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Standard: 1 atom 12 C = 12 amu H = ______amu O = ______amu Atomic mass the mass of an atom in atomic mass units (amu) Micro World atoms & molecules Macro World grams

Average atomic mass (6.941)

SO 2 1S32.07 amu 2O+ 2 x amu SO amu 1 molecule SO 2 = amu Formula mass is the sum of the atomic masses of all the atoms in a compound, in atomic mass units (amu) Calculate the formula mass for SO 2

Calculate the formula mass for: H 2 O = ________ amu CO 2 = ________ amu

NO!! Need to be able to convert to macroscale (grams)

Discovery of the MOLE Amedeo Avogadro Avogadro’s number (N A ) 6.02 × — is the number of particles in exactly one mole of a pure substance.

How do we measure an amount?

mole (mol) is the amount of a substance that contains as many particles as there are atoms in exactly grams of 12 C 1 mol = x Avogadro’s number (N A ) The mole is the SI unit for the amount of substance. 1 dozen = 12

Scientists use the mole as a unit of measure AMU is not practical to use Can’t measure individual atoms very easily Measuring in grams is desirable Ratio of atoms is transferred between amu and grams

For any element atomic mass (amu) = molar mass (grams) 1 mole C atoms = x atoms 1 C atom = amu 1 mole C atoms = g C

Molar mass is the mass of 1 mole of in grams eggs shoes marbles atoms Molar mass for H = ______________ g Molar mass for O = ______________ g

Relating Mass to Numbers of Atoms Units: g/mol Molar mass is the sum of the atomic masses (in grams) in a molecule.

SO 2 1S32.07 grams 2O+ 2 x grams SO grams 1 molecule SO 2 = amu 1 mole SO 2 = g SO 2

For any molecule formula mass (amu) = molar mass (grams) Calculate the molar mass for: H 2 O = ________ g/mol CO 2 = ________ g/mol

Mole Conversions

11.2 g NaCl x Mass to Mole Conversions 1 mol NaCl g NaCl mol NaCl Na: 1 X = Cl: 1 X = g/mol MOLAR MASS

3.2 mol Zn(NO 3 ) 2 x Mole to Mass Conversions g Zn(NO 3 ) 2 1 mol Zn(NO 3 ) g Zn(NO 3 ) 2 Zn: 1 X = N: 2 X = O: 6 X = g/mol MOLAR MASS

8.74 x atoms CaCO 3 x Particle to Mole Conversions 1 mol CaCO x atoms CaCO mol CaCO 3 AVOGADRO’S NUMBER

0.36 mol Al x Mole to Particle Conversions 6.02 x atoms Al 1 mol Al 2.2 x atoms Al AVOGADRO’S NUMBER

Moles and Gases Define Molar Volume The conditions 0 0 C and 1 atm are called standard temperature and pressure (STP). Experiments show that at STP, 1 mole of an ideal gas occupies 22.4 L

1.0 L CO 2 X Volume to Moles.045 mol CO 2 Volume of Gas Molar Volume 22.4 L/ mol 1 mol CO L CO 2

.38 mol He x Moles to Volume 8.5 L He Volume of Gas Molar Volume 22.4 L/ mol 22.4 L He 1 mol He

Mole Map PARTICLES MASSVOLUME MOLE MOLAR MASS 22.4 L 6.02 x What conversion factor do I use???

250 g C 12 H 22 O 11 X Multistep Conversions 1 mol C 12 H 22 O g C 12 H 22 O 11 C: 12 X = H: 22 X 1.01 = O: 11 X = g/mol

4.4 x molecules C 12 H 22 O x molec. C 12 H 22 O 11 1 mol C 12 H 22 O 11 1 mol C 12 H 22 O 11 X g C 12 H 22 O 11

Any sample of compound has many atoms and ions… the formula gives a ratio of those atoms or ions. Formulas Express Composition

Percent Composition: The mass of each element in a compound compared to the entire mass of the compound and multiplied by 100 percent.

There are 2 ways to determine percent composition of a compound: From a chemical formula From experimental data

Percent composition of an element in a compound = n x molar mass of element molar mass of compound x 100% n = the number of moles of the element in 1 mole of the compound

H: 2 x 1.01 = 2.02 O: 1 x = g/mol 2 x1.01 g/mol H g/mol H 2 O X 100 = % H g/mol O g/mol H 2 O X 100 = % O Find the percent composition of the elements in water (H 2 O). H : O :

Calculate Total Mass: Al: 2 x = S: 3 x = O: 12 x = g/mol Find the percent composition of the elements in Al 2 (SO 4 ) 3.

Find the percent composition of the elements in Al 2 (SO 4 ) 3. - continued g/mol Al g/mol Al 2 (SO 4 ) 3 X 100 = % Al g/mol O g/mol Al 2 (SO 4 ) 3 X 100 = % O g/mol S g/mol Al 2 (SO 4 ) 3 X 100 = % S Al : S : O :

Find the percent composition of a compound that contains g of C, g of O, and g H. C = g O = g H = g g

g g C : X 100 = % g g O : X 100 = % g g H : X 100 = 1.45 %

Remember Empirical Formula: A chemical formula that gives the simplest whole-number ratio of the elements in the formula. - Subscripts are used for these ratios.

Determine the empirical formula of a compound found to have 13.5 g of Ca, 10.8 g of O, and g of H. Example Problem 1.If given %, assume 100 g; % to mass 2.Convert mass to mole 3.Divide by the smallest to get ratio 4.Multiply ‘til whole if less than.9 or greater than.1

1 mol g Ca : 1 mol g O : 1 mol 1.01 g H : 13.5 g Ca x g H x 10.8 g O x = mol Ca = mol H = mol O

.337mol Ca : mol mol O : mol mol H : = 1 Ca = 2 H = 2 O

Writing the complete formula: a)Round to whole numbers if less than.1 and/or greater than.9 b)Put parentheses around polyatomic ions c)Re-write the final formula. Ca(OH) 2

Remember Molecular Formula: A chemical formula that gives the actual number of the elements in the molecular compound. Example:C 2 H 4 NOT CH 2 C 6 H 12 O 6 NOT CH 2 O

Comparing Empirical and Molecular Formulas

There is a direct relationship between empirical and molecular formulas. There is a direct relationship between the empirical formula mass and the molecular formula mass. FIND THE COMMON MULTIPLE! The correct ratio can be found by: dividing the experimental formula mass by the empirical formula mass

The empirical formula of a compound of phosphorus and oxygen was found to be P 2 O 5. Experimentation shows that the molar mass of this actual compound is g/mol. What is the compound’s molecular formula? Example Problem

P: 2 x = O: 5 x = g/mol 1.Find Molar Mass of empirical formula. 2. Divide the experimental formula mass by the empirical formula mass g/mol g/mol = 2.00

3. Multiply the subscripts by the common multiple. 2 × (P 2 O 5 ) = P 4 O 10