Mass percent?
Q. I need help with this please. I have to determine the mass percent of K2C2O4 in a sample Unknown B, given the following info. A sample of 1.103 g Unknwon B was dissolved in 25mL DI water and 25 mL 3 M H2SO4. The sample was titrated with 0.04311 M KMnO4 solution and it took 32.77 mL to reach the endpoint. Someone please help.
Asked by summitgirl06 - Sat Apr 5 12:06:43 2008 - - 1 Answers - 0 Comments
A. 1.calculate no.of moles of H2SO4 from the relation n=MV 2.calculate no.of moles of KMnO4 from the relation n=MV 3.write a balanced equation for the reaction between KMnO4 & H2SO4 so you will get moles of H2SO4 reacted with KMnO4 4. Then subtract moles of H2SO4 from (1) so you will get moles of H2SO4 reacted with K2C2O4 5.write a balanced equation for the reaction between K2C2O4& H2SO4 so you will get moles of K2C2O4 reacted with H2SO4 6.then calculate the mass of K2C2O4 from the relation m=n*MW 7.to get the mass % divide the mass in (6) by the total mass of the unknown sample $#$
Answered by Fadia - Sat Apr 5 16:43:59 2008
Q. I need help with this please. I have to determine the mass percent of K2C2O4 in a sample Unknown B, given the following info. A sample of 1.103 g Unknwon B was dissolved in 25mL DI water and 25 mL 3 M H2SO4. The sample was titrated with 0.04311 M KMnO4 solution and it took 32.77 mL to reach the endpoint. Someone please help.
Asked by summitgirl06 - Sat Apr 5 12:06:43 2008 - - 1 Answers - 0 Comments
A. 1.calculate no.of moles of H2SO4 from the relation n=MV 2.calculate no.of moles of KMnO4 from the relation n=MV 3.write a balanced equation for the reaction between KMnO4 & H2SO4 so you will get moles of H2SO4 reacted with KMnO4 4. Then subtract moles of H2SO4 from (1) so you will get moles of H2SO4 reacted with K2C2O4 5.write a balanced equation for the reaction between K2C2O4& H2SO4 so you will get moles of K2C2O4 reacted with H2SO4 6.then calculate the mass of K2C2O4 from the relation m=n*MW 7.to get the mass % divide the mass in (6) by the total mass of the unknown sample $#$
Answered by Fadia - Sat Apr 5 16:43:59 2008
What is the mass percent composition of sodium carbonate using stoichiometry?
Q. A 9.052g mixture of sodium carbonate and sodium chloride contains 3.7130g of sodium. What is the mass percent composition of the mixture? The answer to it is 41.58% sodium carbonate. What I don't know is how to get to the answer. Please help!
Asked by anonymous - Tue Sep 8 12:33:25 2009 - - 1 Answers - 0 Comments
A. let x = mass Na2CO3 let y = mass NaCl x + y = 9.052 moles Na = 3.7130 g/ 22.9898 g/mol=0.1615 2x/ 121.9866 + y / 58.4428 = 0.1615 y= 9.052 - x 2x / 121.9866 + 9.052 - x/ 58.4428= 0.1615 solve for x = mass Na2CO3 % = mass Na2CO3 x 100/ 9.052
Answered by Dr.A - Tue Sep 8 13:15:31 2009
Q. A 9.052g mixture of sodium carbonate and sodium chloride contains 3.7130g of sodium. What is the mass percent composition of the mixture? The answer to it is 41.58% sodium carbonate. What I don't know is how to get to the answer. Please help!
Asked by anonymous - Tue Sep 8 12:33:25 2009 - - 1 Answers - 0 Comments
A. let x = mass Na2CO3 let y = mass NaCl x + y = 9.052 moles Na = 3.7130 g/ 22.9898 g/mol=0.1615 2x/ 121.9866 + y / 58.4428 = 0.1615 y= 9.052 - x 2x / 121.9866 + 9.052 - x/ 58.4428= 0.1615 solve for x = mass Na2CO3 % = mass Na2CO3 x 100/ 9.052
Answered by Dr.A - Tue Sep 8 13:15:31 2009
What is the mass percent of iron in the compound?
Q. 0.2716g of an iron-containing compound yields 0.06136g Fe2O3. What is the mass percent of iron in the compound? I was instructed to calculate from the Mass of Fe2O3 to Moles of Fe2O3... from Moles of Fe2O3 to Moles of Fe... from Moles of Fe to Mass of Fe... to the mass percent of Fe. How can I find the moles of Fe without knowing the molar ratio that would otherwise be given in the chemical equation?
Asked by David - Mon Jun 28 21:13:27 2010 - - 2 Answers - 0 Comments
A. moles = mass/molar mass .06126 g Fe2O3 / ((2 x 55.8)+(3 x 16))g/mol = .00038383 mol Fe2O3 3.8383 x 10^-4 x 2molFe/1molFe2O3 = 7.6766 x 10^-3 mol Fe mass = (moles)(molar mass) mass = (7.6766 x 10^-3) x 55.8 = 4.283543 x 10^-2 g mass percent = (mass substance)/(total mass) 4.283543 x 10^-2 g / .2716 g = .1577 = .158 = 15.8%
Answered by Adespota - Wed Jun 30 12:04:23 2010
Q. 0.2716g of an iron-containing compound yields 0.06136g Fe2O3. What is the mass percent of iron in the compound? I was instructed to calculate from the Mass of Fe2O3 to Moles of Fe2O3... from Moles of Fe2O3 to Moles of Fe... from Moles of Fe to Mass of Fe... to the mass percent of Fe. How can I find the moles of Fe without knowing the molar ratio that would otherwise be given in the chemical equation?
Asked by David - Mon Jun 28 21:13:27 2010 - - 2 Answers - 0 Comments
A. moles = mass/molar mass .06126 g Fe2O3 / ((2 x 55.8)+(3 x 16))g/mol = .00038383 mol Fe2O3 3.8383 x 10^-4 x 2molFe/1molFe2O3 = 7.6766 x 10^-3 mol Fe mass = (moles)(molar mass) mass = (7.6766 x 10^-3) x 55.8 = 4.283543 x 10^-2 g mass percent = (mass substance)/(total mass) 4.283543 x 10^-2 g / .2716 g = .1577 = .158 = 15.8%
Answered by Adespota - Wed Jun 30 12:04:23 2010
How do I go about determining the mass percent in this question?
Q. 2.500 g of a solid mixture contains barium chloride and sodium chloride. The solid is dissolved in water and an aqueous solution containing an excess of sulfate ions is added. A precipitate is formed, collected, dried and is determined to weigh 2.123 grams. What was the mass percent of barium chloride in the original sample?
Asked by themaeyu - Wed Oct 7 00:27:16 2009 - - 1 Answers - 0 Comments
A. first find the difference between the mixture and the precipitate to figure out the mass of the sample then do mass of sample/mixture X 100
Answered by hatorihansotanaka - Wed Oct 7 00:43:05 2009
Q. 2.500 g of a solid mixture contains barium chloride and sodium chloride. The solid is dissolved in water and an aqueous solution containing an excess of sulfate ions is added. A precipitate is formed, collected, dried and is determined to weigh 2.123 grams. What was the mass percent of barium chloride in the original sample?
Asked by themaeyu - Wed Oct 7 00:27:16 2009 - - 1 Answers - 0 Comments
A. first find the difference between the mixture and the precipitate to figure out the mass of the sample then do mass of sample/mixture X 100
Answered by hatorihansotanaka - Wed Oct 7 00:43:05 2009
How can mass percent be found from molarity?
Q. We have a molarity of like 0.033M and we need to determine the percentage of sodium that is in the solution. We got 0.509 g of sodium and placed it in 250ml of sulfuric acid. We had to determine the concentration of the sodium solution and got the 0.0033M. How do you determine mass percent from this data?
Asked by ThE UlTiMaTe MaLe - Thu Mar 25 10:15:36 2010 - - 1 Answers - 0 Comments
A. I do not think that there is a lot of complexity in this problem. Assuming that nothing is lost, and that the dilute solution has density = 1.00g/cm , you have 0.509g of Na in a total mass 250g . mass % Na = 0.509/250*100 = 0.204% Na in the solution. If you go the long way round, and calculate the mass of Na2SO4 in 250ml of a 0.033M ( or is it 0.0033M) solution. You can then calculate the mass of Na in this amount of Na2SO4, which should work out at 0.509g Na. Then you are back to saying that 0.509g Na was dissolved in a volume of250ml, and so on exactly as I did previously.
Answered by Trevor H - Thu Mar 25 11:46:55 2010
Q. We have a molarity of like 0.033M and we need to determine the percentage of sodium that is in the solution. We got 0.509 g of sodium and placed it in 250ml of sulfuric acid. We had to determine the concentration of the sodium solution and got the 0.0033M. How do you determine mass percent from this data?
Asked by ThE UlTiMaTe MaLe - Thu Mar 25 10:15:36 2010 - - 1 Answers - 0 Comments
A. I do not think that there is a lot of complexity in this problem. Assuming that nothing is lost, and that the dilute solution has density = 1.00g/cm , you have 0.509g of Na in a total mass 250g . mass % Na = 0.509/250*100 = 0.204% Na in the solution. If you go the long way round, and calculate the mass of Na2SO4 in 250ml of a 0.033M ( or is it 0.0033M) solution. You can then calculate the mass of Na in this amount of Na2SO4, which should work out at 0.509g Na. Then you are back to saying that 0.509g Na was dissolved in a volume of250ml, and so on exactly as I did previously.
Answered by Trevor H - Thu Mar 25 11:46:55 2010
Trying to determine the mass percent of copper in a copper sulfate + sodium hydroxide reaction?
Q. Is there a better method other than filtering to determine the mass percent of copper in a copper sulfate + sodium hydroxide (plus heat) reaction? Will the filtered copper still contain a lot of impurities? Do I just need to heat it more or is there a different/better method to obtain a more pure copper?
Asked by Ion TT - Sun Apr 4 10:31:40 2010 - - 1 Answers - 0 Comments
A. see the Pearson's chi square test
Answered by Etienne de Quercy, in God Mode - Thu Apr 8 06:36:48 2010
Q. Is there a better method other than filtering to determine the mass percent of copper in a copper sulfate + sodium hydroxide (plus heat) reaction? Will the filtered copper still contain a lot of impurities? Do I just need to heat it more or is there a different/better method to obtain a more pure copper?
Asked by Ion TT - Sun Apr 4 10:31:40 2010 - - 1 Answers - 0 Comments
A. see the Pearson's chi square test
Answered by Etienne de Quercy, in God Mode - Thu Apr 8 06:36:48 2010
What is the mass percent of sodium chloride?
Q. A 0.8000 g sample of impure sodium choride is dissolved in water, and the chloride is precipitated with excess silver nitrate, producing 1.000 g of silver chloride. What is the mass percent of sodium chloride in the original impure sample? a. ) 40.77 b. ) 24.74 c. ) 50.97 d. ) 80.00 e. ) 0.6977 I worked it out and I got e). Is that correct?
Asked by secondaryhoper - Sat Sep 22 19:29:39 2007 - - 1 Answers - 0 Comments
A. Atomic weights: Ag=108 Cl=35.5 Na=23 AgCl=143.5 NaCl=58.5 NaCl + AgNO3 ===> AgCl + NaNO3 1.000gAgCl x 1molAgCl/143.5gAgCl x 1molNaCl/1molAgCl x 58.5gNaCl/1molNaCl = 0.4077g NaCl 0.4077/0.8000 x 100% = 50.96% Looks like c.)
Answered by steve_geo1 - Sat Sep 22 19:48:34 2007
Q. A 0.8000 g sample of impure sodium choride is dissolved in water, and the chloride is precipitated with excess silver nitrate, producing 1.000 g of silver chloride. What is the mass percent of sodium chloride in the original impure sample? a. ) 40.77 b. ) 24.74 c. ) 50.97 d. ) 80.00 e. ) 0.6977 I worked it out and I got e). Is that correct?
Asked by secondaryhoper - Sat Sep 22 19:29:39 2007 - - 1 Answers - 0 Comments
A. Atomic weights: Ag=108 Cl=35.5 Na=23 AgCl=143.5 NaCl=58.5 NaCl + AgNO3 ===> AgCl + NaNO3 1.000gAgCl x 1molAgCl/143.5gAgCl x 1molNaCl/1molAgCl x 58.5gNaCl/1molNaCl = 0.4077g NaCl 0.4077/0.8000 x 100% = 50.96% Looks like c.)
Answered by steve_geo1 - Sat Sep 22 19:48:34 2007
What is the mass percent of MgCl2 in the solid?
Q. A solid mixture contains MgCl2 and NaCl. When 0.5000 g of this solid is dissolved in enough water to form 1.000 L of solution, the osmotic pressure at 25 C is observed to be 0.4030 atm. What is the mass percent of MgCl2 in the solid? (Assume ideal behavior for the solution.) ___% The answer is not 46%
Asked by KickBacknChillax - Wed Dec 30 13:20:57 2009 - - 1 Answers - 1 Comments
A. where i is the dimensionless van 't Hoff factor M is the molarity R=0.08206 L atm mol-1 K-1 is the gas constant T is the thermodynamic (absolute) temperature dropping the "i" monmentarially 0.4030 atm = MRT 0.4030 atm = M (0.08206) (298) M = 0.01648 Molar this 1 litre solution has, then a total of 0.01648 moles of ions of Mg+3, Cl- & Na+ === assume that the grams of MgCl2 = "X" then the garms of NaCl = (0.5000 -X ) grams using molar masses , find moles of each: X grams MgCl2 @ 95.211 g/mol = 0.010503 X moles of MgCl2 times three to find moles of ions released:= 0.03151X mol of ions from MgCl2 (0.5000 -X ) gramsNaCl @ 58.44 g/mol = (0.008556 - 0.01711X) moles of NaCl times two to find moles of ions released: (0. [cont.]
Answered by Steve O - Wed Dec 30 16:17:46 2009
Q. A solid mixture contains MgCl2 and NaCl. When 0.5000 g of this solid is dissolved in enough water to form 1.000 L of solution, the osmotic pressure at 25 C is observed to be 0.4030 atm. What is the mass percent of MgCl2 in the solid? (Assume ideal behavior for the solution.) ___% The answer is not 46%
Asked by KickBacknChillax - Wed Dec 30 13:20:57 2009 - - 1 Answers - 1 Comments
A. where i is the dimensionless van 't Hoff factor M is the molarity R=0.08206 L atm mol-1 K-1 is the gas constant T is the thermodynamic (absolute) temperature dropping the "i" monmentarially 0.4030 atm = MRT 0.4030 atm = M (0.08206) (298) M = 0.01648 Molar this 1 litre solution has, then a total of 0.01648 moles of ions of Mg+3, Cl- & Na+ === assume that the grams of MgCl2 = "X" then the garms of NaCl = (0.5000 -X ) grams using molar masses , find moles of each: X grams MgCl2 @ 95.211 g/mol = 0.010503 X moles of MgCl2 times three to find moles of ions released:= 0.03151X mol of ions from MgCl2 (0.5000 -X ) gramsNaCl @ 58.44 g/mol = (0.008556 - 0.01711X) moles of NaCl times two to find moles of ions released: (0. [cont.]
Answered by Steve O - Wed Dec 30 16:17:46 2009
What is the mass percent of H2SO4 in the battery acid?
Q. A sample of battery acid is to be analyzed for its sulfuric acid content. A 1.00- mL sample weighs 1.228 g. This 1.00-mL sample is diluted to 250.0 mL, and 10.00 mL of this diluted acid requires 31.48 mL of 4.780 10 3 M Ba(OH)2 for its titration. What is the mass percent of H2SO4 in the battery acid?
Asked by zerodood - Fri Mar 13 20:50:49 2009 - - 1 Answers - 0 Comments
A. moles Ba(OH)2 = 0.03148 L x 4.780 x 10^-3 M =0.0001505 Ba(OH)2 + H2SO4 = BaSO4 + 2 H2O moles H2SO4 = 0.0001505 in 10.00 mL moles in 250.00 mL = 0.0001505 x 250.00/ 10.00 = 0.003763 mass H2SO4 = 0.003763 x 98.078 g/mol=0.3691 % = 0.3691 x 100 / 1.228=30.06
Answered by Dr.A - Sat Mar 14 09:42:30 2009
Q. A sample of battery acid is to be analyzed for its sulfuric acid content. A 1.00- mL sample weighs 1.228 g. This 1.00-mL sample is diluted to 250.0 mL, and 10.00 mL of this diluted acid requires 31.48 mL of 4.780 10 3 M Ba(OH)2 for its titration. What is the mass percent of H2SO4 in the battery acid?
Asked by zerodood - Fri Mar 13 20:50:49 2009 - - 1 Answers - 0 Comments
A. moles Ba(OH)2 = 0.03148 L x 4.780 x 10^-3 M =0.0001505 Ba(OH)2 + H2SO4 = BaSO4 + 2 H2O moles H2SO4 = 0.0001505 in 10.00 mL moles in 250.00 mL = 0.0001505 x 250.00/ 10.00 = 0.003763 mass H2SO4 = 0.003763 x 98.078 g/mol=0.3691 % = 0.3691 x 100 / 1.228=30.06
Answered by Dr.A - Sat Mar 14 09:42:30 2009
In finding the mass percent of acetic acid, how do you find the mass of solution?
Q. So to find the mass percent of acetic acid, you would take the mass of acetic acid (.405g) divided by mass of solution times by 100. My question is how do you figure out the mass of solution, what do you add up to get the mass of the whole solution? Formula is NaOH + CH3CO2H --> H2O + C2H3NaO2
Asked by Joel - Sun Oct 18 18:12:41 2009 - - 2 Answers - 0 Comments
A. you need to calculate the mass of all parts of the equation, using your periodic table, then take the Part/Whole. and then convert it to a percent.
Answered by Kepeli - Sun Oct 18 18:19:10 2009
Q. So to find the mass percent of acetic acid, you would take the mass of acetic acid (.405g) divided by mass of solution times by 100. My question is how do you figure out the mass of solution, what do you add up to get the mass of the whole solution? Formula is NaOH + CH3CO2H --> H2O + C2H3NaO2
Asked by Joel - Sun Oct 18 18:12:41 2009 - - 2 Answers - 0 Comments
A. you need to calculate the mass of all parts of the equation, using your periodic table, then take the Part/Whole. and then convert it to a percent.
Answered by Kepeli - Sun Oct 18 18:19:10 2009
What is the composition (by mass percent) of the mixture?
Q. Help! A .1586 g sample of the mixture was dissolved in water. It took 22.90 mL of .1000 M AgNO3 to completely precipitate all the chloride present. What is the composition (by mass percent) of the mixture?
Asked by Jenny - Thu Oct 8 09:09:11 2009 - - 1 Answers - 0 Comments
A. The equation of reaction is: AgNO3 + Cl(-)-->AgCl(s) + NO3(-). All one can tell from that is, there is 22.9 X 0.1 mmol Cl(-) present in the sample. The atomic mass of Cl is 35,453 g/mol. We calculate the mass in grams: M=22.9 x 0.0001 mol x 35.453 g/mol. M = 0,01028 g Per 100 g (by mass percent) 0.65 % Cl(-)
Answered by picus48 - Fri Oct 9 18:31:26 2009
Q. Help! A .1586 g sample of the mixture was dissolved in water. It took 22.90 mL of .1000 M AgNO3 to completely precipitate all the chloride present. What is the composition (by mass percent) of the mixture?
Asked by Jenny - Thu Oct 8 09:09:11 2009 - - 1 Answers - 0 Comments
A. The equation of reaction is: AgNO3 + Cl(-)-->AgCl(s) + NO3(-). All one can tell from that is, there is 22.9 X 0.1 mmol Cl(-) present in the sample. The atomic mass of Cl is 35,453 g/mol. We calculate the mass in grams: M=22.9 x 0.0001 mol x 35.453 g/mol. M = 0,01028 g Per 100 g (by mass percent) 0.65 % Cl(-)
Answered by picus48 - Fri Oct 9 18:31:26 2009
How in the heck to I go about this MASS PERCENT problem?
Q. A hydrocarbon sample with a mass of 6 grams underwent combustion, producing 11 grams of carbon dioxide. If all of the carbon initially present in the compound was converted to carbon dioxide, what was the percent of carbon, by mass, in the hydrocarbon sample? Please, the setup is more crucial than the answer. i need to learn how to do these problems. I have finals soon. Thank you.
Asked by too_cool_4wrds - Tue Nov 27 13:15:05 2007 - - 2 Answers - 0 Comments
A. It's all based on moles. First calculate how many moles of CO2 is represented by 11 grams of CO2. That will tell you how many moles of starting material you had, since one mole of hydrocarbon will presumably result in one mole of CO2. Then, knowing that 12 grams of carbon is 1 mole, you should be able to figure out the mass percent of the original sample.
Answered by Simonizer1218 - Tue Nov 27 14:04:31 2007
Q. A hydrocarbon sample with a mass of 6 grams underwent combustion, producing 11 grams of carbon dioxide. If all of the carbon initially present in the compound was converted to carbon dioxide, what was the percent of carbon, by mass, in the hydrocarbon sample? Please, the setup is more crucial than the answer. i need to learn how to do these problems. I have finals soon. Thank you.
Asked by too_cool_4wrds - Tue Nov 27 13:15:05 2007 - - 2 Answers - 0 Comments
A. It's all based on moles. First calculate how many moles of CO2 is represented by 11 grams of CO2. That will tell you how many moles of starting material you had, since one mole of hydrocarbon will presumably result in one mole of CO2. Then, knowing that 12 grams of carbon is 1 mole, you should be able to figure out the mass percent of the original sample.
Answered by Simonizer1218 - Tue Nov 27 14:04:31 2007
Calculate the mass percent of a solution?
Q. Calculate the mass percent of a solution that is prepared by adding 27.4 g of NaOH to 257 g of H2O .
Asked by Carmen - Sun Nov 15 22:38:54 2009 - - 1 Answers - 0 Comments
Q. Calculate the mass percent of a solution that is prepared by adding 27.4 g of NaOH to 257 g of H2O .
Asked by Carmen - Sun Nov 15 22:38:54 2009 - - 1 Answers - 0 Comments
how do I figure out the mass percent of Tl2SO4 in the sample of TlI?
Q. The thallium (present as Tl2SO4) in a 6.177 g pesticide sample was precipitated as thallium(I) iodide. Calculate the mass percent of Tl2SO4 in the sample if 0.3381 g of TlI was recovered.
Asked by getyoursmarton - Thu Jan 28 19:16:06 2010 - - 1 Answers - 0 Comments
A. moles TlI = 0.3381 g/ 331.29 g/mol=0.001021 moles Tl2SO4 = 0.001021 /2 =0.000510 mass Tl2SO4 = 0.000510 mol x 504.83 g/mol=0.2576 % = 0.2576 x 100 / 6.177=4.170
Answered by Dr.A - Fri Jan 29 09:07:40 2010
Q. The thallium (present as Tl2SO4) in a 6.177 g pesticide sample was precipitated as thallium(I) iodide. Calculate the mass percent of Tl2SO4 in the sample if 0.3381 g of TlI was recovered.
Asked by getyoursmarton - Thu Jan 28 19:16:06 2010 - - 1 Answers - 0 Comments
A. moles TlI = 0.3381 g/ 331.29 g/mol=0.001021 moles Tl2SO4 = 0.001021 /2 =0.000510 mass Tl2SO4 = 0.000510 mol x 504.83 g/mol=0.2576 % = 0.2576 x 100 / 6.177=4.170
Answered by Dr.A - Fri Jan 29 09:07:40 2010
What is the mass percent of oxygen in kaolinite?
Q. Kaolinite, a clay mineral with the formula Al4Si4O10(OH)8, is used as a filler in slick-paper for magazines and as a raw material for ceramics. Analysis shows that 14.35 g of kaolinite contains 8.009 g of oxygen. Calculate the mass percent of oxygen in kaolinite. A. 1.792 mass % B. 24.80 mass % C. 30.81 mass % D. 34.12 mass % E. 55.81 mass %
Asked by Jessica - Wed Sep 12 11:55:06 2007 - - 2 Answers - 0 Comments
A. Analysis shows that14.35g of kaolinite contains 8.009g of oxygen. So E) 55.81%
Answered by danksquish - Wed Sep 12 12:05:06 2007
Q. Kaolinite, a clay mineral with the formula Al4Si4O10(OH)8, is used as a filler in slick-paper for magazines and as a raw material for ceramics. Analysis shows that 14.35 g of kaolinite contains 8.009 g of oxygen. Calculate the mass percent of oxygen in kaolinite. A. 1.792 mass % B. 24.80 mass % C. 30.81 mass % D. 34.12 mass % E. 55.81 mass %
Asked by Jessica - Wed Sep 12 11:55:06 2007 - - 2 Answers - 0 Comments
A. Analysis shows that14.35g of kaolinite contains 8.009g of oxygen. So E) 55.81%
Answered by danksquish - Wed Sep 12 12:05:06 2007
What is the mass percent of NaCl in a solution obtained by mixing 50g of this solution with 215g of H20?
Q. An aqueous solution is 10.5% by mass. What is the mass percent of NaCl in a solution obtained by mixing 50g of this solution with 215g of H20?
Asked by Mikey Walnuts - Thu May 6 11:08:23 2010 - - 1 Answers - 0 Comments
Q. An aqueous solution is 10.5% by mass. What is the mass percent of NaCl in a solution obtained by mixing 50g of this solution with 215g of H20?
Asked by Mikey Walnuts - Thu May 6 11:08:23 2010 - - 1 Answers - 0 Comments
Can you describe an experiment to determine the mass percent of nitrate?
Q. an experiement is to be performed to determine the mass percent of nitrate in the solid mixture of iron (II) nitrate, iron (III) chloride, and iron (III) sulfate. Explain how please!
Asked by hotrod3248705 - Thu Apr 26 00:30:40 2007 - - 2 Answers - 0 Comments
A. I would also try to run a Mass Spectroscopy (Will give you instant answers) or a NMR, or IR. These will tell you exactly what you have and how much you have.
Answered by jcann17 - Thu Apr 26 12:23:26 2007
Q. an experiement is to be performed to determine the mass percent of nitrate in the solid mixture of iron (II) nitrate, iron (III) chloride, and iron (III) sulfate. Explain how please!
Asked by hotrod3248705 - Thu Apr 26 00:30:40 2007 - - 2 Answers - 0 Comments
A. I would also try to run a Mass Spectroscopy (Will give you instant answers) or a NMR, or IR. These will tell you exactly what you have and how much you have.
Answered by jcann17 - Thu Apr 26 12:23:26 2007
How could I possibly solve this mass percent problem?!?
Q. A compound has the formula CH3 (CH2)n MgBr. Calculate the value of n if the mass percent of carbon is 38.1%? I'm not so much interested in the answer as I am in understanding what mental tools I need to know to solve it. Thanks for any help.
Asked by schase10353 - Mon Sep 7 18:19:44 2009 - - 1 Answers - 0 Comments
A. You just need to add up the total mass of the molecule, and then add up the mass of the carbon in the molecule. Then mass percent of carbon is just: Mass of Carbon / Mass of total molecule Of course, since you don't know 'n', it requires a bit of algebra. So the total amount of carbon will be: Total Carbon mass = Mass of single carbon * (1 + n) And the total molecule mass will then be: Molecule mass = Mass of C* (1 + n) + Mass of H* (3 + 2n) + mass of Mg + Mass of Br You get those numbers by literally adding up the number of 'letters' of each element in the molecular formula. And now you just sub these values into the first equation I posted, and it has to equal 38.1%: Mass of C * (1 + n) / [ C*(1+n) + H*(3 + 2n) + Mg + Br] = 0.381… [cont.]
Answered by Call me Batman - Mon Sep 7 18:38:01 2009
Q. A compound has the formula CH3 (CH2)n MgBr. Calculate the value of n if the mass percent of carbon is 38.1%? I'm not so much interested in the answer as I am in understanding what mental tools I need to know to solve it. Thanks for any help.
Asked by schase10353 - Mon Sep 7 18:19:44 2009 - - 1 Answers - 0 Comments
A. You just need to add up the total mass of the molecule, and then add up the mass of the carbon in the molecule. Then mass percent of carbon is just: Mass of Carbon / Mass of total molecule Of course, since you don't know 'n', it requires a bit of algebra. So the total amount of carbon will be: Total Carbon mass = Mass of single carbon * (1 + n) And the total molecule mass will then be: Molecule mass = Mass of C* (1 + n) + Mass of H* (3 + 2n) + mass of Mg + Mass of Br You get those numbers by literally adding up the number of 'letters' of each element in the molecular formula. And now you just sub these values into the first equation I posted, and it has to equal 38.1%: Mass of C * (1 + n) / [ C*(1+n) + H*(3 + 2n) + Mg + Br] = 0.381… [cont.]
Answered by Call me Batman - Mon Sep 7 18:38:01 2009
what is the mass percent of strontium?
Q. What is the mass percent of strontium in 3.65 g sample of an alloy if it takes 32.5 ml of .547 M H2SO4 to precipitate all of the strontium?
Asked by Chase - Wed Dec 17 00:41:45 2008 - - 1 Answers - 0 Comments
A. Chase, Strontium sulfate, SrSO4, is pretty insoluble. Notice there is a 1:1 ratio of Sr^2+ : SO4^2-. The moles of Sr^2+ = moles SO4^2- = moles H2SO4 = 0.547 mole/L * 0.0325 L = 0.01778 moles Grams Sr = 0.01778 moles * 87.62 g/mole = 1.56 grams Percent Sr = 1.56/3.65 x 100% = 42.7% Hope that helped!
Answered by Dr. Buzz - Wed Dec 17 01:25:47 2008
Q. What is the mass percent of strontium in 3.65 g sample of an alloy if it takes 32.5 ml of .547 M H2SO4 to precipitate all of the strontium?
Asked by Chase - Wed Dec 17 00:41:45 2008 - - 1 Answers - 0 Comments
A. Chase, Strontium sulfate, SrSO4, is pretty insoluble. Notice there is a 1:1 ratio of Sr^2+ : SO4^2-. The moles of Sr^2+ = moles SO4^2- = moles H2SO4 = 0.547 mole/L * 0.0325 L = 0.01778 moles Grams Sr = 0.01778 moles * 87.62 g/mole = 1.56 grams Percent Sr = 1.56/3.65 x 100% = 42.7% Hope that helped!
Answered by Dr. Buzz - Wed Dec 17 01:25:47 2008
An experiment is to be performed to determine the mass percent of carbonate in an unknown soluble carbonate sa?
Q. An experiment is to be performed to determine the mass percent of carbonate in an unknown soluble sulfate salt? An experiment is to be performed to determine the mass percent of carbonate in an unknown carbonate salt. Would .2 molar NaCL be an acceptable subsitute for the BaCl2 solution provided for this experiment? Explain
Asked by JTennis - Sat Jan 3 00:24:39 2009 - - 1 Answers - 0 Comments
A. What you are doing here is a gravimetric determination of the carbonate. The chemistry involves: Taking K2CO3 as an example: K2CO3 + BaCl2 BaCO3 + 2KCl The barium carbonate precipitates because it is insoluble, you can filter this off, dry it and determine its dry mass. From this mass you determine the quantity of K2CO3 and so calculate the %. This process depends on the reaction between the barium ion and the carbonate ion. If you try and use sodium chloride in place of the BaCl2, you will not get a precipitate because no double replacemenmt reaction will occur, all the ions being soluble. So the simple answer is: No you cannot replace the BaCl2 with NaCl. You first reference to the determination of the carbonate % in a soluble… [cont.]
Answered by Trevor H - Sat Jan 3 04:18:28 2009
Q. An experiment is to be performed to determine the mass percent of carbonate in an unknown soluble sulfate salt? An experiment is to be performed to determine the mass percent of carbonate in an unknown carbonate salt. Would .2 molar NaCL be an acceptable subsitute for the BaCl2 solution provided for this experiment? Explain
Asked by JTennis - Sat Jan 3 00:24:39 2009 - - 1 Answers - 0 Comments
A. What you are doing here is a gravimetric determination of the carbonate. The chemistry involves: Taking K2CO3 as an example: K2CO3 + BaCl2 BaCO3 + 2KCl The barium carbonate precipitates because it is insoluble, you can filter this off, dry it and determine its dry mass. From this mass you determine the quantity of K2CO3 and so calculate the %. This process depends on the reaction between the barium ion and the carbonate ion. If you try and use sodium chloride in place of the BaCl2, you will not get a precipitate because no double replacemenmt reaction will occur, all the ions being soluble. So the simple answer is: No you cannot replace the BaCl2 with NaCl. You first reference to the determination of the carbonate % in a soluble… [cont.]
Answered by Trevor H - Sat Jan 3 04:18:28 2009
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