Chapter 6 Chemical Quantities.

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Presentation transcript:

Chapter 6 Chemical Quantities

How you measure how much? You can measure mass, or volume, or you can count pieces. We measure mass in grams. We measure volume in liters. We count pieces in MOLES.

Moles Defined as the number of carbon atoms in exactly 12 grams of carbon-12. 1 mole is 6.02 x 1023 particles. Treat it like a very large dozen 6.02 x 1023 is called Avagadro’s number.

Representative particles The smallest pieces of a substance. For a molecular compound it is a molecule. For an ionic compound it is a formula unit. For an element it is an atom.

Types of questions How many oxygen atoms in the following? CaCO3 Al2(SO4)3 How many ions in the following? CaCl2 NaOH

Types of questions How many molecules of CO2 are the in 4.56 moles of CO2 ? How many moles of water is 5.87 x 1022 molecules? How many atoms of carbon are there in 1.23 moles of C6H12O6 ? How many moles is 7.78 x 1024 formula units of MgCl2?

Measuring Moles Remember relative atomic mass? The amu was one twelfth the mass of a carbon 12 atom. Since the mole is the number of atoms in 12 grams of carbon-12, the decimal number on the periodic table is also the mass of 1 mole of those atoms in grams.

Gram Atomic Mass The mass of 1 mole of an element in grams. 12.01 grams of carbon has the same number of pieces as 1.008 grams of hydrogen and 55.85 grams of iron. We can right this as 12.01 g C = 1 mole We can count things by weighing them.

Examples How much would 2.34 moles of carbon weigh? How many moles of magnesium in 24.31 g of Mg? How many atoms of lithium in 1.00 g of Li? How much would 3.45 x 1022 atoms of U weigh?

What about compounds? in 1 mole of H2O molecules there are two moles of H atoms and 1 mole of O atoms To find the mass of one mole of a compound determine the moles of the elements they have Find out how much they would weigh add them up

What about compounds? What is the mass of one mole of CH4? 1 mole of C = 12.01 g 4 mole of H x 1.01 g = 4.04g 1 mole CH4 = 12.01 + 4.04 = 16.05g The Gram Molecular mass of CH4 is 16.05g The mass of one mole of a molecular compound.

Gram Formula Mass The mass of one mole of an ionic compound. Calculated the same way. What is the GFM of Fe2O3? 2 moles of Fe x 55.85 g = 111.70 g 3 moles of O x 16.00 g = 48.00 g The GFM = 111.70 g + 48.00 g = 159.70g

Molar Mass The generic term for the mass of one mole. The same as gram molecular mass, gram formula mass, and gram atomic mass.

Examples Calculate the molar mass of the following and tell me what type it is. Na2S N2O4 C Ca(NO3)2 C6H12O6 (NH4)3PO4

Finding moles of compounds Counting pieces by weighing Using Molar Mass Finding moles of compounds Counting pieces by weighing

Molar Mass The number of grams of 1 mole of atoms, ions, or molecules. We can make conversion factors from these. To change grams of a compound to moles of a compound.

For example How many moles is 5.69 g of NaOH?

For example How many moles is 5.69 g of NaOH?

For example How many moles is 5.69 g of NaOH? need to change grams to moles

For example How many moles is 5.69 g of NaOH? need to change grams to moles for NaOH

For example How many moles is 5.69 g of NaOH? need to change grams to moles for NaOH 1mole Na = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.01 g

For example How many moles is 5.69 g of NaOH? need to change grams to moles for NaOH 1mole Na = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.01 g 1 mole NaOH = 40.00 g

For example How many moles is 5.69 g of NaOH? need to change grams to moles for NaOH 1mole Na = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.01 g 1 mole NaOH = 40.00 g

For example How many moles is 5.69 g of NaOH? need to change grams to moles for NaOH 1mole Na = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.01 g 1 mole NaOH = 40.00 g

Examples How many moles is 4.56 g of CO2 ? How many grams is 9.87 moles of H2O? How many molecules in 6.8 g of CH4? 49 molecules of C6H12O6 weighs how much?

Gases and the Mole

Gases Many of the chemicals we deal with are gases. They are difficult to weigh. Need to know how many moles of gas we have. Two things effect the volume of a gas Temperature and pressure Compare at the same temp. and pressure.

Standard Temperature and Pressure 0ºC and 1 atm pressure abbreviated STP At STP 1 mole of gas occupies 22.4 L Called the molar volume Avagadro’s Hypothesis - at the same temperature and pressure equal volumes of gas have the same number of particles.

Examples What is the volume of 4.59 mole of CO2 gas at STP? How many moles is 5.67 L of O2 at STP? What is the volume of 8.8g of CH4 gas at STP?

Density of a gas D = m /V for a gas the units will be g / L We can determine the density of any gas at STP if we know its formula. To find the density we need the mass and the volume. If you assume you have 1 mole than the mass is the molar mass (PT) At STP the volume is 22.4 L.

Examples Find the density of CO2 at STP. Find the density of CH4 at STP.

The other way Given the density, we can find the molar mass of the gas. Again, pretend you have a mole at STP, so V = 22.4 L. m = D x V m is the mass of 1 mole, since you have 22.4 L of the stuff. What is the molar mass of a gas with a density of 1.964 g/L? 2.86 g/L?

All the things we can change

We have learned how to change moles to grams moles to atoms moles to formula units moles to molecules moles to liters molecules to atoms formula units to atoms formula units to ions

Mass Moles

Mass PT Moles

Mass Volume PT Moles

Mass Volume 22.4 L PT Moles

Mass Volume 22.4 L PT Moles Representative Particles

Mass Volume 22.4 L PT Moles 6.02 x 1023 Representative Particles

Mass Volume 22.4 L PT Moles 6.02 x 1023 Representative Particles Atoms

Mass Volume Moles 6.02 x 1023 Representative Particles Ions Atoms PT Moles 6.02 x 1023 Representative Particles Ions Atoms

Percent Composition Like all percents Part x 100 % whole Find the mass of each component, divide by the total mass.

Example Calculate the percent composition of a compound that is 29.0 g of Ag with 4.30 g of S.

Getting it from the formula If we know the formula, assume you have 1 mole. Then you know the pieces and the whole.

Examples Calculate the percent composittion of C2H4? Aluminum carbonate.

From percentage to formula Empirical Formula From percentage to formula

The Empirical Formula The lowest whole number ratio of elements in a compound. The molecular formula the actual ration of elements in a compound. The two can be the same. CH2 empirical formula C2H4 molecular formula C3H6 molecular formula H2O both

Calculating Empirical Just find the lowest whole number ratio C6H12O6 CH4N It is not just the ratio of atoms, it is also the ratio of moles of atoms. In 1 mole of CO2 there is 1 mole of carbon and 2 moles of oxygen. In one molecule of CO2 there is 1 atom of C and 2 atoms of O.

Calculating Empirical Means we can get ratio from percent composition. Assume you have a 100 g. The percentages become grams. Can turn grams to moles. Find lowest whole number ratio by dividing by the smallest.

Example Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N. Assume 100 g so 38.67 g C x 1mol C = 3.220 mole C 12.01 gC 16.22 g H x 1mol H = 16.09 mole H 1.01 gH 45.11 g N x 1mol N = 3.219 mole N 14.01 gN

Example The ratio is 3.220 mol C = 1 mol C 3.219 molN 1 mol N The ratio is 16.09 mol H = 5 mol H 3.219 molN 1 mol N C1H5N1 A compound is 43.64 % P and 56.36 % O. What is the empirical formula? Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula?

Empirical to molecular Since the empirical formula is the lowest ratio the actual molecule would weigh more. By a whole number multiple. Divide the actual molar mass by the the mass of one mole of the empirical formula. Caffeine has a molar mass of 194 g. what is its molecular mass?

Example A compound is known to be composed of 71.65 % Cl, 24.27% C and 4.07% H. Its molar mas is known (from gas density) is known to be 98.96 g. What is its molecular formula?

Chapter 8

Stoichiometry Greek for “measuring elements” The calculations of quantities in chemical reactions based on a balanced equation. We can interpret balanced chemical equations several ways.

In terms of Particles Atom - Element Molecule Formula unit Molecular compound (non- metals) or diatomic (O2 etc.) Formula unit Ionic Compounds (Metal and non-metal)

2H2 + O2 ® 2H2O Two molecules of hydrogen and one molecule of oxygen form two molecules of water. 2 Al2O3 ® 4Al + 3O2 2 formula units Al2O3 form 4 atoms Al and 3 molecules O2 2Na + 2H2O ® 2NaOH + H2

2Na + 2H2O ® 2NaOH + H2

Look at it differently 2H2 + O2 ® 2H2O 2 dozen molecules of hydrogen and 1 dozen molecules of oxygen form 2 dozen molecules of water. 2 x (6.02 x 1023) molecules of hydrogen and 1 x (6.02 x 1023) molecules of oxygen form 2 x (6.02 x 1023) molecules of water. 2 moles of hydrogen and 1 mole of oxygen form 2 moles of water.

In terms of Moles 2 Al2O3 ® 4Al + 3O2 The coefficients tell us how many moles of each kind

In terms of mass The law of conservation of mass applies We can check using moles 2H2 + O2 ® 2H2O 2.02 g H2 2 moles H2 = 4.04 g H2 1 moles H2 32.00 g O2 1 moles O2 = 32.00 g O2 1 moles O2 36.04 g reactants 36.04 g reactants

In terms of mass 2H2 + O2 ® 2H2O 2H2 + O2 ® 2H2O 18.02 g H2O 2 moles H2O = 1 mole H2O 2H2 + O2 ® 2H2O 36.04 g (H2 + O2) = 36.04 g H2O

Your turn Show that the following equation follows the Law of conservation of mass. 2 Al2O3 ® 4Al + 3O2

Mole to mole conversions 2 Al2O3 ® 4Al + 3O2 every time we use 2 moles of Al2O3 we make 3 moles of O2 2 moles Al2O3 3 mole O2 or 3 mole O2 2 moles Al2O3

Mole to Mole conversions How many moles of O2 are produced when 3.34 moles of Al2O3 decompose? 2 Al2O3 ® 4Al + 3O2 3.34 moles Al2O3 3 mole O2 = 5.01 moles O2 2 moles Al2O3

Your Turn 2C2H2 + 5 O2 ® 4CO2 + 2 H2O If 3.84 moles of C2H2 are burned, how many moles of O2 are needed? How many moles of C2H2 are needed to produce 8.95 mole of H2O? If 2.47 moles of C2H2 are burned, how many moles of CO2 are formed?

Mole to Mole Conversions 2C2H2 + 5 O2 ® 4CO2 + 2 H2O How many moles of C2H2 are needed to produce 8.95 mole of H2O? If 2.47 moles of C2H2 are burned, how many moles of CO2 are formed?

We can’t measure moles!! What can we do? We can convert grams to moles. Periodic Table Then use moles to change chemicals Balanced equation Then turn the moles back to grams. Periodic table

Periodic Table Balanced Equation Periodic Table Mass g A MolesA MolesB Mass g B Decide where to start based on the units you are given and stop based on what unit you are asked for

Conversions 2C2H2 + 5 O2 ® 4CO2 + 2 H2O How many moles of C2H2 are needed to produce 8.95 g of H2O? If 2.47 moles of C2H2 are burned, how many g of CO2 are formed?

For example... If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how much solid copper would form? Fe + CuSO4 ® Fe2(SO4)3 + Cu 2Fe + 3CuSO4 ® Fe2(SO4)3 + 3Cu 1 mol Fe 63.55 g Cu 10.1 g Fe 3 mol Cu 55.85 g Fe 2 mol Fe 1 mol Cu = 17.3 g Cu

2Fe + 3CuSO4 ® Fe2(SO4)3 + 3Cu 3 mol Cu 0.272 mol Cu 0.181 mol Fe = 2 mol Fe 63.55 g Cu 0.272 mol Cu = 17.3 g Cu 1 mol Cu

Could have done it 1 mol Fe 63.55 g Cu 10.1 g Fe 3 mol Cu 55.85 g Fe = 17.3 g Cu

More Examples To make silicon for computer chips they use this reaction SiCl4 + 2Mg ® 2MgCl2 + Si How many moles of Mg are needed to make 9.3 g of Si? 3.74 mol of Mg would make how many moles of Si? How many grams of MgCl2 are produced along with 9.3 g of silicon?

For Example The U. S. Space Shuttle boosters use this reaction 3 Al(s) + 3 NH4ClO4 ® Al2O3 + AlCl3 + 3 NO + 6H2O How much Al must be used to react with 652 g of NH4ClO4 ? How much water is produced? How much AlCl3?

How do you get good at this?

Gases and Reactions

We can also change Liters of a gas to moles At STP 0ºC and 1 atmosphere pressure At STP 22.4 L of a gas = 1 mole If 6.45 moles of water are decomposed, how many liters of oxygen will be produced at STP?

For Example If 6.45 grams of water are decomposed, how many liters of oxygen will be produced at STP? H2O ® H2 + O2 2H2O ® 2H2 + O2 1 mol H2O 1 mol O2 22.4 L O2 6.45 g H2O 18.02 g H2O 2 mol H2O 1 mol O2

Your Turn How many liters of CO2 at STP will be produced from the complete combustion of 23.2 g C4H10 ? What volume of oxygen will be required?

Example How many liters of CH4 at STP are required to completely react with 17.5 L of O2 ? CH4 + 2O2 ® CO2 + 2H2O 22.4 L O2 1 mol O2 1 mol CH4 22.4 L CH4 1 mol O2 1 mol CH4 22.4 L CH4 17.5 L O2 22.4 L O2 2 mol O2 1 mol CH4 = 8.75 L CH4

Avagadro told us Equal volumes of gas, at the same temperature and pressure contain the same number of particles. Moles are numbers of particles You can treat reactions as if they happen liters at a time, as long as you keep the temperature and pressure the same.

Example How many liters of CO2 at STP are produced by completely burning 17.5 L of CH4 ? CH4 + 2O2 ® CO2 + 2H2O 1 L CO2 17.5 L CH4 = 17.5 L CO2 1 L CH4

Particles We can also change between particles and moles. 6.02 x 1023 Molecules Atoms Formula units

Limiting Reagent If you are given one dozen loaves of bread, a gallon of mustard and three pieces of salami, how many salami sandwiches can you make? The limiting reagent is the reactant you run out of first. The excess reagent is the one you have left over. The limiting reagent determines how much product you can make

How do you find out? Do two stoichiometry problems. The one that makes the least product is the limiting reagent. For example Copper reacts with sulfur to form copper ( I ) sulfide. If 10.6 g of copper reacts with 3.83 g S how much product will be formed?

If 10. 6 g of copper reacts with 3. 83 g S If 10.6 g of copper reacts with 3.83 g S. How many grams of product will be formed? 2Cu + S ® Cu2S Cu is Limiting Reagent 1 mol Cu 1 mol Cu2S 159.16 g Cu2S 10.6 g Cu 63.55g Cu 2 mol Cu 1 mol Cu2S = 13.3 g Cu2S = 13.3 g Cu2S 1 mol S 1 mol Cu2S 159.16 g Cu2S 3.83 g S 32.06g S 1 mol S 1 mol Cu2S = 19.0 g Cu2S

How much excess reagent? Use the limiting reagent to find out how much excess reagent you used Subtract that from the amount of excess you started with

Your turn Mg(s) +2 HCl(g) ® MgCl2(s) +H2(g) If 10.1 mol of magnesium and 4.87 mol of HCl gas are reacted, how many moles of gas will be produced? How much excess reagent remains?

Your Turn II If 10.3 g of aluminum are reacted with 51.7 g of CuSO4 how much copper will be produced? How much excess reagent will remain?

Yield The amount of product made in a chemical reaction. There are three types Actual yield- what you get in the lab when the chemicals are mixed Theoretical yield- what the balanced equation tells you you should make. Percent yield = Actual x 100 % Theoretical

Example 6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate. 2Al + 3 CuSO4 ® Al2(SO4)3 + 3Cu What is the actual yield? What is the theoretical yield? What is the percent yield? If you had started with 9.73 g of Al, how much copper would you expect?

Details Practice Percent yield tells us how “efficient” a reaction is. Percent yield can not be bigger than 100 %. How would you get good at this? Practice