"You can't fake chemistry"

## The Basics

## Significant figures

Only some values in a number are considered significant. Significant figures change with the type of calculation in the problem.
Rules are: 1. All non-zero integers are significant. ie: 123 has THREE significant figures 2. Zeros after a decimal are significant if they do not begin a number. ie: 123.00 has FIVE sig. figs. 0.00123 has THREE sig. figs. 3. Zeros BETWEEN two non-zero integers are significant. ie: 1023 has FOUR sig.figs 4. If there is no decimal, a case of trailing zeros, the number of sig. figs is ambiguous. This is fixed by writing the number in scientific notation. ie: 123000 may have 6,5, 4, or 3 significant figures. We can show the true number of significant figures using scientific notation: 1.23 x10^5 means 3 sig. figs and 1.230 x10^4 means 4 and 1.2300 means 5 significant figures and 1.23000 means 6. |
- When multiplying/dividing/raising a number to a power/ taking a root, your answer should contain the same amount of significant figures as the value with the lowest significant figures.
ie: 2.0 X 123= 250 - When adding or subtracting a number, your answer should contain the same amount of significant figures as the number with the least decimal places.
ie: 123.45+ 123= 246 |

## The Metric System

Tera = 10^12
Giga= 10^9 Mega= 10^6 Kilo 10^3 Hect 10^2 Deca 10 Unit Deca 10^-1 Centi 10^-2 Milli 10^-3 Micro 10^-6 Nano 10 ^-9 Pico 10^-12 Femto 10 ^-15 |
* A way to remember this
"Thank God My Karate Helps Ditch [Unkind] Dudes 'Cause Mean Men Need Punching Forreal"1 Degree Celsius = 274.15 degrees Kelvin 9/5 * Degrees C +32 = 1 degree Fahrenheit |

## Elements and the Periodic Table

## Elements

What is an element?!

- The simplest substance with distinct properties that can be broken down by chemical means.
- Each element has a distinct number of protons.
- Elements that have more or less electrons than their neutral state (state of no charge) are called ions.
- Elements that have more or less neutrons than their common state are called isotopes.

## The Periodic Table

ARTSY, GENERALIZED VIDEO:

The periodic table is organized by trends of its elements.

Effective nuclear charge (Z*):

- the net charge experienced by a particular electron resulting from a balance of the attractive force of the nucleus and the repulsive forces of other electrons
- Z* creates a shielding effect of the electrons in the valance shell from the nucleus due the the electrons closer to the nucleus
- Z effective = Z- S where Z is the number of protons (atomic number) and S is the number of electrons screening the nucleus

- the amount of energy required to remove 1 electron from a neutrally charged atom in its ground state, gas phase.
- measured in kilojoules/mol
- for s and p blocks (main group elements) ionization energy usually increases across a period and decreases down a group
- IE increase as more electrons are removed because the atom, now ion, becomes increasingly positive.
- IE is highest to the right of the periodic table as the electrons in each element are most stable
- Always positive and forms cations

- the amount of energy required to add 1 electron to a neutrally charged atom
- measured in KJ/mol
- Electrons are more attracted to a less negative ion therefore as Z* causes the IE to increase, EA increases

- Depends on size of atoms
- Is determined by the distance of the nucleus of the atom (protons and neutrons) to its outermost electron shell
- The periodic table is arranged by its atomic radius

## Electron Configuration

## Compounds

## Pure Substances and Mixtures

Pure Substance: a classification of matter chemically combined with definite properties and composition no matter what quantity exists. Any substance composed of 1 element or 1 type of compound may be considered a pure substance.

- Common Examples: NaCl (table salt), pure water, diamond

- Heterogeneous mixture: A mixture where the constituent substances can be easily distinguished.

- Homogeneous mixture: Also called a solution. A mixture where the constituent substances cannot be easily distinguished and which has a uniform phase of matter and composition throughout.

## Percent Composition

The formula for percent composition is (mass of specific component/total molar mass) X 100%

In the problem:

1. You can determine the value by plugging in values and re-writing the equation. For copper, .95 = mass of specific component (copper)/3.11g) which is .95 X 3.11 g= mass of specific component(copper) which equals 2.9545 grams copper.

Perform this procedure on zinc. .05= (mass of zinc)/3.11 g or .05 X 3.11g = mass of zinc which equals .1555 grams zinc.

2. Take this mass value and convert grams to pounds. The conversion rate is 453.592 grams = 1 pound. So 2.9545g/453.592 g = .00651 lbs copper. And .1555g/453.592g = .0003428 lbs zinc.

3. Multiply these values by the amount per pound. .00651 lbs X $2.42= $.0157542; .0003428 lbs X $1.07= $.000366816.

4. Add these values together to equal a total of $.016121016 which rounded to the correct amount of sig. figs gives the value of $.0161.

Only step 1 of the problem required the percent composition rule. The other steps were just conversions.

In the problem:

*Pennies made in 1972 had a mass of 3.11 grams. Their composition was 95.00% copper and 5.00% zinc. If the current scrap prices of copper and zinc are $2.42/lb and $1.07/lb, respectively, what is the scrap value of a penny?*

1. You can determine the value by plugging in values and re-writing the equation. For copper, .95 = mass of specific component (copper)/3.11g) which is .95 X 3.11 g= mass of specific component(copper) which equals 2.9545 grams copper.

Perform this procedure on zinc. .05= (mass of zinc)/3.11 g or .05 X 3.11g = mass of zinc which equals .1555 grams zinc.

2. Take this mass value and convert grams to pounds. The conversion rate is 453.592 grams = 1 pound. So 2.9545g/453.592 g = .00651 lbs copper. And .1555g/453.592g = .0003428 lbs zinc.

3. Multiply these values by the amount per pound. .00651 lbs X $2.42= $.0157542; .0003428 lbs X $1.07= $.000366816.

4. Add these values together to equal a total of $.016121016 which rounded to the correct amount of sig. figs gives the value of $.0161.

Only step 1 of the problem required the percent composition rule. The other steps were just conversions.

## Percent Abundance

## Density

Note the formula D= M/V. Density is mass divided by volume. Density should be given in units of grams/cm^3. Mass is therefore in the unit of grams. Volume is in the unit of cm^3.

In the problem:

1. Standardize the units. If there is a measurement that is not a unit in the equation we're using (D=M/V) convert the value to a unit present in the equation. Here, 50 FEET squared is in feet, not cm as in the equation D=M/V. So by the conversion rate 30.48 cm equals 1 foot. BUT we have feet squared in this problem so we must square both values. 929.0304 cm^2 therefore equals 1ft^2. 50 ft^2 = 46451.52 cm^2.

2. Determine what known variables exist. The mass of aluminum given is 230.0 grams of aluminum. The density is 2.70 g/cm^3. These units match the equation.

3. Rewrite the equation and plug in values. Since we have 2 of the 3 variables, we can shift the original equation to get our unknown variable, density. D= M/V becomes M/D =V and with values plugged in 230.0 g/ (2.70 g/cm^3). The density is therefore 85.18518519 cm^3.

4. This requires another equation: V= L * W * H. The volume of a cube (foil is a very flattened cube) is equal to length * width * height. Since we have 1524 cm^2 of aluminum, we can infer that this is equal to length * width. Or any 2 of those 3 dimensions multiplied together to get the unit cm^2. So width in cm * length in cm = width*length cm^2. Or width * height = width*height cm^2. We want the thickness which is the height of the flattened, aluminum cube. So volume which is 85.18518519 ft ^3 = 46451.52 cm^2 (length * width) * height. V= L * W * H. Therefore H = V/L*W. H= 85.18518519 cm^3/ 46451.52 cm^2. Height(thickness) is .00183 cm in thickness.

In the problem:

*The aluminum in a package containing 50 ft^2 of kitchen foil weighs approximately 230.0 grams. Aluminum has a density of 2.70 g/cm^3. What is the approximate thickness of the aluminum foil in centimeters?*1. Standardize the units. If there is a measurement that is not a unit in the equation we're using (D=M/V) convert the value to a unit present in the equation. Here, 50 FEET squared is in feet, not cm as in the equation D=M/V. So by the conversion rate 30.48 cm equals 1 foot. BUT we have feet squared in this problem so we must square both values. 929.0304 cm^2 therefore equals 1ft^2. 50 ft^2 = 46451.52 cm^2.

2. Determine what known variables exist. The mass of aluminum given is 230.0 grams of aluminum. The density is 2.70 g/cm^3. These units match the equation.

3. Rewrite the equation and plug in values. Since we have 2 of the 3 variables, we can shift the original equation to get our unknown variable, density. D= M/V becomes M/D =V and with values plugged in 230.0 g/ (2.70 g/cm^3). The density is therefore 85.18518519 cm^3.

4. This requires another equation: V= L * W * H. The volume of a cube (foil is a very flattened cube) is equal to length * width * height. Since we have 1524 cm^2 of aluminum, we can infer that this is equal to length * width. Or any 2 of those 3 dimensions multiplied together to get the unit cm^2. So width in cm * length in cm = width*length cm^2. Or width * height = width*height cm^2. We want the thickness which is the height of the flattened, aluminum cube. So volume which is 85.18518519 ft ^3 = 46451.52 cm^2 (length * width) * height. V= L * W * H. Therefore H = V/L*W. H= 85.18518519 cm^3/ 46451.52 cm^2. Height(thickness) is .00183 cm in thickness.

## Naming

## Molecular vs Empirical Formula

The empirical formula is the formula of a molecule with the most simplified ratio. The molecular formula is the formula of a molecule as it occurs.

A compound has an empirical formula of C3H4O and a molar mass of 112 g/mol. What is the molecular formula?

1. Determine the molar mass of each constituent element. Here there are 3 Carbons, 4 Hydrogen, and 1 Oxygen atoms.

2. Add all the molar masses. 3 * 12.01 amu + 4 * 1.008 amu + 1* 16.00 amu= 56.062 g/mol

3. Divide the given molecular molar mass by the empirical molar mass to give the ratio between the molecular to empirical.

112 g/mol divided by 56.062 g/mol

4. Round the answer to the nearest whole number digit. 112g/mol divided by 56.062 g/mol equals 1.998 = 2

5. Multiply the molecular formula by multiplying the amount of each element by the ratio. 3* 2 = 6 Carbons, 4*2=

8 Hydrogen and 1*2= 2 Oxygen

Answer: C6H8O2

Acetic acid, which is present in vinegar, was found to consist of 40.00% C, 6.71% H, and 53.29% O by mass. Find the empirical formula for acetic acid.

1. Assume 100 grams so 40 grams Carbon, 6.71 grams Hydrogen and 53.29 grams Oxygen.

2. Determine the ratio of mols by dividing the masses of each element by its molar mass. 40/12.001= 3.33 , 6.71/1 = 6.71 and 53.29/16= 3.33. The smallest resulting number is the mol ratio.

3. Divide all numbers found in the previous step by the mol ratio to get the amount. Here, the smallest ratio is 3.33 so for Carbon there is 1, for Hydrogen 2 and Oxygen 1.

Answer: CH2O

A compound has an empirical formula of C3H4O and a molar mass of 112 g/mol. What is the molecular formula?

1. Determine the molar mass of each constituent element. Here there are 3 Carbons, 4 Hydrogen, and 1 Oxygen atoms.

2. Add all the molar masses. 3 * 12.01 amu + 4 * 1.008 amu + 1* 16.00 amu= 56.062 g/mol

3. Divide the given molecular molar mass by the empirical molar mass to give the ratio between the molecular to empirical.

112 g/mol divided by 56.062 g/mol

4. Round the answer to the nearest whole number digit. 112g/mol divided by 56.062 g/mol equals 1.998 = 2

5. Multiply the molecular formula by multiplying the amount of each element by the ratio. 3* 2 = 6 Carbons, 4*2=

8 Hydrogen and 1*2= 2 Oxygen

Answer: C6H8O2

Acetic acid, which is present in vinegar, was found to consist of 40.00% C, 6.71% H, and 53.29% O by mass. Find the empirical formula for acetic acid.

1. Assume 100 grams so 40 grams Carbon, 6.71 grams Hydrogen and 53.29 grams Oxygen.

2. Determine the ratio of mols by dividing the masses of each element by its molar mass. 40/12.001= 3.33 , 6.71/1 = 6.71 and 53.29/16= 3.33. The smallest resulting number is the mol ratio.

3. Divide all numbers found in the previous step by the mol ratio to get the amount. Here, the smallest ratio is 3.33 so for Carbon there is 1, for Hydrogen 2 and Oxygen 1.

Answer: CH2O

## Common Polyatomic Ions

## Solubility Rules

1. All elements in Group 1 (Alkali metals) are soluble and the ammonium (NH4+) ion are soluble

2. All common acetates (CH3CO2^-) and nitrates (NO3^-) are soluble.

3. All compounds of Group 17 (except fluorine (f2)) are soluble except with silver (Ag), mercury (Hg), and lead (Pb).

4. All sulfates are soluble except with calcium (Ca), strontium (Sr), barium (Ba), silver (Ag), mercury (Hg) and lead (Pb).

5. Carbonates (CO3^2-), Hydroxides (OH-), Oxides (O2), Sulfides (S^2-), Chromates (CrO4^2-), and Phosphates (PO4^3-) are insoluble except with those stated in #1

2. All common acetates (CH3CO2^-) and nitrates (NO3^-) are soluble.

3. All compounds of Group 17 (except fluorine (f2)) are soluble except with silver (Ag), mercury (Hg), and lead (Pb).

4. All sulfates are soluble except with calcium (Ca), strontium (Sr), barium (Ba), silver (Ag), mercury (Hg) and lead (Pb).

5. Carbonates (CO3^2-), Hydroxides (OH-), Oxides (O2), Sulfides (S^2-), Chromates (CrO4^2-), and Phosphates (PO4^3-) are insoluble except with those stated in #1

## Thermodynamics (The Study of Processes Involving Heat)

## Heat Transfer:

## Heat Transfer Problems (taken from Chemistry & Chemical Reactivity)

Calculation-based
QUESTIONS: 1. How much energy must be transferred to raise the temperature of a cup of coffee (250 mL) to 20.5 degrees celsius to 95.6 degrees celsius? Assume water and coffee have the same density (1.00 g/mL) and specific heat capacity (4.184 J/g*K) 2. An 88.5 g piece of iron whose temperature is 78.8 degrees celsius is placed in a beaker containing 244 grams of water at 18.8 degrees celsius. When thermal equilibrium is reached, what is the final temperature? (Assume no energy is lost to warm the beaker and its surroundings) 3. You heat a 7.951 sample of gold and silver mixture to 95.51 degrees C and place it in 12.8 g of water initially at 24.93 degrees C. Thermal equilibrium is reached and the final temperature is 26.51 degrees C. Find the percent compositions of gold and silver. The specific heat of water is 4.184 J/g*C. The specific heats of gold and silver are .129 J/g*C and .240 J/g*C respectively. 4. The production of ammonia (NH3) under standard conditions at 25°C is represented by the following thermo-chemical equation: N2( g) + 3 H2(g) → 2 NH3(g); ΔH = −91.8 kJHow much heat is released when 1.463 ✕ 10^4 grams of ammonia is produced? 5. In the recent ice storm the baseball field at Foley Field was covered with 3.78 cm of ice. If the area of the baseball field at Foley Field is 6575 m2, what is the volume of ice, in cubic centimeters? What is the mass, in grams, of ice covering the baseball field at Foley Field? How much heat, in kilojoules, would be needed to melt the layer of ice covering the baseball field at Foley Field? The heat of fusion of ice is 333.5 J/g. 6. Calculate the enthalpy change (sign is important) for the reaction in which 38.00 g of Al(s) reacts with oxygen to form Al2O3(s) at 25°C and one atmosphere: 4 Al(s) + 3 O2(g) → 2 Al2O3(s) The ΔHf° for Al2O3(s) is -1675.7 kJ/mol 7. For each reaction calculate ΔH° at 25°C. - 2 Al(
*s*) + 3 Cl2(*g*) → 2 AlCl3(*s*) - Na2CO3(
*s*) + 2 HCl(*aq*) → 2 NaCl(*aq*) + H2O(*l*) + CO2(*g*)
8. Calculate the enthalpy change (sign is important) for the reaction in which 42.00 g of Fe reacts with 12.50 g of O2 to form Fe2O3( s) at 25°C and one atmosphere:4 Fe(s) + 3 O2(g) → 2 Fe2O3(s)The ΔHf° for Fe2O3 is -824.2 kJ/mol ANSWERS: 1. 78554 or 78,600 J or 79 KJ - The question asks for an amount of energy, meaning the answer will be in joules.
- 1 milliliter (mL) equals 1 gram of water based on density
- The final temperature is 95.6 degrees, and the initial temperature is 20.5 degrees. The change is temperature is the final temp minus the initial: 75.1 degrees celsius
- Now simply plug-in to equation q=mc(delta)t to get amount of heat energy, q
- q= (250)(4.184)(75.1)
2. 21 degrees Celsius - The question asks for temperature, so the answer will be in celsius.
- We realize there are 2 substances and we are given 2 masses with 2 corresponding temperatures. This means we must calculate q values for 2 temperatures. These q values should add to 0 according to the law of conservation of energy.
- The final temperature is the equilibrium temperature, meaning that it has the same value for both water and aluminum, which has a specific heat capacity of .449 J/g*K
- MC(Final Temperature- Initial Temperature) + MC(Final Temperature- Initial Temperature) = 0
- Now simply plug-in to equations q=mc(delta)t to get amount of heat energy, q
- Water: q= (244)(4.184)(Temperature Final- 18.8)
- Aluminum: q= (88.5)(.449)(Temperature Final- 78.8)
- q(water) + q(aluminum)= 0
- Use algebra to solve for the Final Temperature
3. 77.0 % silver; 23.0% gold - In order to do this question, you must realize that an amount of gold and and an amount of silver must combine to make the 7.951 g sample. This means that we must subtract some part of that 7.951 g total to get grams of gold. In algebraic terms: 7.951- x where "x" is the grams of gold.
- Then you plug in the values for the three heat energy equations: one for gold, one for silver, and one for water.
- Gold: q= (x)(26.51 - 95.51)(.129)
- Silver: q= (7.951- x)(26.51- 95.51)(.240)
- Water: q=(12.8)(26.51 - 24.93)(4.184)
- Add these q-values together and solve for x to get the grams of gold
- Place the grams of gold over the total grams of the substance and multiply by 100% to get a percentage. Subtract the percentage of gold from 100% to get the percentage of silver.
4. -3.95 X 10^4 KJ - First, from the grams of ammonia given, we must find the amount, in mols, of ammonia. This is done by dividing the given amount of ammonia by the molar mass (grams) of ammonia. The molar mass of ammonia is 17.031 calculated by adding the individual masses of each element in the ammonia compound. Therefore: 1.463 x 10^4 g divided by 17.031 g = 858.87 mol.
- When we get the value for mols of ammonia, we can multiply that by the 1 mol of reaction divided by the 2 mols of ammonia. 858.87 mol x 1/2.
- Multiply this mol value by the heat of the reaction which -91.8 kj per mol.
5. 2.49 E8; 2.27E8; 7.60E10 6. -1180 kj 7.-906 |
conceptual Questions: |

## Changes of State:

A "Change of state" occurs as a substance goes from 1 phase to another. The more energy an object has, the more likely it will go from solid to liquid to gas to plasma phase. |
heat of fusion: the heat energy required to convert a substance from a solid to a liquid.
heat of vaporization: the heat energy required to convert a substance from a liquid to a gas. sublimation: the process of heat transfer that takes a substance directly from a solid to a gas. Equation 1:
change in u = q + w |