¨The Mole: A specific measurement that is used to help quantify chemical reactions. Its relationships describe how and why different elements and compounds interact.

– Avogadro’s Number: In every element sample, there are always 6.02 X 10²³ molecules present regardless of the type of that particular sample. This amount is equal to one MOLE of a substance.

– Mole is very similar to the word DOZEN as a dozen can refer to 12 of anything. Moles refers to HOW MANY not to SIZE of an element/compound.

Some common substances and their molar amounts:

1. Aluminum - Al: 1 mol, 1 gram atomic mass = 27 grams and contains 6.02 X 10²³ particles.

2. Oxygen - O₂: 1 mol, 1 gram formula mass = 32 grams and contains 6.02 X 10²³ molecules.

3. Water - H₂O: 1 mol, 1 gram formula mass = 18 grams and contains 6.02 X 10²³ molecules.

4. Dry ice - CO₂: 1 mol, 1 gram formula mass = 44 grams and contains 6.02 X 10²³ molecules.

¨Mole Relationships: THESE can be used as CONVERSION FACTORS to solve problems!!!

Molar mass for each element is on the periodic table Any molar mass can be a conversion factor

True for any gas at STP

True for anything 1 mol of eggs, 1 mole of Gold, 1 mole of sugar

¨The sum of the atomic masses of all the atoms represented by the formula of the substance.

– This formula is referring to the molecular or compound formula (O₂ or NaCl).

– It is also referred to as the MOLAR MASS.

– The molar mass is useful as it can be quantified in terms of grams (as opposed to atoms) which is a value that can be measured out

Other names for Molar Mass: YOU NEED TO KNOW THESE!!!

1. Formula Mass

2. Gram Atomic Mass

3. Gram Formula Mass

4. Gram Molecular Mass

Formula Mass Example

¨What is the formula mass of water, H₂O ?

(2 molecules H) X 1.0 amu = 2.02 amu

(1 molecule O) X 16.0 amu = 16.00 amu

Formula mass of H₂O = 18.02 amu

Gas Volume

¨One mole of gas occupies 22.4 L (at STP)

– the volume is dependent on the temperature and the pressure.

– Because of this, chemists often state the temperature and pressure as 0E C and 1.00 atmosphere (101.3 kPa, or 760 millimeter of Hg).

• This is referred to as STP or STANDARD TEMPERATURE AND PRESSURE.

Moles and Conservation of Mass

¨Coefficients in a balanced equation provide number of moles of a particular compound that are required for the chemical reaction to occur or describe the amount of product.

¨Molar Conversions: The mole is often used as a stepping stone when trying to calculate amounts in reference to reaction amounts (both before and after)

Moles and molecules

¨How many molecules in 1.5 moles of S?

– 1.5 moles of S x 6.02 x 10²³ particles of S = 9.0 x 10²³ particles of S

1 mole S

¨How many moles are in 3.5 x 10²⁵ particles of oxygen?

– 3.5 x 10²³ particles O x 1 mole oxygen = 0.58 moles of O

6.02 x 10²³ particles of O

–

Moles and grams

¨How many grams of CO₂ are present in 1.5 moles of CO₂?

– 1.5 moles CO₂ X 44.0098 g CO₂ = 66 grams CO₂

1 mole CO₂

¨How many moles are in 30.0 grams of lithium?

– 30.0 grams Li X 1 mole Li = 4.32 mol Li

6.941 g Li

Moles and liters

¨ How many liters are there in 2 moles of O₂?

2.0 mol O₂ X 22.4 liters = 45 liters O₂

1 mol O₂

¨ How many moles are there in 350 liters of N₂

350 L N₂ X 1 mole N₂ = 16 moles N₂

22.4 liters

¨For any mole calculation, report your final answer in the correct number of significant figures.

– The number of significant figures you need to use should always be equal to the number of sig figs that were used in the original question, NEVER in the conversion factors

¨Percentage Composition: The percentage by mass of each of the elements in a compound

Percentage Composition Example: A sample of a compound containing carbon and oxygen had a mass of 88 g. Experimental procedures showed that 24 g of this sample was carbon, and the remaining 64 g was oxygen. What was the percent composition of this compound?

% carbon = mass of carbon in sample X 100% = 24 g / 88 g X 100 % = 27 %

mass of sample

% oxygen = mass of oxygen in sample X 100% = 64 g / 88 g X 100 % = 73%

mass of sample

27% + 73% = 100%.

¨Empirical Formulas show the simplest whole-number ratio in which the atoms of the elements are present in the compound. The empirical formula can be found in a couple of ways

– the mass of elements

– the percentage composition

– a chemical analysis

Empirical Formulas from Mass

¨A sample of a brown colored gas that is a major air pollutant is found to contain 2.34 g of N and 5.34 g of 0. What is the simplest formula of the compound?

1. Begin by converting the masses of the indicated elements to moles

2.34 g N * (1mol of N) = 0.167 mol N 5.34 g O * (1 mol O) = 0.334 mol O

14.0 g N 16.0 g O

2. Next, to find the whole number ratio, divide by the smaller molar amount. In this example, nitrogen has less moles than oxygen.

0.167 mol N = 1 0.334 mol 0 = 2

0.167 mol N 0.167 mol N

– The whole number ratio are the values that become the subscripts in the compound so the empirical formula for this compound is NO₂.

Empirical Formula from Percentage Composition

¨What is the empirical formula of a compound composed of 43.7% P and 56.3% O?

1. Assume that you have 100 total grams of the compound. By making this assumption, the amount of P is now 43.7 g and the amount of O is 56.3 g.

2. Once this mass is known, find the number of moles present in this mass.

3. 43.7 g P * (1 mol P) = 1.41 mol P 56.3 g O * (1 mol O) = 3.52 mol O

31.0 g P 16.0 g O

4. Next, divide by the smaller molar amount. In this example, P.

1.41 mol P = 1 3.52 mol O = 2.50

1.41 mol P 1.41 mol P

5. To find the whole number, multiply the decimal until a whole number is reached. In this case, multiply both values by 2 so that the empirical formula for this compound is P₂O₅

Empirical Formula from Chemical Analysis

¨A 1.025 g sample of a compound that contains only carbon and hydrogen was burned in oxygen to give carbon dioxide and water vapor products. These products were trapped separately and weighed. It was found that 3.007 g of CO₂ and 1.845 g H₂O were formed. What is the empirical formula?

1. Find the mass of C and H in this sample

3.007 g CO₂ X 12.011 g C = 0.8207 g C 1.845 g H₂O X 1.00794 g H = 0.2064 g H

44.0098 g CO₂ 44.0098 g CO₂

2. Convert Grams to mols

0.8207 g C X 1 mol C = 0.06833 mol C 0.2064 g H X 1 mol H = 0.2048 mol H

12.011 g C 1.00794 g H

3. Determine the subscripts by dividing by the lower mole amount (here its C)

0.06833 mol C = 1 0.2048 mol H = 3

0.06833 mol C 0.06833 mol C

¨Final answer: CH₃

Molecular Formulas: Molecular formulas give both the atom ratios and also describes the actual number of atoms of each kind of element. Two different compounds can have identical empirical formulas but different molecular formulas.

Determining the molecular formula of a compound

¨A colorless liquid used in rocket engines, whose empirical formula is NO₂, has a mass of 92.0. What is its molecular formula?

1. Calculate the molar mass of the empirical formula to be 46.0

2. Divide the molecular formula mass by the empirical formula mass to determine how many times the empirical formula occurs in this molecular formula:

92/46 = 2

3. Distribute this value through the empirical formula to find a final answer of N₂O₄

Molecular Formula from percentages and molar mass

¨The difference in this type of calculation is that you must first find the actual mass of each kind of element. This is completed by multiplying the element (in decimal form) by the molar mass. The element is now converted to moles which will give a whole number ratio

nStoichiometry: the study of the quantitative, or measurable relationships that exist in chemical formulas and chemical reactions

nCoefficients in the balanced equations can be read as Particles, moles or volume of gas at STP!!!:

2H₂ + O₂ à 2H₂O

nRead as : Two moles of hydrogen gas react with one mole of oxygen gas to produce two moles of water.

n Use the factor label method to solve:

– Write the grams of known product and place this value over one.

– Use molar mass (of the known substance) to convert the grams of the known substance to moles

• The moles should be on top to cancel grams.

– Use the coefficients of the balanced chemical equation to convert from the moles of the known substance to moles of the unknown substance

• The moles of unknown compound should be on top so the known moles will cancel out.

– Use molar mass (of the unknown substance) to convert the moles of the unknown substance to grams

• Grams should be on top as that is the unit you want

NH₄NO₃ ® N₂O + 2H₂O

nMole to Mole Example: How many moles of N₂O and H₂O are produced from 2.25 moles of NH₄NO₃?

2.25 moles NH₄NO₃ x 1 mol N₂O = 2.25 mol N₂O

1 mol NH₄NO₃

2.25 moles NH₄NO₃ x 2 mol H₂O = 4.5 mol H₂O

1 mol NH₄NO₃

n The coefficients from the balanced equation are used to convert from moles of one substance to moles of another substance

NH₄NO₃ ® N₂O + 2H₂O

nMass to Mass Example: How many grams of N₂O and H₂O are produced from 10.0 g of NH₄NO₃?

10.0 g NH₄NO₃ x 1 mol NH₄NO_{3 x} 1 mol N₂O x 44.0 g N₂O = 5.50 g N₂O

80.1 g NH₄NO₃ 1 mol NH₄NO₃ 1 mol N₂O

10.0 g NH₄NO₃ x 1 mol NH₄NO_{3 x} 2 mol H₂O x 18.02 g H₂O = 4.50 g N₂O

80.1 g NH₄NO₃ 1 mol NH₄NO₃ 1 mol H₂O

nStoichiometry can be used to verify the law of conservation of mass

NH₄NO₃ ® N₂O + 2H₂O

nWe calculated that 10.0 g NH₄NO₃ of decomposed to form 5.50 g N₂O of and 4.50 g of H₂O.

10 g = 5.50 g + 4.50 g Mass was neither created nor destroyed!!!

2 NaN₃ ® 2 Na + 3 N₂

nMass – Volume Example: How many liters of N₂ gas are produced from 125 g of NaN₃?

125 g NaN₃ x 1 mol NaN_{3 x} 3 mol N₂ x 22.4 L N₂ = 64.6 L N₂ (STP)

65.0 g NaN₃ 2 mol NaN₃ 1 mol N₂

Limiting Reactant Problems : In real life, quantities of reactants are not usually available in the exact ratio described by the balanced equation.

• Limiting Reactant: the reactant that runs out first – determines how much product you get!

– The quantities of products formed in a reaction are always determined by the limiting reactant

• Excess Reactant: is the reactant that is leftover after the reaction has ended.

n Limiting Reactant Example: Wood burning in air. The oxygen in the air is considered the reactant in excess and the wood is considered the limiting reactant. When the wood is all burned, there will no longer be a reaction occurring (fire).

Write the balanced equation, calculate product for each reactant, the one that makes the least is limiting!!!

3.5 g Cu is added to a solution containing 6.0 g of silver (I) nitrate. Determine the limiting reactant.

Cu + 2 AgNO₃ ® Cu(NO₃)₂ + 2 Ag

3.5 g Cu x 1 mol Cu _x 2 mol Ag = 0.11 mol Ag

63.5 g Cu 1 mol Cu

6.0 g AgNO₃ x 1 mol AgNO₃ x 2 mol Ag = 0.035 mol Ag Silver nitrate is the limiting Reactant

169.9 g AgNO₃ 2 mol AgNO₃

Volume – Volume Problems: are much like mole-mole problems. Use Coefficients of balanced equation to convert from moles of one substance to moles of another substance.

N₂ + 3 H₂ ® 2 NH₃

Example: Calculate the volume of H₂ gas that reacts with 15.5 liters of N₂ (at STP).

15.5 L N₂ x 3 L H₂ = 46.5 L H₂

1 L N₂

Percent Yield

Expected Yield: the amount of product that should be obtained based on the calculations

Percent Yield: the percent of the expected yield that was actually obtained

Percent yield = actual yield x 100%

expected yield

Solving Percent Yield Problems

1. Write the balanced chemical equation

2. Calculate the expected yield

3. Calculate the percent yield

Percent Yield Example Problem

A 5.00 g piece of Cu is placed in excess silver (I) nitrate. 15.2 g of silver metal is produced. What is the percent yield?

Cu + 2 AgNO₃ ® Cu(NO₃)₂ + 2 Ag

5.0 g Cu x 1 mol Cu _x 2 mol Ag x 107.9 g Ag = 17.0 g Ag

63.5 g Cu 1 mol Cu 1 mol Ag

Percent Yield = 15.2 g Ag x 100 % = 89.4%

17.0 g Ag

Answers to 6A Packet:

Percentage Composition: 1. 24.74% K, 34.74% Mn, 40.50% O 2. 2.77% H, 97.23 % Cl

3. 16.25% Mg, 18.89%N, 64.72% O 4. 28.19% N, 8.13% H, 20.77% P, 42.92% O

5. 15.77% Al, 28.12% S, 56.11% O 6. 39.17 g Oxygen, 17.49 g Fe, 108.82 g Ag

Mixed Mole Problems: 1. 1100 g CO₂ 2. 560 L CO₂(g) 3. 2.02 x 10²⁴ molecules NH₃(g)

4. 57.1 g NH₃ 5. 3.9x 10²² atoms NO₂ 6. 3.5 L O₂(g) 7. 9.4 x10²² molecules O₂ 1 x10²³ atoms O

Determining Empirical Formulas

1. CH₄ 2. KCl 3. AlPO₄ 4. MgBr₂ 5. Na₂SO₄ 6. CuSO₄·5H₂O

Determining Molecular Formulas: 1. N₂O₄ 2. C₅H₁₀ 3. C₂H₄O₂ 4. C₄H₁₀O 5. C₂H₄O

Composition of Hydrates: 1. 36.1% 2. 67.51% 3. 14% 4. 32% 5. 2.56g 6. 3

The Mole and You: 1. 1.2 x 10²⁴ molecules 2. 9.0 x 10²³ molecules 3. 4.5 x 10²³ molecules

4. 3.04 x 10²⁴ molecules 5. 2.11 x 10²³ molecules 6. 1.00 mole 7. 2.00 moles

8. 2.5 x10^-4 mole 9. 5.6 x 10² mole 10. 1.2 x 10^-4 mole 1. 158.04 g/mol 2. 74.55 g/mol 3. 142.07 g/mol

4. 164.1 g/mol 5. 342.21 g/mol 6. 149.12 g/mol 7. 249.72 g/mol 8. 262.87 g/mol 9. 350.31g/mol

10. 458.19 g/mol 11. 62.03 g/mol 12. 617.18 g/mol 13. 304 g/mol 14. 351.91 g/mol 15. 77.16 g/mol

Moles and mass: 1. 0.43 mol 2. 1.27 mol 3. 0.6 mol 4. 0.99 mol 5. 0.14 mol

6. 146.125 g NaCl 7. 49. g H₂SO₄ 8. 269 g KmnO₄ 9. 19 g KCl 10. 8.0 x 10² g CuSO₄·5H₂O