ap chemistry chapter 3 mr. solsman. goals of this chapter include: relating mass to the number of...

AP Chemistry

Chapter 3

Mr. Solsman

• Goals of this chapter include:

• Relating mass to the number of entities

• Converting mass into a chemical equation

• In short—developing an understanding of the mole

• We typically measure objects by counting individual units, counting in units, or weighing them.

• However atoms and molecules are too small to count individually.

• A unit called the mole has been devised to count chemical entities by weighing them.

• The mole (mol) is the SI unit for the amount of substance.

• The amount of a substance that contains the same number of entities as there are atoms in exactly 12 g of carbon-12.

• This number is also called Avogadro’s number.

• One mole contains 6.022 x 1023 entities.

• 1 mol carbon-12 contains 6.022 x 1023 atoms.

• 1 mol H2O contains 6.022 x1023 molecules.

• 1 mol NaCl contains 6.022 x1023 formula units.

• How big is a mole???

• How big is a mole???

• 1 mole of periods lined up equals the radius of our galaxy!

• How big is a mole???

• 1 mole of periods lined up equals the radius of our galaxy!

• One mole of marbles would cover the USA 70 miles deep!

• How big is a mole???

• 1 mole of periods lined up equals the radius of our galaxy!

• One mole of marbles would cover the USA 70 miles deep!

• One mole of water (18 mL) can easily be swallowed in one gulp!

• The atomic mass of an element expressed in amu’s is the same as the mass of 1 mole of atoms of the element expressed in grams.

• 1 Fe atom has a mass of 55.85 amu and 1 mol of Fe atoms has a mass of 55.85 grams.

• 1 Cl atom has a mass of 35.45 amu and 1 mol of Cl atoms has mass of 35.45 grams.

• Also 1 mol of Fe atoms contains exactly 6.022 x 1023 atoms.

• The same relationship holds for compounds:

• 1 molecule of H2SO4 has a mass of 98.08 amu and 1 mol of H2SO4 has a mass of 98.08 grams and contains 6.022 x 1023 molecules.

• Two Key Points

• 1. The mole maintains the same mass relationship between macroscopic samples as exists between individual chemical entities.

• 2. The mole relates the number of chemical entities to the mass of a sample of those entities.

• The Molar Mass (MM) is the mass per mole of its entities (atoms, molecules, or formula units). Units are grams per mole (g/mol).

• Juglone, a dye, is produced from the husks of black walnuts. The formula for Juglone is C10H6O3. Calculate the molar mass of Juglone.

• Juglone, a dye, is produced from the husks of black walnuts. The formula for Juglone is C10H6O3. Calculate the molar mass of Juglone.

• 10(12.01) + 6(1.008) + 3(16.00) = 174.1 g

• Silver is used in jewelry, tableware, and coins. How many grams of silver are in 0.0342 mol of Ag?

• Silver is used in jewelry, tableware, and coins. How many grams of silver are in 0.0342 mol of Ag?

• 3.69 g Ag

• How many iron atoms are in 95.8 g Fe?

• How many iron atoms are in 95.8 g Fe?

• 1.03 x 1024 atoms

• A crystalline form of carbon called graphite is used in “lead” pencils. How many moles of carbon are in 31.5 mg of graphite?

• A crystalline form of carbon called graphite is used in “lead” pencils. How many moles of carbon are in 31.5 mg of graphite?

• 2.63 x 10-3 mole carbon

• The transition element manganese is essential for bone growth. What is the mass in grams of 3.22 x1020 Mn atoms found in 1 kg of bone?

• The transition element manganese is essential for bone growth. What is the mass in grams of 3.22 x1020 Mn atoms found in 1 kg of bone?

• 0.0294 g Mn

• Mass Percent

• The molecular or formula mass and the chemical formula are used to find the mass percent of any element present in a compound.

• Mass % of element X =

• (grams of X divided by

• mass of 1 mol of compound in grams) times 100

• (The individual mass percents of the elements in the compound must add up to 100%.)

• Glucose (C6H12O6) is used by living cells to generate chemical energy. What is the mass percent of each element in glucose?

• Glucose (C6H12O6) is used by living cells to generate chemical energy. What is the mass percent of each element in glucose?

• Mass % C = 40.00 %

• Mass % H = 6.714 %

• Mass % O = 53.29 %

• How many grams of carbon are in 16.55 grams of glucose?

• How many grams of carbon are in 16.55 grams of glucose?

• 6.620 g C

• Fleming was never able to isolate penicillin in its pure form. It has the formula C14H20N2SO4. Compute the mass percent of each element.

Mass Percent II

• Fleming was never able to isolate penicillin in its pure form. It has the formula C14H20N2SO4. Compute the mass percent of each element.

• C = 53.82%• H = 6.453%• N = 8.969 %• S = 10.27%• O = 20.49%

Amount (mol) of element Xin 1 mol of compound

Mass (g) of X in 1 mol of compound

Divide by mass of 1 mol compound

Mass fraction of X

X 100 %

Mass % of X

• Empirical Formulas

• The simplest whole-number ratio of the moles of each element in a compound.

Mass (g) of each element

Divided by MM (g/mol)

Amount (mol) of each element

Use moles as subscripts

Preliminary formula

Change to integer subscripts

Empirical Formula

• Elemental analysis of a sample of an ionic compound gave the following results:

• 2.82 g Na, 4.25 g Cl, and 7.83 g O. What is the empirical formula and the name of this compound?

• Elemental analysis of a sample of an ionic compound gave the following results:

• 2.82 g Na, 4.25 g Cl, and 7.83 g O. What is the empirical formula and the name of this compound?

• NaClO4 sodium perchlorate

• An unknown metal M reacts with sulfur to form a compound M2S3. If 3.12 g of M reacts with 2.88 g S, what are the names of M and M2S3?

• An unknown metal M reacts with sulfur to form a compound M2S3. If 3.12 g of M reacts with 2.88 g S, what are the names of M and M2S3?

• M is chromium

• Cr2S3 chromium(III) sulfide

• One of the most widespread environmental carcinogens is benzo[a]pyrene (MM = 252.30 g/mol). It is found in coal dust, cigarette smoke, and even charcoal-grilled meats. Analysis shows 95.21 mass % C and 4.790 mass % H. What is the molecular formula of this hydrocarbon?

• One of the most widespread environmental carcinogens is benzo[a]pyrene (MM = 252.30 g/mol). It is found in coal dust, cigarette smoke, and even charcoal-grilled meats. Analysis shows 95.21 mass % C and 4.79 mass % H. What is the molecular formula of this hydrocarbon?

• C20H12

• Empirical formulas tell the relative number of each type of atom.

• Molecular formulas tell the actual number of each type of atom.

• The catch is that different compounds can also have the same molecular formula. That is why structural formulas are important.

• Suppose a chemist uses combustion analysis to determine the EF and MF of vitamin C (MM = 176.12 g/mol).

• The following data was taken using a 1.000 g sample of vitamin C.

• Mass of CO2 absorber after combustion was 85.35 g and before combustion was 83.85 g

• Mass of H2O absorber after combustion was 37.96 g and before combustion was 37.55 g.

• What is Vitamin C’s molecular formula?

Chemical Equations

• A chemical change involves a reorganization of the atoms in one or more substances.

• In a chemical reaction, atoms are neither created or destroyed. All atoms present as reactants must be accounted for among the products.

• Making sure this rule is obeyed is called balancing a chemical equation for a reaction..

• 1. The coefficients in chemical equations represent numbers of molecules, not masses of molecules.

• 2. Counting is done in masses since we can’t count molecules directly.

• 3. The mole ratio is the ‘bridge” which allows us to determine a quantity of one substance given info about another.

• In the destruction of marble statuary by acid rain aqueous nitric acid reacts with calcium carbonate to form carbon dioxide, water, and aqueous calcium nitrate. Write the balanced chemical equation.

• In the destruction of marble statuary by acid rain aqueous nitric acid reacts with calcium carbonate to form carbon dioxide, water, and aqueous calcium nitrate. Write the balanced chemical equation.

• CaCO3(s) + 2 HNO3(aq) CO2(g) + H2O(l)

• + Ca(NO3)2(aq)

• In an engine, octane (C8H18) mixes with oxygen from the air and burns to form CO2 and water. Write a balanced chemical equation for this reaction.

• In an engine, octane (C8H18) mixes with oxygen from the air and burns to form CO2 and water. Write a balanced chemical equation for this reaction.

• 2 C8H18 + 25 O2 16 CO2 + 18 H2O

• Stoichiometry

• If you know the number of moles of one of the substances in a reaction, the balanced chemical equation tells you the number of moles of all the other substances in the reaction.

• 1. Balance the equation for the reaction.• 2. Convert the known mass of the reactant or

product to moles of that substance.• 3. Use the balanced equation to set up the

appropriate mole ratios.• 4. Use the appropriate mole ratios to calculate the

number of moles of the desired reactant or product.

• 5. Covert the mole ratio back to grams, if required.

• Copper is obtained from copper(I) sulfide in a multistage process. The first step is to roast the ore with oxygen gas to form powered copper(I) oxide and gaseous sulfur dioxide. (a) How many moles of O2 are needed to completely roast 10.0 mol of copper(I) sulfide?

• Copper is obtained from copper(I) sulfide in a multistage process. The first step is to roast the ore with oxygen gas to form powered copper(I) oxide and gaseous sulfur dioxide. (a) How many moles of O2 are needed to completely roast 10.0 mol of copper(I) sulfide? 15.0 mol O2

• (b) How many grams of sulfur dioxide are formed when 10.0 mol of copper(I) sulfide are roasted?

• (b) How many grams of sulfur dioxide are formed when 10.0 mol of copper(I) sulfide is roasted? 641 g SO2

• (c) How many kilograms of O2 are required to form 2.86 kg of Copper(I) oxide?

• (b) How many grams of sulfur dioxide are formed when 10.0 mol of copper(I) sulfide is roasted? 641 g SO2

• (c) How many kilograms of O2 are required to form 2.86 kg of Copper(I) oxide?

• 0.959 kg O2

• Thermite is a mixture of iron(III) oxide and aluminum powders used in welding. (It undergoes a spectacular reaction to yield solid aluminum oxide and molten iron.)

• (a) How many grams of iron form when 135 g of aluminum reacts?

• (b) How many atoms of Al react for every 1.00 g of aluminum oxide that is formed?

• Fe2O3(s) + 2 Al(s) Al2O3(s) + 2 Fe(l)

• Fe2O3(s) + 2 Al(s) Al2O3(s) + 2 Fe(l)

• 279 g Fe

• Fe2O3(s) + 2 Al(s) Al2O3(s) + 2 Fe(l)

• 279 g Fe

• 1.18 x 1022 Al atoms

• Baking soda (NaHCO3) is often used as an antacid. It neutralizes excess hydrochloric acid in the stomach:

• NaHCO3 + HCl H2O + CO2 + NaCl

• Milk of magnesia is also used as an antacid. Mg(OH)2 + 2HCl 2H2O + MgCl2

• Which is the most effective per gram?

Antacid

• 1.00 g of Mg(OH)2 will neutralize 3.42 x 10-2 mol HCl. It is the better choice.

• Limiting Reactants (Reagents)

• So far we have assumed that there is enough reactants present that they are all used up in reactions. In practice however, there is usually one in excess and so the other one(s) limit how much product can be made from the one in excess.

Limiting Reagents

• The limiting reactant, therefore, limits or determines the amount of product that can be formed in a reaction.

• Excess reagents are not completely reacted.

• When working problems, all quantities must be converted to moles first before the amount of product, based on the moles of the limiting reagent can be calculated.

• Suppose we mix hydrazine (N2H4) and dinitrogen tetraoxide to form nitrogen gas and water vapor. How many grams of nitrogen gas form when 1.00 x 102 grams of N2H4 and 2.00 x 102 grams N2O4 are mixed?

• Suppose we mix hydrazine (N2H4) and dinitrogen tetraoxide to form nitrogen gas and water vapor. How many grams of nitrogen gas form when 1.00 x 102 grams of N2H4 and 2.00 x 102 grams N2O4 are mixed?

• 2 N2H4(l) + N2O4(l) 3 N2(g) + 4 H2O(g)

• Suppose we mix hydrazine (N2H4) and dinitrogen tetraoxide to form nitrogen gas and water vapor. How many grams of nitrogen gas form when 1.00 x 102 grams of N2H4 and 2.00 x 102 grams N2O4 are mixed? 131 g N2

• 2 N2H4(l) + N2O4(l) 3 N2(g) + 4 H2O(g)

• How many grams of solid aluminum sulfide can be prepared by the reaction of 10.0 g of aluminum and 15.0 g of sulfur? How much of the non-limiting reagent is in excess?

• How many grams of solid aluminum sulfide can be prepared by the reaction of 10.0 g of aluminum and 15.0 g of sulfur? How much of the non-limiting reagent is in excess?

• 16 Al(s) + 3 S8(s) 8 Al2S3(s)

• How many grams of solid aluminum sulfide can be prepared by the reaction of 10.0 g of aluminum and 15.0 g of sulfur? How much of the non-limiting reagent is in excess?

• 16 Al(s) + 3 S8(s) 8 Al2S3(s)

• 23.4 g Al2S3 1.59 g Al left over

Calculating Percent Yield

• In theory, we always hope for a 100% yield.

• But in practice this is nearly impossible.

• The theoretical yield is the maximum amount of a product that could be formed from the given amounts of reactants.

• The actual yield is the amount that actually is formed.

Percent Yield

• The percent yield is the ratio of the actual yield to the theoretical yield expressed as a percent.

• % Yield = (actual yield/theory yield) x 100%

Reasons for Yield

• We get less than 100 % for the following reasons:

• 1. Reactions don’t always go to completion.

• 2. Impure reactants and side reactions• 3. Loss of product during filtration,

transferring, evaporation, etc• 4. Human measuring error

• Marble (calcium carbonate) reacts with hydrochloric acid to form calcium chloride solution, water, and carbon dioxide. What is the percent yield of CO2 when 3.65 g of the gas is collected when 10.0 g of marble reacts?

• Marble (calcium carbonate) reacts with hydrochloric acid to form calcium chloride solution, water, and carbon dioxide. What is the percent yield of CO2 when 3.65 g of the gas is collected when 10.0 g of marble reacts?

• CaCO3(s) + 2 HCl(aq) CaCl2(aq) + H2O(l) + CO2(g)

• Marble (calcium carbonate) reacts with hydrochloric acid to form calcium chloride solution, water, and carbon dioxide. What is the percent yield of CO2 when 3.65 g of the gas is collected when 10.0 g of marble reacts?

• CaCO3(s) + 2 HCl(aq) CaCl2(aq) + H2O(l) + CO2(g) 4.40 g CO2 or 83 %