Stoichiometry Molarity and more

Stoichiometry Molarity and more
Some questions require whiteboards. If you want to draw, you can come after school and draw on the big board. Bonus points may be available for suggesting good wrong answers for future use. So if you were thinking that the answer you really wanted wasn’t among the choices, put it on a piece of paper to submit. Write down the slide # and submit your suggestion.

The following diagram represents the collection of elements formed by the decomposition of a compound. The blue spheres represent nitrogen atoms and the red spheres represent oxygen atoms, what was the empirical formula of the original compound? N6O12 N3O6 NO2

We can not know the molecular formula from the information given.
The following diagram represents the collection of elements formed by the decomposition of a compound. The blue spheres represent nitrogen atoms and the red spheres represent oxygen atoms, what was the empirical formula of the original compound? N6O12 N3O6 NO2 We can not know the molecular formula from the information given. We only know the ratio of N’s to O’s in the original compound.

The following diagram represents the collection of carbon dioxide and water formed by the decomposition of a hydrocarbon. What was the empirical formula of the original hydrocarbon? C4H16 C2H8 CH4

The following diagram represents the collection of carbon dioxide and water formed by the decomposition of a hydrocarbon. What was the empirical formula of the original hydrocarbon? C4H16 C2H8 CH4 While the diagram indicates 4 carbons, and you might think there could have been 1 C4H16, 2 C2H8, or 4 CH4. However, the maximum number of H’s that can attach to C’s is CnH2n+2. Thus to achieve the 1:4 C:H ratio, both the empirical and molecular formula must have been CH4.

The following diagram represents a high-temperature reaction between CH4 and H2O. Based on this diagram, write a balanced chemical equation to represent this reaction.

While the diagram actually represents
The following diagram represents a high-temperature reaction between CH4 and H2O. Based on this diagram, write a balanced chemical equation to represent this reaction. CH4 + H2O → CO2 + 3H2 While the diagram actually represents 2CH4 + 2H2O → 2CO2 + 6H2 It is more appropriate to write chemical equations is the lowest whole number ratio.

Break out the scrap paper to sketch a response.
Nitrogen and hydrogen react to form ammonia (NH3). Consider the model of the mixture shown below. Draw a representation of the product mixture, assuming the reaction goes to completion. Which color sphere best represents nitrogen and which color for hydrogen? Break out the scrap paper to sketch a response.

8N’s, 4N2 require 24 H’s, 12H2 for a complete reaction.
Nitrogen (N2) and hydrogen (H2) react to form ammonia (NH3). Consider the model of the mixture shown below. Blue spheres = N and white spheres = H. Draw a representation of the product mixture, assuming the reaction goes to completion. N2 + 3H2 → 2NH3 8N’s, 4N2 require 24 H’s, 12H2 for a complete reaction. Only 9H2 are present, thus H2 limits. 9H2 require 3N2, one N2 in excess, and 6NH3 are produced.

Break out the whiteboards to sketch a response.
Nitrogen monoxide and oxygen react to form nitrogen dioxide. Consider the model of the mixture shown below. Blue spheres = N and red spheres = O. Draw a representation of the product mixture, assuming the reaction goes to completion. Break out the whiteboards to sketch a response.

8NO require 4O2 for a complete reaction.
Nitrogen monoxide and oxygen react to form nitrogen dioxide. Consider the model of the mixture shown below. Blue spheres = N and white spheres = O. Draw a representation of the product mixture, assuming the reaction goes to completion. 2NO + O2 → 2NO2 8NO require 4O2 for a complete reaction. 5O2 are present, thus O2 is in excess and NO limits. 8NO require 4O2, one O2 in excess, and 8NO2 are produced.

The molarity an aqueous solution made from 4
The molarity an aqueous solution made from 4.2 g of sodium fluoride dissolved to make 500 ml of solution would be 2.1 M 0.20 M 0.05 M 50 M M 8.4 M No calculator

The molarity an aqueous solution made from 4
The molarity an aqueous solution made from 4.2 g of sodium fluoride dissolved to make 500 ml of solution would be 2.1 M 0.20 M 0.05 M 50 M M 8.4 M moles Molarity, M = V (L)

The molarity of the sodium ions in an aqueous solution made from 4
The molarity of the sodium ions in an aqueous solution made from 4.2 g of sodium fluoride dissolved to make 500 ml of solution. No calculator 2.1 M 4.2 M 0.20 M 0.10 M 0.05 M

The molarity of the sodium ions an aqueous solution made from 4
The molarity of the sodium ions an aqueous solution made from 4.2 g of sodium fluoride dissolved to make 500 ml of solution. 2.1 M 4.2 M 0.20 M of the solution the molarity of each ion is present in a 1:1 ratio with the “molecule” and thus has the same molarity, 0.20 M for each ion, Na+ and F− NaF → Na+ + F− 0.10 M 0.05 M

impossible to determine.
In a M solution of K2SO4 the total concentration of all the ions is 0.010 M 0.020 M 0.030 M 0.070 M impossible to determine. No calculator

In a 0.010 M solution of K2SO4 the total concentration of the ions is
When dissolves, 3 ions are produced. K2SO4 → 2K+ + SO42− resulting in 3 ions per particle dissolved 0.020 M solution of K+ ions and M solution of SO42− ions 0.070 M impossible to determine.

When making a 1.0 M aqueous solution of NaCl. Select all that apply.
It is best to dissolve g of NaCl in a ml beaker and add 1000 ml of water with a graduated cylinder. It is best to dissolve g of NaCl in a ml beaker and use a graduated cylinder to add 1000 ml of water. It is best to put g of NaCl in a volumetric flask and then add 1000 ml of water. It is best to dissolve g of NaCl in some water in a volumetric flask and then fill the volumetric flask up to the 1 L mark.

When making a 1.0 M aqueous solution of NaCl. Select all that apply.
It is best to dissolve g of NaCl in a 2000 ml beaker and add 1000 ml of water with a graduated cylinder. It is best to dissolve g of NaCl in a 2000 ml beaker and use a graduated cylinder to add 1000 ml of water. It is best to put g of NaCl in a volumetric flask and then add 1000 ml of water. It is best to dissolve g of NaCl in some water in a volumetric flask and then fill the volumetric flask up to the 1 L mark. This of course will allow for the room that the dissolved salt will take up as part of the total volume of the solution. moles Molarity, M = V (L) This is volume of the solution, not volume of water added.

24 ml of a 0. 10 M solution of Al(NO3)3 is combined with 24 ml of 0
24 ml of a 0.10 M solution of Al(NO3)3 is combined with 24 ml of 0.10 M solution of Na2CO3. Write out balanced “overall” equation on your Mega-white boards. Be sure and indicate the precipitate.

Write out balanced “overall” equation on paper.
24 ml of a 0.10 M solution of Al(NO3)3 is combined with 24 ml of 0.10 M Na2CO3. Write out balanced “overall” equation on paper. 3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) + 6NaNO3 Net ionic equation?

24 ml of a 0.10 M solution of Al(NO3)3 is combined with 24 ml of 0.10 M Na2CO3. Mega-white boards 3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) + 6NaNO3 3CO32− + 2Al3+ → Al2(CO3)3(ppt) sodium and nitrate ions are spectator ions

3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) + 6NaNO3
24 ml of a 0.10 M solution of Al(NO3)3 is combined with 24 ml of 0.10 M Na2CO3. 3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) + 6NaNO3 What is the mass of the dried precipitate? 7.8 g 70.2 g 0.19 g 0.28 g 0.56 g 187 g 842 g 1123 g 1685 g MM (g/mol) 3Na2CO3 = 106 Al(NO3)3 = 213 Al2(CO3)3 = 234 NaNO3 = 85

24 ml of a 0.10 M solution of Al(NO3)3 is combined with 24 ml of 0.10 M Na2CO3. 3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) + 6NaNO3 What is the mass of the dried precipitate? # g 2.4 mmol each, and because of the 3:2 ratio between the reactants, the 3Na2CO3 limits which equals 0.19 g Al2(CO3)3 MM (g/mol) 3Na2CO3 = 106 Al(NO3)3 = 213 Al2(CO3)3 = 234 NaNO3 = 85

24 ml of a 0.10 M solution of Al(NO3)3 is combined with 24 ml of 0.10 M Na2CO3. 3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) + 6NaNO3 What is the molar concentration of each ion in solution after the reaction? Mega-white boards MM (g/mol) 3Na2CO3 = 106 Al(NO3)3 = 213 Al2(CO3)3 = 234 NaNO3 = 85

24 ml of a 0.10 M solution of Al(NO3)3 is combined with 24 ml of 0.10 M Na2CO3. 3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) + 6NaNO3 What is the molar concentration of each ion in solution after the reaction? Remember - the volume of the solution has doubled which affects the concentrations Treat the spectator ion concentrations as completely unreacted in twice the volume. Assume the limiting ion in the precipitate is effectively gone from solution. Calculate the mmoles left over (this means a subtraction) of the excess ion that forms the precipitate and calculate its molarity in the total volume.

24 ml of a 0.10 M solution of Al(NO3)3 is combined with 24 ml of 0.10 M Na2CO3.3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) + 6NaNO3 What is the molar concentration of each ion in solution after the reaction? Assume the CO32− ≃ 0

Calculate the Molarity of NO3− ions in 40 ml of a 0
Calculate the Molarity of NO3− ions in 40 ml of a 0.20 M solution of Mg(NO3)2 after reacting with 60 ml of a 0.10 M solution of Al(NO3)3 No Calculator Type in a numerical answer.

Calculate the Molarity of NO3− ions in 40 ml of a 0
Calculate the Molarity of NO3− ions in 40 ml of a 0.20 M solution of Mg(NO3)2 after reacting with 60 ml of a 0.10 M solution of Al(NO3)3 No Calculator

What is the mass of copper(II) sulfate (molecular weight = 159
What is the mass of copper(II) sulfate (molecular weight = g/mol) in 40. mL of 2.0 M copper(II) sulfate? 3.2 g 5.5 g 8.8 g 13 g 16 g 32 g No Calculator

Learn to estimate using easy math
What is the mass of copper(II) sulfate (molecular weight = g/mol) in 40. mL of 2.0 M copper(II) sulfate? 3.2 g 5.5 g 8.8 g 13 g 0.1 mol would mean 16g, half that or 0.05 mol would be ~8g, thus the 0.08 mol must be in between≃13g 16 g 32 g No Calculator Learn to estimate using easy math

How many grams of zinc nitrate(189 g/mol) contain 48 grams of oxygen atoms?
No Calculator 95 g 125 g 145 g 165 g none of the above

How many grams of zinc nitrate(189 g/mol) contain 48 grams of oxygen atoms?
No Calculator 95 g Look for easy math. Zn(NO3)2 3 O’s = 48g, since 2 NO3‘s per zinc nitrate, you only need 0.5 mol of zinc nitrate = half of molar mass 125 g 145 g 165 g 189 g

A 2.00 gram mixture of calcium carbonate and calcium chloride are treated with an excess of hydrochloric acid and 0.66 grams of carbon dioxide (44 g/mol) are produced. What is the percent of CaCO3 (100 g/mol) by mass in the original mixture? No Calculator 25% 30% 50% 75% 90% Write a net ionic equation to represent the hydrochloric acid + calcium carbonate reaction

A 2.00 gram mixture of calcium carbonate and calcium chloride are treated with an excess of hydrochloric acid and 0.66 grams of carbon dioxide (44 g/mol) are produced. What is the percent of CaCO3 (100 g/mol) by mass in the original mixture? No Calculator 25% 30% 50% 75% thus mol of CaCO3 to start = 1.5 g and 1.5 is 75% of 2 g total 90% CaCO3 + H+ ➙ H2O + CO2 + Ca2+

No Calculators CuS2 CuS Cu2S Cu2S3 Cu3S2
In which compound below is the mass ratio of copper to sulfur closest to 2:1? No Calculators CuS2 CuS Cu2S Cu2S3 Cu3S2

No Calculators CuS2 ~1:1 CuS ~2:1 Cu2S ~4:1 Cu2S3 ~4:3 Cu3S2 ~6:4
In which compound below is the mass ratio of copper to sulfur closest to 2:1? Expect easy math. MM Cu=63.6 and S=32. For calculation purposes assume 60 and 30 CuS2 ~1:1 CuS ~2:1 Cu2S ~4:1 Cu2S3 ~4:3 Cu3S2 ~6:4 No Calculators

No Calculators 460 g/mol 230 g/mol 110 g/mol 55 g/mol
A certain compound contains only one sodium atom and is 5% sodium by mass. What is the molar mass of the compound? No Calculators 460 g/mol 230 g/mol 110 g/mol 55 g/mol none of the above

A certain compound contains only one sodium atom and is 5% sodium by mass. What is the molar mass of the compound? No Calculators 460 g/mol make the math easy. molar mass of Na is 23, so 23 is 10% of 230, and thus 5% of 460. 230 g/mol 110 g/mol 55 g/mol none of the above

Analysis of a tellurium oxide compound indicated 84. 22 % tellurium
Analysis of a tellurium oxide compound indicated % tellurium. The molar mass is between 580 and 610 g/mole. Determine the molecular formula. Yes, Calculators TeO Te2O3 Te2O TeO2 TeO3 Te3O Te4O TeO4 Te4O6

The Molecular Formula is Te4O6. Te2O3 MM = 303.2 thus Te4O6.
Analysis of a tellurium oxide compound indicated % tellurium. The molar mass is between 580 and 610 g/mole. Determine the molecular formula. 46 The Molecular Formula is Te4O6. Te2O3 MM = thus Te4O6.

Determine the empirical formula of a nerve gas that gave the following analysis: 39.10% C, 7.67% H, 26.11% O, 16.82% P, % F. Yes, Calculators The Formula is CvHwOxPyFz. Type your answer in as a number vwxyz (no spaces, no letters, just one number) i.e. for C3H12O4PF2, you would type in

The Empirical Formula is C6H14O3PF
Determine the empirical formula of a nerve gas that gave the following analysis: 39.10% C, 7.67% H, 26.11% O, 16.82% P, % F. 614311 The Empirical Formula is C6H14O3PF

Combustion analysis is used to determine the amount of carbon, hydrogen, and oxygen in a combustible compound. Measure the mass of compound to be combusted. Measure the mass of water produced. Measure the mass of carbon dioxide produced oxygen will make up any remaining mass in original compound. CxHyOz + O2 ➙ H2O + CO2

empirical ratio ➙ then molar mass Calculate for molecular formula
Cumene is a organic compound contains only carbon and hydrogen that is used in the production of acetone and phenol in the chemical industry. Combustion of 47.6 mg of cumene produces mg of carbon dioxide and 42.8 mg of water. The molar mass of cumene is between 115 and 125 g/mole. Determine the molecular formula. Yes, Calculators Determine moles of C Determine moles of H empirical ratio ➙ then molar mass Calculate for molecular formula

Cumene is a organic compound contains only carbon and hydrogen that is used in the production of acetone and phenol in the chemical industry. Combustion of 47.6 mg of cumene produces mg of carbon dioxide and 42.8 mg of water. The molar mass of cumene is between 115 and 125 g/mole. Determine the molecular formula. thus C3H4 thus C9H12

A combustion device was used to determine the empirical formula of an organic compound. A g sample was burned and produced g of carbon dioxide and g of water and no other oxides. Determine the empirical formula for the compound. Compounds that may contain oxygen are a bit trickier. determine mass of C, mass of H check to see if there there is any “missing mass” that would be oxygen back to moles of each element determine empirical formula, molar mass, then molecular formula Yes, Calculators

A combustion device was used to determine the empirical formula of an organic compound. A g sample was burned and produced g of carbon dioxide and g of water and no other oxides. Determine the empirical formula for the compound. thus C7H6O2

Yes, Calculators The Formula is CwHxOyNz.
A confiscated white substance, suspected of being cocaine, was purified by a forensic chemist and subjected to elemental analysis. Combustion of a mg sample yielded mg of carbon dioxide and mg of water. Analysis for nitrogen showed that the compound contained 4.62 % N by mass. Calculate the empirical formula. Yes, Calculators The Formula is CwHxOyNz. Type your answer in as a number wxyz (no spaces, no letters, just one number) i.e. for C3H12O4N2, you would type in 31242

A confiscated white substance, suspected of being cocaine, was purified by a forensic chemist and subjected to elemental analysis. Combustion of a mg sample yielded mg of carbon dioxide and mg of water. Analysis for nitrogen showed that the compound contained 4.62 % N by mass. Calculate the empirical formula. 172141 C17H21O4N

No Calculators CH4 C2H6 C3H4 C3H6 C2H2
50. grams of an unknown hydrocarbon are burned in an excess of oxygen to form 130 grams of carbon dioxide and 54 grams of water. What might this hydrocarbon be? No Calculators CH4 C2H6 C3H4 C3H6 C2H2

No Calculators CH4 C2H6 C3H4 C3H6 C2H2
50. grams of an unknown hydrocarbon are burned in an excess of oxygen to form 130 grams of carbon dioxide and 54 grams of water. What might this hydrocarbon be? CH4 C2H6 C3H4 C3H6 C2H2 No Calculators

The combustion analysis of 19. 8mg of an organic acid produced 39
The combustion analysis of 19.8mg of an organic acid produced 39.6 mg of carbon dioxide and 16.2 mg of water. The molar mass is ~88 g/mole. Determine the molecular formula. Yes, Calculators

Now let’s draw a possible structural formula of this organic acid.
The combustion analysis of 19.8 mg of an organic acid produced 39.6 mg of carbon dioxide and 16.2 mg of water. The molar mass is ~88 g/mole. Determine the molecular formula. thus C2H4O1 MM=44 thus C4H8O2 Now let’s draw a possible structural formula of this organic acid.

O = CH3CH2CH2 -C-O-H All organic acids have a -COOH group.
A possible structural formula of an organic acid with the formula C4H8O2 All organic acids have a -COOH group. Thus the formula below would be an option. Name? -C-O-H O = CH3CH2CH2

This 4-carbon acid would be butanoic acid
A possible structural formula of an organic acid with the formula C4H8O2 All organic acids are carbonroot-oic acid This 4-carbon acid would be butanoic acid -C-O-H O = CH3CH2CH2

Again, type in a numerical answer in order CwHx(and)Oy(if necessary).
Dianabol is one of the anabolic steroids that has been used by some athletes to increase the size and strength of their muscles. It is similar to the male hormone testosterone. Some studies indicate that the desired effects of the drug are minimal, and the side effects, which include sterility, behavior changes, increased risk of liver cancer and heart disease, keep most people from using it. The molecular formula of Dianabol, which consists of carbon, hydrogen, and oxygen, can be determined using the data from two different experiments. In the first experiment, g of Dianabol is burned, and g CO2 and g H2O are formed. In the second experiment, the molecular mass of Dianabol is found to be g/mole. What is the molecular formula for Dianabol Again, type in a numerical answer in order CwHx(and)Oy(if necessary).

The molecular formula of Dianabol, which consists of carbon, hydrogen, and oxygen, can be determined using the data from two different experiments. In the first experiment, g of Dianabol is burned, and g CO2 and g H2O are formed. In the second experiment, the molecular mass of Dianabol is found to be g/mole. What is the molecular formula for Dianabol C20H28O2

Again, type in a numerical answer in order CwHx(and)Oy(if necessary).
One of the additives in unleaded gasoline that replaced tetraethyl lead in leaded gasoline is called MTBE. When g MTBE is burned completely, g CO2 and g H2O form. In a separate experiment the molecular mass of MTBE is found to be What is the molecular formula for MTBE? Again, type in a numerical answer in order CwHx(and)Oy(if necessary).

One of the additives in unleaded gasoline that replaced tetraethyl lead in leaded gasoline is called MTBE. When g MTBE is burned completely, g CO2 and g H2O form. In a separate experiment the molecular mass of MTBE is found to be What is the molecular formula for MTBE? C5H12O

For those of you who have found this last material too easy....
Try these next three challenge problems to stimulate your brain and keep you sharp. These would NOT likely show up on an AP exam.

Type in the atomic number of the element.
Special Challenge Problem #1 When aluminum is heated with an element from Group 6A of the periodic table an ionic compound is formed. When an experiment is performed with an unknownı element of group 6A, the product is 18.56% aluminum by mass. Determine the identity of the reacting element and the formula of the compound. Type in the atomic number of the element.

Special Challenge Problem When aluminum is heated with an element from Group 6A of the periodic table an ionic compound is formed. When an experiment is performed with an unknownı element of group 6A, the product is 18.56% aluminum by mass. Determine the identity of the reacting element and the formula of the compound. Hint #1 Since the element is in group 6A, we know that the formula must be Al2X3

Special Challenge Problem #1 When aluminum is heated with an element from Group 6A of the periodic table an ionic compound is formed. When an experiment is performed with an unknownı element of group 6A, the product is 18.56% aluminum by mass. Determine the identity of the reacting element and the formula of the compound. Hint #2 Set up a ratio to determine the total molar mass of the compound compared to the total mass of aluminum, using the molar masses. 54/MM = 18.56/100% MM = 291

Knowing the molar mass of Al, solve for the MM of X 291 - 54 = 237
Special Challenge Problem #1 When aluminum is heated with an element from Group 6A of the periodic table an ionic compound is formed. When an experiment is performed with an unknownı element of group 6A, the product is 18.56% aluminum by mass. Determine the identity of the reacting element and the formula of the compound. And finally... Knowing the molar mass of Al, solve for the MM of X = 237 X3 = 237 so MM of X = 79 Thus it must be Se

Type in the atomic number of the element.
Special Challenge Problem #2 An element X forms a compound with two chlorine attached (XCl2) and with 4 chlorines attached (XCl4). Treatment of g of XCl2 with excess chlorine forms g of XCl4. Calculate the atomic mass of X and identify which element it is likely to be. Type in the atomic number of the element.

Special Challenge Problem #2 An element X forms a compound with two chlorine attached (XCl2) and with 4 chlorines attached (XCl4). Treatment of g of XCl2 with excess chlorine forms g of XCl4. Calculate the atomic mass of X and identify which element it is likely to be. Hint #1 XCl2 + Cl2 --> XCl4 Determine the mass of Cl2 in XCl4 Then the moles of Cl2 which is the same as the moles of X Which can lead you to the molar mass of X

10 g of XCl2 + Cl2 --> 12.55 g of XCl4
Special Challenge Problem #2 An element X forms a compound with two chlorine attached (XCl2) and with 4 chlorines attached (XCl4). Treatment of g of XCl2 with excess chlorine forms g of XCl4. Calculate the atomic mass of X and identify which element it is likely to be. Hint #2 10 g of XCl2 + Cl2 --> g of XCl4 Thus there must be 2.55 g Cl2 in the XCl4 and thus there must also be 2.55 g Cl2 in the XCl2

Special Challenge Problem #2 An element X forms a compound with two chlorine attached (XCl2) and with 4 chlorines attached (XCl4). Treatment of g of XCl2 with excess chlorine forms g of XCl4. Calculate the atomic mass of X and identify which element it is likely to be. Hint #3 Determine the moles of Cl2 in XCl2, and since there is a 1:1 ratio of X:Cl2, we would know the moles of X. 2.55g of Cl2 * 1 mole Cl2/71 g * 1 mole X/1mole Cl2 = moles X in XCl2,

Special Challenge Problem #2 An element X forms a compound with two chlorine attached (XCl2) and with 4 chlorines attached (XCl4). Treatment of g of XCl2 with excess chlorine forms g of XCl4. Calculate the atomic mass of X and identify which element it is likely to be. and finally... Since 2.55g of the 10g of XCl2 is Cl, 7.45g of the 10g must be X. Calculate the molar mass of X 7.45 g / moles = ~207g/mole. Thus X might be lead.

Type in the % of the CuO compound.
Special Challenge Problem #3 A g sample of a mixture containing only CuO and Cu2O was treated with hydrogen to produce g of pure copper metal. Calculate the percent composition of the mixture. (i.e. What percent of the mixture is each of the two compounds?) Type in the % of the CuO compound.

Special Challenge Problem #3 A 1
Special Challenge Problem #3 A g sample of a mixture containing only CuO and Cu2O was treated with hydrogen to produce g of pure copper metal. Calculate the percent composition of the mixture. (i.e. What percent of the mixture is each of the two compounds?) Hint #1 Solving this problem will require two variables, thus two equations. First equation: The mass of each compound in the mixture will equal the total mass. Second equation: The moles of copper in each compound will equal the moles of copper in the total mixture.

First equation - total mass:
Special Challenge Problem #3 A g sample of a mixture containing only CuO and Cu2O was treated with hydrogen to produce g of pure copper metal. Calculate the percent composition of the mixture. (i.e. What percent of the mixture is each of the two compounds?) First equation - total mass: You could name the mass of Cu2O as x and the mass of CuO as y. The two substances make up the total mass of mixture. xg + yg = 1.500g

Special Challenge Problem #3 A g sample of a mixture containing only CuO and Cu2O was treated with hydrogen to produce g of pure copper metal. Calculate the percent composition of the mixture. (i.e. What percent of the mixture is each of the two compounds?) Second equation - total moles: We know that the moles of copper in the two compounds will add up to the total moles of copper. Xg * 1 mole Cu2O/143.1 g * 2 Cu/1 Cu2O = moles of copper in Cu2O Yg * 1 mole CuO/79.55 g * 1 Cu/1 CuO = moles of copper in CuO 1.252 g Cu * 1 mole/63.55 g = total moles Cu x y =

Special Challenge Problem #3 A g sample of a mixture containing only CuO and Cu2O was treated with hydrogen to produce g of pure copper metal. Calculate the percent composition of the mixture. (i.e. What percent of the mixture is each of the two compounds?) and finally... Simply the last equation x y = 1.112 x + y = 1.567 Subtract the other x y equation. 1.112 x + y = (-x -y = ) equals: x = thus X = 0.6 g Cu2O 0.6 g Cu2O / g = 40% Cu2O in mixture Thus 60% CuO in mixture

Keeping Track of Electrons in Redox Reactions
Oxidation Numbers Keeping Track of Electrons in Redox Reactions

Oxidation Oxidation Numbers aka Oxidation States
A concept devised to keep track of electrons in a redox reaction. Increase in oxidation number = oxidation LEO (Lose Electrons = Oxidation) Decrease in oxidation number = reduction says GER (Gain Electrons = Reduction) The sum of oxidation numbers in neutral compounds must = 0 The sum of oxidation numbers in a polyatomic ion = the charge of the ion

Determining Oxidation #
The charge on a monotomic ion is the Ox # Atoms in elemental form are zero H in H2, atoms in a lump of iron, Fe, P atoms in P4 Nonmetals usually have negative Ox # - but they can be positive. oxygen is −2 in both ionic and molecular compounds except in peroxides in which the oxidation number is -1. hydrogen is always +1 when bonded to nonmetals and always −1 when bonded to metals. fluorine is −1 in all compounds. The other halogens are usually −1 in binary compounds, but when combined with oxygen they are positive and have various oxidation states.

The many oxidation states of manganese, Mn
Mn , Mn = 0 MnCl2 , Mn = +2 MnF3 , Mn = +3 MnO2 , Mn = +4 K3MnO4 , Mn = +5 K2MnO4 , Mn = +6 KMnO4 , Mn = +7 Methylcyclopentadienyl manganese tricarbonyl MnC5H4CH3(CO)3 , Mn = What??

Determine the oxidation number of each element in the following substances
CaCl2 PbO2 Cl2 S8 SO42- SO32- H2O H2O2 hydrogen peroxide AlH3 ClO3-

Determine the oxidation number of each element in the following substances
+2 -1 +4 -2 CaCl2 PbO2 Cl2 S8 SO42− SO32− H2O H2O2 hydrogen peroxide AlH3 ClO3− +4 -2 +1 -2 +1 -1 +3 -1 +6 -2 +5 -2 +6

Determine the oxidation number of phosphorus in Mg2P2O7
Submit a numeric value. If it is negative, put on the - sign. If positive, just leave it.

Determine the oxidation number of phosphorus in Mg2P2O7
+2 +5 -2 Mg2P2O7

Determine the oxidation number of iron in K4Fe(CN)6
Submit a numeric value. If it is negative, put on the - sign. If positive, just leave it.

Determine the oxidation number of iron in K4Fe(CN)6
+1 +2 -1 K4Fe(CN)6 (CN= -1)

Determine the oxidation number of chromium in Na2Cr2O7

Determine the oxidation number of chromium in Na2Cr2O7
+1 +6 -2 Na2Cr2O7

NO N2O NO2 N2H4 NH3 Li3N N2 NO2− NO3−
Nitrogen is the master of multiple oxidation states. Determine the oxidation number of nitrogen in each of the following compounds. NO N2O NO2 N2H4 NH3 Li3N N2 NO2− NO3−

NO N2O NO2 N2H4 NH3 Li3N N2 NO2− NO3−
Nitrogen is the master of multiple oxidation states. Determine the oxidation number of nitrogen in each of the following compounds. +2 -2 +1 -3 NO N2O NO2 N2H4 NH3 Li3N N2 NO2− NO3− +1 -2 +4 -2 +3 -2 -2 +1 +5 -2 -3 +1

Stoichiometry Molarity and more

Similar presentations

Presentation on theme: "Stoichiometry Molarity and more"— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Stoichiometry Molarity and more

Similar presentations

Presentation on theme: "Stoichiometry Molarity and more"— Presentation transcript:

Similar presentations

About project

Feedback