What is the difference between Molecular mass and Formula mass?
Q. I know that they`ve both got the same way of finding the answer but how do i know that i have to find the Molecular mass and Formula mass of... lets say sand(SiO2) or of this thing Na2Co3(i dont lnow the name) plz help me i really stink in Chemistry and this thing is the biggest problem
Asked by Saad A - Sat May 17 04:12:02 2008 - - 5 Answers - 0 Comments
A. Really, they are basically the same thing. Molecular mass is the mass of a molecule, while Formula Mass is the mass of ionic compound. It's really a terminology issue but they are both calculated the same way. In the above, sand (SiO2 or silicon dioxide) is formed by covalent bonds and thus is a molecule. And since it is a molecule, you would be finding its Molecular Mass. On the other hand, Na2CO3 (Sodium Carbonate) has an ionic bond and is an ionic compound. not a molecule. So, you would be finding it's Formula Mass. And I assuming you know how to find the masses, so I won't touch on that.
Answered by Blieux Monkey - Sat May 17 04:28:04 2008
Q. I know that they`ve both got the same way of finding the answer but how do i know that i have to find the Molecular mass and Formula mass of... lets say sand(SiO2) or of this thing Na2Co3(i dont lnow the name) plz help me i really stink in Chemistry and this thing is the biggest problem
Asked by Saad A - Sat May 17 04:12:02 2008 - - 5 Answers - 0 Comments
A. Really, they are basically the same thing. Molecular mass is the mass of a molecule, while Formula Mass is the mass of ionic compound. It's really a terminology issue but they are both calculated the same way. In the above, sand (SiO2 or silicon dioxide) is formed by covalent bonds and thus is a molecule. And since it is a molecule, you would be finding its Molecular Mass. On the other hand, Na2CO3 (Sodium Carbonate) has an ionic bond and is an ionic compound. not a molecule. So, you would be finding it's Formula Mass. And I assuming you know how to find the masses, so I won't touch on that.
Answered by Blieux Monkey - Sat May 17 04:28:04 2008
How to find epirical formula/molar mass/ molecular formula of NO2?
Q. I've struggled on this question for two hours and I just don't know where to start. A sample of NO2 gas weighs .150 g in 36.5 mL. 1.) Find the empirical formula 2.) Find molar mass 3.) Find molecular formula
Asked by Ash57 - Thu Nov 5 21:41:31 2009 - - 1 Answers - 0 Comments
A. 1.the empirical formula is the ratio of the atoms in the molecule in this case N:O = 1 :2 empirical formula is NO2 2.I guess we assume standard temp and pressure 1 mole of gas occupies 22.4 liters at STP we have 36.5 ml so we have 36.5 x 10 ^-3 / 22.4 moles of gas = 0.0016295 moles these moles weigh .150 g so the mass of 1 mole = .15 / 0.0016295 so molecular mass =92.05g in the empirical formula we have N = 14/ 46 = 30.4457 % O = 32/ 46 = 69.5543 % in our actual compound we have N = 30.44% of 92.05= 28g or 2 moles O = 69% of 92.05 = 64g = 4 moles our molecular formula is N2O4 I hope you can understand the solution P:
Answered by unknown - Sat Nov 7 01:55:01 2009
Q. I've struggled on this question for two hours and I just don't know where to start. A sample of NO2 gas weighs .150 g in 36.5 mL. 1.) Find the empirical formula 2.) Find molar mass 3.) Find molecular formula
Asked by Ash57 - Thu Nov 5 21:41:31 2009 - - 1 Answers - 0 Comments
A. 1.the empirical formula is the ratio of the atoms in the molecule in this case N:O = 1 :2 empirical formula is NO2 2.I guess we assume standard temp and pressure 1 mole of gas occupies 22.4 liters at STP we have 36.5 ml so we have 36.5 x 10 ^-3 / 22.4 moles of gas = 0.0016295 moles these moles weigh .150 g so the mass of 1 mole = .15 / 0.0016295 so molecular mass =92.05g in the empirical formula we have N = 14/ 46 = 30.4457 % O = 32/ 46 = 69.5543 % in our actual compound we have N = 30.44% of 92.05= 28g or 2 moles O = 69% of 92.05 = 64g = 4 moles our molecular formula is N2O4 I hope you can understand the solution P:
Answered by unknown - Sat Nov 7 01:55:01 2009
Can someone help me out with percent by mass and figuring out the molecular formula by mass please?
Q. A compound contains 39.17 percent phosphorus and 60.83 percent sulfur. It's experimentally determined mass is 158.12 g/mol. What is the molecular formula of this compound ? A certain metal, M forms two oxides M3O and M2O. If the percent by mass of M in M3O is 73.0 % what is the percent by mass of M in M2O ?
Asked by __A_YAHOO_USER__ - Thu Sep 17 23:28:35 2009 - - 1 Answers - 0 Comments
A. You need to determine the number of moles of each element in a sample of the compound. From this you can determine the empirical formula (lowest whole number ratio of atoms in the molecule). From the empirical formul and molar mass you can determine the actual molecular formula Start with 100 g In 100 g there will be 39.17 g P 60.83 g S moles P = mass / molar mass = 39.17 g / 30.97 g/mol = 1.265 moles moles S = 60.83 g / 32.07 g/mol = 1.897 moles Now arrange in a ratio moles P : moles S = 1.265 : 1.897 Divide each number by the smallest number P : S (1.265/ 1.265) : (1.897 / 1.265) 1 : 1.5 This is not a whole number ratio yet. Multiply by the smallest number that will result in 2 whole numbers, in this case 2 2 x (P : S) 2 x (… [cont.]
Answered by Lexi R - Fri Sep 18 00:08:22 2009
Q. A compound contains 39.17 percent phosphorus and 60.83 percent sulfur. It's experimentally determined mass is 158.12 g/mol. What is the molecular formula of this compound ? A certain metal, M forms two oxides M3O and M2O. If the percent by mass of M in M3O is 73.0 % what is the percent by mass of M in M2O ?
Asked by __A_YAHOO_USER__ - Thu Sep 17 23:28:35 2009 - - 1 Answers - 0 Comments
A. You need to determine the number of moles of each element in a sample of the compound. From this you can determine the empirical formula (lowest whole number ratio of atoms in the molecule). From the empirical formul and molar mass you can determine the actual molecular formula Start with 100 g In 100 g there will be 39.17 g P 60.83 g S moles P = mass / molar mass = 39.17 g / 30.97 g/mol = 1.265 moles moles S = 60.83 g / 32.07 g/mol = 1.897 moles Now arrange in a ratio moles P : moles S = 1.265 : 1.897 Divide each number by the smallest number P : S (1.265/ 1.265) : (1.897 / 1.265) 1 : 1.5 This is not a whole number ratio yet. Multiply by the smallest number that will result in 2 whole numbers, in this case 2 2 x (P : S) 2 x (… [cont.]
Answered by Lexi R - Fri Sep 18 00:08:22 2009
What is molar mass and molecular formula?
Q. A 7.85 g sample of a compound with the empirical formula of C5H4 is dissolved in 301 g of benzene. The freezing point of the solution is 1.05*C below that of pure benzene. What are the molar mass and molecular formula of this compound?
Asked by Fletcher M - Sat Aug 8 04:12:19 2009 - - 1 Answers - 0 Comments
Q. A 7.85 g sample of a compound with the empirical formula of C5H4 is dissolved in 301 g of benzene. The freezing point of the solution is 1.05*C below that of pure benzene. What are the molar mass and molecular formula of this compound?
Asked by Fletcher M - Sat Aug 8 04:12:19 2009 - - 1 Answers - 0 Comments
What are molar mass and molecular formula of sorbitol?
Q. An aqueous solution containing 1gm of sorbitol in 100 g water has a freezing point of -0.102*C. Elemental analysis indicates that sorbitol is 39.56% C, 7.75% H 52.7% O The K(f) of water is 1.86*C/m What are molar mass and molecular formula of sorbitol?
Asked by Austin M - Sun Feb 8 21:21:59 2009 - - 2 Answers - 0 Comments
A. similar to above, but remember the following equations: T(f)=K(f)mi m=moles of solute/kilograms of solvent k(f)=1.86 C/m i=1 because sorbitol is non-dissociating m= molality, what we are trying to find. set everything equal to m. T(f)=difference in freezing point between the solution and the pure solvent, in this case it is -0.102 C - 0 C, or -0.102 C. molality is defined as moles solute over kilograms of solvent. We know the kilograms of solvent (0.100 kg = 1g). in other words we can multiply our molality by 0.1 to give us the number of moles of sorbitol in our solution. You can then divide that number of moles by the mass of the sorbitol to get the molar mass. You can figure out the empirical formula of sorbitol based on the… [cont.]
Answered by dpf90 - Sun Feb 8 21:43:07 2009
Q. An aqueous solution containing 1gm of sorbitol in 100 g water has a freezing point of -0.102*C. Elemental analysis indicates that sorbitol is 39.56% C, 7.75% H 52.7% O The K(f) of water is 1.86*C/m What are molar mass and molecular formula of sorbitol?
Asked by Austin M - Sun Feb 8 21:21:59 2009 - - 2 Answers - 0 Comments
A. similar to above, but remember the following equations: T(f)=K(f)mi m=moles of solute/kilograms of solvent k(f)=1.86 C/m i=1 because sorbitol is non-dissociating m= molality, what we are trying to find. set everything equal to m. T(f)=difference in freezing point between the solution and the pure solvent, in this case it is -0.102 C - 0 C, or -0.102 C. molality is defined as moles solute over kilograms of solvent. We know the kilograms of solvent (0.100 kg = 1g). in other words we can multiply our molality by 0.1 to give us the number of moles of sorbitol in our solution. You can then divide that number of moles by the mass of the sorbitol to get the molar mass. You can figure out the empirical formula of sorbitol based on the… [cont.]
Answered by dpf90 - Sun Feb 8 21:43:07 2009
Calcium carbonate has the formula CaCO3. What is the relative formula mass (relative molecular mass) of calci?
Q. Calcium carbonate has the formula CaCO3. What is the relative formula mass (relative molecular mass) of calcium carbonate? Relative atomic masses: Ca = 40; C = 12; O = 16
Asked by ben h - Sun Sep 28 14:22:57 2008 - - 4 Answers - 0 Comments
A. 40+12+16+16+16=100 g/mol
Answered by Johnny D - Sun Sep 28 14:29:01 2008
Q. Calcium carbonate has the formula CaCO3. What is the relative formula mass (relative molecular mass) of calcium carbonate? Relative atomic masses: Ca = 40; C = 12; O = 16
Asked by ben h - Sun Sep 28 14:22:57 2008 - - 4 Answers - 0 Comments
A. 40+12+16+16+16=100 g/mol
Answered by Johnny D - Sun Sep 28 14:29:01 2008
How to find the molecular formula with molar mass?
Q. Combustion analysis of a 1.893g sample of progesterone produced 5.563 g of CO2 and 1.627 g H2O. The molar mass of progesterone is 314.46 g/mol. Find the molecular formula
Asked by Katherine W - Tue Sep 15 02:47:03 2009 - - 1 Answers - 0 Comments
A. 5.563g CO2 dont worry about the oxygen yet 5.563g CO2 * 1 molCO2 / 44.01g CO2 to get the amt of moles in CO2 .1264 mol CO2 * 1 mol C / 1 mol CO2 gets amt of C moles in 1 CO2 mol .1264 mol C --- 1.627 g H20 1.627 g H20 * 1 mol H20 / 18.016 g H2O gets amt of moles in H2O .0903 mol H2O* 2 mol H/1 mol H2O gets amt of H moles in 1 H20 mol .1806 mol H --- Now in order to find the oxygen you have to .1264 mol C * 12.01 g C/ 1 mol C = 1.518 g C .1806 mol H * 1.008 g H/ 1 mol H = .1820 g H Now subtract these values from the sample 1.893g - (1.518 g C + .1820 g H) = .193 Then Mol O = .193 g O * 1 mol O/ 16.00 g O = .0121 mol O Now divide all of the moles of C,H, and O by lowest value .1264 mol C/ .0121 mol O = 10.45 = 10.5 .1806 mol H/ .0121… [cont.]
Answered by katrose - Wed Sep 16 12:09:57 2009
Q. Combustion analysis of a 1.893g sample of progesterone produced 5.563 g of CO2 and 1.627 g H2O. The molar mass of progesterone is 314.46 g/mol. Find the molecular formula
Asked by Katherine W - Tue Sep 15 02:47:03 2009 - - 1 Answers - 0 Comments
A. 5.563g CO2 dont worry about the oxygen yet 5.563g CO2 * 1 molCO2 / 44.01g CO2 to get the amt of moles in CO2 .1264 mol CO2 * 1 mol C / 1 mol CO2 gets amt of C moles in 1 CO2 mol .1264 mol C --- 1.627 g H20 1.627 g H20 * 1 mol H20 / 18.016 g H2O gets amt of moles in H2O .0903 mol H2O* 2 mol H/1 mol H2O gets amt of H moles in 1 H20 mol .1806 mol H --- Now in order to find the oxygen you have to .1264 mol C * 12.01 g C/ 1 mol C = 1.518 g C .1806 mol H * 1.008 g H/ 1 mol H = .1820 g H Now subtract these values from the sample 1.893g - (1.518 g C + .1820 g H) = .193 Then Mol O = .193 g O * 1 mol O/ 16.00 g O = .0121 mol O Now divide all of the moles of C,H, and O by lowest value .1264 mol C/ .0121 mol O = 10.45 = 10.5 .1806 mol H/ .0121… [cont.]
Answered by katrose - Wed Sep 16 12:09:57 2009
What is the difference between gram molecular mass and gram formula mass?
Q. What is the difference between gram molecular mass and gram formula mass?
Asked by bored - Sat Jan 5 18:24:08 2008 - - 3 Answers - 0 Comments
A. Gram molecular mass is the molecular weight of the compound. For example, glucose: C6H14O6. The gram formula mass is the weight of a compound that is not expressable as the whole, like NaCl: 58.5.
Answered by steve_geo1 - Sat Jan 5 18:32:38 2008
Q. What is the difference between gram molecular mass and gram formula mass?
Asked by bored - Sat Jan 5 18:24:08 2008 - - 3 Answers - 0 Comments
A. Gram molecular mass is the molecular weight of the compound. For example, glucose: C6H14O6. The gram formula mass is the weight of a compound that is not expressable as the whole, like NaCl: 58.5.
Answered by steve_geo1 - Sat Jan 5 18:32:38 2008
What are the molecular formula and the molar mass of the compound?
Q. Pheromones are compounds secreted by the females of many insects to attract males. One of these compounds contain 80.78% C, 13.56% H, and 5.66% O. A solution of 1 g of this pheromone in 8.50 g of benzene freezes at 3.37 degrees Celsius. What are the molecular formula and the molar mass of the compound? ( the normal freezing point of pure benzene is 5.50 degrees celsius )
Asked by Angelical Abby - Wed May 28 21:24:24 2008 - - 2 Answers - 0 Comments
A. imagine you have 100 g 80.78 g of carbon = 6.73 mol 13.56 g of H = 13.45 mol 5.66 g of O = 0.353 mol Divide all by the lowest 6.73 / 0.353 = 19 13.45 / 0.353 = 38 0.353 / 0.353 = 1 Your empirical formula is C19H38O The freezing point depression constant for benzene is 5.12 C kg/mol Your depression (ha!) is 5.50 - 3.37 = 2.13 C That means your concentration is 2.13/5.12 = 0.416 m So in 8.5 g of solvent you have 0.416 m x 0.0085 kg = 0.00354 mol 1 g / 0.00354 mol = 282 g/mol Your empirical formula weighs 12(19) + 38 + 16 = 282 So you have one empirical formula per molecular formula and your molar mass is 282.
Answered by Fly On The Wall - Wed May 28 21:40:16 2008
Q. Pheromones are compounds secreted by the females of many insects to attract males. One of these compounds contain 80.78% C, 13.56% H, and 5.66% O. A solution of 1 g of this pheromone in 8.50 g of benzene freezes at 3.37 degrees Celsius. What are the molecular formula and the molar mass of the compound? ( the normal freezing point of pure benzene is 5.50 degrees celsius )
Asked by Angelical Abby - Wed May 28 21:24:24 2008 - - 2 Answers - 0 Comments
A. imagine you have 100 g 80.78 g of carbon = 6.73 mol 13.56 g of H = 13.45 mol 5.66 g of O = 0.353 mol Divide all by the lowest 6.73 / 0.353 = 19 13.45 / 0.353 = 38 0.353 / 0.353 = 1 Your empirical formula is C19H38O The freezing point depression constant for benzene is 5.12 C kg/mol Your depression (ha!) is 5.50 - 3.37 = 2.13 C That means your concentration is 2.13/5.12 = 0.416 m So in 8.5 g of solvent you have 0.416 m x 0.0085 kg = 0.00354 mol 1 g / 0.00354 mol = 282 g/mol Your empirical formula weighs 12(19) + 38 + 16 = 282 So you have one empirical formula per molecular formula and your molar mass is 282.
Answered by Fly On The Wall - Wed May 28 21:40:16 2008
What is the molecular formula if its molar mass is aproxumately 121g?
Q. the percent composition of cysteine is 29.74% C, 5.82% H, 26.41% O, 11.56% N, 26.47% S. What is the molecular formula it its molar mass is approximately 121g?
Asked by Tim - Mon Oct 26 12:09:44 2009 - - 1 Answers - 0 Comments
A. assume 100g... moles C = 29.74g x (1 mole / 12.01g) = 2.476 moles H = 5.82g x (1 mole / 1.008g) = 5.774 moles O = 26.41g x (1 mole / 16.00g) = 1.651 moles N = 11.56g x (1 mole / 14.01g) = 0.8251 moles S = 26.47g x (1 mole / 32.07g) = 0.8253 now we simply those #'s by dividing all by the smallest. that forces at least 1 # to = 1 and the rest to be > 1 moles C = 2.476 / 0.8251 = 3 moles H = 5.774 / 0.8251 = 7 moles O = 1.651 / 0.8251 = 2 moles N = 0.8251 / 0.8251 = 1 moles S = 0.8253 / 0.8251 = 1 and your empirical formula is C3H7O2NS.. and that has empirical mass = 121 g/unit so there must be 1 unit per molecule. and your molecular formula is C3H7NO2S *** alternately... you could start with 121g... and skip the empirical formula… [cont.]
Answered by m w - Mon Oct 26 12:32:41 2009
Q. the percent composition of cysteine is 29.74% C, 5.82% H, 26.41% O, 11.56% N, 26.47% S. What is the molecular formula it its molar mass is approximately 121g?
Asked by Tim - Mon Oct 26 12:09:44 2009 - - 1 Answers - 0 Comments
A. assume 100g... moles C = 29.74g x (1 mole / 12.01g) = 2.476 moles H = 5.82g x (1 mole / 1.008g) = 5.774 moles O = 26.41g x (1 mole / 16.00g) = 1.651 moles N = 11.56g x (1 mole / 14.01g) = 0.8251 moles S = 26.47g x (1 mole / 32.07g) = 0.8253 now we simply those #'s by dividing all by the smallest. that forces at least 1 # to = 1 and the rest to be > 1 moles C = 2.476 / 0.8251 = 3 moles H = 5.774 / 0.8251 = 7 moles O = 1.651 / 0.8251 = 2 moles N = 0.8251 / 0.8251 = 1 moles S = 0.8253 / 0.8251 = 1 and your empirical formula is C3H7O2NS.. and that has empirical mass = 121 g/unit so there must be 1 unit per molecule. and your molecular formula is C3H7NO2S *** alternately... you could start with 121g... and skip the empirical formula… [cont.]
Answered by m w - Mon Oct 26 12:32:41 2009
how can i find the molar mass and molecular formula from this?
Q. Analysis of a volatile liquid shows that it contains 37.23% C, 7.81% H, and 54.96% Cl by mass. At 150 oC and 1.00 atm, 500 ml of the vapor has a mass of 0.922 g. a) What is the molar mass of the compound? b) What is the molecular formula of the compound? thanks so much :)
Asked by David - Mon Nov 16 00:55:57 2009 - - 1 Answers - 0 Comments
Q. Analysis of a volatile liquid shows that it contains 37.23% C, 7.81% H, and 54.96% Cl by mass. At 150 oC and 1.00 atm, 500 ml of the vapor has a mass of 0.922 g. a) What is the molar mass of the compound? b) What is the molecular formula of the compound? thanks so much :)
Asked by David - Mon Nov 16 00:55:57 2009 - - 1 Answers - 0 Comments
Explain how to find the molecular formula when given the empirical formula and molar mass?
Q. Empirical formula: HgCl Molar Mass: 472.2 g/mol Can you plz show me the steps? Also, what about finding the molecular formula from: 94.1% O and 5.9% H molar mass=34 grams
Asked by Matt Strader - Mon Apr 13 00:51:20 2009 - - 1 Answers - 0 Comments
A. Take the HgCl. If this was the final molecule, what would its molar mass be? Hg = 200.59, Cl = 35.453 , HgCl = 236.043g/mol But you have been told that the actual molar mass is 472.2g/mol 472.2/236.043 = 2.000. This tells you that two lots of the empirical formula is required to make up the molecular formula. The molecular formula is: Hg2Cl2 2)O = 94.1, divide by atomic mass = 94.1/15.99 = 5.885 H = 5.9 divided by atomic mass = 5.9/1.008 = 5.853 The ratio is 1:1, so empirical formula is HO. Molar mass HO = 1.008+15.99 = 19.998 34/16.998 = 2.000 molecular formula is 2(HO) = H2O2
Answered by Trevor H - Mon Apr 13 06:09:16 2009
Q. Empirical formula: HgCl Molar Mass: 472.2 g/mol Can you plz show me the steps? Also, what about finding the molecular formula from: 94.1% O and 5.9% H molar mass=34 grams
Asked by Matt Strader - Mon Apr 13 00:51:20 2009 - - 1 Answers - 0 Comments
A. Take the HgCl. If this was the final molecule, what would its molar mass be? Hg = 200.59, Cl = 35.453 , HgCl = 236.043g/mol But you have been told that the actual molar mass is 472.2g/mol 472.2/236.043 = 2.000. This tells you that two lots of the empirical formula is required to make up the molecular formula. The molecular formula is: Hg2Cl2 2)O = 94.1, divide by atomic mass = 94.1/15.99 = 5.885 H = 5.9 divided by atomic mass = 5.9/1.008 = 5.853 The ratio is 1:1, so empirical formula is HO. Molar mass HO = 1.008+15.99 = 19.998 34/16.998 = 2.000 molecular formula is 2(HO) = H2O2
Answered by Trevor H - Mon Apr 13 06:09:16 2009
What is the difference in calculating formula and molecular mass?
Q. In my book it says the same for both. is there even a difference?
Asked by barnacle - Thu Mar 22 19:32:20 2007 - - 2 Answers - 0 Comments
A. You calculate both the same way, but technically a formula mass refers to an ionic substance which can form large crystals (NaCl). We calculate the mass of one formula unit from the crystal so we don't have to add all the units in the crystal. Molecular mass refers to molecular substances which exist as separate molecules (CO2).
Answered by physandchemteach - Thu Mar 22 19:37:51 2007
Q. In my book it says the same for both. is there even a difference?
Asked by barnacle - Thu Mar 22 19:32:20 2007 - - 2 Answers - 0 Comments
A. You calculate both the same way, but technically a formula mass refers to an ionic substance which can form large crystals (NaCl). We calculate the mass of one formula unit from the crystal so we don't have to add all the units in the crystal. Molecular mass refers to molecular substances which exist as separate molecules (CO2).
Answered by physandchemteach - Thu Mar 22 19:37:51 2007
Chemistry. calculating molecular&formula mass?
Q. calculate the molecular or formula mass of each of the following compounds a. ethyl ether, C4H10O, once used as an anesthetic b. titanium(IV) oxide, TiO2, a covering medium used in house paint c. silver nitrate, AgNO3, used to kill bacteria d. acetone, CH3COCH3, used in nail polisih remover
Asked by ugh - Thu Apr 3 20:41:53 2008 - - 1 Answers - 0 Comments
A. To find the formula mass, you simply add up the molecular weights of all the atoms there are. [The weights can be found on a periodic table] a. There are 4 C atoms at 12 each = 48; 10 H at 1 each = 10; and 1 O at 16 = 16. For the whole thing, 48+10+16=74. b. Follow the same rules. 47.9+32=79.9 c. 107.9+14+48=169.9 d. 12+3+12+16+12+3=58
Answered by Meg - Thu Apr 3 20:53:42 2008
Q. calculate the molecular or formula mass of each of the following compounds a. ethyl ether, C4H10O, once used as an anesthetic b. titanium(IV) oxide, TiO2, a covering medium used in house paint c. silver nitrate, AgNO3, used to kill bacteria d. acetone, CH3COCH3, used in nail polisih remover
Asked by ugh - Thu Apr 3 20:41:53 2008 - - 1 Answers - 0 Comments
A. To find the formula mass, you simply add up the molecular weights of all the atoms there are. [The weights can be found on a periodic table] a. There are 4 C atoms at 12 each = 48; 10 H at 1 each = 10; and 1 O at 16 = 16. For the whole thing, 48+10+16=74. b. Follow the same rules. 47.9+32=79.9 c. 107.9+14+48=169.9 d. 12+3+12+16+12+3=58
Answered by Meg - Thu Apr 3 20:53:42 2008
What is the molecular formula of a substance that has an empirical formula of C2H5 and a molecular mass of 58?
Q. What is the molecular formula of a substance that has an empirical formula of C2H5 and a molecular mass of 58?
Asked by crownmecca - Sat May 10 22:10:53 2008 - - 1 Answers - 0 Comments
A. Mr(C2H5) = 2(12) + 5(1) Mr = 29 58/29 = 2 --> 2(C2H5) ---> Molecular formula is C4H10 hope this helps^_^
Answered by **PiNoY YFC** - Sat May 10 22:16:00 2008
Q. What is the molecular formula of a substance that has an empirical formula of C2H5 and a molecular mass of 58?
Asked by crownmecca - Sat May 10 22:10:53 2008 - - 1 Answers - 0 Comments
A. Mr(C2H5) = 2(12) + 5(1) Mr = 29 58/29 = 2 --> 2(C2H5) ---> Molecular formula is C4H10 hope this helps^_^
Answered by **PiNoY YFC** - Sat May 10 22:16:00 2008
If the substance has a molecular mass of 160 +/-5 g, what is its molecular formula?
Q. If the substance has a molecular mass of 160 +/-5 g, what is its molecular formula?
Asked by boss - Sun Sep 3 22:17:33 2006 - - 2 Answers - 0 Comments
A. Not enough clue. Many combinations of atoms can sum up to such molecular mass.
Answered by dactylifera001 - Sun Sep 3 23:14:18 2006
Q. If the substance has a molecular mass of 160 +/-5 g, what is its molecular formula?
Asked by boss - Sun Sep 3 22:17:33 2006 - - 2 Answers - 0 Comments
A. Not enough clue. Many combinations of atoms can sum up to such molecular mass.
Answered by dactylifera001 - Sun Sep 3 23:14:18 2006
Chemistry Questions. Empirical formula/Molecular Formula/Mass percent! [ Give it a shot! ]?
Q. I thought that i was good in chemistry in the past. When my little sister brought me these formulas, however, it seems I'm not as good as I thought. haha. Oh well. Practice makes perfect. Give it a try. 1. ( cut short ) 35.5% C, 4.77% H, 8.29% N, 13.6% Na, 37.9% O. What is the empirical formula? 2. Ibuprofen yields a molar mass of 206 g/mol and a percent composition of 75.7% C. 8.80% H, 15.5% O. Determine the molecular formula. 3. Cu3(OH)4SO4 and Cu(OH)6SO4. Determine the mass percent of copper in these compounds. Thanks for the help.
Asked by Christian - Sun Nov 4 16:55:04 2007 - - 3 Answers - 0 Comments
A. (I'll edit and update as I go through. 1-3) 1. Convert %mass into grams and find the # of moles. 35.5g Carbon/12.0g= 2.958 moles 4.77g Hydogen/1.0 g=4.77 moles 13.6g Sodium/22.98g = .5918 mol 8.29g Nitrogen/14.0 g = .5921 mol 37.9 g Oxygen/16.0g = 2.3688 mol Divide the # of moles by the lowest factor (in this case .5918) To get... 2.958 mol C/.5918=5 4.77 mol H/.5918=8 .5918 mol Na/.5918=1 .5921 mol N/.5918=1 2.3688 mol O/.5918=4 Empirical formula: C5H8NANO4 2. Same as #1: Convert percent into grams, then moles. CARBON: 75.7g/12=6.31mol HYDROGEN: 8.8g/1g=8.80 mol OXYGEN: 15.5g/16= .968 mol Then, divide by lowest factor (which is .968 mol. to get... ---13 Carbons, 18 hydrogens, and 2 oxygens! ---Add these up (13*12)+(1*18)*(2*16) [cont.]
Answered by Confused Teen - Sun Nov 4 17:05:08 2007
Q. I thought that i was good in chemistry in the past. When my little sister brought me these formulas, however, it seems I'm not as good as I thought. haha. Oh well. Practice makes perfect. Give it a try. 1. ( cut short ) 35.5% C, 4.77% H, 8.29% N, 13.6% Na, 37.9% O. What is the empirical formula? 2. Ibuprofen yields a molar mass of 206 g/mol and a percent composition of 75.7% C. 8.80% H, 15.5% O. Determine the molecular formula. 3. Cu3(OH)4SO4 and Cu(OH)6SO4. Determine the mass percent of copper in these compounds. Thanks for the help.
Asked by Christian - Sun Nov 4 16:55:04 2007 - - 3 Answers - 0 Comments
A. (I'll edit and update as I go through. 1-3) 1. Convert %mass into grams and find the # of moles. 35.5g Carbon/12.0g= 2.958 moles 4.77g Hydogen/1.0 g=4.77 moles 13.6g Sodium/22.98g = .5918 mol 8.29g Nitrogen/14.0 g = .5921 mol 37.9 g Oxygen/16.0g = 2.3688 mol Divide the # of moles by the lowest factor (in this case .5918) To get... 2.958 mol C/.5918=5 4.77 mol H/.5918=8 .5918 mol Na/.5918=1 .5921 mol N/.5918=1 2.3688 mol O/.5918=4 Empirical formula: C5H8NANO4 2. Same as #1: Convert percent into grams, then moles. CARBON: 75.7g/12=6.31mol HYDROGEN: 8.8g/1g=8.80 mol OXYGEN: 15.5g/16= .968 mol Then, divide by lowest factor (which is .968 mol. to get... ---13 Carbons, 18 hydrogens, and 2 oxygens! ---Add these up (13*12)+(1*18)*(2*16) [cont.]
Answered by Confused Teen - Sun Nov 4 17:05:08 2007
Need chemistry help! Calculating epirical formula, molecular formula, and molar mass?
Q. ~Calculate the number of atoms in 2.00g of platinum ~The composition of acetic acid is 40.00% carbon, 6.71% hydrogen and 53.29% oxygen. Calculate the empirical formula for acetic acid ~ The compund borazine consista of 40.29% boron, 7.51% hydrogen and 52.20% nitrogen and its molecular mass is 80.50. Calculate the molecular formula for borazine.
Asked by curiousshortstories - Wed Dec 10 20:13:08 2008 - - 1 Answers - 0 Comments
Q. ~Calculate the number of atoms in 2.00g of platinum ~The composition of acetic acid is 40.00% carbon, 6.71% hydrogen and 53.29% oxygen. Calculate the empirical formula for acetic acid ~ The compund borazine consista of 40.29% boron, 7.51% hydrogen and 52.20% nitrogen and its molecular mass is 80.50. Calculate the molecular formula for borazine.
Asked by curiousshortstories - Wed Dec 10 20:13:08 2008 - - 1 Answers - 0 Comments
HELP - 11th grade organic chemistry : What's the relative molecular mass and the formula of the alcohol?
Q. 1.6 grams of alcohol was found to react with excess of metallic sodium producing 0.56 dm^3 of hydrogen gas measured at standard temperature and pressure (s.t.p). What's the molecular mass of the alcohol and what's the formula?
Asked by PackFan - Thu Nov 20 05:39:26 2008 - - 2 Answers - 1 Comments
A. idea here is ... 2 R-OH + Na ---> 2 R-O-Na + H2 so let's 1) calculate moles H2 from PV = nRT 2) calculate moles R-OH from the balanced equation. 3) calculate molar mass R-OH from moles = mass / molar mass.. 4) then the formula... *** 1 *** PV = nRT n = PV/RT P = 1 atm V = 0.56 dm x (10 cm / 1 dm) x (1 ml / 1 cm ) x (1 L / 1000 mL) = 0.56 L R = 0.0821 L atm/moleK T = 273 K n = (1 atm) x (0.56 L) / [ (0.0821 Latm/moleK) x (273K) ] = 0.0250 moles H2 *** 2 *** from balanced equation, 2 moles alcohol ---> 1 mole H2 0.0250 moles H2 x (2 moles Alcohol / 1 mole H2) = 0.0500 moles R-OH *** 3 *** moles = mass / molar mass molar mass = mass / moles = 1.6 g / 0.0500 moles = 32 g/mole *** 4 *** you know the formula has 32 g/mole. you… [cont.]
Answered by m w - Thu Nov 20 09:32:36 2008
Q. 1.6 grams of alcohol was found to react with excess of metallic sodium producing 0.56 dm^3 of hydrogen gas measured at standard temperature and pressure (s.t.p). What's the molecular mass of the alcohol and what's the formula?
Asked by PackFan - Thu Nov 20 05:39:26 2008 - - 2 Answers - 1 Comments
A. idea here is ... 2 R-OH + Na ---> 2 R-O-Na + H2 so let's 1) calculate moles H2 from PV = nRT 2) calculate moles R-OH from the balanced equation. 3) calculate molar mass R-OH from moles = mass / molar mass.. 4) then the formula... *** 1 *** PV = nRT n = PV/RT P = 1 atm V = 0.56 dm x (10 cm / 1 dm) x (1 ml / 1 cm ) x (1 L / 1000 mL) = 0.56 L R = 0.0821 L atm/moleK T = 273 K n = (1 atm) x (0.56 L) / [ (0.0821 Latm/moleK) x (273K) ] = 0.0250 moles H2 *** 2 *** from balanced equation, 2 moles alcohol ---> 1 mole H2 0.0250 moles H2 x (2 moles Alcohol / 1 mole H2) = 0.0500 moles R-OH *** 3 *** moles = mass / molar mass molar mass = mass / moles = 1.6 g / 0.0500 moles = 32 g/mole *** 4 *** you know the formula has 32 g/mole. you… [cont.]
Answered by m w - Thu Nov 20 09:32:36 2008
How do you find the molecular formula when given the grams and molar mass?
Q. I understand how to do it when I'm given the percentages of the elements, but not with the grams! please help! example: 41.05 g C, 3.45 g H, 20.51 g O molar mass of 152g/mol find the molecular formula
Asked by guitarhero - Thu Sep 24 19:59:28 2009 - - 1 Answers - 0 Comments
A. if we can assume that the compound only has C,H & O Moles of C = 41.05/12 =3.42 Moles of H = 3.45/1= 3.45 moles of O =20.51/16=1.28 dividing by the smallest to get the ratios C:H:O = 2.67:2.67:1 multiply by 3 to get whole numbers 8:8:3 empirical formula is C8H8O3 empirical formula mass = 8*12+8*1+3*16 = 152 which is the same as the molecular mass so the molecular formula is C8H8O3
Answered by unknown - Thu Sep 24 20:22:12 2009
Q. I understand how to do it when I'm given the percentages of the elements, but not with the grams! please help! example: 41.05 g C, 3.45 g H, 20.51 g O molar mass of 152g/mol find the molecular formula
Asked by guitarhero - Thu Sep 24 19:59:28 2009 - - 1 Answers - 0 Comments
A. if we can assume that the compound only has C,H & O Moles of C = 41.05/12 =3.42 Moles of H = 3.45/1= 3.45 moles of O =20.51/16=1.28 dividing by the smallest to get the ratios C:H:O = 2.67:2.67:1 multiply by 3 to get whole numbers 8:8:3 empirical formula is C8H8O3 empirical formula mass = 8*12+8*1+3*16 = 152 which is the same as the molecular mass so the molecular formula is C8H8O3
Answered by unknown - Thu Sep 24 20:22:12 2009
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