6
Answered by m w - Tue Apr 21 17:57:28 2009
What is the empirical formula of a compound that has three times as many hydrogen atoms as carbon atoms....?
Q. question continued: What is the empirical formula of a compound that has three times as many hydrogen atoms as carbon atoms, but only half as many oxygen atoms as carbon atoms? I got C2H6O. Is this right? Thanks all!
Asked by Julie - Fri Sep 25 01:21:16 2009 - - 1 Answers - 0 Comments
A. Yes it looks right !!!
Answered by unknown - Fri Sep 25 02:09:32 2009
Q. question continued: What is the empirical formula of a compound that has three times as many hydrogen atoms as carbon atoms, but only half as many oxygen atoms as carbon atoms? I got C2H6O. Is this right? Thanks all!
Asked by Julie - Fri Sep 25 01:21:16 2009 - - 1 Answers - 0 Comments
A. Yes it looks right !!!
Answered by unknown - Fri Sep 25 02:09:32 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 or 80%Carbon and 20%Hydrogen by mass?
Q. What is the empirical formula or 80%Carbon and 20%Hydrogen by mass?
Asked by flacormrz - Mon Jun 11 16:21:01 2007 - - 4 Answers - 0 Comments
A. Assume you have 100g of material 80g = carbon, 20 g = hydrogen moles carbon = 80g * mole/12g = 6.667 moles moles hydrogen = 20 g * mole/1g = 20 moles divide both by 6.667 6.667/6.667 = 1 C 20/6.667 = 3 H The empirical formula is CH3
Answered by Dr Dave P - Mon Jun 11 16:46:42 2007
Q. What is the empirical formula or 80%Carbon and 20%Hydrogen by mass?
Asked by flacormrz - Mon Jun 11 16:21:01 2007 - - 4 Answers - 0 Comments
A. Assume you have 100g of material 80g = carbon, 20 g = hydrogen moles carbon = 80g * mole/12g = 6.667 moles moles hydrogen = 20 g * mole/1g = 20 moles divide both by 6.667 6.667/6.667 = 1 C 20/6.667 = 3 H The empirical formula is CH3
Answered by Dr Dave P - Mon Jun 11 16:46:42 2007
Methyl salicylate is 63.1% carbon, 5.31% hydrogen, and 31.6% Oxygen. What's the empirical formula?
Q. Methyl salicylate, or oil of wintergreen, is produced by the wintergreen plant. It can also be prepared easily in a laboratory. Methyl salicylate is 63.1% carbon, 5.31% hydrogen, and 31.6% Oxygen. Calculate the empirical formula of methyl salicylate. i got stuck on this Qs. ca anyone do it, and explain it to me, please?
Asked by Zayka - Tue Nov 17 19:57:42 2009 - - 1 Answers - 0 Comments
A. Pick a mass (I ususally use 100g) and work out how many moles of each atom would be in this mass moles = mass / molar mass Carbon = 63.1 %, so mass in 100 g = 63.1 g moles C = 63.1 g / 12.01 g/mol = 5.25 moles Hydrogen = 5.31 %, so in 100 g there are 5.31 g moles H = 5.31 g / 1.008 g/mol = 5.27 moles Oxygen = 31.6 %, so mass in 100 g = 31.6 g moles O = 31.6 g / 16.00 g/mol = 1.975 moles Now you know moles of each arrange in a ratio moles C : moles H : moles O = 5.25 : 5.27 : 1.975 To get it to the simplest whole number ratio start by Dividing each number by the smallest number 5.25/ 1.975 : 5.27/1.975 : 1.975/1.975 = 2.66 : 2.66 : 1 These are not whole number yet. Multipy the whole thing by the smallest number required to get them… [cont.]
Answered by Lexi R - Tue Nov 17 22:42:51 2009
Q. Methyl salicylate, or oil of wintergreen, is produced by the wintergreen plant. It can also be prepared easily in a laboratory. Methyl salicylate is 63.1% carbon, 5.31% hydrogen, and 31.6% Oxygen. Calculate the empirical formula of methyl salicylate. i got stuck on this Qs. ca anyone do it, and explain it to me, please?
Asked by Zayka - Tue Nov 17 19:57:42 2009 - - 1 Answers - 0 Comments
A. Pick a mass (I ususally use 100g) and work out how many moles of each atom would be in this mass moles = mass / molar mass Carbon = 63.1 %, so mass in 100 g = 63.1 g moles C = 63.1 g / 12.01 g/mol = 5.25 moles Hydrogen = 5.31 %, so in 100 g there are 5.31 g moles H = 5.31 g / 1.008 g/mol = 5.27 moles Oxygen = 31.6 %, so mass in 100 g = 31.6 g moles O = 31.6 g / 16.00 g/mol = 1.975 moles Now you know moles of each arrange in a ratio moles C : moles H : moles O = 5.25 : 5.27 : 1.975 To get it to the simplest whole number ratio start by Dividing each number by the smallest number 5.25/ 1.975 : 5.27/1.975 : 1.975/1.975 = 2.66 : 2.66 : 1 These are not whole number yet. Multipy the whole thing by the smallest number required to get them… [cont.]
Answered by Lexi R - Tue Nov 17 22:42:51 2009
A compound contains 50% Magnesium, 24% carbon, 16% oxygen, and 10% hydrogen. What is the empirical formula?
Q. Another problem that has me confused is the following. A compound consists of 40% calcium, 12% carbon, and 48% oxygen by mass. What is the empirical formula by mass? Is there a difference in how you calculate?
Asked by Joanna - Sun Sep 20 16:35:09 2009 - - 2 Answers - 0 Comments
A. This question is all about atomic weight which you can read off the periodic table for each element. The AW for Mg, C, O and H is 24.3,12, 16 and 1 respectively. So number of mols of each is 50/24.3 = 2, 24/12 = 2, 16/16 = 1 and 10/1 = 10. The empirical formula is therefore Mg2C2OH10. There are some addition isotopes of Mg which means that the average atomic weight is slightly higher than 24.3 which makes the no of mols of Mg very close to 2. Your second question is very similar: Atomic Mass of Ca is 40, so mols of each element is: Ca 40/40 = 1, C 12/12 = 1, O 48/12 = 4 so the empirical formula is CaCO4.
Answered by Bertybign - Sun Sep 20 16:56:04 2009
Q. Another problem that has me confused is the following. A compound consists of 40% calcium, 12% carbon, and 48% oxygen by mass. What is the empirical formula by mass? Is there a difference in how you calculate?
Asked by Joanna - Sun Sep 20 16:35:09 2009 - - 2 Answers - 0 Comments
A. This question is all about atomic weight which you can read off the periodic table for each element. The AW for Mg, C, O and H is 24.3,12, 16 and 1 respectively. So number of mols of each is 50/24.3 = 2, 24/12 = 2, 16/16 = 1 and 10/1 = 10. The empirical formula is therefore Mg2C2OH10. There are some addition isotopes of Mg which means that the average atomic weight is slightly higher than 24.3 which makes the no of mols of Mg very close to 2. Your second question is very similar: Atomic Mass of Ca is 40, so mols of each element is: Ca 40/40 = 1, C 12/12 = 1, O 48/12 = 4 so the empirical formula is CaCO4.
Answered by Bertybign - Sun Sep 20 16:56:04 2009
What is the empirical formula of the compound that contains 32.01 percent carbon, 4.03 percent hydrogen, and 6
Q. What is the empirical formula of the compound that contains 32.01 percent carbon, 4.03 percent hydrogen, and 63.96 percent oxygen? (If the subscripts in the empirical formula for the compound are, for example, 3, 2 and 1, you would enter your answer as C3H2O omitting the subscript 1.) first right answer gets best answer
Asked by aks s - Fri Jan 26 22:32:58 2007 - - 2 Answers - 0 Comments
A. Assuming you mean percent by weight, it is C2 H3 O3
Answered by Brian C - Fri Jan 26 22:39:27 2007
Q. What is the empirical formula of the compound that contains 32.01 percent carbon, 4.03 percent hydrogen, and 63.96 percent oxygen? (If the subscripts in the empirical formula for the compound are, for example, 3, 2 and 1, you would enter your answer as C3H2O omitting the subscript 1.) first right answer gets best answer
Asked by aks s - Fri Jan 26 22:32:58 2007 - - 2 Answers - 0 Comments
A. Assuming you mean percent by weight, it is C2 H3 O3
Answered by Brian C - Fri Jan 26 22:39:27 2007
A hydrocarbon contains 90% by mass of carbon. What is its empirical formula?
Q. So this is what the question asks: A hydrocarbon (contains only carbon and hydrogen) contains 90% by mass of carbon. What is its empirical formula?
Asked by IB chem - Sat Nov 14 13:42:03 2009 - - 1 Answers - 0 Comments
A. assume 100g moles C = 90g x (1 mole / 12g) = 7.5 moles H = 10g x (1mole / 1g) = 10 divide both by the smallest.. 7.5 moles C = 1 moles H = 4/3 multiply both by 3.. C3H4
Answered by m w - Sat Nov 14 14:06:07 2009
Q. So this is what the question asks: A hydrocarbon (contains only carbon and hydrogen) contains 90% by mass of carbon. What is its empirical formula?
Asked by IB chem - Sat Nov 14 13:42:03 2009 - - 1 Answers - 0 Comments
A. assume 100g moles C = 90g x (1 mole / 12g) = 7.5 moles H = 10g x (1mole / 1g) = 10 divide both by the smallest.. 7.5 moles C = 1 moles H = 4/3 multiply both by 3.. C3H4
Answered by m w - Sat Nov 14 14:06:07 2009
What is the empirical formula for of a compound that contains 40.9% carbon, 4.58% Hydrogen, 54.52% oxygen?
Q. and has a molecular W of 264g/mol?
Asked by A A - Mon Aug 17 20:34:20 2009 - - 3 Answers - 0 Comments
A. Mole C = 40.9 g C ( 1mole C/12 g C) =3.41 mole mole H = 4.58 g H ( 1mole H/1.008 g H) = 4.54 mole Mole O = 54.52 g O (1 mole O/16 g O) = 3.41 mole Divide all the moles by smallest mole (3.41) C = 3.41 mole/3.41 mole = 1 H = 4.54 mole/3.41 mole = 1.33 O = 3.41 mole/3.41 mole = 1 The empirical formula = C1H1.33O1 Empirical formula cannot have decimals. Multiply an integer to make it a whole number. Multiply the formula by 3. (C1H1.33O1) x 3 C3H4O3 Empirical formula is C3H4O3.
Answered by Pushpa Padmanabhan - Mon Aug 17 21:30:12 2009
Q. and has a molecular W of 264g/mol?
Asked by A A - Mon Aug 17 20:34:20 2009 - - 3 Answers - 0 Comments
A. Mole C = 40.9 g C ( 1mole C/12 g C) =3.41 mole mole H = 4.58 g H ( 1mole H/1.008 g H) = 4.54 mole Mole O = 54.52 g O (1 mole O/16 g O) = 3.41 mole Divide all the moles by smallest mole (3.41) C = 3.41 mole/3.41 mole = 1 H = 4.54 mole/3.41 mole = 1.33 O = 3.41 mole/3.41 mole = 1 The empirical formula = C1H1.33O1 Empirical formula cannot have decimals. Multiply an integer to make it a whole number. Multiply the formula by 3. (C1H1.33O1) x 3 C3H4O3 Empirical formula is C3H4O3.
Answered by Pushpa Padmanabhan - Mon Aug 17 21:30:12 2009
The ratio of carbon to hydrogen to nitrogen atoms in nicotine is 5 to 7 to 1. What is the molecular formula of
Q. Heres the question: The ratio of carbon to hydrogen to nitrogen atoms in nicotine is 5 to 7 to 1. What is the molecular formula of nicotine if its gram molecular mass is 162 g? The empirical formula is C5H7N. how do i work this problem out (please step by step)
Asked by katiereading - Sun Dec 16 16:15:04 2007 - - 2 Answers - 0 Comments
A. Mass empirical formula = ( 12x5 ) + ( 7 x 1 ) + 14 = 81 g/mol Divide 162 by 81 = 2 Multiply by 2 the empirical formula C10H14N2 ( molecular formula)
Answered by Dr.A - Sun Dec 16 16:20:28 2007
Q. Heres the question: The ratio of carbon to hydrogen to nitrogen atoms in nicotine is 5 to 7 to 1. What is the molecular formula of nicotine if its gram molecular mass is 162 g? The empirical formula is C5H7N. how do i work this problem out (please step by step)
Asked by katiereading - Sun Dec 16 16:15:04 2007 - - 2 Answers - 0 Comments
A. Mass empirical formula = ( 12x5 ) + ( 7 x 1 ) + 14 = 81 g/mol Divide 162 by 81 = 2 Multiply by 2 the empirical formula C10H14N2 ( molecular formula)
Answered by Dr.A - Sun Dec 16 16:20:28 2007
What is the empirical formula of a compound that is 3.05% carbon, 0.26% hydrogen, and 96.69% iodine?
Q. A- C2HI7 B- CH2I5 C- C3H2I11 D- CHI3
Asked by vikaskindacool - Wed Feb 27 12:27:46 2008 - - 3 Answers - 0 Comments
A. B I think
Answered by yupyup - Wed Feb 27 12:31:13 2008
Q. A- C2HI7 B- CH2I5 C- C3H2I11 D- CHI3
Asked by vikaskindacool - Wed Feb 27 12:27:46 2008 - - 3 Answers - 0 Comments
A. B I think
Answered by yupyup - Wed Feb 27 12:31:13 2008
What is the empirical formula of a compound that contains 80.0% carbon and 20.0% hydrogen by mass?
Q. What is the empirical formula of a compound that contains 80.0% carbon and 20.0% hydrogen by mass?
Asked by pimpdaddy696949006 - Tue Mar 24 21:40:28 2009 - - 1 Answers - 0 Comments
A. 3*80/12 = 20; 3*20/1 = 60 The empirical formula of this compound is CH3.
Answered by Hahaha - Wed Mar 25 00:10:05 2009
Q. What is the empirical formula of a compound that contains 80.0% carbon and 20.0% hydrogen by mass?
Asked by pimpdaddy696949006 - Tue Mar 24 21:40:28 2009 - - 1 Answers - 0 Comments
A. 3*80/12 = 20; 3*20/1 = 60 The empirical formula of this compound is CH3.
Answered by Hahaha - Wed Mar 25 00:10:05 2009
empirical formula?
Q. #1)Combustion analysis of toluene, a common organic solvent, gives 8.84 mg of CO2 and 2.06 mg of H2O. If the compound contains only carbon and hydrogen, what is its empirical formula? #2) Menthol, the substance we can smell in mentholated cough drops, is composed of C, H, and O. A 0.1105mg sample of menthol is combusted, producing 0.3112 mg of CO2 and 0.1275 mg of H2O. What is the empirical formula for menthol? #3) If the compound has a molar mass of 156 g/mol, what is its molecular formula? Please explain the answer..
Asked by helpme77 - Thu Sep 27 16:18:43 2007 - - 1 Answers - 0 Comments
A. #2) 1st, let's calculate how many mg of C we have.. we know that C in this case comes only from CO2 so we can do the following: 0.3112mg CO2 x 1mm CO2/ 44.01mg CO2 x 1mmol C/1mmol CO2 x 12.01mg C/1mmol C = 0.0849 mg CO2 similiarly, H comes from H2O, then: 0.1275mg H2O x 1mmol H2O/ 18.016mg H2O x 2mmol H/1mmol H2O x 1.008 mg H/1 mmol H = 0.01426 mg H % C= (0.0849 mg C/ 0.1105mg Sample) x 100 = 76.8% % H= (0.01426 mg H/ 0.1105mg Sample) x 100= 12.9% therefore %O= 100-% C-% H =10.29% Now: 76.8/12.01=6.39 12.9/1.008=12.79 10.20/16= 0.64 ===> devide by smallest ratio you get: 6.39/0.64= 10 mol C 12.79/0.64= 20 mol H 0.64/0.64= 1 mol O The emperical formula of menthol is C 10H 20 O #3) MW= 10x12.01 + 20x1.008 + 16 = 156 g/mol meaning, … [cont.]
Answered by ZikZak - Mon Oct 1 01:28:12 2007
Q. #1)Combustion analysis of toluene, a common organic solvent, gives 8.84 mg of CO2 and 2.06 mg of H2O. If the compound contains only carbon and hydrogen, what is its empirical formula? #2) Menthol, the substance we can smell in mentholated cough drops, is composed of C, H, and O. A 0.1105mg sample of menthol is combusted, producing 0.3112 mg of CO2 and 0.1275 mg of H2O. What is the empirical formula for menthol? #3) If the compound has a molar mass of 156 g/mol, what is its molecular formula? Please explain the answer..
Asked by helpme77 - Thu Sep 27 16:18:43 2007 - - 1 Answers - 0 Comments
A. #2) 1st, let's calculate how many mg of C we have.. we know that C in this case comes only from CO2 so we can do the following: 0.3112mg CO2 x 1mm CO2/ 44.01mg CO2 x 1mmol C/1mmol CO2 x 12.01mg C/1mmol C = 0.0849 mg CO2 similiarly, H comes from H2O, then: 0.1275mg H2O x 1mmol H2O/ 18.016mg H2O x 2mmol H/1mmol H2O x 1.008 mg H/1 mmol H = 0.01426 mg H % C= (0.0849 mg C/ 0.1105mg Sample) x 100 = 76.8% % H= (0.01426 mg H/ 0.1105mg Sample) x 100= 12.9% therefore %O= 100-% C-% H =10.29% Now: 76.8/12.01=6.39 12.9/1.008=12.79 10.20/16= 0.64 ===> devide by smallest ratio you get: 6.39/0.64= 10 mol C 12.79/0.64= 20 mol H 0.64/0.64= 1 mol O The emperical formula of menthol is C 10H 20 O #3) MW= 10x12.01 + 20x1.008 + 16 = 156 g/mol meaning, … [cont.]
Answered by ZikZak - Mon Oct 1 01:28:12 2007
What is the empirical formula of a sample that contains 23.4g Carbon, 3.9g Hydrogen and 20.8g Oxygen?
Q. What is the empirical formula of a sample that contains 23.4g Carbon, 3.9g Hydrogen and 20.8g Oxygen?
Asked by marshmallow - Thu Apr 23 01:15:33 2009 - - 2 Answers - 0 Comments
A. Do exactly what we did in your last question: 1) divide each mass by atomoic mass: C = 23.4 / 12.011 = 1.9482 H = 3.9/1.008 = 3.869 O = 20.8/15.99 = 1.3008 Divide through by smallest C = 1.9482/1.3008 = 1.4977 H = 3.869/1.3008 = 2.9743 O = 1.3008/1.3008 = 1.0 You cannot have a molecule C1.5H3O, so multiply through by 2 to bring to whole integers: Empirical formula = C3H6O2 Compound is probably CH3CH2COOH propanoic acid.
Answered by Trevor H - Thu Apr 23 04:57:52 2009
Q. What is the empirical formula of a sample that contains 23.4g Carbon, 3.9g Hydrogen and 20.8g Oxygen?
Asked by marshmallow - Thu Apr 23 01:15:33 2009 - - 2 Answers - 0 Comments
A. Do exactly what we did in your last question: 1) divide each mass by atomoic mass: C = 23.4 / 12.011 = 1.9482 H = 3.9/1.008 = 3.869 O = 20.8/15.99 = 1.3008 Divide through by smallest C = 1.9482/1.3008 = 1.4977 H = 3.869/1.3008 = 2.9743 O = 1.3008/1.3008 = 1.0 You cannot have a molecule C1.5H3O, so multiply through by 2 to bring to whole integers: Empirical formula = C3H6O2 Compound is probably CH3CH2COOH propanoic acid.
Answered by Trevor H - Thu Apr 23 04:57:52 2009
how can i Find the empirical formula of a compound that is 92.3% carbon and 7.7% Hydrogen. ?
Q. how can i Find the empirical formula of a compound that is 92.3% carbon and 7.7% Hydrogen. ?
Asked by yudyzz - Mon Oct 27 22:55:14 2008 - - 1 Answers - 0 Comments
Q. how can i Find the empirical formula of a compound that is 92.3% carbon and 7.7% Hydrogen. ?
Asked by yudyzz - Mon Oct 27 22:55:14 2008 - - 1 Answers - 0 Comments
What is the empirical formula of a substance that contains 0.150 moles of carbon, 0.300 moles of hydrogen, and?
Q. 0.150 of oxygen? please help!!!
Asked by Unidentified - Mon Sep 22 19:21:38 2008 - - 1 Answers - 0 Comments
A. You have to divide each number by the smallest number of moles. The smallest number here is 0.150 Carbon: 0.150 moles / 0.150 = 1 Hydrogen: .300 moles / 0.150 = 2 Oxygen: 0.150 moles / 0.150 = 1 So the empirical formula is CH2O
Answered by marylynac1988 - Mon Sep 22 19:28:56 2008
Q. 0.150 of oxygen? please help!!!
Asked by Unidentified - Mon Sep 22 19:21:38 2008 - - 1 Answers - 0 Comments
A. You have to divide each number by the smallest number of moles. The smallest number here is 0.150 Carbon: 0.150 moles / 0.150 = 1 Hydrogen: .300 moles / 0.150 = 2 Oxygen: 0.150 moles / 0.150 = 1 So the empirical formula is CH2O
Answered by marylynac1988 - Mon Sep 22 19:28:56 2008
dtermine the empirical formula for a compound that has 9.09g carbon, 1.52g hydrogen, and 14.4g fluorine.?
Q. dtermine the empirical formula for a compound that has 9.09g carbon, 1.52g hydrogen, and 14.4g fluorine.?
Asked by ash_rash_13 - Wed May 30 22:24:53 2007 - - 2 Answers - 0 Comments
A. moles (Carbon) = 9.09g/12.01 = 0.757 mol moles (Hydrogen) = 1.52g/1.0079g.mol-1 = 1.51 mol moles (Fluorine) = 14.4g/18.9984g.mol-1 = 0.758 mol Divide all numbers by 0.757, giving the approximate ratio of 1:2:1 (C:H:F) as the empyrical formula.
Answered by Lab Monkey #31 - Wed May 30 22:35:28 2007
Q. dtermine the empirical formula for a compound that has 9.09g carbon, 1.52g hydrogen, and 14.4g fluorine.?
Asked by ash_rash_13 - Wed May 30 22:24:53 2007 - - 2 Answers - 0 Comments
A. moles (Carbon) = 9.09g/12.01 = 0.757 mol moles (Hydrogen) = 1.52g/1.0079g.mol-1 = 1.51 mol moles (Fluorine) = 14.4g/18.9984g.mol-1 = 0.758 mol Divide all numbers by 0.757, giving the approximate ratio of 1:2:1 (C:H:F) as the empyrical formula.
Answered by Lab Monkey #31 - Wed May 30 22:35:28 2007
A compound has: carbon, 38.44%; hydrogen, 4.84%; chlorine, 56.73%. Calculate the empirical formula ?
Q. A compound has: carbon, 38.44%; hydrogen, 4.84%; chlorine, 56.73%. Calculate the empirical formula ?
Asked by liseerocks - Sat Sep 13 22:43:33 2008 - - 3 Answers - 0 Comments
A. Assume you have 100g of this compound, you would then have 38.44g Carbon 4.84g hydrogen 56.73g Chlorine converting each to mols (get Molar mass from a periodic table) 3.2 mol C 1.6 mol Cl 4.8 mol H divide each by the lowest of the numbers to get each in the lowest possible terms, 2 mol C 1 mol Cl 3 mol H (you can round here, if you are a whole number +/- .1 or if your brave .2 but never more, if they don't come to whole numbers multiply by whole numbers until you cna get them in whole numbers) so your formula would be C2H3Cl
Answered by Chemguy - Sat Sep 13 23:02:23 2008
Q. A compound has: carbon, 38.44%; hydrogen, 4.84%; chlorine, 56.73%. Calculate the empirical formula ?
Asked by liseerocks - Sat Sep 13 22:43:33 2008 - - 3 Answers - 0 Comments
A. Assume you have 100g of this compound, you would then have 38.44g Carbon 4.84g hydrogen 56.73g Chlorine converting each to mols (get Molar mass from a periodic table) 3.2 mol C 1.6 mol Cl 4.8 mol H divide each by the lowest of the numbers to get each in the lowest possible terms, 2 mol C 1 mol Cl 3 mol H (you can round here, if you are a whole number +/- .1 or if your brave .2 but never more, if they don't come to whole numbers multiply by whole numbers until you cna get them in whole numbers) so your formula would be C2H3Cl
Answered by Chemguy - Sat Sep 13 23:02:23 2008
Empirical Formula Homework help?
Q. 2 problems Compound has 1.121 g nitrogen, .161 g hydrogen, .48g Carbon, .64 g Oxygen. Empirical Formula? Compund containing carbon, hydrogen, and oxygen w/ 48.38% carbon and 8.12% oxygen. Empirical Formula? For number 1, I got C H(4) N(20) O For number 2 I got C(8) O H(87) What did you guys get?
Asked by wikipediaftw - Wed Apr 9 20:40:31 2008 - - 1 Answers - 0 Comments
A. Yup, your right. Calculated it myself
Answered by slik_dog1 - Wed Apr 9 20:44:58 2008
Q. 2 problems Compound has 1.121 g nitrogen, .161 g hydrogen, .48g Carbon, .64 g Oxygen. Empirical Formula? Compund containing carbon, hydrogen, and oxygen w/ 48.38% carbon and 8.12% oxygen. Empirical Formula? For number 1, I got C H(4) N(20) O For number 2 I got C(8) O H(87) What did you guys get?
Asked by wikipediaftw - Wed Apr 9 20:40:31 2008 - - 1 Answers - 0 Comments
A. Yup, your right. Calculated it myself
Answered by slik_dog1 - Wed Apr 9 20:44:58 2008
From Yahoo Answer Search: 'empirical formula carbon hydrogen'
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