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Atomic Mass

How heavy is an atom of oxygen? There are different kinds of oxygen

atoms. More concerned with average atomic

mass. Based on abundance of each element in

nature. Don’t use grams because the numbers

would be too small

Measuring Atomic Mass

Unit is the Atomic Mass Unit (amu) One twelfth the mass of a carbon-12 atom. = 1.660540 * 10 -27 Amu is the number on the periodic table Refers to the mass of one atom of an

element Each isotope has its own atomic mass we

need the average from percent abundance.

Calculating averages

You have five rocks, four with a mass of 50 g, and one with a mass of 60 g. What is the average mass of the rocks?

Total mass = 4 x 50 + 1 x 60 = 260 g Average mass = 4 x 50 + 1 x 60 = 260 g

5 5 Average mass = 4 x 50 + 1 x 60 = 260 g

5 5 5

Calculating averages

Average mass = (4/5) x 50 + (1/5) x 60 = 260 g

Average mass = .8 x 50 + .2 x 60 80% of the rocks were 50 grams 20% of the rocks were 60 grams Average = % as decimal x mass + % as

decimal x mass + % as decimal x mass

Atomic Mass

Is not a whole number because it is an average.

These are responsible for the decimal numbers on the periodic table.

The MoleThe Mole

6.02 X 6.02 X 10102323http://www.teachertube.com/viewVideo.php?title=Happy_Mole_Day_to_You_Chemistry_

Song&video_id=41557

Molar Mass

Molecular Mass/Molecular Weight:

The molecular mass/weight is the amount of mass in 1 mole of the molecule.

Molar Mass

Molar mass of atoms: 1 mole of Br atom = 79.9 g / mole. 1 mole of Sn atom = 118.7 g / mole.

Molar masses for compounds or Molecules are the sum of molar masses of each atom. 1 Mole of CaCl2 = 111.1 g / mole {40.1 +

2(35.5)}

Percent composition

Percent of each element in a compound.

TO Calculate:1. Find the mass of each element2. Divide that by the molar mass3. Multiply by a 100.

Percent Composition

Find the percent composition of CH4

C= 1x 12g= 12g H= 4 x 1g = 4g

16g is the molar mass of CH4

% of C = 12/16 = 75% % of H = 4/16 = 25%

Empirical Formulas

Empirical Formula =Empirical Formula = the lowest ratio of atoms in a molecule.

Ex. A sample is 59.53% C, 5.38%H, 10.68%N, and 24.40%O what is its empirical formula?

Formula from percentage

1. Assume 100 g of sample2. Divide by the molar mass for each

element.3. Find the smallest approximate

whole number ratio of smallest number to each of the other element

4. Determine formula of the compound.

Example 1. Determine the empirical formula of a compound that contains 36.5% sodium, 25.4% sulfur, and 38.1% oxygen.

Assume 100 g of sample

Solution:  By assuming that we can study a 100g sample of the compound, we can change % to grams. so:

So the problem now reads: Determine the empirical formula of a compound that contains 36.5g sodium, 25.4g sulfur, and 38.1g oxygen.

Divide by the molar mass for each element

Now we can solve them by finding the molar ratio by which the elements combine.

       mass of that element in the sample

 moles of an element = ---------------------Molar mass of the

element                                            

Divide by the molar mass for each element     36.5 g

# of moles of sodium = ------------- = 1.59 moles                                     23.0g/mole

                                     25.4g# of moles of sulfur = --------------- = 0.791 moles                                    32.1 g/mole

                                       38.1 g# of moles of oxygen = --------------- = 2.38 moles                                       16.0 g/mole         

Find the smallest approximate whole number ratio of smallest number to each of the other element # of moles of Na  = 1.59 moles

                              -------------- = 2.01                              0.791 moles        

# of moles of S   = 0.791 moles                              -------------- = 1                              0.791 moles

# of moles of O   = 2.38  moles                              -------------- = 3.01                              0.791 moles

Determine formula of the compound The ratio shows that 2 atoms of

sodium combine with 1 atom of sulfur and 3 atoms of oxygen, so our answer is Na2SO3.

Answer: Na2SO3

Mass to moles(mass is already given)

1. Determine mass of each of the elements.2. Divide by the molar mass for each element3. Find the smallest approximate whole

number ratio of smallest number to each of the other element

4. Determine formula of the compound.

Solution:  Find the mass of the empirical formula (CH2):

C = 12.0 x 1 atom = 12.0 gH = 1.01 x 2 atoms = 2.02 g                               ----------                                 14.0 g

next, divide that number into the molecular mass:

56.0 g-------  = 414.0 g

Now use that number, 4, as a multiplier for the subscripts in the empirical formula:

CH2 x 4 = C4H8 Answer = C4H8

Another Way to Determine Molecular Formulas

Caffeine a stimulant found in coffee, contains, 49.48% carbon, 5.15% hydrogen, 28.87% nitrogen, and 16.49% oxygen by mass and has a molar mass of 194.2 g/mol determine molecular formula

1. Assume 100 total grams the element for each 100 grams of compound

2. Multiply by molar mass of the compound.

3. Divide by molar mass of the element.

49.48 g C x 194.2g x 1 mol C = 8.01 mol C100g caffeine 1 mol 12.01g C

5.15 g H x 194.2g x 1 mol H = 9.92 mol H100g caffeine 1 mol 1.01g H

28.87g N x 194.2g x 1 mol N = 4.002 mol N100g caffeine 1 mol 14.01g N

16.49 g O x 194.2g x 1 mol O = 2.001 mol O100g caffeine 1 mol 16.0g O

Molecular Formula C8H10N4O2

Law of Conservation of Mass and Balancing Chemical Equations

Matter is neither created nor destroyed during a chemical reaction. Therefore, all the atoms that were present at the start of the reaction must be present at the end of the reaction.

Steps to Balance Chemical Equations

1. Write the formula equation with the correct symbols and formulas.

Na + Cl2 NaCl2. Count the number of atoms of each

element on each side of the arrow.3. Balance atoms by using coefficients.

2Na + Cl2 2NaCl4. Check your work by counting atoms of

each element.

Mole/Mole Problems

1. A balanced chemical equation2. Determine the mole ratio

3. Set-Up problem….start with the given

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Li + O2 Li2O

1. Balance equation: 4 Li + O2 2 Li2O

2. Mole ratio of lithium to lithium oxide: 4 mol lithium to 2 mol lithium oxide

2. Set-Up problem, start with given.

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2 mol Li x 2 mol Li2O =

4 mol Li

Answer: 1 mol Li2O

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Mass/Mole and Mole/Mass Problems Must have 3 pieces of information:1. Balanced chemical equation2. Mole ratio3. Molar mass

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Problem 1

What mass in grams of magnesium oxide is produced from 2.00 mol of magnesium?

Mg + O2 MgO

2 Mg + O2 2 MgO Molar Mass of MgO 40.31 g/mol2.00 mol Mg x 2 mol MgO x 40.31g MgO = 2 mol Mg mol MgO

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Mass/Mass Problems

1. Balanced equation2. Mole ratio3. Molar Masses

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Problem 1

Tin (II) fluoride, SnF2 is used in toothpaste. How many grams of tin (II) fluoride are produced from the reaction of 30.0 g of HF with Sn? (Hydrogen is also a product.)

Sn + HF SnF2 + H2

Sn + 2 HF SnF2 + H2

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Sn + 2 HF SnF2 + H2

How many grams of tin (II) fluoride are produced from the reaction of 30.0 g of HF with Sn?

Mole ratio? 2 mole HF : 1 mole SnF2

Molar masses of HF and SnF2? HF: 20.01 g/mol SnF2: 156.71 g/mol

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How many grams of SnF2 are produced from the reaction of

30.0 g of HF with Sn?30.0gHF x mol HF x 1 mol SnF2 x

156.71gSnF2

20.01gHF 2 mol HF mol SnF2

Answer: 117 g SnF2

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Limiting Reactant Problems1. Balanced chemical equation.2. Determine the number of moles of

each reactant.3. Divide each number of moles of

each reactant by the coefficient in the balanced equation. The smallest number is the limiting reactant.

4. Proceed with solving problem with original numbers of limiting reactant.

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Limiting Reactant ProblemIf 20.5 g of chlorine is reacted with

20.5 g of sodium, which is the limiting reactant? How much salt is formed?

Cl2 + Na NaCl

Cl2 + 2 Na 2 NaCl

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Cl2 + 2 Na 2 NaCl

If 20.5 g of chlorine is reacted with 20.5 g of sodium, which is the limiting reactant?

20.5 g Cl2 x 1 mol Cl2 = 0.289 mol

70.9 g20.5 g Na x 1 mol Na = 0.892 mol 22.99 gCl2 : 0.289 ÷ 1 = 0.289

Na : 0.892 ÷ 2 = 0.445

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How much salt is formed?

Limiting Reactant? Chlorine0.289 mol Cl2 x 2 mol NaCl x 58.44 g

NaCl 1 mol Cl2 1 mol NaCl

33.8 g NaCl

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Percent Yield

Theoretical Yield: the maximum amount of product that can be produced from a given amount of reactant.

Actual Yield: the measured amount of product obtained from a reaction.

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Percent Yield =

Actual yield x 100 Theoretical yield

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