Chemistry 101 - Chapter 3

Products

Substances produced in a chemical reaction; it appears to the right of the arrow in a chemical equation

The unbalanced equation for the reaction between methane and bromine is: __CH4(g) + ))Br2(l) --> CBR4(s) + ___HBr)g) Once this equation is balances what is the value of the coefficient in front of Bromine Br2? a) 4 b) 2 c) 6 d) 3 e) 1

Answer: A

Write the balanced equation for the reaction that occurs when ethylene glycol, C2H4(OH)2 burns in air. (check image)

Answer: B

Calculate the number of moles of glucose (C6H12O6) in a 5.380 g sample

C6H12O6 total mass is = 180 g/mol 5.380 g x (1 mol/ 180 g) = 0.0299 mol

Excess Reactants

Reactant leftovers when a reactions stops.

Formula Weight

(FW); also called the Molecular Weight (MW) The sum of atomic weights (AW) of the atoms in the chemical formula of the substance Ex: H2SO4 H = 1.008 S = 32.06 O = 16.00 = (2 x 1.008) + 32.06 + (4 x 16.00) = 98.08 or 98.1 amu

Symbols for States of Matter

(l) = Liquids (s) = Solids (g) = Gasses (aq) = Aqueous

Isopropyl alcohol, sold as rubbing alcohol, is composed of C, H, and O. Combustion of 0.255 g of isopropyl alcohol produces 0.561 g of CO2 and 0.306 g of H2O. Determine the empirical formula of isopropyl alcohol.

0.561 g CO2 x (1 mol CO2/ 44.0 g CO2) x (1 mol C/1 mol CO2) x (12.0 g C/ 1 mol C) = 0.153 g C 0.306 g H2O x (1 mol H2O/18.0 g H2O) x (2 mol H/1 mol H2O) x (1.01 g H/1 mol H) = 0.0343 g H 0.255 g - 0.0343 g H - 0.153 g C = 0.068 0.152 g C x (1 mol/12.01 g) = 0.0127 mol C 0.0343 g H x (1 mol H/1.008 g) = 0.034 mol H 0.068 g O x (1 mol/16.00 g O) = 0.0043 mol O 0.0127 mol C/0.0043 mol O = 3 0.034 mol H/0.0043 mol O= 8 0.0043 mol O/0.0043 mol O= 1 C3H8O

Without using a calculator, arrange these samples in order of increasing numbers of O atoms: 1 mol H2O 1 mol CO2 3 x10^23 molecules of O3

1 mol H2O = 6.022x10^23 O 1 mol CO2 = 2(6.022x10^23) O = 12x10^23 O 3x10^23 O3 is that number times 3 = 9x10^23 1 mol H2O then 3x10^23 O3 then 1 mol CO2

The reaction 2 H2 (g) + O2 (g) --> 2H2O (g) is used to produce electricity in a hydrogen fuel cell. Suppose a fuel cell contains 150 g of H2 (g) and 1500 g O2 (g) (each measured to two significant figures). How many grams of water can form?

150 g H2 x (1 mol H2/ 2.02 g) x (1 mol O2/2 mol H2) = 37.5 mol O2 original mol of H2 would be 75 mol H2 (following first multiplication/division step) 1500 g O2 x (1 mol O2/ 32 g O2) x (2 mol H2/1 mol O2) = 93.75 mol H2 original mol of O2 would be 46.88 mol O2 (following first multiplication/division step) 75 mol H2 - 93.75 mol H2 = -18.75 mol H2 46.88 mol O2 - 37.5 mol O2 = 9.38 mol O2 Limiting Reagent is H2 of 150 g 150 g H2 x (1 mol H2/ 2.02 g H2) x (2 H2O/2 mol H2) x (18 g/ 1 mol H2O) = 1350 g H2O or 1.3x10^2 H2O

When 24 mol of methanol and 15 mol of oxygen combine in the combustion reaction 2 CH3OH (l) + 3 O2 (g) --> 2 CO2 (g) + 4 H2O (g) What is the excess reactant and how many moles of it remain at the end of the reaction?

24 mol CH3OH x (3 mol O2/2 mol CH3OH) = 36 mol O2 15 - 36 mol O2 = -21 mol O2 15 mol O2 x (2 mol CH3OH/3 mol O) = 10 mol CH3OH 24 mol CH3OH - 10 mol CH3OH = 14 mol CH3OH Limiting Reactant = Oxygen Excess Reactant = Methanol by 14 mol

Methanol, CH3OH, reacts with oxygen from air in a combustion reaction to form water and carbon dioxide. What mass of water is produced in the combustion of 23.6 g of methanol?

2CH3OH + 3O2 -> 2CO2 + 4H2O Molar Mass CH3OH 12 + 4 + 16 = 32 g/mol H2O 2 + 16 = 18 g/mol 23.6 g CH3OH x (1 mol C3HOH/32 g CH3OH) x (4 mol H2O/ 2 mol CH3OH) x (18 g H2O/1 mol H2O) = 26.55 g H2O

Write the balanced equation for the reaction that occurs when methanol, CH3OH (l) is burned in air CH3OH(l) + O2(g) --> CO2(g) + H2O (g)

3 CH3OH(l) + 3 O2(g) --> 2 CO2(g) + 4 H2O(g)

A 5.325-g sample of methyl benzoate, a compound used in the manufacture of perfumes, contains 3.758 g of carbon, 0.316 g of hydrogen, and 1.251 g of oxygen. What is the empirical formula of this substance?

3.758 g C x (1 mol/12.01 g) = 0.313 mol C 0.316 g H x (1 mol/ 1.008 g) = 0.313 mol H 1.251 g O x (1 mol/16.00 g) = 0.078 mol O 0.313 mol C/0.078 mol O = 4 0.313 mol H/0.078 mol O = 4 0.078 mol O/0.078 mol O = 1.0 C:H:O is 4:4:1 Empirical Formula = C4H4O

Molten gallium reacts with arsenic to form the semiconductor, gallium arsenide, GaAs, used in light-emitting diodes and solar cells: Ga(l) + As(s) --> GaAs (s) If 4.00 g gallium is reacted with 5.50 g of arsenic, how many grams of the excess reactant are left at the end of the reaction? a) 1.20 g As b) 1.50 g As c) 4.30 g As d) 8.30 g As

4.00 g Ga x (1 mol Ga/69.72 g Ga)= 0.05737 mol Ga 0.05737 mol Ga x (1 mol As/1 mol Ga) = 0.05737 mol As 5.50 g As x (1 mol As/74.92 g As) = 0.07341 mol Ga 0.07341 mol Ga x (1 mol As/1 mol Ga) = 0.07341 mol Ga *no further work is needed, because they are a 1:1 ration of Ga:As, the Gallium will be used up first, and Arsenic will be the excess by (5.50-4.00=1.50 g) 1.50 g As Answer: B

The compound dioxane, which is used as a solvent in various industrial processes, is composed of C, H, and O atoms. Combustion of a 2.203-g sample of this compound produces 4.401 g CO2 and 1.802 g H2O. A separate experiment shows that it has a molar mass of 88.1 g/mol. Which of the following is the correct molecular formula for dioxane?

4.401 g CO2 x (1 mol CO2/44.0 g CO2) x (1 mol C/1 mol CO2) x (12.01 g C/1 mol C) = 1.2013 g C 1.802 g H2O x (1 mol H2O/18.0 g H2O) x (2 mol H/1 mol H2O) x (1.008 g H/ 1 mol H) = 0.1009 g H 2.203 g - 1.2013 g - 0.1009 g = 0.9008 g O 1.2013 g C x (1 mol C/12.01 g) = 0.10002 mol C 0.1009 g H x (1 mol H/1.008 g) = 0.10009 mol H 0.9008 g O x (1 mol O/ 16.0 g O) = 0.0563 mol O 0.10002 mol C/0.10002 mol C = 1 0.10009 mol H/0.10002 mol C = 1 0.0563 mol O/ 0.10002 mol C = 0.56 2(1:1:0.5)= (2:2:1) C2H2O Empirical Weight = 2(12.01) + 2 + 16 = 42.02 88.1/42.02 = 2 2(C2H2O) = C4H4O2 Answer: B

Which of the following is the balanced chemical reaction for the combination reaction between Lithium metal and Oxygen gas? a) Li(s) + O(g) -> LiO(s) b) 4 Li(s) + O2(g) -> 2Li2O(s) c) Li(s) + O2(g) -> LiO2(s) d) 2Li(s) + O2(g) _> 2LiO(s)

A combination between a metal and nonmetal produces a solid. It takes 2 Li to combine with 1 O because Li forms an Li+ cation, and oxygen forms an O2- anion. Answer: B

Decomposition Reaction

A reaction in which a single substance is broken into two or more substances

Combination Reactions

A reaction in which two or more substances react to form one product

When the following equation is balanced with the smallest set of whole numbers, what is the coefficient of O2 (g)? ___C3H8 (g) + ___ O2(g) --> __CO2(g) + ___H2O(g)

Answer: C3H8(g) + 5 O2(g) --> 3 CO2(g) + 4 H2O(g)

In the following diagram, the white spheres represent hydrogen atoms and the blue spheres represent nitrogen atoms (check image). The two reactants combine to form a single product, ammonium, NH3, which is not shown. Based on the contents of the left (reactants) box, how many NH3 molecules should be shown in the right (products) box? a) 2 b) 3 c) 4 d) 6 e) 9

Answer: D

How many sulfur atoms are in 1.10 mol of Aluminum Sulfide

Aluminum Sulfide = Al2S3 1.10 mol of Al2S3 x (3 mol S/ 1 mol Al2S3) x (6.022x10^23 S/1 mol of S) = 1.99x10^24 atoms of S

Write the balanced equation for the reaction that occurs when ethanol, C2H5OH (l), burns in air. C2H5OH(l) + O2(g) --> CO2(g) + H2O(g)

C2H5OH(l) + 3 O2(g) --> 2 CO2(g) + 3 H2O(g)

What is the percentage of nitrogen, by mass, in calcium nitrate? a) 8.54% b) 17.1% c) 13.7% d) 24.4% e) 82.9%

Ca(NO3)2 % mass = ((# of X)(mass of X))/Total mass of Ca(NO3)2 40.1 + 2(14) + 6(16) = 164.1 amu Ca 40.1/164.1 = 0.244 x 100 = 24.4% N 2(14)/164.1 = 0.171 x 100 = 17.1% O 100 - 24.4 - 17.1 = 58.5% or 6(16)/164.1 = 0.585 x 100 = 58.5% Answer: B

Calculate the mass, in grams , of 0.433 mol of calcium nitrate

Ca(NO3)2 40.1 + 2(14) + 6(16) = 164.1 g/mol 0.433 mol x (164.1 g/mol) = 71.1 g

Which of the following is the correct formula weight for Calcium Phosphate? a) 278.2 amu b) 135.1 amu c) 175.1 amu d) 182.2 amu e) 310.2 amu

Ca3(PO4)2 3(40.1) + 2(31.0) + 8(16) = 310.3 amu Answer: E

A sample of an ionic compound containing iron and chlorine is analyzed and found to have a molar mass of 126.8 g/mol. What is the charge of the iron in this compound? a) 1+ b) 2+ c) 3+ d) 4+

Fe = 55.85 g/mol Cl = 35.45 g/mol 126.8 g/mol Fe is 2+ Cl is - 126.8 - 35.45 + 55.85 = 35.5 FeCl2 Answer: B

When 1.57 mol O2 reacts with H2 to form H2O, how many moles of H2 are consumed in the process?

First write a balanced equation: O2 + 2H2 -> 2H2O 1.57 mol O2 x (2 mol H2/1 mol O2) = 3.14 mol H2

Solid lithium hydroxide is used in space vehicles to remove the carbon dioxide gas exhaled by astronauts. The hydroxide reacts with the carbon dioxide to form solid lithium carbonate and liquid water. How many grams of carbon dioxide can be absorbed by 1.00 g of lithium hydroxide?

LiOH(g) + CO2(g) --> Li2CO3 (s) + H2O (l) First balance the equation: 2LiOH (g) + CO2 (g) --> Li2CO3 (s) + H2O (l) Molar Mass LiOH 6.94 + 16 + 1 = 23.97 g/mol Molar Mass CO2 12.01 + 2(16) = 44.01 g/mol 1.00 g LiOH x (1 mol LiOH/23.97 g LiOH) x (1 mol CO2/2 mol LiOH) x (44.01 g CO2/ 1 mol CO2) = 0.92 g CO2

a) How many nitric acid molecules are in 4.20 g of HNO3? b) How many O atoms are in this sample?

Nitric Acid = HNO3 1 + 14 + 3(16) = 63 g/mol 4.20 g HNO3 x (1 mol HNO3/63 g) x (6.022x10^23 molecules/ 1 mol) = 4.02 x 10^22 molecules of HNO3 b) 4.01x10^22 molecules HNO3 x (3 Oxygen atoms/ 1 molecule HNO3) = 1.2x10^23 O atoms

Could the empirical formula determined from chemical analysis be used to tell the difference between acetylene, C2H2 and benzene, C6H6?

No, because both have the empirical formula of CH

In the molecular level views shown in the figure, how many C, H, and O atoms are present as reactants? Are the same number of each type of atom present as products?

Reactants C = 2 H = 8 O = 8 Products C = 2 H = 8 O = 8

Actual Yield

The amount of product actually obtained from reaction

Elemental Composition

The percentage composition of an element in a substance % mass = ((number of atoms of element) x (atomic weight of element))/ Formula weight of substance

Theoretical Yield

The quantity of product formed when all of the limiting reagent reacts

Law of Conservation of Mass

The total mass of materials present after a chemical reaction is the same as the total mass present before the reaction

Avogadro's Number

The value in which the number of "X" in one mole of the substance/object 6.022x10^23 ex: 1 mol 12-C atoms = 6.022x10^23 12-C atoms 1 mol H2O molecules = 6.022x10^23 molecules 1 mol NO3(-) ions = 6.022x10^23 ions

If 3.00 g of titanium metal is reacted with 6.00 g of chlorine gas, Cl2, to form 7.7 g of titanium (IV) chloride in a combination reaction, what is the percent yield of the product? a) 65% b) 96% c) 48% d) 86%

Ti (s) + 2 Cl2 (g) -> TiCl4 (s) 3.00 g Ti x (1 mol/47.87) = 0.0627 mol Ti 0.0627 mol Ti x (2 mol Cl2/1 mol Ti) = 0.125 mol Cl2 6.00 g Cl2 x (1 mol Cl2/71 g Cl2) = 0.0845 mol Cl2 0.0845 mol Cl2 x (1 mol Ti/2 mol Cl2) = 0.0423 mol Ti 0.0627 - 0.0423 mol Ti = 0.0204 mol Ti 0.0845 - 0.125 mol Cl2 = -0.0405 mol Cl2 6.00 g Cl2 x (1 mol Cl2/71 g Cl2) x (1 mol TiCl4/2 mol Cl2) x (189.9 g TiCl4/1 mol TiCl4) = 8.02 g TiCl4 (7.7 g/8.02g) x 100 = 96% Answer: B

Calculate the percentage of potassium, by mass, in K2PtCl6

Total Mass = 2(39.1) + 195.1 + 6(35.5) = 486.3 amu 2(39.1)/486.3 = 0.161 x 100 = 16.1%

A 508 g sample of sodium bicarbonate (NaHCO3) contains how many moles of Sodium Bicarbonate?

Total mass 23 + 1 + 12 + 3(16) =84 g/mol 508 g x (1 mol/ 84 g) = 6.05 mol

Molecular Formula from Empirical Formula Tips n' Tricks

We can obtain the molecular formula for any compound from its empirical formula if we know either the molecular weight or the molar mass of the compound. The subscripts in the molecular formula of a substance are always whole-number multiples of the subscripts in its empirical formula Ex: Ascorbic Acid C3H4O3 Empirical formula Weight = 3(12.0 amu) + 4(1.0 amu) + 3(16.0 amu) = 88.0 amu The experimentally determined molecular weight is 176 amu SO 176 amu/88.0 amu = 2 Therefore we multiple the subscripts in the empirical formula by this multiple - giving the molecular formula of C6H8O6

Does this reaction produce or consume thermal energy (heat)? C3H8(g) + 5O2(g) --> 3 CO2(g) + 4H2O (g)

Yes, because it is a combustion reaction

Ethylene glycol, used in automobile antifreeze, is 38.7% C, 9.7% H, and 51.6% O by mass. Its molar mass is 62.1 g/mol. a) What is the empirical Formula of Ethylene Glycol?? b) what is its molecular formula?

a) 62.1 g x 0.387 = 24.03 g C 62.1 g x 0.097 = 6.02 g H 62.1 g x 0.516 = 32.04 g O 24.03 g C x (1 mol/12.01 g) = 2.001 mol C 6.03 g H x (1 mol/1.008 g) = 5.98 mol H 32.04 g O x (1 mol/16.00 g) = 2.003 mol O 2.001 mol C/2.001 mol C = 1 5.98 mol H/2.001 mol C = 3 2.003 mol O/2.001 mol C = 1 1:3:1 Empirical Formula = CH3O b) CH3O 12 + 3 + 16 = 31 amu 62.1 (experimental weight)/31 (empirical weight) = 2 2(CH3O) Molecular Formula = C2H6O2

Balance these equations by providing the missing coefficients: a)___Fe(s) + ___O2(g) --> ___Fe2O3(s) b) ___Al(s) + ___HCl(aq) --> ___AlCl3(aq) + ___H2(g) c) ___CaCO3(s) + ___ HCl(aq) --> ___CaCl2(aq) + ___CO2(g) + ___H2O(l)

a) Fe = 4; O2 = 3; Fe2O3 = 2 b) Al = 2; HCl = 6; AlCl3 = 2; H2 = 3 c) CaCO3 = 3; HCl = 6; CaCl2 = 3; CO2 = 3; H2O = 3 simplified: 1, 2, 1, 1, 1,

The following diagram represents a chemical reaction in which the red spheres are oxygen atoms and the blue spheres are nitrogen atoms. a) Write the chemical formulas for the reactants and products. b) Write a balanced equation for the reaction. c) Is the diagram consistent with the law of conservation of mass?

a) NO(g) + O2(g) --> NO2(g) b) 8 NO(g) + 4 O2(g) --> 8 NO2(g) Simplified: 2 NO(g) + O2(g) --> 2 NO2(g) c) Yes

Which of the following samples contains the fewest sodium atoms? a) 1 mol sodium oxide b) 45 g sodium fluoride c) 50 g sodium chloride d) 1 mol sodium nitrate

a) Na2O b) NaF c) NaCl d) NaNO3 answer: D

What is the mass, in grams of a) 6.33 mol of NaHCO3 b) 3.0x10^-5 mol of Sulfuric Acid

a) NaHCO3 23 + 1 + 12 + 3(16) = 84 g/mol 6.33 mol x (84 g/1 mol) = 532 g b) H2SO4 2 + 32.1 + 4(16) = 98.1 g/mol 3.0x10^-5 mol x (98.1 g/ 1 mol) = 2.9x10^-3 g

a) Which has more mass, a mole of water (H2O) or a mole of glucose? (C6H12O6) b) Which contains more molecules, a mole of water or a mole of glucose?

a) The mole of glucose has more mass b) Both of them contain the same amount of molecules (6.022x10^23)

Percentage Composition

the percentage by mass contributed by each element in the substance

Percent Yield

the ratio of the actual (experimental) yield of a product to its theoretical (calculated) yield, multiplied by 100 Check Image for Equation

Calculate the percentage of Carbon, Hydrogen and Oxygen (by mass) in C12H22O11

% mass = ((# of X)(mass of X))/Total mass of C12H22O11 Total mass = 12(12) + 22(1) + 11(16) = 342.0 amu C 12(12.0)/342.0 = 0.42 x 100 = 42.1% H 22(1)/342.0 = 0.064 x 100 = 6.4% O 100 - 42.1 - 6.4 = 51.5 % or 11(16)/342 = 0.515 x 100 = 51.5%

Molecular Weight

(MW) The sum of the atomic weights (AQ) of the atoms represented by their chemical formula for a molecule

Calculate the molar mass of Ca(NO3)2

= 40.1 + 2(14) + 6(16) = 164.1 g/mol

Chemical Equation

A representation of a chemical reaction using the chemical formulas of the reactants and products; a balanced chemical equation contains equal numbers of atoms of each element on both sides of the equation

Mole

The amount of matter that contains as many "X" (atoms, molecules or whatever the object consists of) at the number of in exactly 12 g of isotopically 12-C.

Mass Number and a Mole

The atomic weight of an element in atomic mass units is numerically equal to the mass in grams of 1 mol of that element ex: H has an amu of 1.008 which is also 1.008 g/mol

Molar Mass

The mass of 1 mole of substance in grams; it is numerically equal to the Formula weight in AMU units Ex: NaCl FW 30.0 + 35.5 = 65.5 Molar Mass (g/mol) = FW so it is 65.5 g/mol

Limiting Reactant

The reactant that is completely consumed in a reaction; the amount of product that can form is limited by the complete consumption of the limiting reactant

Stoichiometrically Equivalent Quantities

The relationship between the coefficients and the number of molecules in the BALANCED reaction are the relative numbers of moles

Stoichiometry

The relationships among the quantities of reactants and products involved in chemical reactions

How many oxygen atoms are in a) 0.25 mol Ca(NO3)2 and b) 1.50 mol of Sodium Carbonate?

a) 0.25 mol Ca(NO3)2 x (6 mol O/ 1 mol Ca(NO3)2) x (6.022x10^23 atoms O/1 mol O) = 9.03x10^23 atoms O b) 1.50 mol of Na2CO3 x (3 mol O/1 mol Na2CO3) x (6.022x10^23 atoms O/ 1 mol O) = 2.71x10^24 atoms O

a) Caproic acid, responsible for the odor of dirty socks, is composed of C, H, and O atoms. Combustion of a 0.225-g sample of this compound produces 0.512 g CO2 and 0.209 g H2O. What is the empirical formula of caproic acid? b) Caproic acid has a molar mass of 116 g/mol. What is its molecular formula?

a) 0.512 g CO2 x (1 mol/ 44.0 g CO2) x (1 mol C/1 mol CO2) x (12.01 g/1 mol C) = 0.1397 g C 0.209 g H2O x (1 mol H2O/18.0 g H2O) x (2 mol H/1 mol H2O) x (1.008 g H/1 mol H) = 0.0234 g H 0.225 - 0.1397 - 0.0234 = 0.0619 g O 0.1397 g C x (1 mol/12.01 g) = 0.01163 mol C 0.0234 g H x (1 mol H/1.008 g) = 0.0232 mol H 0.0619 g O x (1 mol O/16 g O) = 0.0039 mol O 0.01163 mol C/0.0039 mol O = 3 0.0232 mol H/0.0039 mol O = 6 0.0039 mol O/ 0.0039 mol O = 1 Empirical Formula is C3H6O b) C3H6O 3(12.01) + 6 + 16 = 58.03 g/mol (116 g/mol)/ (58.03 g/mol) = 2 2(C3H6O)= C6H12O2

a) When 1.50 mol of Al and 3.00 mol of Cl2 combine in the reaction 2 Al(s) + 3 Cl2(g) --> 2 AlCl3 (s), which is the limiting reactant? b) How many moles of AlCl3 are formed? c) How many moles of the excess reactant remain at the end of the reaction?

a) 1.50 mol Al x (3 mol Cl2/ 2 mol Al) = 2.25 mol Cl2 3.00 mol Cl2 x (2 mol Al/3 mol Cl2) = 2 mol Al 1.5 Al - 2 Al = -0.5 Al 3 Cl2 - 2.25 Cl2 = 0.75 mol Cl2 Limiting Reagent is Al b) 1.50 mol Al x (2 AlCl3/2 mol Al) = 1.5 mo AlCl3 c) 0.75 mol Cl2

Calculate the Formula Weight of a) Sucrose, C12H22O11 (table sugar) b) Calcium Nitrate, Ca(NO3)2

a) 12(12.01) + 22(1.01) + 11(16) = 342.3 amu b) 40.1 + 2(14.01) + 6(16) = 164.1 amu

Imagine you are working on ways to improve the process by which iron ore containing Fe2O3 is converted into iron: Fe2O3(s) + 3 CO(g) --> 2 Fe(s) + 3 CO2(g) a) If you start with 150 g of Fe2O3 as the limiting reactant, what is the theoretical yield of Fe? b) If your actual yield is 87.9 g, what is the percent yield?

a) 150 g Fe2O3 x (1 mol/ 159.7 g Fe2O3) x (2 mol Fe/1 mol Fe2CO3) x (55.85 g Fe/ 1 mol Fe) = 105 g Fe(s) b) (87.9 g/105 g) x 100 = 83.7%

When a 2.00 g strip of zinc metal is placed in an aqeous solution containing 2.50 g of silver nitrate, the reaction is Zn(s) + 2 AgNO3 (aq) --> 2 Ag (s) + Zn(NO3)2 (aq) a) Which reactant is limiting? b) How many grams of Ag form? c) How many grams of Zn(No3)2 form? d) How many grams of the excess reactant are left at the end of the reaction?

a) 2.00 g of Zn x (1 mol Zn/65.38 g) = 0.0306 mol Zn 0.0306 mol Zn x (2 mol AgNo3/1 mol Zn) = 0.0612 mol AgNO3 2.50 g AgNO3 x (1 mol AgNO3/169.91 g AgNO3) = 0.0147 mol AgNO3 0.0147 mol AgNO3 x (1 mol Zn/ 2 mol AgNO3) = 0.0074 mol Zn 0.0306 mol Zn - 0.0074 mol Zn = 0.0232 mol Zn 0.0147 mol AgNO3 - 0.0612 mol AgNO3 = -0.0465 mol AgNO3 Limiting Reactant is AgNO3 b) 2.50 AgNO3 x (1 mol AgNO3/169.91 g) x (2 mol Ag/ 2 mol AgNO3) x (107.9 g Ag/1 mol Ag) = 1.588 g Ag c) 2.50 AgNO3 x (1 mol AgNO3/169.91 g AgNO3) x (1 mol Zn(NO3)2/2 mol AgNO3) x (189.4 g Zn(NO3)2/1 mol Zn(NO3)2) = 1.393 g Zn(NO3)2 d) 2.50 AgNO3 x (1 mol AgNO3/169.91 g AgNO3) x (1 mol Zn/2 mol AgNO3) x (65.38 g Zn/1 mol Zn) = 0.481 g Zn used 2.00 g Zn - 0.481 g used Zn = 1.519 g Zm

Adipic acid, H2C6H8O4, used to produce nylon, is made commercially by a reaction between cyclohexane (C6H12) and O2: 2 C6H12(l) + 5 O2(g) --> 2 H2C6H8O4(l) + 2 H2O (g) a) Assume that you carry out this reaction with 25.0 g of cyclohexane and that cyclohexane is the limiting reactant. What is the theoretical yield of adipic acid? b)If you obtain 33.5 g of adipic acid, what is the percent yield for the reaction?

a) 25.0 g C6H12 x (1 mol C6H12/84 g C6H12) x (2 mol H2C6H8O4/2 mol C6H12) x ( 146 g H2C6H8O4/1 mol H2C6H8O4) = 43.5 g H2C6H8O4 b) (33.5 g/43.5 g) x 100 = 77.0%

Calculate the formula weight of a) Al(OH)3 b) CH3OH c) TaON

a) 27.0 + 3(16) + 3(1.0) = 78.0 amu b) 12.0 + 4(1.0) + 16 = 32.0 amu c) 180.9 + 16 + 14 = 210.9 amu

In this reaction, there are two reactants, ethylene, C2H4, which is shown, and oxygen, O2, which is not shown; and two products, CO2 and H2O, both of which are shown. a) Write a balanced chemical equation for the reaction b) Determine the number of O2 molecules that should be shown in the left (reactants box).

a) 3 C2H4(g) + 9 O2(g) --> 6 CO2(g) + 6 H2O(g) Simplest: C2H4(g) + 3O2(g) --> 2CO2(g) + 2H2O(g) b) 9 O2 molecules should be on the left

a) How many glucose molecules are in 5.23 g of C6H12O6? b) How many oxygen atoms are in this sample?

a) 6(12) + 12(1) + 6(16) = 180 g/mol 5.23 g (1 mol/ 180 g) x (6.022x10^23 molecules/ 1 mol) = 1.75x10^22 molecules of Glucose b) 1.75x10^22 molecules of glucose x (6 atoms O/ 1 molecule of glucose) x = 1.05x10^23 atoms of O

What is the mass, in grams, of a) 0.50 mol of diamond (C) b) 0.155 mol of ammonium Chloride

a) C = 12 g/mol 0.50 mol x (12 g/ 1 mol) = 6 g b) NH4Cl 14 + 4(1) + 35.5 = 53.5 g 0.155 mol x (53.5 g/ 1 mol) = 8.29 mol

Mesitylene, a hydrocarbon found in crude oil, has an empirical formula of C3H4 and an experimentally determined molecular weight of 121 amu. What is its molecular formula?

C3H4 = 3(12.01 amu) + 4(1.008 amu) = 40.062 amu for empirical weight 121 amu (molecular weight)/40.062 (empirical weight) = 3 3(C3H4)= C9H12

How many chlorine atoms are in 12.2 g of CCl4? a) 4.77 x 10^22 b) 7.34 x 10^24 c) 1.91 x 10^23 d) 2.07 x 10^23

CCl4 12 + 4(35.5) = 154 g/mol 12.2 g CCl4 x (1 mol/154 g CCl4) x (4 Cl/ 1 CCl4) x (6.022x10^23 atoms Cl/ 1 mol Cl) = 1.91x10^23 atoms of Cl Answer: C

Procedure for Calculating an Empirical Formula from Percentage Composition

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Combustion Analysis

Commonly used for compounds containing principally Carbon and Hydrogen When they are completely combusted, C turns into CO2 and H turns into H2O From the masses of CO2 and H2O, we can calculate the number of moles of C and H in the original sample (empirical formulas). If a third element is present, then its mass can be determined by subtracting the measured masses of C and H from the original sample mass

Balance the equation

2 Na(s) + 2 H2O(l) --> 2 NaOH(aq) + H2(g)

A 2.144-g sample of phosgene, a compound used as a chemical warfare agent during World War I, contains 0.260 g of carbon, 0.347 g of oxygen, and 1.537 g of chlorine. What is the empirical formula of this substance? a) CO2CL6 b) COCL2 c) C(0.022)O(0.022)Cl(0.044) d) C2OCl2

2.144 g Phosphogene 0.260 g C 0.347 g O 1.527 g Cl 0.260 g C x (1 mol/ 12.01 g) = 0.023 mol C 0.347 g O x (1 mol/ 16.00 g) = 0.022 mol O 1.527 g Cl x (1 mol/ 35.45 g) = 0.043 mol Cl 0.023 mol C/0.022 mol O = 1.0 0.022 mol C/0.022 mol O = 1.0 0.043 mol Cl/0.022 mol O = 1.95 --> 2 Empirical Formula 1:1:2 COCL2 Answer: B

The most important commercial process for converting N2 from the air into nitrogen-containing compounds is based on the reaction of N2 and H2 to form ammonia (NH3)L N2(g) + 3H2(g) --> 2NH3 (g) How many moles of NH3 can be formed from 3.0 mol of N2 and 6.0 mol of H2?

3.0 mol N2 x (3 mol H2/ 1 mol N2) = 9.0 mol H2 6.0 mol O2 - 9.0 mol O2 = -3 mol O2 **Limiting reactant 6.0 mol H2 x (1 mol N2/3 mol H2) = 2 mol N2 3 mol N2 - 3 mol N2 = 1 mol N2 H2 is our limiting reactant because it will be used up before the N2 is able to entirely react, (lowest number) Moles NH3 = 6.0 mol H2 x (2 mol NH3/3 mol H2) = 4.0 mol NH3

What is the molar mass of glucose? C6H12O6

6(12) + 12(1) + 6(16) = 180 g/mol

Calculate the number of H atoms in 0.350 mol of C6H12O6

0.350 mol C6H12O6 x (12 H mol/ 1 mol C6H12O6) x (6.022x10^23 atoms/1 mol H) = 2.53x10^24 H atoms

How many sulfur atoms are in 0.45 mol BaSO4?

0.45 mol BaSO4 x (1 mol S/1 mol BaSO4) x (6.022x10^23 atoms/1 mol S) = 2.71x10^23 atoms of S

How many moles of water are in 1.00 L of water, whose density is 1.00 g/mL?

1 L = 1000 mL H2O = 2(1) + 16 = 18 g/mol 1000 ml x (1.00 g/ 1 mL) x (1 mol/ 18 g) = 55.56 mol

In the previous example, 1.00 g of C4H10 reacts with 3.59 g of O2 to form 3.03 g of CO2. Using only addition and subtraction, calculate the amount of H2O produced.

1.00g C4H10 + 3.59g O2 = 4.59 g total to reach 4.59 g - 3.03 g CO2 = 1.56 g of H2O produced

Without using a calculator, arrange these samples in order of increasing numbers of carbon atoms: 12 g 12-C, 1 mol C2H2, 9x10^23 molecules of CO2

12g of 12-C then 1 mol C2H2 9x10^23 then 12 g of C is 1 mol of C = 12x10^23 1 mol of C2H2 contains 6.02x10^23 C2H2 molecules; because there are 2 Carbon, there are double the amount = ~12x10^23 C atoms)

Ascorbic acid (vitamin C) contains 40.92% C, 4.58% H, and 54.50% O by mass. What is the empirical formula of ascorbic acid?

C 40.92% = 40.92 g C H 4.58% = 4.58 g H O 54.50% = 54.50 g O 40.92 g x (1 mol/ 12.01 g) = 3.407 mol C 4.58 g x (1 mol/ 1.008 g) = 4.54 mol H 54.50 g x (1 mol/ 16.00 g) = 3.406 mol O 4.54 mol H/3.406 O = 1.3 mol or 1:1 for H:O 3.307 mol C/3.306 O = 1.0 mol or 1:1 for C:O 3.306 mol O/3.306 mol O = 1.0 Because the ratio of H:O is 1.3, we have to multiple each by 3 (1.3 is too far from 1 to account for error so we can't round it to 1) Therefore: C:H:O = (3x1): (1.3 x 3): (3x1) = (3:4:3) C3H4O3

Cyclohexane, a commonly used organic solvent, is 85.6% C and 14.4% H by mass with a molar mass of 84.2 g/mol. What is its molecular formula?

C = 12.01 amu H = 1.008 amu 84.2 x 0.856 = 72.07 g of C 84.2 x 0.144 = 12.1 g of H 72.07 g C x (1 mol/12.01) = 5.99 mol C 12.1 g H x (1 mol/ 1.008 g) = 12.00 mol 12.00 mol H/6.00 mol C = 2 CH2 12 + 2 = 14 empirical weight (84.2 g/mol)/(14 g/mol) = 6 Molecular Formula is 6(CH2) = C6H12 Answer: D

Reactants

Starting substances in a chemical reaction; appears to the left of the arrow in a chemical reaction

Propane, C3H8, is a common fuel used for cooking and home heating. What mass of O2 is consumed in the combustion of 1.00 g of propane? a) 5.00 g b) 0.726 g c) 2.18 g d) 3.63 g

First Balance Equation - C3H8 + 5 O2 --> 3 CO2 + 4 H2O Molar Mass C3H8 3(12.01) + 8 = 44.03 g/mol O2 2(16) = 32 g/mol 1.00 g C3H8 x (1 mol C3H8/ 44.03 g C3H8) x (5 mol O2/ 1 mol C3H8) x (32 g O2/ 1 mol O2) = 3.63 g Answer: D

If the amount of H2 is doubled, how many moles of H2 would have formed?

If originally the H2 amount was 10 mol, and 7 mol of O2. When the H2 is doubled (20 mol H2) Moles of O2 = 20 mol H2 x (1 mol O2/2 mol H2) = 10 mol O2 7 mol O2 - 10 mol O2 = -3 mol O2 All Oxygen would react, but there would be excess Hydrogen

How many atoms of Mg, O, and H are represented by the notation 3 Mg(OH)2?

Mg = 3 O = 6 H = 6

Determine how many grams of water are produced in the oxidation of 1.00 g of glucose, C6H12O6: C6H12O6 (s) + 6 O2 (g) --> CO2 (g) + 6 H2O (l)

Molar Mass 6(12) + 12 + 6(16) = 180 g/mol 1.00 g C6H12O6 x (1 mol C6H12O6/180 g) x (6 mol H2O/1 mol C6H12O6) x (18.0 g H2O/1 mol H2O) = 0.60 g H2O

Decomposition of KClO3 is sometimes used to prepare small amounts of O2 in the laboratory: 2 KClO3 (s) + 2 KCl (s) + 3 O2 (g) How many grams of O2 can be prepared from 4.50 g KClO3?

Molar Mass KClO3 39.1 + 35.5 + 3(16) = 122.6 g/mol Molar Mass O2 2(16) = 32 g/mol 4.50 g KClO3 x (1 mol KClO3/ 122.6 g) x (3 mol O2/ 2 mol KClO3) x (32 g/ 1 mol O2) = 1.77 g O2

Sodium Hydroxide reacts with carbon dioxide to form sodium carbonate and water: 2 NaOH(s) + CO2 (g) --> Na2CO3(s) + H2O (l) How many grams of Na2CO3 can be prepared from 2.40 g of NaOH? a) 3.18 g b) 6.36 g c) 1.20 g d) 0.0300 g

Molar Mass NaOH 22.99 + 16 + 1 = 39.99 g/mol Molar Mass Na2CO3 2(22.99) + 12.01 + 3(16) = 105.99 g/mol 2.40 g NaOH x (1 mol NaOH/39.99 g NaOH) x (1 mol Na2CO3/ 2 mol NaOH) x (105.99 g/1 mol Na2CO3) = 3.18 g Na2CO3 answer: A

What units would you put under "molar mass" and "Avogadro's number" on this diagram?

Molar Mass would be g/mol; Avogadro's number is normally the unit for the smallest quantifiable substance (atoms, etc)

Combustion Reactions

Rapid reactions that produce a flame; most combustion involves a reactions with oxygen Note: The combustion of Hydrocarbons always result in CO2 and H2O