 #  Dalton used the percentages of elements in compounds and the chemical formulas to deduce the relative masses of atoms  Unit is the amu(atomic mass.

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 Dalton used the percentages of elements in compounds and the chemical formulas to deduce the relative masses of atoms  Unit is the amu(atomic mass unit) ◦ 1 amu = 1.66 x 10 -24 g  We define the masses of atoms in terms of atomic mass units ◦ 1 Carbon atom = 12.0 amu, ◦ 1 Oxygen atom = 16.0 amu ◦ 1 O 2 molecule = 2(16.00 amu) = 32.00 amu

 Atomic masses allow us to convert weights into numbers of atoms If a sample of carbon weighs 3.00 x 10 20 amu how many atoms of carbon are present?

 The mass of 1 Al atom = 27.0 amu  Use the relationship as a conversion factor Calculate the Mass (in amu) of 75 atoms of Al

 A mole is a counting unit, just like a dozen, except that a mole is much larger.  A mole (mol) is a number that represents the number of representative particles of a substance.  It is always equal to 6.02 x 10 23 and is referred to as Avogadro’s number.

 Atom - single element, for example: Na, C  Diatomic molecule – H 2, N 2, O 2, F 2, Cl 2, Br 2, I 2  Molecule – contains no metals, for example: H 2 O, C 12 H 22 O 11  Ion – only ions, for example, Ca 2+, P 3-  Formula Unit – always starts with a metal, for example: CaF 2, CuSO 4

 Avogadro’s # can be used as a conversion factor where 1 mol= 6.02 x 10 23 atoms, ions, molecules, or formula units depending on the type of substance you have.

 Convert 4.27 x 10 23 molecules of CO 2 to moles  How many formula units are there in 2.78 moles of CaCl 2 ?

 How many molecules are there in 0.85 moles of octane, C 8 H 18 ?  How many moles are there in 7.6 x 10 22 Na + ions?

 When converting between grams and moles, 1 mole of an element equals the average atomic mass of that element in grams, rounded to one number after the decimal. ◦ 1 mol Ca = 40.1 g Ca1 mol P = 31.0 g P ◦ 1 mol Cl = 35.5 g Cl1 mol K = 39.1 g K ◦ 1 mol S = 32.1 g S1 mol Cu = 63.6 g Cu ◦ 1 mol Br = 79.9 g Br1 mol Si = 28.1 g Si

¬ How many moles are there in a 10.0 gram sample of Al? ¬ How many grams are there in a 0.643 mol sample of Fe?

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 How many formula units of Al 2 O 3 are there in 0.750 moles of Al 2 O 3 ? How many atoms of each element are in the sample of Al 2 O 3 ?

 Molar mass – mass of 1 mol of a pure substance usually written in grams per mole (g/mol)  The subscript next to each element in a formula tells you how many moles of that particular element are in the compound.

 When finding molar mass of a compound if there are parentheses in the formula, you must multiply each element in the parentheses times the number outside to determine the total number of moles of each element.

 Determine the molar mass for:  Fe 2 (SO 4 ) 3  Cu(ClO 3 ) 2  Ni(C 4 H 4 O 6 ) 2

 Determine the number of moles in a 22.3 gram sample of Al 2 (CO 3 ) 3

 How many grams of ammonium phosphate, (NH 4 ) 3 PO 4, are present in a 0.082 mole sample of (NH 4 ) 3 PO 4 ?

 How many grams of gold, Au, are present in a 8.21 x 10 22 atom sample of Au?

 Determine the number of formula units are there in a 47.9 gram sample of Mg(NO 3 ) 2

 To determine percent composition 1.Determine molar mass of your compound (g/mol). 2. Divide the total mass for each element by the molar mass of the compound. 3. Multiply each element’s % by 100. If you add up the percentages for all of the elements, you should get very close to 100%.

 Determine the percent composition for each element in the following compounds: ◦ K 2 SO 4 ◦ Ba 3 (PO 4 ) 2

 Empirical formula – lowest whole-number ratio of the moles in a compound.  Molecular formulas are simple multiples of empirical formulas that can be reduced if each element in the compound is divisible by the same whole number.  Empirical formulas cannot be reduced any further and still have whole numbers for each element in the compound.

C6H9O3C6H9O3  C 12 H 22 O 11 C4H6O2C4H6O2 C3H8N4C3H8N4  Ca 3 (S 2 O 3 ) 2

1. If given percentages for each element, convert each one directly to grams. 2. Divide the grams of each element by its molar mass to get moles of each element. 3. Divide each number of moles by the smallest number of moles.

4. All answers must be within 0.1 of a whole number. If not, then all of the elements must be multiplied by a small whole number so that all elements are whole numbers. If 0.5, then multiply all elements by 2

 A compound is composed of 50.0% S and 50.0% O. What is the empirical formula of this compound?

 A compound contains 33.88% Cu, 14.94% N, and the remaining % is Oxygen. What is the empirical formula of this compound?

Determine the Empirical Formula of a compound that contains 55.2% Carbon, 2.3% Hydrogen, 18.4% Oxygen and 24.1% Nitrogen

1. Determine the empirical formula for the compound. 2. Determine the molar mass of the empirical formula. 3. Divide the molar mass of the molecular formula over the molar mass of the empirical formula. 4. Multiply the answer from #3 by the empirical formula.

Molar Mass of Molecular Formula ---------------------------- = X Molar Mass of Empirical Formula Molecular Formula = (Empirical formula) X  To check to see if your molecular formula is correct, determine the molar mass of your molecular formula. It should be equal to the molar mass of the molecular formula you were given in the problem.

 Example: A compound has an empirical formula of C 2 H 3 O. What is the molecular formula if the molecular formula mass is 172.0 g/mol?

 Determine the empirical and molecular formulas for a compound containing 94.1%O and 5.9% H. The molar mass of the molecular formula equals 34.0 g/mol.

 Determine the empirical and molecular formulas for a compound containing 26.1%C, 50.7% N and 23.2% O. The molar mass of the molecular formula equals 828.0 g/mol.

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